Question

Calculate the pH of the resulting solution when 50.0ml of .350 M H2CO3 is added to 25.0mL of .400 M KOH

Answer 1

pKa1 = 6.37

pKa2 = 10.33

Ka1 = 10^-pKa1 = 4.266 x 10^-7

Ka2 = 10^-pKa2 = 4.677 x 10^-11

Since Ka1 >> Ka2, pH will be governed by H₂CO₃ dissociation.

Moles of H₂CO₃ added = V1 * M1

= 50 * 0.35 = 17.5 mMoles

Moles of KOH added = V2 * M2

= 25 * 0.4 = 10 mMoles

Total volume of solution, V = 50 + 25 mL = 75 mL

H₂CO₃ <--> H+ + HCO₃^-

(17.5 – x) <--> x + x

After 10 mMoles of base addition:

mMoles of H₂CO₃ in solution = 17.5 – 10 – x

mMoles of H+ in solution = x

mMoles of HCO₃^- in solution = 10 + x

[H₂CO₃] = (17.5 – 10 – x) / V

[H+] = x / V

[HCO₃^-] = (10 + x) / V

Ka1 = [H+] [HCO₃-] / [H₂CO₃]

x * (10 + x) / (V * (17.5 – 10 – x)) = 4.266 x 10^-7

x * (10 + x) / (75 * (17.5 – 10 – x)) = 4.266 x 10^-7

Assume x << 1

Solving we get,

x = 2.4 x 10^-5 mMoles

[H+] = x /V = 3.2 x 10^-7 M

pH = -log(3.2 x 10^-7)

= 6.49