text:
The reason behind the steep rise in pH in the acid base titration curve
Most books refer to a steep rise in pH when a titration reaches the equivalence point. However, I do not understand why … I mean I am adding the same drops of acid to the alkali but just as I near the correct volume (i.e. the volume required to neutralize the alkali), the pH just suddenly increases quickly.

I've decided to tackle this question in a somewhat different manner. Instead of giving the chemical intuition behind it, I wanted to check for myself if the mathematics actually work out. As far as I understand, this isn't done often, so that's why I wanted to try it, even though it may not make the clearest answer. It turns out to be a bit complicated, and I haven't done much math in a while, so I'm kinda rusty. Hopefully, everything is correct. I would love to have someone check my results.
My approach here is to explicitly find the equation of a general titration curve and figure out from that why the pH varies quickly near the equivalence point. For simplicity, I shall consider the titration to be between a monoprotic acid and base. Explicitly, we have the following equilibria in solution
$$\ce{HA <=> H^+ + A^-} \ \ \ → \ \ \ K_\text{a} = \ce{\frac{[H^+][A^-]}{[HA]}}$$
$$\ce{BOH <=> B^+ + OH^-} \ \ \ → \ \ \ K_\text{b} = \ce{\frac{[OH^-][B^+]}{[BOH]}}$$
$$\ce{H2O <=> H^+ + OH^-} \ \ \ → \ \ \ K_\text{w} = \ce{[H^+][OH^-]}$$
Let us imagine adding two solutions, one of the acid $\ce{HA}$ with volume $V_\text{A}$ and concentration $C_\text{A}$, and another of the base $\ce{BOH}$ with volume $V_\text{B}$ and concentration $C_\text{B}$. Notice that after mixing the solutions, the number of moles of species containing $\ce{A}$ ($\ce{HA}$ or $\ce{A^-}$) is simply $n_\text{A} = C_\text{A} V_\text{A}$, while the number of moles of species containing $\ce{B}$ ($\ce{BOH}$ or $\ce{B^+}$) is $n_\text{B} = C_\text{B} V_\text{B}$. Notice that at the equivalence point, $n_\text{A} = n_\text{B}$ and therefore $C_\text{A} V_\text{A} = C_\text{B} V_\text{B}$; this will be important later. We will assume that volumes are additive (total volume $V_\text{T} = V_\text{A} + V_\text{B}$), which is close to true for relatively dilute solutions.
In search of an equation
To solve the problem of finding the final equilibrium after adding the solutions, we write out the charge balance and matter balance equations:
Charge balance: $\ce{[H^+] + [B^+] = [A^-] + [OH^-]}$
Matter balance for $\ce{A}$: $\displaystyle \ce{[HA] + [A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}$
Matter balance for $\ce{B}$: $\displaystyle \ce{[BOH] + [B^+]} = \frac{C_\text{B} V_\text{B}} {V_\text{A} + V_\text{B}}$
A titration curve is given by the pH on the $y$-axis and the volume of added acid/base on the $x$-axis. So what we need is to find an equation where the only variables are $\ce{[H^+]}$ and $V_\text{A}$ or $V_\text{B}$. By manipulating the dissociation constant equations and the mass balance equations, we can find the following:
$$\ce{[HA]} = \frac{\ce{[H^+][A^-]}}{K_\text{a}}$$ $$\ce{[BOH]} = \frac{\ce{[B^+]}K_\text{w}}{K_\text{b}\ce{[H^+]}}$$ $$\ce{[A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{a}}{K_\text{a} + \ce{[H^+]}}\right)$$ $$\ce{[B^+]} = \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{b}\ce{[H^+]}}{K_\text{b}\ce{[H^+]} + K_\text{w}}\right)$$
Replacing those identities in the charge balance equation, after a decent bit of algebra, yields:
$$\ce{[H^+]^4} + \left(K_\text{a} + \frac{K_\text{w}}{K_\text{b}} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^3} + \left(\frac{K_\text{a}}{K_\text{b}}K_\text{w} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} K_\text{a} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}K_\text{a} - K_\text{w}\right) \ce{[H^+]^2} - \left(K_\text{a} K_\text{w} + \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\frac{K_\text{a}}{K_\text{b}} K_\text{w} + \frac{K^2_\text{w}}{K_\text{b}}\right) \ce{[H^+]} - \frac{K_\text{a}}{K_\text{b}} K^2_\text{w} = 0$$
Now, this equation sure looks intimidating, but it is very interesting. For one, this single equation will exactly solve any equilibrium problem involving the mixture of any monoprotic acid and any monoprotic base, in any concentration (as long as they're not much higher than about $1~\mathrm{\small M}$) and any volume. Though it doesn't seem to be possible to separate the variables $\ce{[H^+]}$ and $V_\text{A}$ or $V_\text{B}$, the graph of this equation represents any titration curve (as long as it obeys the previous considerations). Though in its full form it is quite daunting, we can obtain some simpler versions. For example, consider that the mixture is of a weak acid and a strong base. This means that $K_\text{b} \gg 1$, and so every term containing $K_\text{b}$ in the denominator is approximately zero and gets cancelled out. The equation then becomes:
Weak acid and strong base:
$$\ce{[H^+]^3} + \left(K_\text{a} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\ce{A} + V_\ce{B}} K_\ce{a} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}K_\ce{a} - K_\ce{w}\right) \ce{[H^+]} - K_\ce{a} K_\ce{w} = 0$$
For a strong acid and weak base ($K_\text{a} \gg 1$), you can divide both sides of the equation by $K_\text{a}$, and now all terms with $K_\text{a}$ in the denominator get cancelled out, leaving:
Strong acid and weak base:
$$\ce{[H^+]^3} + \left(\frac{K_\ce{w}}{K_\ce{b}}+\frac{C_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}\right) \ce{[H^+]^2} - \left(K_\ce{w} + \frac{C_\text{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}} \frac{K_\ce{w}}{K_\ce{b}}\right) \ce{[H^+]} - \frac{K^2_\ce{w}}{K_\ce{b}} = 0$$
The simplest case happens when adding a strong acid to a strong base ($K_\ce{a} \gg 1$ and $K_\ce{b} \gg 1$), in which case all terms containing either in the denominator get cancelled out. The result is simply:
Strong acid and strong base:
$$\ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]} - K_\ce{w} = 0$$
It would be enlightening to draw some example graphs for each equation, but Wolfram Alpha only seems to be able to handle the last one, as the others require more than the standard computation time to display. Still, considering the titration of $1~\text{L}$ of a $1~\ce{\small M}$ solution of a strong acid with a $1~\ce{\small M}$ solution of a strong base, you get this graph. The $x$-axis is the volume of base added, in litres, while the $y$-axis is the pH. Notice that the graph is exactly as what you'll find in a textbook!
Now what?
With the equations figured out, let's study how they work. We want to know why the pH changes quickly near the equivalence point, so a good idea is to analyze the derivative of the equation and figure out where they have a very positive or very negative value, indicating a region where $\ce{[H^+]}$ changes quickly with a slight addition of an acid/base.
Suppose we want to study the titration of an acid with a base. What we need then is the derivative $\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}}$. We will obtain this by implicit differentiation of both sides of the equations by $\displaystyle \frac{\ce{d}}{\ce{d}V_\ce{B}}$. Starting with the easiest case, the mixture of a strong acid and strong base, we obtain:
$$\frac{\ce{d[H^+]}}{\ce{d} V_\ce{B}}=  \frac{K_\ce{w} - C_\ce{B} \ce{[H^+] - [H^+]^2}}{2(V_\ce{A} + V_\ce{B}\left) \ce{[H^+]} + (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}\right)}$$
Once again a complicated looking fraction, but with very interesting properties. The numerator is not too important, it's the denominator where the magic happens. Notice that we have a sum of two terms ($2(V_\ce{A} + V_\ce{B})\ce{[H^+]}$ and $(C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A})$). The lower this sum is, the higher $\displaystyle \frac{\mathrm{d}\ce{[H^+]}}{\mathrm{d} V_\ce{B}}$ is and the quicker the pH will change with a small addition of the base. Notice also that, if the solutions aren't very dilute, then the second term quickly dominates the denominator because while adding base, the value of $[H^+]$ will become quite small compared to $C_\ce{A}$ and $C_\ce{B}$. Now we have a very interesting situation; a fraction where the major component of the denominator has a subtraction. Here's an example of how this sort of function behaves. When the subtraction ends up giving a result close to zero, the function explodes. This means that the speed at which $\ce{[H^+]}$ changes becomes very sensitive to small variations of $V_\ce{B}$ near the critical region. And where does this critical region happen? Well, close to the region where $C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$ is zero. If you remember the start of the answer, this is the equivalence point!. So there, this proves mathematically that the speed at which the pH changes is maximum at the equivalence point.
This was only the simplest case though. Let's try something a little harder. Taking the titration equation for a weak acid with strong base, and implicitly differentiating both sides by $\displaystyle \frac{\ce{d}}{\ce{d} V_\ce{B}}$ again, we get the significantly more fearsome:
$$\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}} = \frac{ -\frac{V_\ce{A}}{(V_\ce{A} + V_\ce{B})^2} \ce{[H^+]} (C_\ce{B}\ce{[H^+]} - C_\ce{B} K_\ce{a} + C_\ce{A} K_\ce{a})}{3\ce{[H^+]^2 + 2[H^+]}\left(K_\ce{a} + \frac{C_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}}\right) + \frac{K_\ce{a}}{V_\ce{A} + V_\ce{B}} (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}) -K_\ce{w}}$$
Once again, the term that dominates the behaviour of the complicated denominator is the part containing $C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$, and once again the derivative explodes at the equivalence point.

@SatwikPasani It took a bit of effort (about 6-7 hours from start to finish, including checks and writing up the answer), but I was too curious to let it go. Though I've never done this calculation, I quickly saw the general solution outline, it just took a while to actually do the algebra and make sure I didn't screw up somewhere. Perseverance paid off, though I wish I could explore the equations numerically a bit more. It would also be fantastic to consider a diprotic weak acid and a strong base, but that would require another truckload of algebra and time. I think this is enough for now.

If anyone is interested, Chem.SE user PH13 has far more expertise in pH titimetry, and has produced wonderful related answers such as this.

Hi @NicolauSakerNeto would you mind explaining how you obtained the mathematical equation you put in WolframAlpha from the strong acid-strong base chemical equation?

@Razorlance The biggest difference is that while here I deduced a formula involving the concentration of hydrogen ions, the graph has pH as the $y$-axis rather than $\mathrm{[H^+]}$. To get this, all that is needed is to use the identity  $\mathrm{[H^+]=10^{-pH}}$, which is where the $10^{-y}$ terms come from. The $(1+x)$ factors multiplied in the first and last terms of the Wolfram Alpha equation are a consequence of removing the original $(1+x)$ denominator in the middle term, which comes from $(V_A+V_B)$.

