Question

25.000 mL of formic acid (HCHO2) is titrated to its equivalence point with 20.13 mL of 0.1192 M potassium hydroxide. Calculate the pH of the mixture at the acid’s equivalence point.

Answer 1

HCOOH+ NaOH----> HCOONa + H2O

1 mole of HCOOH requires 1mole of NaOH

Moles of HCOO- ( from HCOONa)= Volume(L)* Molarity of NaOH =(20.13/1000)*0.1192=0.002399

volume after reaction = 25 (HCOOH)+ 20.13 (NaOH)= 45.13 ml = 45.13/1000 = 0.04513L

concentration of HCOO- =0.002399/0.04513= 0.053169

HCOO- + H2O --------->HCOOH+ OH-

Kb= [HCOOH] [OH-]/ [HCOO-]

Ka for HCOOH= 1.8*10-4 and Kb= 10^-14/(1.8*10^-4)= 5.555*10-11

let x= drop in concentration of HCOO- to reach equilibrium

At equilibrium, [OH-] = [HCOOH] =x

x²/(0.053169-x)= 5.555*10^-11

when solved using excel, x= 1.72*10^-6

[OH-] =1.72*10-6, pOH= -log[OH-] = 5.76, pH= 14-5.76= 8.24