A 29.2 mL aliquot of carbonic acid that has a concentration of 0.637 M will be titrated with 0.358 M NaOH. Calculate the pH of the solution upon the addition of 8.52 mL of NaOH.

Question

The Ka of the acid at the temperature of the experiment is 4.95 x 10^-7

Answer 1

Carbonic acid = H2CO3

H2CO3 + 2NaOH ==> Na2CO3 + H2O

moles H2CO3 present = 29.2 ml x 1 L / 1000 mls x 0.637 mol/L = 0.0186 mols

moles NaOH added = 8.52 ml x 1 L / 1000 mls x 0.358 mol/L = 0.00305 mols

mols H2CO3 neutralized = 0.00305 mol NaOH x 1 mol H2CO3 / 2 mol NaOH = 0.00153 mols

mols H2CO3 remaining = 0.0186 mol - 0.00153 mol = 0.01707 mols H2CO3 remaining

Final volume = 29.2 mls + 8.52 mls = 37.72 mls = 0.03772 L

[H2CO3] = 0.01707 mols / 0.03772 L = 0.453 M

H2CO3 <==> H⁺ + HCO₃^-

Ka = 4.95x10^-7 = [H⁺][HCO₃^-] / [H₂CO₃]

4.95x10^-7 = (x)(x) / 0.453

x² = 2.24x10^-7

x = [H⁺] = 4.74x10^-4

pH = -log [H⁺] = 3.32