Can you give some reference (English) of a book that does this calculation? I haven't really seen such a calculation in any introductory physical chemistry books or anywhere else (quick google search).

@ApoorvPotnis Sorry I don't really have a reference. I think some graduate-level analytical chemistry textbooks on pH titimetry may be the best place to start.

What would you call this equation of SA-SB in a few words (to describe it) e.g. derives pH value?

@DialFrost The equations in the "in search of an equation" section are just called titration curves.

Quick question what is the "d" in the derivative of concentration of H+ ions over volume?

@DialFrost $\frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}}$ stands for "the derivative of the concentration of $\ce{[H^+]}$ with respect to volume B (the volume of base added)". $\frac{\ce{d}}{\ce{d}V_\ce{B}}$ represents the differentiation operation with respect to the variable $V_\ce{B}$. This is standard notation in calculus.

sorry if I am bothering you too much, but last clarification (hopefully), what does $\ce{K_\text{a}}$ mean in your first equation given

@DialFrost $K_\text{a}$ is the acid dissociation/ionizaton constant for the acid $\ce{HA}$, $K_\text{b}$ is the base dissociation/ionizaton constant for the base $\ce{BOH}$, and $K_\text{w}$ is the autoionization constant of water. To be perfectly frank, if you're not familiar with even these terms, I'm not too sure how much you can get out of this answer...

I'm just clarifying what they mean, but I can understand your answer well enough :)

A PRACTICAL DEMONSTRATION WITH PYTHON
Nicolau answer is an excellent analysis of the function that describes the titration but I would like to follow the Ben Norris answer (that I think to tackle better the problem at its origin)  using Python hoping that this could be useful to become more confident and explore titration on your own.

Acidify pure water and calculate its pH
I think acidify water should be the first experiment you should do to understand pH. Imagine you have 1 L of water and imagine to add drop by drop HCl 0.1M. Assuming the volume of a drop is 1 mL you are adding 0.0001 mol of $H^{+}$ to the solution. With every drop, furthermore, you are adding 1 mL of water to the solution, so the concentration of $H^{+}$ after the first drop should be:
$$\ce{[ H^{+} ]= \frac{mol~H^{+}_{init} + (mol~H^{+}_{drop} \times n)}{V_{init} + (V_{drop} \times n)}}$$
Where "init" indicate the initial amount of $H^{+}$ and $V$olume of water. and n is the number of drops you added.
$H^{+}_{init}$
Before you add the first drop of acid there are already $\ce{H3O^{+}}$ because water dissociate in $\ce{H3O+ ~, OH- }$ and the product of their concentration ($\ce{[H3O+] [OH- ]}$ at 25 °C) is $10^{-14}$. In pure water there is the same amount of $\ce{[H3O+]}$ and $\ce{[OH- ]}$ so $\ce{[H3O+] = [OH- ]}$ and we can write $\ce{[H3O+]^{2} = 10^{-14}}$ that is $\ce{[H3O+] = 10^{-7}}$.
The following python code simulates the addition of the drops and plots the concentration:
import numpy as np
import matplotlib.pyplot as plt

dropVolume=0.001#L
molH=0.0001#mol (you add this mol of H+ for every drop)
molOH=0.12#molt
intVolume=1#L
intHmol=10**(-7)#mol
n = np.arange(0, 30, 0.1)#here we create the drops.
#Plot concentration of H3O+
fig = plt.figure()
ax = fig.add_subplot(111)
cH=(intHmol+molH*n)/(intVolume+dropVolume*n)
plt.ylabel(r'$ [H_{3}O^{+}]$ ')
plt.xlabel(r'$mL$ of acid added')
plt.plot(n,cH)

The result is this:

You can notice that the result is quite obvious: adding acid the $[H_3O^{+}]$ increase proportionally. Note also that $H^{+}_{init}$ is negligible.
-log$[H_3O^{+}]$
Now you can calculate the pH.
$$\ce{pH = -log[H3O+ ]}$$
#Plot -log [H3O+]
fig1 = plt.figure()
ax1 = fig1.add_subplot(111)
pH=np.log10((intVolume+dropVolume*n)/(intHmol+molH*n))
pOH=np.log10((intVolume+dropVolume*n)/((10**(-14))/(intHmol+molH*n)))
plt.plot(n,pOH,'--',color='red',label='pOH   $-log[OH^{-}]$')
plt.plot(n,pH,label="pH   $-log[H_{3}O^{+}]$")
plt.ylabel(r'pH or pOH')
plt.xlabel(r'$mL$ of acid added')
plt.xlim(-1,30)
plt.grid(color='red')
plt.legend(loc=10)
plt.tick_params(length=4)
plt.xticks(np.arange(30))

This is more interesting. You can see that the pH suddenly decreases from 7 to 4 after the first mL of acid added. This is, in fact, something similar what Ben Norris noted in his answer and is only related to the logarithmic operation. After this introduction, we can deal more easily with a real titration. If you want you can try to increase the pH of pure water adding $OH^{-}$ and then calculating the hydronium concentration $\ce{[H3O+]}$ from $\ce{[H3O+] [OH- ] = 10^{-14}}$ in this way $\ce{[H3O+]  = \frac{10^{-14}}{[OH- ]}}$ and plot the pH. In the example, I've done the inverse with pOH.

Titration of HCl with NaOH
This is a simulation of a  titration of a solution of HCl with NaOH the system at the beginning has a lower pH that will increase adding drop by drop a solution of  NaOH. In this case, there are already present $H^{+}$ from HCl and the following neutralization reaction will occur until all the HCl is neutralized:
$$\ce{H+ + Cl- + Na+ + OH- -> H2O + NaCl}$$
So we should divide the process into three steps:

$$\ce{H+ > OH-}$$ until this preposition is true the number of moles of $H^{+}$ will lower down by the same amount of $OH^{-}$ added due to the neutralization reaction: $$\ce{[ H_{3}O^{+} ]= \frac{mol~H^{+}_{HCl} - (mol~OH^{-}_{drop} \times n)}{V_{init} + (V_{drop} \times n)}}$$

$$\ce{H+ = OH-}$$ at the equivalence point $\ce{[H3O+] = [OH- ]}$ so $\ce{[H3O+] = 10^{-7}}$.

$$\ce{H+ < OH-}$$ After the equivalence point we are actually "basifying" brine (solution of water and salts), so we can get $\ce{[H3O+]}$ from $\ce{[H3O+]  = \frac{10^{-14}}{[OH- ]}}$ we will use absolute value to avoid negative concentration: $$\ce{[ H_{3}O^{+} ]= \frac{ \frac{10^{-14}}{\lvert mol~H^{+}_{HCl} - (mol~OH^{-}_{drop} \times n )\rvert}}{V_{init} + (V_{drop} \times n)}}$$

Here is the code:
import numpy as np
import matplotlib.pyplot as plt

dropVolume=0.001#L
molOH=10**(-4)#mol
molH=10**(-2)#mol
intVolume=1#L
n = np.arange(0, 200, 0.1)
y=[]
for i in n:
    if (molH-molOH*i)>0:
        y.append(np.log10((intVolume+dropVolume*i)/(molH-molOH*i)))
    if  (molH-molOH*i)==0:
        y.append(7)
    if (molH-molOH*i)<0:
        y.append(np.log10((intVolume+dropVolume*i)/(((10**(-14))/np.absolute(molH-molOH*i)))))

fig = plt.figure()
ax = fig.add_subplot(111)
plt.plot(n,y)
plt.ylabel(r'pH ($-log[H_{3}O^{+}]$)')
plt.xlabel(r'$mL$ of acid added')
plt.grid(color='red')

With these conditions we obtain the usual titration curve:

You can easily see that derives directly from the previous example we can also plot the $[OH^{-}], ~ [H_3O^{+}]$.

$[H_3O^{+}]$ simply goes lower down to $10^{-14}$ at the equivalence point, and then continue to goes down while $[OH^{-}]$ increase.
Conclusion
So, in fact, the answer to your question is that relative to the Self-ionization of water equilibrium $\ce{[H3O+] [OH- ] = 10^{-14}}$ at equivalence point you are switching between a solution with an excess of $\ce{[H3O+] }$ to a solution with an excess of $\ce{[OH- ]}$ and are these excesses that are determinant. Of course, if you plot the pH you have a steep rise if you plot $[H_3O^{+}]$ you got only a little step. This is why the usage a logarithmic function (precisely a p function) is more convenient even if maybe more tricky, otherwise, you can zoom a little bit with Python to the equivalence point, this gives you an idea of what I mean with "excess"...

It took a long time for me to realise, but looking at your code, the steep part of the curve can simply be attributed to logarithms. I just want to emphasise that for others. Your code and middle two graphs helped emphasise that, thank you

Here's a desmos graph.

The pH titration curve is shaped as it is because pH is a logarithmic scale. pH is the negative base ten logarithm of hydrogen ion concentration:
$$\text{pH}=-\log_{10}{\ce{[H+]}}$$
Thus, each change of 1 pH unit is a 10-fold change in concentration, and linear of value that 10-fold change gets smaller as the pH increases. For example, going from pH 1 to pH 2 means changing the concentration of $\ce{H+}$ from $10^{-1}$ to $10^{-2}$ or $0.100... -0.010... =0.099...$
Going from pH 2 to pH 3 means means changing the concentration of $\ce{H+}$ from $10^{-2}$ to $10^{-3}$ or $0.0100... -0.0010... =0.0099...$ If you are added fixed amounts of base (which is what you do during a titration), then that fixed amount has a larger effect on the concentration of $\ce{H+}$ as the pH goes up.

