pH headache?

Question

A 0.20 M sodium chlorobenzoate (NaC7H4ClO2) solution has a pH of 8.65. Calculate the pH of a 0.13 M chlorobenzoic acid (HC7H4ClO2) solution.

Answer 1

Let's substitute the group C7H4ClO2 with A for the sake of simplicity

Rationale of the solution:

NaA is the salt of the weak acid HA with the strong base NaOH. Therefore it will hydrolyse and it will have Kb=Kw/Ka.From there we will find Ka and then we will use that to find the pH of the other solution.

In detail:

.. .. .. .. .. .. A- +H2O<=> HA + OH-
Initial .. .. .. C
React .. .. . x
Produce .. .. .. .. .. .. .. .. .. x .. .. x
At Equil .. C-x .. .. .. .. .. .. .x .. .. x

[H2O] is considered constant in such cases so it is not included in the equilibrium constant expressions and the ICE table.

Kb= [HA][OH-]/[A-] and since Kb=Kw/Ka we have
Kw/Ka= x^2/(C-x) =>
Ka= Kw(C-x)/x^2
but C=0.2 and x=[OH-] = 10^-pOH= 10^-(14-pH)= 10^-(14-8.65)= 10^-5.35
So Ka= (10^-14)*(0.2-10^-5.35)/(10^-5.35)^2 = 10^-4

.. .. .. .. .. .. HA <=> H+ + A-
Initial .. .. .. .0.13
Dissociate .. x
Produce .. .. .. .. .. .. x .. .. x
At equil .. 0.13-x .. .. x .. .. x

Ka= [H+][A-]/[HA] = x^2/(0.13-x)

Let's assume that 0.13 >> x so that 0.13-x=0.13
Then the equation is simplified to x^2/0.13= Ka =>
x= squareroot( 0.13*Ka) =SQRT(0.13*10^-4) = 3.6*10^-3

pH= -log(x) =-log(3.6*10^-3) = 2.44

To be honest 3.60*10^-3 marginally fits our assumption.
If we decided that the assumption is not valid then we would solve the quadratic (x^2+Ka-Ka*0.13=0) and get x=3.56 *10^-3 and pH= 2.45

Answer 2

Hey, your question is just like my question! Except I use HA instead of HC7H3ClO2. If I get my answer then I can answer your question.