pH at the equivalence point

by gkangelexa
Tags: equivalence, point
 P: 81 My book says that you cannot use the Henderson-Hasselbalch equation to find the pH at the equivalence point when titrating a weak acid with a strong base.... I was wondering why not? it says that instead you must use the Kb of the conjugate base and that we find that from the Kw and the Ka. Why? and how do we find the Ka?
 P: 5,462 The H-H equation is derived for weak acids/bases and buffers on the assumption that [A-] >> [OH-] Which is true for the initial solution of weak acid. So you can use H-H to calculate the intitial pH. However once you start adding strong alkali you are adding (lots of) [OH-] so the assumption no longer holds. There are, in fact, four different distinct calculations needed to derive the neutralisation curve for a strong base v a weak acid.
P: 21,727
 Quote by Studiot The H-H equation is derived for weak acids/bases and buffers on the assumption that [A-] >> [OH-]
Care to elaborate? As far as I can tell there is no such assumption. Henderson-Hasselbalch equation is not different from acid dissociation reaction quotient definition, it is just rearranged. And it holds always, as long as you put in equilibrium concentrations of [A-] and [HA].

 Which is true for the initial solution of weak acid. So you can use H-H to calculate the intitial pH.
Sorry, but you can't. Please show how you are going to use H-H equation:

$$pH = pK_a + \log \frac {[A^-]}{[HA]}$$

to calculate pH of a pure acetic acid. Say, 0.1M.

Edit: to clarify, by "pure, 0.1M acetic acid" I mean 0.1M solution of acetic acid that doesn't contain anything but water and acetic acid.

P: 5,462

pH at the equivalence point

So why did the OP's book say the same as mine about the H-H equation?
P: 21,727
 Quote by Studiot So why did the OP's book say the same as mine about the H-H equation?
It is true that you can't use HH equation to calculate pH at equivalence point, this statement is correct. But you can't use it to calculate initial pH either, for exactly the same reasons.

Please show how is the HH equation derived using assumption that you listed, and please explain how you are going to use HH equation to calculate pH of the pure weak acid solution.
 P: 5,462 Since this is a working day and it will take me some while to produce the derivation of the equations I will defer that exercise until this evening. In the meantime, since you apparently disagree with my explanation of the original question and since I was the only respondent how about posting your explanation?
 P: 5,462 As promised here is a derivation of the H-H equation, establishing why the OH- concentration is crucial and using it to calculate the initial pH of ).1M acetic acid as requested. The calculation is compared with another method of achieving the pH. This method will only work because at the outset the only source of OH- is the water and this is 9 orders of magnitude less than the H+ from the acetic acid. The formula can be adapted for the equilibrium point on titrating with 0.1M sodium hydroxide. as follows pH = 0.5pKw + 0.5pKa + 0.5logC pH = 7 + 2.37 + (-0.65) = 8.72 where C is the acetic acid initial concentration. Attached Thumbnails
P: 21,727
 Quote by Studiot As promised here is a derivation of the H-H equation, establishing why the OH- concentration is crucial and using it to calculate the initial pH of ).1M acetic acid as requested. The calculation is compared with another method of achieving the pH. This method will only work because at the outset the only source of OH- is the water and this is 9 orders of magnitude less than the H+ from the acetic acid. The formula can be adapted for the equilibrium point on titrating with 0.1M sodium hydroxide. as follows pH = 0.5pKw + 0.5pKa + 0.5logC pH = 7 + 2.37 + (-0.65) = 8.72 where C is the acetic acid initial concentration.
Sorry, but - first of all - equation you derived:

$$[H^+] = \sqrt{[HA] K_a}$$

or

$$pH = \frac 1 2 pK_a + \frac 1 2 \log{[HA]}$$

is not a Henderson-Hasselbalch equation. Henderson-Hasselbalch equation is the one used for calculation of a buffer pH:

$$pH = pK_a + \log \frac {[A^-]}{[HA]}$$

See for example:

http://www.biology.arizona.edu/bioch...ets/ph/HH.html
http://www.wiley.com/college/pratt/0..._equation.html
http://www.chemteam.info/AcidBase/HH-Equation.html
http://people.rit.edu/pac8612/webion.../ione89wu.html
http://www.lsbu.ac.uk/biology/biolchem/acids.html

and so on.

Now to the derivation:

As far as I know equation you listed has no name, and I am afraid there are several problems with your derivation and calculations.

Your final equation (bottom of the first page) uses [HA] which you used to denote equilibrium concentration of HA (undissociated acid). You can't use it to calculate pH, as you don't know it. What you usually know is the analytical concentration Ca and that's the information your final result should depend on. Ca should be introduced into the problem through the mass balance of the acid:

$$C_a = [HA] + [A^-]$$

You calculated correct pH value only because you entered analytical concentration of the acid where your equation shows equilibrium concentration of HA. This can be done after using another crucial approximation, that you have not mentioned - if the acid is not dissociated more than 5% we can assume:

$$[HA] \approx C_a$$

$$[H^+] = \sqrt{C_a K_a}$$