Let's walk through a simple titration:

You have 10 mL of 0.100 M $\ce{HCl}$ that you want to titrate with 0.100 M $\ce{NaOH}$. We will consider the change to the solution after each mL of base is added. We have to remember to keep track of moles of acid and base, since the volume is changing (we are adding base) and we need to remember that autoionization of water. The concentrations of $\ce{H+}$ and $\ce{OH-}$ never go to zero because of the autoionization:
$$\ce{H2O <=> H+ + OH-} \ \ \ \ K_w=[\ce{H+}][\ce{OH-}]=10^{-14}$$
Let $V_b$ be the volume of base added. At $V_b=0$, we have our original acid solution, which contains $\left(\dfrac{10\  \text{mL}}{1000 \ \frac{\text{mL}}{\text{L}}}\right)\left(0.100\ \dfrac{\text{M}}{\text{L}}\right)=0.001 \ \text{moles of HCl}$. Each time we increase $V_b$ by 1 mL, we add $\left(\dfrac{1\  \text{mL}}{1000 \ \frac{\text{mL}}{\text{L}}}\right)\left(0.100\ \dfrac{\text{M}}{\text{L}}\right)=0.0001 \ \text{moles of NaOH}.$ We decrease the moles of $\ce{H+}$present by this amount and recalculate concnetration for the new volume. Then we recalculation pH. We do this until...
When we reach $V_b =10 \ \text{mL}$, we will have added exactly as much base as there was acid originally. The concentrations of $\ce{H+}$ and $\ce{OH-}$ are both $10^{-7}\ \text{M}$, controlled by the autoionization. This is the equivalence point. As we add more base, we increase the concentration of $\ce{OH-}$. We can use the $K_w$ to calculate the concentration of $\ce{H+}$ and then the pH.
 V_b     V_T     moles H+     moles OH-     [H+]        [OH-]        pH
 0 mL    10 mL   1.00E-3      1.00E-15      1.00E-1     1.00E-13     1.00
 1 mL    11 mL   9.00E-4      1.34E-15      8.18E-2     1.22E-13     1.09
 2 mL    12 mL   8.00E-4      1.80E-15      6.67E-2     1.50E-13     1.18
 3 mL    13 mL   7.00E-4      2.41E-15      5.38E-2     1.86E-13     1.27
 4 mL    14 mL   6.00E-4      3.27E-15      4.29E-2     2.33E-13     1.37
 5 mL    15 mL   5.00E-4      4.50E-15      3.33E-2     3.00E-13     1.48
 6 mL    16 mL   4.00E-4      6.40E-15      2.50E-2     4.00E-13     1.60
 7 mL    17 mL   3.00E-4      9.63E-15      1.76E-2     5.67E-13     1.75
 8 mL    18 mL   2.00E-4      1.62E-15      1.11E-2     9.00E-13     1.95
 9 mL    19 mL   1.00E-4      3.61E-15      5.26E-3     1.90E-12     2.28
10 mL    20 mL   2.00E-9      2.00E-9       1.00E-7     1.00E-7      7.00
11 mL    21 mL   4.41E-14     1.00E-4       2.10E-12    4.76E-3     11.68
12 mL    22 mL   2.42E-14     2.00E-4       1.10E-12    9.09E-3     11.96
13 mL    23 mL   1.76E-14     3.00E-4       7.67E-13    1.30E-2     12.12
14 mL    24 mL   1.44E-14     4.00E-4       6.00E-13    1.67E-2     12.22
15 mL    25 mL   1.25E-14     5.00E-4       5.00E-13    2.00E-2     12.30
16 mL    26 mL   1.13E-14     6.00E-4       4.33E-13    2.31E-2     12.36
17 mL    27 mL   1.04E-14     7.00E-4       3.86E-13    2.59E-2     12.41
18 mL    28 mL   9.80E-15     8.00E-4       3.50E-13    2.86E-2     12.45
19 mL    29 mL   9.34E-15     9.00E-4       3.22E-13    3.10E-2     12.49
20 mL    30 mL   9.00E-15     1.00E-3       3.00E-13    3.33E-2     12.52

Notice that between $V_b =9$ and $V_b = 11$, the change in $\ce{[H+]}$ and $\ce{[OH-]}$ are very large compared to their changes before and after the equivalence point. This is because near the equivalence point, the concentrations are very low and controlled by the autoionization and not by the amount of acid or base present. Even if we expand the range around the equivalence point and determine the pH between $V_b =9$ and $V_b = 11$ after 0.1 mL, we still see similar behavior.

https://www.quora.com/Whats-the-reason-behind-the-steep-rise-in-pH-in-the-acid-base-titration-curve
is pretty intuitive:
(This answer assumes that you are titrating acid with base. You see the same effect, in reverse, when titrating base with acid.)
You are losing the buffering effect of a high acid concentration.
Adding a drop (about 0.03 mL) of 1 M NaOH adds 0.03 mmol of NaOH to your solution. (There are about 30 drops per mL, 1/30 = 0.033333...)
Suppose you begin with a solution of 0.1 M acid (pH =1 for a strong acid); that means that you have, in a typical 10-mL aliquot, 1 mmol of acid. It takes about 33 drops of your 1M NaOH solution to to neutralize that amount, at 0.03 mmol per drop.
Let's assume that you diluted your aliquot to 20 mL (so that the volume can be assumed to be constant, that simplifies things). That makes your concentration 0.05 M, for a pH of  1.30.
The first drop changes the H+ concentration from 0.05 M (1 mmol in 20 mL) to 0.0485 M (0.97 mmol in 20 mL), which corresponds to a pH of 1.31.


text:
How does mass spectroscopy of nicotine work?
Take this nicotine molecule:

And this mass spectrum:

I'm not sure how to explain all the peaks.
This might explain the peak at 42:

But can this peek als be explained by something else in the molecule?
Like the decomposition of the ring with the double bonds:

This might explain the peak at 84:

And I'm clueless on the peak at 133

The peak at 133 corresponds to a loss of 29 mass units (nicotine is 162).  N-CH3 has a mass of 29

How about the others? Do you think my guesses were correct?

How can you know whether the peak at 42 is the decomposition of the ring with the double bonds or the ring without the double bonds?

Yes, I think your other guesses are reasonable.  As Klaus points out, you would need to look at the mass spectrum of some model compounds and/or some isotopically labeled compounds to know for sure.  Also, especially in larger molecules, if you have a high resolution mass spectrometer that can help, because if you calculate the mass to 6 decimal places, different fragments with the same mass - but different formula - will give rise to different exact masses.  As to which ring is fragmenting, the pyridine ring is aromatic and therefor has more stability then the saturated

pyrolidine ring, making the pyridine less likely to crack.

"pyrolidine ring, making the pyridine less likely to crack" How can you explain the fact that altghouh stable, there is no peak in the MS for 78... I would expect it to be the base peak.

I think that your assignment is reasonable.
As ron has already pointed out, cleavage of an endocyclic $\ce{C-N}$ bond (= opening of the pyrolidine ring) in nicotine is likely. The resulting open chain fragment then loses $\ce{NCH3}$.
The peak at 84 represents a N-methylpyrolidine fragment and indicates that the bond between the two heterocycles is cleaved. This also makes sense.
In general, it it a good idea to work from higher to lower masses and not to spend to much brain on the very small fragments.

How can you know whether the peak at 42 is the decomposition of the ring with the double bonds or the ring without the double bonds?

@Pickering You might be able to rule that out by comparison with the MS of pyridine. Further experimental proof could involve the specific isotopic labelling of the starting material.

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Smelly matters in urine - do they have lower or higher boiling point than water?
I don't know what exactly are the compounds that causes urine to be smelly, but does it (or do they) have higher or lower boiling points than water? If I have both liquid and exposed them to open air, will the water dry up first or will the smell disappear first?

There are several reasons that a smell vanishes.
One obvious reason is that a smelly component has a very low boiling point and simply evaporates. Another reason might be a chemical reaction of the "stinker" when exposed to air, which converts it to something untraceable.
On the other hand, other reactions might result in quite the opposite but I rather doubt that you want to perform a test run with open bowls of urine over the next week at home ;)
Here is an older article from 1971 on the gaschromatographic analysis of volatile organic compounds in human urine after extraction with diethylether.
On a first view, however, the ketones that they found usually have a rather nice smell.
In order to get a hand on the components in the vapour phase above the urine, others have performed a headspace GC-MS analysis, see this application note
and this article.
Again, the ketones usually have a pleasant smell, but they also identified some of the highly volatile bad boys, such as trimethylamine, methanthiol, and dimethylsulfide. These have boiling points of ~5, ~6 and ~40 °C, respectively.

It is important to note that unless the urine is sterile, its composition will change significantly over a few hours, as bacteria will grow feeding of the urea and producing ammonia. Therefore, even though other smelly volatiles are removed with time, ammonia is produced continuously and is tied with the exponential growth of bacteria, so older urine gets progressively more smelly.

@NicolauSakerNeto do you mean that left in the open, it will grow more smelly (because of the bacteria activity) instead of less smelly (because of evaporation)?

@OlivierBeaumont Precisely. Though I wouldn't really recommend you check it out yourself!

@NicolauSakerNeto oh god, why Now that explains what happened =)) I thought it was related to the boiling points.

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Is a diamond a single molecule?
Assuming a perfect diamond with no impurities. Would this diamond considered to be a single large molecule?
Browsing the interweb I found several opinions about this but did not find a clear Yes or No.
From what if gathered so far I'm pretty sure a diamond is a monocrystalline solid. Since all its parts are connected with covalent bonds I would assume that it should be considered a molecule even though it has no fixed number of atoms in its molecule formula.

"[...] even though it has no fixed number of atoms in its molecule formula" That exactly would be my argument against it.

Diamond is a covalent network solid, like a number of other common materials (quartz, graphite, glass, and a whole bunch of stuff).
Because they are not discrete molecules - there is no 'diamond' molecule the same way there are molecules of caffeine, benzoic acid, citric acid, N,N-dimethylaminopyridine, etc. - network solids form one of the two main classes of macromolecules, the other being polymers. You'll note that the Wikipedia article on macromolecules seems to imply that 'macromolecule' and 'polymer' are synonymous. They are not, at least not to those chemists working in the field.

The Wikipedia Article about network solids contains a key phrase: "In a network solid there are no individual molecules and the entire crystal may be considered a macromolecule." For me this means: Diamonds are macromolecules. Thanks!

@ChristophGrimmer-Dietrich Giant molecule is another term used often.

It isn't helpful to describe network solids as molecules, so we probably shouldn't
The problem here is that there in't a strict definition of molecule in chemistry. So we have to decide whether particular uses of the term are helpful or not.
We could argue that any entity where all the bonds are covalent is a "molecule". This would include diamond and polymers. But this is not helpful in practice. polymer scientists don't, for example, find it helpful to describe the components of a polymer as a molecule (they might use the more specific term macro-molecule). Most non-crosslinked polymers don't consist of single "molecules" but a mixture of many entities with different chain lengths. Talking about "molecules" isn't much use in describing or characterising the overall polymer. Many properties depend on the statistical distribution of chain length and thinking individually about each specific chain is unhelpful and irrelevant.
For many substances it is helpful to talk about the molecules making up the substance. Usually this is because the molecules that make up the substance are well defined. We can exactly describe the atomic makeup of each molecule making up the substance. Every molecule of aspirin is $$\ce{C9H8O4}$$ and no matter how much we have that describes the substance in a useful way. Where a substance is made up from more than one molecule, this is still true.
The problem with network solids is that there is no unique molecule. Every single crystal contains a different number of atoms; every diamond is different. So the idea of being able to describe the common feature of the substance as being the properties of the components (molecules) making it up goes out the window. It is better to call them network solids and think about the properties as a product of how the atomic components are connected. In mineralogy, for example, there are a lot of network solids. Silicate minerals, for example, can contain silica 3D networks (like diamond, in fact), flat silica sheets, linear silica chains and small silica units. Describing these components as "molecules" adds nothing useful; recognising the structures as components of a network is clearer and more useful.
In conclusion describing the components of network solids as "molecules" doesn't help so it isn't a useful use of the term.

