Question

A 25.0 mL sample of 0.30 M HCOOH is titrated with 0.20 M KOH. What is the pH of the solution after 25.0 mL of KOH has been added to the acid? Ka = 1.8x10^-4

Answer 1

HCOOH is a weak acid which undergoes partial dissociation as follows:
HCOOH(aq) ↔ H⁺(aq) + HCOO^-(aq)

When KOH is added, the OH- from KOH react with HCOOH as follows:
HCOOH(aq) + OH^-(aq) --> HCOO^-(aq)+H₂O(l)

No of moles of HCOOH
= [HCOOH] x (volume of HCOOH in L)
= 0.30 x 25.0 x 10^-3
= 7.5 x 10^-3 moles

No of moles of OH^- = [KOH] x (volume of KOH in L)
= 0.20 x 25.0 x 10^-3
= 5.00 x 10^-3

The required ratio of the no of moles of HCOOH to OH^- is 1:1, but since there is more HCOOH than required, OH- is the limiting reagent, implying that all of it will react.

No of moles of HCOOH remaining after the reaction
= (7.5 - 5.00) x 10^-3
= 2.5 x 10^-3

[HCOOH] after the reaction
= no of moles of HCOOH after the reaction/volume of the solution after the reaction in L
= (2.5 x 10^-3) / ((25 + 25) x 10^-3)
= 0.05 M

No of moles of HCOO^- formed after KOH was added
= no of moles of OH^- that reacted
= 5.00 x 10^-3

[HCOO^-] after KOH was added
= no of moles of HCOO^-/volume of the solution in L
= (5.00 x 10^-3) / ((25 + 25) x 10^-3)
= 0.100 M

K_a= [H⁺][HCOO^-]/HCOOH]
[H⁺] after the addition of KOH
= (K_a)([HCOOH])/([HCOO^-])
= (1.8 x 10^-4)(0.05)/(0.100)
= 9.00 x 10^-5 M

pH = -log[H+]
= -log(9.00 x 10^-5)
= 4.04