Question

What is the pH in a titration of 125 mL of 0.45 F sulfurous acid (H2SO3) with 1.23 M solution of sodium hydroxide after 15 mL of base have been added?

Answer 1

H2SO3 + 2NaOH --> Na2SO3 + 2H2O

Mmoles of H2SO3 = 125mL*0.45M = 56.25mmoles

Mmoles of NaOH = 15mL*1.23M = 18.45mmoles

ICE table-

H2SO3 + 2NaOH --> Na2SO3 + 2H2O

Initial:56.25. 36.9. 0 . 0

Change: -36.9 . -36.9. +36.9 . +73.8

Equilbrm: 19.35 . 0 36.9. 73.8

So, after adding 15mL of NaOH-

Mmoles of H2SO3 = 19.35mmoles

Mmoles of Na2SO3= 36.9mmoles

Volume = 125mL + 15mL = 140mL

So, [H2SO3] = 19.35mmoles/140mL = 0.138M

[Na2SO3] = 36.9mmoles/140mL = 0.264M

Since an acid and its conjugate base are in equilibrium we will use Henderson-Hasslebalch equation to find the pH

Ka for H2SO3 = 1.54*10^-2

pKa = -log(Ka) = 1.81

pH = pKa + log[Na2SO3/H2SO3]

pH = 1.81 + log[0.264/0.138]

pH = 1.81 +0.282 = 2.09