We have a definition of molecule: https://goldbook.iupac.org/html/M/M04002.html

a fixed link  to the link jianrga mentioned https://web.archive.org/web/20190416235713/https://goldbook.iupac.org/html/M/M04002.html

Single crystal diamond is literally a single molecule.  Every carbon is covalently bonded to the whole.  A modest level of impurities makes no difference.  Consider epoxy.  A well compounded, well measured, well mixed sample will crosslink into a single molecule with every atom bonded to the whole body.  Ionic crystals are collections of interacting formula units.
Diamond, graphite, and buckeyballs are not organic compounds, for they are not compounds.

Buckyballs are discrete molecules and therefore, by any definition, molecules. Diamond and graphite are better called covalent network solids.

A diamond is not considered  as a molecule because each carbon atom is covalently bonded with four other carbon atoms. This is what makes diamond a network solid. Since it's a whole network of covalently bonded atoms(carbons), diamond is not considered  to be one molecule.


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Splitting of $d$ orbitals when ligands approach central metal ion
In my high school chemistry book, it is written that when ligands approach the central metal ion (transition metal ion) to form dative bonds, the $3d$ orbitals split into two: two which are in higher energy level and the other 3 in lower energy orbital...
One thing that doesn't make intuitive sense to me is that why 3 orbitals move to a lower energy level... if the ligands repel the electrons in the $3d$ orbitals, how can some of the orbitals possibly lose energy and be at a lower enrgy state than they were currently at before...

The picture you pose is only a part. The full picture is posted below.
This solves your doubt. The splitting does not occur about the original level, i.e. not about the energy level the d-orbital has in complete absence of the ligand's electrostatic field. But, the splitting occurs about a hypothetical Barycentre. This is the energy level of the d-orbitals (still degenerate) in a hypothetical spherical electric field (with spherical symmetry). Due to the symmetry of the hypothetical barycentre field, the d-orbitals remain degenerate but at an energy higher than in the absence of the ligands. Now, the $t_{2g}$ group which is lowered in case of an octahedral field, is still above the original free ion energy level in the complete absence of the ligands.
EDIT
By conservation of energy, we mean that the energy is conserved during splitting. The very process of splitting occurs symmetrically about the barycentre, so that the net amount by which $t_{2g}$ is lowered ($3\times 0.4\Delta_o$) is equal to the amount by which $e_g$ is raised ($2\times 0.6\Delta_o$). But the net energy in absence of field and in presence of field need not be same for energy conservation to hold. Whatever the rise in the net energy (from free ionic to barycentre) will be taken from the kinetic energy (or potential) of the approaching ligands.

That makes more sense, so all the orbitals have a higher energy level than the original.However, in my book, it says that: "from the point of view of conservation of energy, it is necessary that this split happens in such a way that the total energy of the 3d orbitals remains same "... doesn't that contradict the diagram you have given (i.e. how is energy conserved?)

@Eliza Please see the edit in my answer

so as the ligand approaches the 3d orbitals, the KE of the ligands is converted to the energy of the 3d orbitals... the "barycentre" just shows what would have happened if splitting was symmetrical... so the energy of the higher and lower energy 3d orbitals will just equal to the total energy of the barycentre.. is that it? I just want to check :)

@Eliza To the best of my knowledge, yes. Just a minor clarification, saying that it is the kinetic energy of the ligands converting to the energy of the orbitals would not always be correct. The energy is a mixture of that released by the electrostatic attraction of the overall positive ion and the negative ligands, kinetic energy of either species, or any other energy involved in the interaction.


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The last element's atomic number
I was just thinking what can be the last atomic number that can exist within the range of permissible radioactivity limit and considering all other factors in quantum physics and chemical factors.

Nobody really knows. Using the naive Bohr model of the atom, we run into trouble around $Z=137$ as the innermost electrons would have to be moving above the speed of light. This result is because the Bohr model doesn't take into account relativity. Solving the Dirac equation, which comes from relativistic quantum mechanics, and taking into account that the nucleus is not a point particle, then there seems to be no real issue with arbitrarily high atomic numbers, although unusual effects start happening above $Z \approx 173$. These results may be overturned by an even deeper analysis with current quantum electrodynamics theory, or a new theory altogether.
As far as we can tell, however, we will never get anywhere close to such atomic numbers. Very heavy elements are extremely unstable with respect to radioactive decay into lighter elements. Our current method of producing superheavy elements is based on accelerating a certain isotope of an relatively light element and hitting a target made of an isotope of a much heavier element. This process is extremely inefficient, and it takes many months to produce significant amounts of material. For the heaviest elements, it takes years to detect even a handful of atoms. The very short lifetime of the heaviest targets and the very low collision efficiency between projectile and target mean that it will be extremely difficult to go much further than the current 118 elements. It is possible that we may find somewhat more stable superheavy isotopes in the islands of stability around $Z=114$ and $Z=126$, but the predicted most stable isotopes (which even then are not expect to last more than a few minutes) have such a huge amount of neutrons in their nuclei that we have no idea how to produce them; we may be condemned to merely skirt the shores of the islands of stability, while never climbing them.
EDIT: Note that the best calculation presented above is based on quantum electrodynamics alone, i.e. only electromagnetic forces are taken into account. Obviously, to predict how nuclei will behave (and therefore how many protons you can stuff into a nucleus before it's impossible to go any further), one needs detailed knowledge of the strong and weak nuclear forces. Unfortunately, the mathematical description of nuclear forces is still an incredibly tough problem in physics today, so no one can hope to provide a rigorous answer from that angle.
There must be some limit, as the residual nuclear forces are very short-ranged. At some point there will be so many protons and neutrons in the nucleus (and the resulting nucleus will have become so large) that the diametrically opposite parts of the nucleus won't be able to "detect" each other, as they are too far away. Each additional proton or neutron produces a weaker stabilization via the strong nuclear force. Meanwhile, the electrical repulsion between protons has infinite range, so every additional proton will contribute repulsively just the same. This is why heavier elements need higher and higher neutron-to-proton ratios to remain stable.
Thus, at some atomic number, possibly not much higher than our current record of $Z=118$, the electrical repulsion of the protons will always win against the strong nuclear attractions of protons and neutrons, no matter the configuration of the nucleus. Hence, all sufficiently heavy atomic nuclei will suffer spontaneously fission almost immediately after coming into existence, or all the valid reaction pathways to reach an element will require events which are so fantastically unlikely that if even all the nucleons in the entire observable Universe were being collided with each other since the Big Bang in an attempt to synthesize the heaviest element possible, we would statistically expect some sufficiently heavy atom to not have been produced even once.

Using the naïve Bohr model of the atom, we run into trouble around $Z = 2$...

@leftaroundabout Only with respect to the accuracy of energy levels, not the stability of the atom itself!

With respect to any property these atoms have. The Bohr model simply doesn't work out for anything but 2-body systems, so it can't really apply to atoms other than hydrogen (though it can well apply to $\ce{He}^+$ etc.).

@leftaroundabout Fair enough. I guess Bohr's model is just often mentioned for historical reasons, to show that models can set limits (even if wrong) and because $v^{1s}_e=Z\alpha c$ is a very simple result. Of course, the Dirac equation itself is also an approximation (a much better one, no doubt). We don't even need a new theory to overturn its conclusions; at some point even more subtle QED effects will become appreciable, and how they will alter the final picture is still unknown, as far as I understand.

We could describe the Bohr model as a model for the ion produced when only one electron is left.  That is the model which, having no Einsteinian relativity, renders the (1s) electronic orbital speed greater than that of light.

An "element" must be defined as the set of all atomic nuclei having a specified number of protons.  Definitions based on electrons (or other leptons) can't be used because how many electrons are associated with an element changes with the atom's environment.
Defining an "atomic nucleus" as a set of protons and neutrons, in a common nuclear potential well, whose mean life is large with respect to the time it took the set to form.  (A nuclear interaction takes place over a span of time on the order of $1\times10^{-23}$ sec.)
If you add neutrons to a nucleus, each is more weakly bound than the last.  Eventually, the last neutron added is unbound, so it comes right back out.  Usually, this happens within a time comparable to $1\times10^{-23}$ sec.  For each proton number, Z, there is a maximum number of neutrons, call it Nd, which can be in a nucleus with Z protons.  The set of nuclides $(Z,Nd)$ is a curve on a Z,N plane known as the neutron dripline.  The neutron dripline defines the maximum size a nucleus with a given number of protons may have.
If a nucleus with Z protons has too few neutrons, one of two things will happen: It may eject a proton or it may fission.  Large nuclei will almost invariably fission, though, so that's the important criterion.  The simplest workable model of an atomic nucleus is the "liquid drop model".  Since its charges are trying to push it apart, though, thinking of a nucleus as a tiny, highly stressed balloon gives a better idea of the forces in play.  Electric repulsion varies as $(Z^2 / r_{eff})$ where $r_{eff}$ is distance between equivalent point charges.  What pulls the nucleus together is what amounts to surface tension – unbalanced nuclear cohesion – and the total "surface energy" stored varies as $(r^2)$, where r is nuclear radius. The ratio between Coulomb and surface energies is defined by $(Z^2 / r_{eff})*(1/ r^2) = K$.  Set $r_{eff} = r$.  Nuclear volume is proportional to the total number of particles, $A = Z+N$, in a collection.  That means r varies as $A^{1/3}$, so $(Z^2 / r^3) = K = (Z^2)/A$.  K is called a "fissility parameter."  A given value of K defines a set of nuclei which have similar liquid-drop-model barriers against spontaneous fission.  For specified value of K, $N(Z) = (1/K)*(Z^2) - Z$ defines a curve of constant fission barrier height on the $(Z,N)$ plane.  One particular curve defines the line dividing sets of nucleons for which a fission barrier exists and sets of nucleons which do not.  In other words, it defines the minimum number of neutrons which a nucleus of given Z may have.
At least one nuclear model includes nuclei with up to $330$ neutrons and $175$ protons(1).  An equation for the neutron dripline as a function of Z can be extracted from their dripline.  A second equation for $N/Z$ as $f(Z)$ can be used to construct an alternate dripline curve.  KUTY's neutron dripline does not show any dramatic changes below $N=330$.  Still, when extrapolating into the unknown, it seems prudent to consider the upper limit to neutron count in a nucleus to be $1/4$ order of magnitude ($1.77$) times larger.

Liquid-drop theory predicts immediate fission for $K>50$; however, the liquid drop model does not account for the additional binding produced by nuclear structure.  The maximum K value for any nucleus in the KUTY model can be used as a guide to how large K must be to overcome these corrections.  Taking that value as the geometric mean between $K=50$ and the value of K to be used gives $K=102$.  (This was the highest of three techniques tried.)
For large Z, the fission curve rises faster than the dripline curve.  The point at which they meet is the largest possible nucleus.  Anything larger will immediately decay by neutron emission or fission.  Nominally, the largest nucleus $Z=592$, $N=2846$ - but that is way too much precision for this sort of calculation.  It's reasonable to say that the largest possible nucleus has $Z <600$ and $N < 3000$.
It's entirely possible that my remarks are completely wrong.  I hope so, because that would mean somebody who knows more about it than I do has come up with a better answer.

