Question

An acid with a single acidic proton has a pKa of 11.6. An aqueous solution of the acid is prepared, with the pH adjusted by the addition of strong acid or base. The acid is found to be 36 % ionized in this solution. What is the pH of the solution? Answer to two decimal places. Just provide a number as your answer, without pH in front.

Answer 1

Let the acid is HA

It will dissociate as with % dissociation = 36%

LEt the inital concentration = [HA]

                                  HA            -->           H+         +    A-

Inital concentration       a    0 0

Change                      -0.36a                      +0.36a                      0.36a

At equilibriub            a-0.36a                      0.36a                        0.36a

The Ka = [H+] [ A-] / {HA]

pKa = 11.6 = -logKa

So Ka = -antilog of pKa

2.51 X 10^-12 = Ka = 0.36a X 0.36a / (a-0.36a)

we can ignore 0.36a in denominator as Ka is very samll

So, 2.51 X 10^-12 = 0.36a² / a

2.51 X 10^-12 = 0.36a

So , a = 0.069 X 10^-12

so [H+] = 0.36 X a = 0.069 X 10^-12 X 0.36 = 0.025 X 10^-12

now the total concentration of H+ in the solution will be due to acid dissociation and as well as water

[H+] due to water = 10^-7

So total concentration of [H+] = 2.5 X 10^-10 + 10^-7

[H+] = 10^-7 [2.5 X 10^-3 + 1] = 10^-7 X 1.0025

pH = -log[H+] = 6.99