How much 0.1M NaOH must be added to 45.0 mL of 0.1M HOAc to give a solution whose pH is 4.30?

Question

How much 0.1 M NaOH must be added to 45.0 mL of 0.1 M HOAc (Ka for acetic acid = 1.8 x 10^-5) to give a solution whose pH is 4.30?

Answer 1

let acetic acid = HAc

OH- + HAc = Ac- + H2O

Initially when NaOH is added.

V is in L

mol Ac- = 0

mol NaOH = 0.1 V mol in (0.045+V) L

mol HAc = 0.0045 mol in (0.045+V) L

At equilibrium

mol NaOH = 0.1 V - 0.1 V = 0

mol HAc = 0.0045 - 0.1 V in 0.045+V L

mol Ac- = 0.1 V in 0.045+V L

[H+] = 10^-4.30 = 5.01 * 10^-5 M

Ka = 1.8 x 10^-5 M

Ka = [H+] . [salt] / [acid]

1.8 x 10^-5 = 5.01 * 10^-5 .  0.1 V / 0.0045 - 0.1 V

0.000000081 - 0.0000018V = (5.01 *  10^-6)V

8.1*10^-8 - 1.8*10^-6V = 5.01*10^-6V

(5.01+1.8)*10^-6V = 8.1*10^-8

6.81*10^-6 V = 8.1*10^-8

V = 0.01189 L = 11.89 L