"Decay Modes and a Limit of Existence of Nuclei"; Hiroyuki Koura; http://tan11.jinr.ru/pdf/10_Sep/S_2/05_Koura.pdf

I like your last sentence ;) This is nice analysis, but you go too far with "Definitions based on electrons (or other leptons) can't be used" While you can't define element by  number of electrons, they are actually needed for these nuclei to become part of element. Nucleus with half-time too short to bind electrons around it isn't enough to get element. This is a matter of definition which might be changed, though.

Naively, the nuclear electric field at Z ~ 137 or greater, reciprocal of the Fine Structure constant, would "spark the vacuum."  Vacuum would be torn into electron-positron pairs.  The electrons go in to convert protons to neutrons plus neutrinos.  As stated above, non-classical treatment suggests we'll never get near a cold nucleus that sparks the vacuum.  RHIC and LHC rip the vacuum hot by colliding deeply relativistic gold or lead nuclei.
The big problem in making new heavy elements is getting enough neutrons in there while colliding the heaviest nucleus with the lightest to do the job.  Fusion charge repulsion is the product of the two charges.  It must be minimized.  Ca-48 is stable for a footnote, and it's run its course for fusion with available transuranics.  The isotopes produced are way too neutron deficient to stick around. One might do something clever with a big custom-configured H-bomb - lots of compression and neutron density there - but sample retrieval is problematic.

The real thing is that we do NOT know for sure. Accepting Quantum Mechanics, we have some parts of the answer, but we can not know the rest until we test it. Curiously, the higher Z is, the more relativistic the atom is. So, relativistic quantum mechanics do matter in the Periodic Table.
Option 1. Something avoids to allow elements with Z>118. That is quite unlikely. Pekka Pykko did simulations until 173...
Option 2. The maximum Z is somewhere between 122-173. It seems unlikely too!
The only caveat is that Z=137 is special because the Dirac equation implies that the electron 1s has a speed greater than light for Z>137 due to the value of the fine structure constant. However, this is done neglecting the finite extension of the nuclei. So, here we realize that the ultimate fate of the atom depends on the stability of the nuclei...
Option 3. Elements are stable until 173, nuclear physics acts to avoid higher elements some point between 137 and 173...This is more likely, but nobody knows...
Option 4. Elements are allowed in principle, even at supercritical Z or above until the nuclei can not support more shells. We don't know really what happens here due to limited simulations. Maybe quantum computers can help us here (I hope so) to simulate elements and atoms in these quantum realm.
Option 5. There is no limit on Z. I consider this thing unlikely unless something is discovered beyond the current quark-lepton periodic table of fundamental particles.
I have written about the Bohr model at my blog, and written about the last element too. Here: http://www.thespectrumofriemannium.com/2013/06/30/log113-bohrs-legacy-i/, http://www.thespectrumofriemannium.com/2013/07/10/log114-bohrs-legacy-ii/, http://www.thespectrumofriemannium.com/2013/07/10/log115-bohrs-legacy-iii/, http://www.thespectrumofriemannium.com/2013/12/31/log150-bohr-and-doctor-who-amc%c2%b3/, http://www.thespectrumofriemannium.com/2014/05/26/log151-bohrlogy-i/ , http://www.thespectrumofriemannium.com/2014/05/26/log152-bohrlogy-ii/, http://www.thespectrumofriemannium.com/2015/07/04/log171-from-bohrlogy-to-dualities/, and http://www.thespectrumofriemannium.com/2017/07/11/log183-bohrlogy-some-pocket-formulae/

Upper limit of Z is very large indeed but there should be a gap starting at at least 118 and ending somewhere around 10^57.

@joshua you mean a neutron star?  If that is an element, it would be a very strange one in which the nucleons are held together by gravity instead of nuclear forces and most of the electrons are trapped inside!

How high atomic number can go is a nuclear question.  Any lump of positively-charged matter will find and bind as many electrons as it needs.
While it is true that the limiting value of Z is unknown an unpredictable, the same is not true for an outer boundary to the region of a (Z,N) or (Z,A) plane within which atomic nuclei can exist.  Since an outer boundary is desired, only fission and neutron emission (but not beta + neutron decay) need be considered.
For each Z, there is a maximum number of neutrons which can bind together into a nucleus.  The locus of these points is the neutron dripline.  That dripline has only been studied experimentally at low Z, but has been studied theoretically up to rather high Z by a few authors.  These predictions can be combined and multiplied by a factor of safety to allow for errors and by a second factor of safety to account for the possibility of unknown stabilizing effects.
For each Z, there is also a minimum number of neutrons required to prevent immediate (under 10^-20 sec) fission.  During fission, a nucleus distorts until its surface energy is insufficient to prevent Couloumb repulsion from separating the nucleus into two parts.  The activation energy needed to distort the nucleus is called binding energy.  An analytical expression for fission barrier magnitude is possible with a liquid-drop approximation.  This expression for fission barrier as a function of Z and A can take negative values.  A must-fission curve Nmin(Z) can be constructed from this by setting fission barrier equal to -1 * strongest structural correction energy * a safety factor.  That way, even a stabilized nucleus will have a fission barrier less than zero, so must fission on a nuclear timescale.
The must-fission curve rises faster than the neutron dripline, so the two intersect at a specific Z and N.  This is as large as a nucleus can be; anything bigger, either in Z or N, will explode or spray neutrons long before it can find its first electron.
When quantified, the maximum Z must be less than 340 and the maximum N must be less than 1300.  That does not mean Z or N can actually be anywhere near that large, only that they can't be bigger.  As a personal guess, I'd predict maximum Z will probably be around 220.
The analysis described has not been published in reviewed journal.  It has, however, been posted to the "Transuranic Elements Wiki" as an article: "The Final Element".

You have had at least 2 accounts with the same user name, which is a violation of the code of conduct. I have deleted one of them. Please do not create multiple accounts with the same name. I also encourage you to register this particular account.

I did so by accident, and still have no idea how I did it.

Turns out I was wrong in re: code of conduct. That said, you should stick with one AND register it.

It might be possible that they can keep on going to find new elements by using cycloterons and fusing atoms together till they reach a blackhole. So  I believe that may be the limit for atomic number can be calculated by the singularity of Kerr Newman metric. But by assuming periodic table pattern if one other element could be found there comes at least 32 others to complete one other row of periodic table to reach may be 150 atomic number.

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Epsom salts - heavy metal contamination myth?
This recipe for home-made handwash (liquid soap) consists of bar soap, water, Epsom salts, and coconut oil.
The author specifically warns against using 'artificial' Epsom salts, preferring the naturally occurring type, because the 'artificial' type is supposedly contaminated by heavy metals incl. arsenic.

Note – Be sure to get natural salts as some are man made and a full of
  unwanted chemicals. A lot of Epsom salts are made from synthetic
  materials, they are manufactured in factories from neutralized
  Sulfuric Acid, which is very high in heavy metals and arsenic. The
  REAL Epsom salt is naturally occurring mineral deposit in spring.

According to Wikipedia, Epsom salts are made by reacting sulfuric acid with magnesium oxide or magnesium carbonate.
I find it difficult to believe that 'artificial' Epsom salts would have appreciable amounts of heavy metal contamination, as neither sulfuric acid nor magnesium oxide are known to be frequently contaminated with heavy metals. A quick search on Google Scholar gives no results on the topic of heavy metal contamination in Epsom salts.
The only way I am aware that heavy metals might enter the process is if the sulfuric acid was produced from smelter stack gases, which do have appreciable heavy metal content. This is done at Mount Isa Mines, where $\ce{SO2}$ from the copper smelter stack gas is scrubbed out to make sulfuric acid, but the entire output of this plant goes straight to fertilizer production.
Is the above claim about heavy metal contamination of Epsom salts credible, or just an urban myth?

I've never heard about these claims before and I had too look it up too. It seems that the "arsenic" contamination is mostly discussed on sites with little to no scientific background. I think that your  analysis is reasonable and agree that the contamination story is just an urban myth.

I wouldn't be surprised to find that "natural" salts had higher levels of contamination.

There is a lot of tramp sulfuric acid which contains all sorts of impurities everywhere- its an easy way get rid of sulfur dioxide produced from smelting and other places.

Here in the US, they sell different grades of Epsom salt, from USP down to plant food.  I'd presume that depending upon what the product will ultimately be used for, the manufacturer may use MgO with different purity levels - the lower the purity, the cheaper the product.  Even the "bath salts" grade contain trace amounts of arsenic and selenium (see the first link below).  Among the "medicinal" offerings, some are labeled "for external use" and others are labeled "for external and internal use."  I suspect that anything with the "for external and internal use" label would be more than safe enough to use in handwash.
Here are some links you might find helpful:
http://k.b5z.net/i/u/2182313/f/Tech%20Data%20Sheets/Tech_Data_Sheet-bath_salt-epsom_USP-3.pdf
http://www.drugs.com/otc/104235/pepsom-flying-p-epsom-salt.html

As seen on a 25kg bag of Epsom Salts:

I did XRF for years. A mg/kg is a part per million (ppm). When you start looking at ppm levels of contaminants in salts you will find all sorts of impurities. Even ACS reagent grade chemicals are full of junk at the ppm level. It is really really hard to get 99.9999% pure of anything. // Not disputing label, just pointing out that these would be trace level impurities...

Chemicals from chemical makers have labels about hazards and any kind of minor components and well as the grade of purity. I don't know Epson salt specifically in your country, but most probably there is a cosmetic or food grade version, which must be safe for you. If a salt is pure enough for bath salt, it is pure enough for washing hands (you don't drink soap, so food grade is not necessary). UNLIKE many natural, organic or whatever shaman-branded products, the "artificial" chemicals has a strong quality and safety control and they actually make all those info public.
Just a note:
Home-soap makers are notoriously bad in chemistry and bad in safety. I have seen many recipes that make no sense, extremely hazardous and you produce a mediocre-to-bad quality soap. From the wording of the inserted quote I would expect he/she is a nice, but uninformed amateur at best. Be careful with those recipes.

home-made handwash (liquid soap) consists of bar soap, water, Epsom
  salts, and coconut oil.

Bar soap - OK.  Might be sodium or potassium fatty acid carboxlates, or syndet.
Epson salts - not OK.  Bivalent cations turn literal soap into insoluble slime.  Synets don't much like it either.
Coconut oil - not OK.  Why are you adding grease to soap?
Supermarket, soap aisle, find something nice.  I once washed my hands with Dove bar soap.  It took a day to get them clean again.  Maybe that's what you want.

The question is about heavy metal contamination of Epsom salts, not so much about the particular home-made soap recipe. (I buy mine from the supermarket, same as everyone else.)

The question is about how to recognise nonsense writing. And Uncle Al's answer shows that the writer of that recipe sells snake oil.

