Question

For the solution of 0.020 moles of formic acid (HCOOH) and 0.010 moles of sodium formate (pka = 3.7545 at 25 C) obtain:
a) the reactions involved in the process

b) the derivation of the Henderson-Hasselbalch expression from equilibrium for the dissolution of the formic acid.

c) the pH of the system and the pOH

d) the concentration of H₃O⁺

e) the concentration of OH^- of the system

Answer 1

a) Reactions involved

0.01 HCOONa + 0.01 H₂O --> 0.01 HCOOH + OH^- + Na⁺

0.01 HCOOH + 0.01 OH^- + 0.01 Na⁺ + 0.02 HCOOH + 0.03 H₂O --> 0.03 H₃O⁺ + 0.03 HCOO^- + 0.01 OH^- + 0.01 Na⁺

b) Henderson-Hasselbalch expression for dissolution of Formic Acid

Reaction involved in dissolution of Formic Acid is:

HCOOH + H₂O --> H₃O⁺ + HCOO^-

Ka = { [H₃O⁺] [HCOO^-] } / [HCOOH]

[H₃O⁺] = Ka [HCOOH] / [HCOO^-]

Taking negative log on both sides

- log [H₃O⁺] = - log Ka - log { [HCOOH] / [HCOO^-] }

pH = - log Ka - log { [HCOOH] / [HCOO^-] }

pH = pKa + log { [HCOO^-] / [HCOOH] }

c) pH of the system = 3.7545 + log {0.03/0.02} = 3.7545 + log (1.5) = 3.7545 + 0.176 = 3.9305

d) The concentration of H3O⁺ (as from the balanced chemical equation) = 0.03 moles/volume (the volume of the system is not defined in the question)

e) Concentration of OH^- (as from the balanced chemical equation) = 0.01 moles/volume (the volume of the system is not defined in the question)