Carbonic acid is a weak, monoprotic acid with a pKa of 6.38. Calculate the follo
ID: 953801 • Letter: C
Question
Carbonic acid is a weak, monoprotic acid with a pKa of 6.38. Calculate the following pH values. For any buffer solution, assume the assumptions will be valid and simplify by using the Henderson-Hasselbalch equation. Show your work. Attach additional pages if necessary.
a. Determine the pH of 25 mL of a 0.10 M solution of carbonic acid.
b. Determine the pH after 20.0 mL of a 0.10 M NaOH solution is added to the 25 mL of 0.10 M carbonic acid.
c. Determine the pH at the equivalence point.
d. Determine the pH after 30 mL of 0.10 M NaOH solution is added to the 25 mL of 0.10 M carbonic acid.
Explanation / Answer
a. H2CO3 <==> H+ + HCO3-
Ka = [H+][HCO3-]/[H2CO3]
4.17 x 10^-7 = x^2/0.1
x = [H+] = 2.04 x 10^-4 M
pH = -log[H+] = 3.69
b. moles of NaOH added = 0.1 M x 20 ml = 2.0 mmol
[H2CO3] remaining = 0.1 M x 5 ml/45 ml = 0.011 M
[HCO3-] formed = 0.1 M x 20 ml/45 ml = 0.044 M
pH = pKa + log([base]/[acid])
= 6.38 + log(0.044/0.011)
= 6.98
c. pH at equivalence point
Volume of base added = 25 ml
[HCO3-] formed = 0.1 M x 25 ml/50 ml = 0.05 M
HCO3- + H2O <==> H2CO3 + OH-
Kb = 1 x 10^-14/4.17 x 10^-7 = x^2/0.05
x = [OH-] = 3.46 x 10^-5 M
pOH = -log[OH-] = 4.46
pH = 14 - pOH = 9.54
d. Excess NaOH = 0.1 M x 5 ml/55 ml = 9.10 x 10^-3 M
pOH = -log(9.10 x 10^-3) = 2.04
pH = 14 - pOH = 11.96