Well I'm no expert but I wanted to do a liver flush and bought some Epsom salts (magnesium sulfate heptahydrate), apparently safe for ingestion and after seeing this page further searching led to this:
http://www.fao.org/fileadmin/templates/agns/pdf/jecfa/cta/68/Magnesium_Sulfate.pdf
Section 4.2 is relevant and the key phrase is "limited to those that originate from the starting material".

LOL - I'd agree. Anyone who wants to do a "liver flush" is no expert.


text:
Nature of transition metals
The transition metals have valence and penultimate shell incompletely filled . so does this mean that the transition metals first fill up their valence shell and then penultimate shell?

You should change you title in something like this "Can transition metals first fill up their valence shell and then penultimate shell?" although I think your question will be closed...

The short answer to your question is, usually, but not always.  The energy differences between these levels are very small, so sometimes other factors come into play that can cause exceptions.  Here are a few exceptions:
 Predicted configuration   Actual configuration

Cr    [Ar] 3d4 4s2  ...........  [Ar] 3d5 4s1
Cu    [Ar] 3d9 4s2  ...........  [Ar] 3d10 4s1
Ag    [Kr] 4d9 5s2  ...........  [Kr] 4d10 5s1
Au    [Xe] 4f14 5d9 6s2  .. [Xe] 4f14 5d10 6s1
There is a special stability associated with a full shell, in fact that's probably why you suspected that "transition metals first fill up their valence shell and then penultimate shell."  There also appears to be a stability advantage associated with a half-filled shell and this can explain many of the anomalies.  The reason that a half-filled shell may have some extra stability is that there are no paired electrons (Hund's rule) in such a configuration.


text:
Why are cations ordinarily smaller than anions?
I'm reading on ionic crystal structures and am confused by this from my textbook:

Because the metallic elements give up electrons when ionized, cations are ordinarily smaller than anions

I understand that having a positive charge will pull the electrons closer, but shouldn't the atom from the larger period (row on periodic table), have more electron shells and be the larger atom?

It isn't true that any cation will be smaller than any anion.
You intuition that atoms with larger electron counts from lower rows of the periodic table will be bigger is correct. Size for atoms is determined by the highest occupied electron shell or orbital as the interactions with other things that define size are, crudely, caused by interactions between the electrons in their outer orbitals.
So a Rb+ cation radius is about 166pM but an F- anion is smaller at about only 119pm. There is a good table of such values in the Wikipedia article on ionic radii which also explains some of the trends and some of the difficulties in actually defining size for ions.

Generally we say that cation is smaller than anion in case of iso-electronic species (see the fig). If you consider ions which are not iso-electronic, the statement may not be true always. As you say, if you consider cations consisting of many shells (say 6) and if you compare them with anions consisting of less shells (say 2), in that case cations will be larger than anion.


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What does the negative superscript in the term symbol for ground state Oxygen stand for?
I think that I understand most of the parts of the symbol shown for ground state oxygen $\ce{{}^{16}O2}$, which is ${}^{3}\Sigma^{-}_{g}$.
However, I was hoping that someone could tell me what exactly the negative superscript sign means? And how do you identify between a positive and negative state? I think it has something to do with a mirror plane.

The ground state term symbol doesn't depend on the isotope - it is also $^3\Sigma_g^-$ for $\ce{^{17}O ^{16}O}$, $\ce{^{18}O2}$ etc. However, the resulting nuclear spin statistics for rotational states will of course be different for different isotopomers.

You're right that it has to do with a mirror plane. The negative sign means that the overall electronic wavefunction changes sign when reflected in a mirror plane that includes the internuclear axis.
For ground-state oxygen, reflecting in a mirror plane will change the sign of one unpaired $\pi$ electron (the one that sticks out of the mirror plane) but not the other, resulting in an overall change in sign. (This is slightly simplified, a more rigorous explanation would take into account the complex-valued orbitals rather than their real-valued linear combinations.)

If you look at the point group $D_{\infty h} $ you will see how the symmetry works


text:
Is acetate a weak base due to its resonance structure?
I just read that acetic acid is an exception to the general idea that the conjugate base to a weak acid is a strong base.  The example says that the conjugate base of a weak acid will only be strong if the acid is a weaker acid than water.
What I'm curious about though, is if acetate is a weak base due to its resonance structure.  Does the resonance cause acetate to be more stable than hydrogen acetate?
Is this true in the more general case where some substituent can bond to the atoms involved in the resonance?
Thanks.

Please be more specific in the "general idea that the conjugate base to a weak acid is a strong base". How general it is in your opinion? All carboxylic acids, e.g.?

@ssavec Well, it's true in some cases, such as with $\mathrm{NH_3}$ and its conjugate base $\mathrm{NH_2^-}$, but it's not always true such as in the case of hydrogen acetate and acetate.  I'm just simply curious about how much resonance affects how basic a given compound is.

Maybe there is some confusion on what the weak acid means. Just calculate the equilibrium concentration of the species involved and guess yourself, what is weak and what is strong. To call NH3 a weak acid is technically possible, but nonsense, similar to call a boling water "weakly cold". For sure, it is cold compared to e.g. lava, but for common use, it is damn hot. And this acid/base scale is even muuuuch wider, than temperature.

@ssavec I'm completely aware that the idea of a strong/weak acid is completely relative, but it's a well-established notion.  In Carey's Organic Chemistry text, 7th ed., he offers as a definition that a strong acid is an acid stronger than $\mathrm{H_3O^+}$ and a weak acid as an acid weaker than $\mathrm{H_3O^+}$.  And you're really missing the point - I just want to know how the basicity of a species is affected by the presence of a resonance structure.

I can agree with the cited definition. Find one also for the basicity.
And I am not missing the point, I am openly ignoring it. Whatever stabilizing effect on acid or it's conjugated base affects the pKa and pKb of the reverse reaction by the very same amount, but in opposing sense. Thermodynamics.

Consider this generic acid-base reaction in water:
$$\ce{HA <=>[K_\text{a}] A- + H+ }$$
The (thermodynamic) dissociation constant $K_\text{a}$ depends mainly on two things:

The stability of the product $\ce{A-}$
The instability of the starting material $\ce{HA}$

The higher one of those factors is, the higher the dissociation constant. So in effect, what a chemist states is the following:
The carboxylate anion is the conjugated base of the carboxylic acid, whose strength is determined by the stability of said carboxylate ion and the instability of the acid. It may be reasoned that carboxylic acids would be much less strong if it was not for the additional stability via conjugation.


text:
Solubility, pressure and Henry's law

Determine the solubility of $\ce{N2}$ in water exposed to air at $\pu{25^\circ C}$, if the atmospheric pressure is $\pu{1.2 atm}$, assume that the mole fraction of nitrogen is $0.78$ in air and the Henry's Law constant for nitrogen in water at this temperature is $\pu{6.1 \times 10^{-4} M/atm}$.

I have the answer, but i want to know how to solve for it. I've been reading through my notes and the book trying to find a reference through which to approach this problem and i have found none. If anyone can guide me in the right direction that would suffice.

What are the units of your answer? You need a solubility, which is an amount per volume. If your amount is moles, the your solubility is in moles/liter: $\text{M}$.
That unit appears in the Henry's Law constant! $6.1\times 10^{-4} \frac{\text{M}}{\text{atm}}$. What can you do to get rid of the atmospheres in the denominator? You probably need the pressure and the mole fraction.

Henry's law is given by the formula $p=K_\mathrm h\chi$. Here $p$ is partial pressure of the gas $K_\mathrm h$ is Henry's constant for that gas and $\chi$ is mole fraction.Using this you can find it.

I downvoted this answer because there are several forms of henry's law (http://en.wikipedia.org/wiki/Henry's_law) but the mole fraction form does not answer this question.


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What happens to electrons in metal complexes after excitation by visible light?
My book writes:

When white light shines through a solution of a complex ion of a transition metal, photons of a particular frequency are absorbed and their energy promotes an electron from lower energy level to upper energy level. We perceive the complementary color to the light absorbed.

However, I have a question about this. Once the electron is promoted to a higher energy level, won't it fall back down and give off a photon?
Surely the electron cannot remain at the higher energy level forever, as that would mean that no further electronic transitions could taking place, and the solution would turn colourless.

Absorption of a photon typically results in a vibrationally excited higher electronic state of the same multiplicity.
$$\ce{S_0 ->[$h\nu_\mathrm{ex}$] S_1}$$
In most cases, the excited state deactivates through internal conversion in a radiationless process via vibrational energy exchange with solvent molecules. No light is emitted here, but the solvent is actually heated up, although it might not be measurable.
There are however cases, in which the initially formed excited singlet state species first relaxes to its lowest vibrational mode and then emits a photon to reach the ground state.
$$\ce{S_1 -> S_0 + $h\nu_\mathrm{f}$}$$
This deactivation via radiation is known as fluorescence. Fluorescence lifetimes are typically rather short (in the ns range) and the emitted photon has a lower energy than the absorbed one. This means that the emission spectrum is shifted to the red end of the uv spectrum ($\lambda_\mathrm{f} > \lambda_\mathrm{ex}$). The effect is known as Stokes shift.
A classical case for a fluorescent transition metal complex is $\ce{Ru(bpy)3^{2+}}$.
In acetonitrile, it absorbs around $\lambda_\mathrm{ex} = 460~\mathrm{nm}$. The fluorescence is observed around $\lambda_\mathrm{em} = 620~\mathrm{nm}$.

Other bidentate ligands - more rigid than 2,2'-bipyridine (bpy) - often used are 1,10-phenanthroline (phen), dipyrido-[3,2-a:2',3'-c]phenazine (dppz), and benzo[i]dipyrido-[3,2-a:2',3'-c]phenazine (dppn).

In other cases, the excited singlet state undergoes a change in the spin multiplicity (Intersystem crossing) to a triplet state.
$$\ce{S_1 ->[ISC] T_1}$$
These excited triplet states are usually much longer-lived. When they eventually deactivate via emission of a photon, this process is known as phosphorescence.
$$\ce{T_1 -> S_0 + $h\nu_\mathrm{p}$}$$

Keeping all these processes in mind, one question arises - particular from younger students or pupils:
If our metal complex absorbs a photon and forms an excited state with such a short lifetime, why can we see a nevertheless observe permanent colour?
The answer is in the enormous numbers!
Continuous irradiation with white light means that we send a huge number of photons. Moreover, the solution of our transition metal complexes in a cuvette (or a test tube) also contains more than one molecule. In fact, much more!
As a result, we have a huge number of incidents

The complex absorbs a photon of a certain energy, which means that one colour is "swallowed". The complex forms a short-lived "excited state"
the solution reflects the remaining light
the excited state complex deactivates again

Since this happens all the time while we shine the light on our sample and for a lot of molecules, we observe a summary of the events: a coloured solution.

I am really sorry to say this but this answer was too complicated for a high school student like me. If you do have the time, could you please add a more simple answer as to why we seen the solutions of complex ions of transition metals to be continuously colorful...I learned that color is absorbed from white light during a $d$-$d$ transition... but once this transition occurs and complementary color seen, how do we continue to see a color as the electron stays in the upper energy level...

