Question

A solution is made by mixing 400 mL of 0.156 M KOH with 0.53 L of 0.095 M HCOOH. Calculate the equilibrium concentrationsof H^+, HCOOH, HCOO^-, OH^- and Na^+ in solution. Show work.

Answer 1

mmol of KOH added= M*V= 0.156 M * 400 mL = 62.4 mmol
mmol of HCOOH added= M*V= 0.095 M * 530 mL = 50.35 mmol

KOH + HCOOH ----> HCOOK + H2O
50.35 mmol of each will react with each other to form 50.35 mmol of HCOOK and 50.35 mol of H2O

mmol of KOH remaining = 62.4-50.35 = 12.05 mmol
Total volume = 400 mL + 530 mL = 930 mL
[K+] = [OH-] =number of moles of KOH/volume of water
= 12.05 mmol/930 mL
=0.013 M

[HCOOH] = 0 M

[H+] = 10^-14/[OH-]
= 10^-14 / (0.013)
= 7.69*10^-13 M

HCCO- will not exist as it is. It will exist as HCOOK
[HCOOK] = number of moles of HCOOK /volume of solution
= 50.35 mmol / 930 mL
=0.054 M

How come Na+ came. It is not even there in solution