Acids & Base Chemistry Math Question

Question

A solution of carbonic acid has its pH adjusted to 8.19 by addition of strong base. What is the equilibrium concentration ratio of carbonate ion to carbonic acid in the final solution? The acid dissociation constants for carbonic acid are pK₁=6.35, pK₂=10.33

[CO₃^2-]/[H₂CO₃]=

Answer 1

Carbonic acid has 2 titratable protons therefore it can dissociate twice:

H₂CO₃ <=> H⁺ + HCO₃^- <=> H⁺ + CO₃^2-

1) Begin by finding the ratio of H₂CO₃ to HCO₃. Use Henderson-Hasselbalch equation and appropriate pKa.

pH = pKa + log [base]/[acid]

8.19 = 6.35 + log [HCO₃]/[H₂CO₃]

1.84 = log [HCO₃]/[H₂CO₃]

10^1.84 = [HCO₃]/[H₂CO₃]

[HCO₃]/[H₂CO₃] = 69.18

2) Use the same calculations to find the ratio of HCO₃^- to CO₃^2-.

pH = pKa + log [base]/[acid]

8.19 = 10.33 + log [CO₃^2-]/[HCO₃^-]

-2.14 = log [CO₃^2-]/[HCO₃^-]

10^-2.14 = [CO₃^2-]/[HCO₃^-]

[CO₃^2-]/[HCO₃^-] = 0.0072

3) Now that you have both ratios you need to combine them to get the ratio of CO₃^2- to H₂CO₃.

Since [HCO₃]/[H₂CO₃] = 69.18

We can solve for [HCO₃] = 69.18 * [H₂CO₃]

Then we can plug it in to the other ratio

[CO₃^2-] / 69.18 * [H₂CO₃] = 0.0072

[CO₃^2-] / [H₂CO₃] = 0.0072 * 69.18

[CO₃^2-] / [H₂CO₃] = 0.5