the titration of 261 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) with 1.1 M KOH.

Question

How many mL of the 1.1 M KOH are needed to raise the pH of the carbonic acid solution to a pH of 6.942?

Answer 1

Carbonic is a weak diacid and it reacts with a strong base, KOH. When H2CO3 is titrated by KOH, the curve will show 2 equivalence points hence the 2 Ka values.

H₂CO₃ + KOH → KHCO₃ + H₂O (Ka = 4.3 x 10^-7) H₂CO_{3 -->} HCO₃^-

KHCO₃ + KOH → K₂CO₃ + H₂O (Ka = 5.6x10^-11) HCO₃^- _--> CO₃^2-

Use the first Ka value to calculate the ratio of [HCO3-]/[H2CO3] at pH 5.935 with the Henderson-Hasselbalch equation:

pKa = -log Ka

-log (4.3 x10^-7)= 6.366

pH = pKa + log [HCO3-]/[H2CO3]

6.942 = 6.366 + log [HCO3-]/[H2CO3]

[HCO3-]/[H2CO3] = 3.767

In other words: [HCO3-] = 3.767[H2CO3]

From the initial solution, we know that [HCO3-] + [H2CO3] = 0.501 M. So:

3.767[H2CO3] + [H2CO3] = 0.501

4.767[H2CO3] = 0.501

[H2CO3] = 0.105 M

0.501-0.105= 0.396 M = [HCO3-]

Moles HCO3- = 0.346 M X 0.261 L = 0.09 moles

This is equal to the moles of KOH added in the titration.

So, to get the volume of KOH:

0.09 mol / 1.1 M = 0.082 L = 82 mL KOH added