They say it's possible to find the growth-rate ie the second derivative of a titration curve on wikipedia. I am interested in finding & deriving such a formula, because I need to know why the growth-rate is so small in the beginning of the titration but it gets so much bigger as you add larger volumes of the titrant.. Shouldn't it be the opposite, if it isn't linear?
I don't get it, if a strong acid is titrated gradually by a strong base, shouldn't the pH growth-rate until the equivalence point be constant? Ie the function should be linear until the equivalence point?
If I would guess the reason, would it be because the water is kind of "resistant" to titration? Let's take the picture below as an example: there are kind of "reserves" of un-protolyzed acids which act as a buffer vs the titrating base?
[PLAIN]http://www.files.chem.vt.edu/chem-ed/titration/graphics/titration-strong-acid-35ml.gif[/QUOTE]
The curve represents the pH and in turn the pH is given by the -log[H+]
So you would expect the curve to follow a logarithmic curve as the hydrogen ion concentration falls by an even amount.
BUT the hydrogen ion concentration does not fall by an even amount as the volume of the mixture is constantly increasing while the molar amount of hydrogen is constantly decreasing.
The result of the three factors above is the complex curve shown. The best thing to do is calculate a few examples:
For the addition of NaOH (0.1M) to 25 ml HCl (0.1M)
Initially there is only 0.1M HCl so the pH = 1
Add 1ml of NaOH
-----------------
Moles of NaOH added = 0.001 x 0.1 = 0.0001
Moles of Hydrogen ions used up = 0.0001
Moles of H+ remaining = (0.025 x 0.1) - 0.0001 = 0.0024
New volume = 0.026 ml
New [H+] = 0.0024/0.026 = 0.0923
pH = 1.035
So, addition of 1 ml increases the pH by only 0.035 units
Add another 9 ml NaOH (total 10 ml)
----------------------------------
Moles of NaOH added = 0.01 x 0.1 = 0.001
Moles of Hydrogen ions used up = 0.001
Moles of H+ remaining = (0.025 x 0.1) - 0.001 = 0.0015
New volume = 0.035 ml
New [H+] = 0.0015/0.035 = 0.0429
pH = 1.37
So, addition of 10 ml total NaOH increases the pH by only 0.37 units
Add another 10 ml NaOH (total 20 ml)
----------------------------------
Moles of NaOH added = 0.02 x 0.1 = 0.002
Moles of Hydrogen ions used up = 0.002
Moles of H+ remaining = (0.025 x 0.1) - 0.002 = 0.0005
New volume = 0.045 ml
New [H+] = 0.0005/0.045 = 0.0111
pH = 1.95
So, addition of 20 ml total NaOH increases the pH by only 0.95 units
Add another 4 ml NaOH (total 24 ml)
----------------------------------
Moles of NaOH added = 0.024 x 0.1 = 0.0024
Moles of Hydrogen ions used up = 0.0024
Moles of H+ remaining = (0.025 x 0.1) - 0.0024 = 0.0001
New volume = 0.049 ml
New [H+] = 0.0001/0.049 = 0.00204
pH = 2.69
So, addition of 24 ml total NaOH increases the pH by only 1.69 units
But notice that the last 4 ml increased the pH by more than the prevous 10 ml
Add another 0.9 ml NaOH (total 24.9 ml)
----------------------------------
Moles of NaOH added = 0.0249 x 0.1 = 0.00249
Moles of Hydrogen ions used up = 0.00249
Moles of H+ remaining = (0.025 x 0.1) - 0.00249 = 0.00001
New volume = 0.0499 ml
New [H+] = 0.00001/0.0499 = 0.0002
pH = 3.70
So, addition of 24.9 ml total NaOH increases the pH by 2.7 units
But, the last 0.9ml changed the pH by over 2 pH units.
Now at neutralisation the pH is 7 so an addition of a further 0.1ml is going to change the pH from 3.7 to 7.0.
After the neutralisation point the pH is calculated by finding the excess [OH-] and using kw to get pH.
The result is that the curve is almost symmetrical, but not quite so as the volume keeps increasing.
