Question

Consider a titration of 100-mL of .25 M formic acid (Ka-1.8 x 10^-4) with .2 M KOH

What is the pH after 100 mL of base has been added?

Answer 1

Given:
M(HCOOH) = 0.25 M
V(HCOOH) = 100 mL
M(KOH) = 0.2 M
V(KOH) = 100 mL


mol(HCOOH) = M(HCOOH) * V(HCOOH)
mol(HCOOH) = 0.25 M * 100 mL = 25 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.2 M * 100 mL = 20 mmol


We have:
mol(HCOOH) = 25 mmol
mol(KOH) = 20 mmol

20 mmol of both will react

excess HCOOH remaining = 5 mmol
Volume of Solution = 100 + 100 = 200 mL
[HCOOH] = 5 mmol/200 mL = 0.025M

[HCOO-] = 20/200 = 0.1M

They form acidic buffer
acid is HCOOH
conjugate base is HCOO-


Ka = 1.8*10^-4

pKa = - log (Ka)
= - log(1.8*10^-4)
= 3.745

use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.745+ log {0.1/2.5*10^-2}
= 4.35


Answer: 4.35