What is the pH of a 0.100 M solution of NaHCO3 at 25∘C, given that the Ka of H2CO3 is 4.3×10−7?

Question

 What is the pH of a 0.100 M solution of NaHCO3 at 25∘C, given that the Ka of H2CO3 is 4.3×10−7?

Answer 1

H2CO3 <==> H+ + HCO3^- Ka = 4.3x10^-7

HCO3^- + H2O ==> H2CO3 + OH^- Kb = 1x10^-14/4.3x10^-7 = 2.3x10^-8

Kb = 2.3x10^-8 = [H2CO3][OH^-]/[HCO3^-] = (x)(x)/0.100

x² = 2.3x10^-9

x = 4.8x10^-5 = [OH^-]

pOH = 4.3

pH = 9.7