My bad, I didn't realize that you attend school. Keep the difficult explanations for later in your scientific career as a photochemist ;-). I added a bit to my answer to make it useful for now :)

Thank you so much. This was much easier to understand. Just one thing, when the excited electron falls back to the lower energy level, is the energy of the electron released as heat?I know this is a stupid question but the thing is that I mix up this concept with the concept of flame colors of different metal ions :/

Yes, exactly :)

Ok. Also, do the electrons fall back to ground state and release heat as that state is unstable?

Yes, you got it :)

Flame colours are nice! But there's no absorption of photons here. It's the other way around. With the heat of the flame, an excited state is generated! This one deactivates to the ground state by releasing the energy as light. So, flame colours result from emission of light.

Here's a simple explanation of why that happens:-
First the electron gets promoted to the higher level after absorbing one photon corresponding to the visible light. This makes you see the complementary color, as you already said.
BUT when the electron comes back to the unexcited state, it does not do that in a single step! The electron comes back to the unexcited state taking smaller intermediate steps. Each of these smaller transitions corresponds to the emission of a low frequency photon belonging to the Infra Red region. This is outside the human visible range and thus we don't "see" these photons. But these Infra Red photons do heat up the solution though the change may not be measurable.


text:
Ligand exchange reactions... are they one way or reversible?
Are ligand exchange reactions one-way  reactions or reversible? I know this is a very silly question but its not said outright in any place...
For example, in my high school chemistry book, these two ligand exchange reactions are shown:
$$\ce{[Cu(H2O)6]^{2+} (aq) + 4Cl- (aq) <=> [CuCl_4]^{2-} (aq) + 6H2O (l)}$$
$$\ce{[Cu(H2O)6]^{2+} (aq) + 4NH3 (aq)-> [Cu(NH3)4(H2O)2]^{2+} (aq) + 4H2O (l)}$$
The first is reversible but the second is one-way
Also, my teacher stated that ligand exchange reactions are irreversible... this confused me further
Can someone please tell me what is right...

Let's take the binding of $\ce{O2}$ and $\ce{CO}$ to hemoglobin as example. $\ce{CO}$ does bind much better than $\ce{O2}$!
We can consider the reaction of oxygen-loaded hemoglobin with $\ce{CO}$ a ligand exchange reaction. If that would be completely irreversible, the logical consequences in the case of an intoxication would be: [...] six feet under ;)
Instead, intoxicated patients are treated with oxygen at higher pressure.
To me, that sounds like Le Chatelier in the ICU and it only makes sense in the case of equilibria.

so ligand exchange reactions involving transition metal complexes are actually all reversible...?

@Eliza Yes, in principle. But the affinity of some metals to particular (polydentate) ligands can be very strong. So you have a bad time trying to replace these ligands with some others.

in that case, would it still be a reversible reaction but the equilibrium would almost be to the right (i.e. the side of the reaction with the complex ion with polydentate ligand)...?

From a theoretical point of view, I'd still say that it's reversible. Practically, cleavage (and re-building) of the complexes might be achieved by playing with the pH. Under acidic conditions, nitrogen atoms in the ligand would be protonated and the ligand thus cannot "bite".


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DFT optimization of a molecule in gaussian and find electrophilic sides
I am not a pro, but I have some experience with gaussian. For example, I can optimize a structure, calculate the frequency and rotational energies.
In the lab I made a nucleophilic substitution reaction where the Nucleophile $\ce{X}$ reacted with $\ce{R-\text{alkyl}-CH2-I}$ to form $\ce{R-\text{alkyl}-CH2-X}$ (iodide was eliminated). However, my molecule has on another position a benzylic bond "$\ce{R-Ph-CH2-O-R}$" as well and I noticed that a side reaction can occur at that position (depends on the nucleophile $\ce{X}$) where the nucleophile $\ce{X}$ reacted with "$\ce{R-Ph-CH2-O-R}$" to form "$\ce{R-Ph-X}$".
So there are two reactions: (1) The normal reaction with the alkyl $\ce{R-\text{alkyl}-CH2-I}$ to form $\ce{R-\text{alkyl}-CH2-X}$ and the side reaction (2) where the benzylic position in $\ce{R-Ph-CH2-O-R}$ is cleaved by the nucleophile $\ce{X}$ to form $\ce{R-Ph-X}$.
I am now trying to understand why the nucleophile $\ce{X}$ reacted with a benzylic $\ce{CH2}$ group at all even when its just a 15% conversion rather than the $\ce{R-\text{alkyl}-CH2-\text{Halogen}}$ group which is the favoured reaction. I think this could be illustrated by MOs, what do you think?
I already optimized the structure of my molecule and generated the formatted checkpoint file to display the molecular orbitals with GaussView. When I actually look at the HOMO and LUMO I don't get any useful information. At the benzylic position there is not really a high coefficient (no big MO) and at the $\ce{CH2}$ atom of "$\ce{R-CH2-I}$" there is even a smaller coefficient.
I had expected to see at least a big MO at the carbon atom of the alkyl $\ce{R-CH2-I}$ - where the nucleophilic reaction is favoured - but thats not the case.
So my question is, how can I analyze the reaction behaviour with DFT calculations, especially with Gaussian? I think calculating transition states is to much. It would be nice to just get some useful information about the molecules reactive sides (electrophilic sides where nucleophiles can react). How can I do that? Which calculation should I make?
By the way I used this B3LYP basis set:
b3lyp/6-31g(d,p) nosymm geom=connectivity
Thanks.

"So my question is, how can I analyze the reaction behaviour with DFT calculations, especially with Gaussian?"  -- To truly do this, you would have to map out the reaction pathway which involves characterizing transition states and computing the intrinsic reaction coordinate.  This would require a huge amount of effort than you would be willing to commit.  Try looking up how to compute "Electrostatic Potentials" in G09.  I think this is what you are looking for judging by your post.  What integration grid are you using? (don't say 'default').  Also Cartesian or spherical harmonic functions?

Have you tried to calculate Fukui function for your molecules? It may help, and it is relatively little effort.


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Why is there one octahedral and two tetrahedral positions for every anion in a close packed structure?
I'm reading about ionic crystal structures with planes of close-packed anions and the interstitial sites for cations. There are two types of interstitial sites: tetrahedral and octahedral. That makes perfect sense. What confuses me is this:

for each of these anion spheres, one octahedral and two tetrahedral positions exist.

Where does this come from? It would seem that for every new anion sphere added, there is a single new tetrahedral position and a single new octahedral position.

Do you know how many octahedral sites and tetrahedral sites is a single "sphere" in contact with, when arranged in a closed packed structure with 12 co-ordinating spheres?


text:
Comparision of donation power
There is an R an organic group connected to either Oxygen or Nitrogen. R is such a group which has almost zero electronic effects (Resonance effect and inductive effect). I want to ask if N or O are to donate a lone pair then which one will have higher donating capacity?
According to my opinion  it should be Oxygen as it is more electronegative than Nitrogen but I'm still not confident and want some proof considering other factors also and neither do I have an answer.

...According to my opinion it should be Oxygen as it is more electronegative than Nitrogen...

It's correct that oxygen is more electronegative than nitrogen. However, this makes an oxygen group a poorer electron donating group than the corresponding nitrogen group. Electronegativity means that the atom is less willing to give up the electrons. Nitrogen is more willing to give up (donate) its electrons because it is less electronegative.


text:
Are highly nitrated cyclopentadiene compounds possible?
Has anybody synthesized di- or higher nitrated cyclopentadienes possibly with methyl or other groups in the 1 position (analogs of TNT)?

Mono-nitrated cyclopentadienes and their benzo-analogues (indene, fluorene) have been prepared in the context of studies on nitro-aci-tautomerism.
As far as higher substitution is concerned, there seems to be at least a crystal structure of potassium dinitrotrichlorocyclopentadienide reported in Acta Cryst., B
In the abstract, the authors (Y. Otaka, F. Marumo, and Y. Saito) mention the preparation of this compound by Haga in 1969, (sorry, reference is behind paywall) via

nitration of hexachlorocyclopentadiene with alkali nitrite in an aqueous or ethanolic solution

NOTE: Hexachlorocyclopentadiene once was a favoured starting material in the synthesis of insecticides via Diels-Alder reaction. However, it is a potent teratogen and the "mother" of least five members of the dirty dozen (persistent organic pollutants) nowadays banned for good reason.
You might want to read the MSDS and then decide that you'd rather prefer to work with something else.

Thank you K.W. for the references.  No need by me for MSDS ;-) - not in the biz -- just a curiosity of mine in analogy to TNT!  I am sure that even if the more highly substituted CPD compounds could be made they would be horrendously explosive; maybe as stable as nitroglycerine itself!  Thanks again!  No reply needed.


text:
Which is better leaving group
How can I compare the leaving group ability of $\ce{Br^+}$ and $\ce{Cl^+}$ ions for the electrophilic substitution reactions for aromatic compounds?

I've never heard about halonium atoms as leaving groups in electrophilic aromatic substitutions. Would you mind to provide a sketch of a reaction you're thinking of?

Yes.This is not possible.$Br^+$ and$Cl^+$ cation cannot exist as free ions leaving the aromatic ring in electrophilic substitution

@KamleshPatel please substantiate your answers with sources, facts, reasonings…

I'm voting to close this question as off-topic because it is based on wrong premises.

http://onlinelibrary.wiley.com/doi/10.1002/hlca.19720550633/abstract

I think it would be nice to write sth about ipso electrophilic aromatic substitutions, not close this.

Could this possibly be referring to leaving group ability in terms of metal-halogen exchange ?

I presume you're referring to $\ce{Br-}$ and $\ce{Cl-}$ in aromatic nucleophilic substitution reactions. In that case, $\ce{Br-}$ is typically the superior leaving group to $\ce{Cl-}$, due to the former's larger size lending it greater polarizability (i.e., the ability to disperse the negative charge over a greater area). In general, when comparing elements in the same group, the dominant factor in determining anion stability is the larger size as one moves down the group. This largely accounts for the order of acidity of the halogen-containing binary acids (i.e., $\ce{HI} > \ce{HBr} > \ce{HCl}$), the greater acidity of thiols over alcohols, etc. Solvent effects are relevant, since smaller anions are more easily and thoroughly solvated in protic solvents. However, as far as I'm aware, this effect is rarely decisive, and $\ce{Br-}$ usually remains the better leaving group.
That said, the rate-determining step in nucleophilic aromatic substitution reactions is not the loss of the leaving group, but rather the formation of the negatively charged Meisenheimer complex resulting from the initial nucleophilic attack. This intermediate is extremely high in energy due to the temporary loss of aromaticity it entails. Thus, the reaction is largely controlled by kinetics, and therefore dependent upon the stability of the intermediates and transition states. $\ce{Cl-}$ is more electronegative than $\ce{Br-}$, and therefore better stabilizes the negative charge in the intermediate complex by induction ($-I$ effect). Consequently, we can expect the chlorobenzene to be more reactive in practice under most conditions, with the disparity in leaving group quality being less important. A cursory search yielded one reference confirming this order, but unfortunately I don't have access at the moment to more authoritative primary sources.
In summary, while $\ce{Br-}$ is the superior leaving group, its loss is not the rate-determining step. Hence, chlorobenzenes will in fact be more reactive due to chlorine's superior ability to inductively stabilize the intermediate anionic complex.

