What is the pH in a titration of 125 mL of 0.45 F sulfurous acid (H2SO3)...
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?
Homework Answers
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
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