No.I meant $Br^+$ and $Cl^+$.

@DevgeetPatel, apologies, I came to the wrong conclusion. That said, I think your question requires clarification/rephrasing. The halogens do not form isolated cations under typical reaction conditions, and certainly not when acting as leaving groups. In electrophilic aromatic substitution reactions, the halogens react with Lewis acid catalysts to form highly reactive zwitterionic complexes. While these complexes may contain halogen atoms with positive formal charges, at no point do said atoms act as cationic leaving groups. Given that, it's unclear to me what you're really asking.

If possible(hypothetical case considered) which one will be better?


text:
Can two precipitates form in a precipitation reaction that starts with only two compounds in water?
Can two precipitates form in a precipitation reaction that starts with only two compounds in water?

Could you add a little bit of discussion why you think it should or should not be possible? Judging by the phrasing of your question I think you know reactions that form 1 precipitate in such a situation? How much do you know about the thermodynamics of these kind of reactions?

I think his question is: can two salts react to precipitate two new salts?

Actually, I'm confused by the title. Does the OP mean "only two molecules"? This means exactly two molecules. Or does the OP mean "only two compounds", with the amount of each compound open to interpretation?

I don't see why this should be closed - it seems a perfectly reasonable question to me

I suggested a long clarification because I understood it and I wasn't sure why other people didn't so I thought I should be thorough, but I just realized that it had already been edited and, based on Ben Norris's comment, was probably much more confusing originally. I suspect Gaurang Tandon's or whoever else's edits were enough and many people already understood the question.

The number of examples is considerably great.
Consider the following:
$$\ce{Ba(OH)2(aq) + MgSO4(aq)  -> BaSO4(s) + Mg(OH)2(s)} $$
Above reactants are reasonably soluble in water.
From Wikipedia, $\ce{Ba(OH)2} $ has a solubility of 1.67 g / 100 mL at 20 °C and $\ce{MgSO4} $ a solubility of 25.5 g / 100 mL at 20 °C

Yes is the answer to your question provided that reactant two molecules are salts.Two salts can react with eact other and form two new stable salts depending on their $\mathrm{K}_{sp}$ values.


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How do I create a concentrated solution of singlet oxygen (in the dark)?
Without appealing to reactions that require photons / light, how do chemists achieve a very high concentration of continually produced, though quite short lived, singlet oxygen in aqueous solution (i.e. $\ce{{}^1O_2}$ where dioxygen is typically in the triplet $\ce{{}^3O_2}$ state)?  I suppose I'll also ask the same question if we have the freedom to select an organic solvent of choice for the buffer?  By "high concentration" I mean at concentration from $\approx 10~\mathrm{\mu{}M} - 100(+)~\mathrm{\mu{}M}$.
My guess would be that should be a way to do this starting with a solution containing some amount of $\ce{H_2O_2}$, but I'm not quite sure what a reasonable process might be.
Here's the wikipedia link for singlet oxygen ($\ce{{}^3O_2}$)
This reaction should happen at or around room temperature.
The reason I'm specifying that I'd like to do this in the dark, and to have the singlet oxygen continually produced at room temperature (given its short lifetime, which is $\leq 100~\mathrm{\mu{}s}$ under any conditions I've heard of), is because I'm curious if I could easily set up some buffer that lets me dose a sample with singlet oxygen on a lab bench with a pipette.  Given the short lifetime of singlet oxygen, having to use a continuous radiation source presents difficulties that, while surmountable, lead me to ask this question.
Provided the above, I'd like to avoid reactions that vent gases like $\ce{Cl_2}$ and require fume hoods in the sense of "... well, if you value your life ..." vs. "... OSHA says so ...".

Potassium ferricyanide (and other ion complexes, like hemoglobin) will decompose hydrogen peroxide into water and (presumably) singlet oxygen. This process is necessary for the luminol reaction.
$$\ce{2H2O2 \overset{Fe}{->} 2H2O + ^1O2}$$

Thanks, do you know what temperature is required for this reaction?  Also, how do we know that singlet oxygen is being produced?

I'd always go by the old photochemical Schenck method, but since you insist:
The thermal decomposition of

Phosphonite ozonides or
9,10-diphenylanthracene endoperoxide

yield $\ce{^1O2}$.
Technically, $\ce{^1O2}$, may also be generated at low oxygen pressure (i a vacuum line) by microwave radiation. (Source)
As far as aqueous solutions are concerned, one should keep in mind that the lifetime of singlet oxygen ($^1\Delta$) is around 2 µs.
For detection, one typically relies on the luminescence at $\lambda$ = 1270 nm.

I think this is getting close to what I'm looking for... what temperatures are required?  I'm looking for something that happens at or around room temperature (e.g. $25C$)?

I would have to look that up.

Unraveling a stoichiometric organic adduct in situ is the way to go.  However,
$\ce{H2O2 + NaOCl → O2(a1Δg) + NaCl + H2O}$
It is one thing to mix streams of supermarket $\ce{H2O2}$ and laundry bleach to get a dimer nice red glow in a darkened room.  It is quite another thing to mix higher concentrations of each.  Concentrated $\ce{H2O2}$ is exceptionally dangerous on skin!  It diffuses in until it hits living tissue with peroxidase.  It then pops into subcutaneous gas bubbles attended with remarkable pain.  Concentrated hypochlorite is evil, including conproportionation with chloride to chlorine gas.
Full safety equipment: face shield, gloves, apron - and remember to shield your neck.  In a running fume hood with the door down.  Get literature references before you even think about doing it.  A hefty dual barrel syringe pump feeding a static mixer might barely pull it off. Keep the reacting volume small as you rocket through the juices.  You might take singlet oxygen somewhere outside the reaction zone.
100 mph = 4.47 mm/100 microsecond half-life

This sounds awesome.   Are you aware of any literature references for the kinetics of singlet oxygen production over time (e.g. as monitored by absorbance @ 1268 nm) as a function of $\ce{H2O2}$ and $\ce{NaOCl}$?

I'm slightly worried though, since it seems like this would also yield chlorine gas?  Consider that $\ce{H2O2 + 2NaOCl -> 2NaOH + O2 + Cl2}$?

Hypochlorite is already ~pH 13, then Le Chatelier.  Any generated chlorine will disproportionate in strong alkali.  Do it with ventilation.

In my experience, concentrated hydrogen peroxide is painful on skin, but leave no chemical burns, and whitened skin returns to normal quite soon. On the other hand, 30% hydrogen peroxide may decompose with high heat output if a proper catalyst (manganese or iron salt) is poured into it, and highly concentrated hydrogen peroxide (>50%) may be explosive.

There is an excellent article on singlet oxygen on Wikipedia. Singlet oxygen can be generated by the action of light on Rose Bengal dye. The brighter the light, the more generated. It is reported to have a half-life of microseconds in solvents and 72 minutes in air at ambient temperature.


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What is the (conceptual) difference between binding energy and adsorption energy?
I know the equations, but I have some difficulty getting the difference between the two concepts of binding energy and adsorption energy.

For solids, the binding energy is what makes the crystal (if you will) hold together. It is what makes the individual constituents of the compound bind together.
Adsorption however is an entirely different process: Here, some molecules interact with the surface of another compound. Since (strictly speaking) this is no chemistry but mere physics (there is no formation or breaking of bonds) there are other interactions at play, mainly van der Waals-type interactions.
As these are weaker (it's easier to wipe off the water from your windshield than rip out a silicon atom) and aren't based on the same principles as the binding energy, they get a different name.
I hope that cleared things up for you.

Chemisorption is adsorption where an actual covalent bond is formed. Just to throw that out there.

I have the same question. Though the answer from tschoppi is only partially correct. Chemists have divided adsorption processes on at least two classes: physisorbtion and chemisorbtion.  The tschoppi's answer is relevant for physisorption, though it does not answers the question. Exampple of chemisorbtion: carbon monoxide on platinum. In this case CO is realy chemically bonded to top platinum atoms. The difference between the adsorbtion energy and the binding energy depends on the person you ask. It is a matter of definition. I would call binding energy is the enrgy of a chemical bond between a particular adsorbed molecule and a particlar atom of the surface. This energy should be weakly influence by the neightbouring adsorbates. However, adsorbate do interact. For strongly covered surface the energy required to pull out a molecul is less then for an almost clean surface. This energy, I believe, is the adsorbtion energy, which is coverage dependednt. At low coverages the adsorbtion energy and binding energy are the same. That is the difference is due to the intercation of adsorbates with each other.


text:
Does isotretinoin affects contraception hormonal pills - Ethinylestradiolum, Drospirenonum
Does isotretinoin affects contraception hormonal pills - Ethinylestradiolum, Drospirenonum? Asking as a pharmacologist not a patient.

My professional knowledge of vitamin A derivatives is limited to preparative aspects and some photochemical properties.
But it seems that the German Bundesinstitut für Arzneimittel und Medizinprodukte (BfArM) provides some information in german and english.
I didn't know whether a link to the search would survive the session, so the link above is just to their website and you would need to search yourself in case you're interested.

Note: I had a look at the Fachinformationen (RTF in German) among the documents "linked".
In this publication, it is very explicitly pointed out that Isotretinoin is a TERATOGEN and efficient contraception therefore is absolutely vital (in the truest sense of the word).
It is suggested that patients use two independent methods of contraception, of which one should be a "barrier method", e.g. the use of condoms.

In the link you provided I have found: http://www.bfarm.de/SharedDocs/Downloads/EN/Drugs/vigilance/PSURs/csp/g-j/isotretinoin.pdf?__blob=publicationFile&v=3
There is no mentioning of interaction between hormonal contraception and isotretinoin.
I found(link at the BOTTOM) "Isotretinoin may reduce the efficacy of these medicines in some women, and they are not 100% reliable. Subcutaneous hormone implants or Mirena® may also be considered." but this is very inexact http://www.dermnetnz.org/treatments/isotretinoin.html

Could you have a look on that: https://www.ncbi.nlm.nih.gov/pubmed/15116059

Please note that I'm not a pharmacologist, I only did some therapeutic drug monitoring in the last millenium. Results and conclusion sound to me like "[...] actually, our data are a bit shaky and we can't really prove a correlation", but better safe than sorry. The "Fachinformationen" (maybe you can run it through a translator) comes to similar conclusions: We're not certain if there's a pharmacological interaction, but if something goes wrong, the sh*t certainly hits the fan.