Question

A 100.0 mL solution containing 0.8366 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.3203 M KOH. Calculate the pH of the solution after the addition of 45.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.

Answer 1

Mass of maleic acid = 0.8366 g

Moles of maleic acid = 0.8366 / 116.072 = 7.21 x 10^-3 moles

Conc. of KOH solution = 0.3203 M

Volume of KOH solution = 45 mL = 0.045 L

=> Moles of KOH added = 0.3203 x 0.045 = 0.01441 moles

H2M + NaOH -----> HM- + H2O

HM- + NaOH -------> M2- + H2O

Overall: H2M + 2NaOH -----> M2- + 2H2O

According to the stoichiometry of the reaction 1 mole of H2M reacts with 2 moles of KOH

7.21 x 10^-3 moles of H2M reacts with 7.21 x 10^-3 x 2 = 0.01442 moles of KOH

Moles of M2- produced after reaction = 7.21 x 10^-3 moles

Total volume of solution = 100 + 45 = 145 mL = 0.145 L

=> [M2-] = 7.21 x 10^-3 / 0.145 = 0.04971 M

M2- + H2O --------> HM- + OH-

pKb2 for this reaction = 14 - 6.27 = 7.73

=> Kb2 = 1.86 x 10^-8

HM- + H2O --------> H2M + H2O

pKb1 for this reaction = 14 - 1.92 = 12.08

Kb1 = 8.32 x 10^-13

M2- + H2O --------> HM- + OH-

0.04971 -X.................X........X+Y

HM- + H2O --------> H2M + OH-

X-Y..........................Y.......X+Y

Kb2 = 1.86 x 10^-8 = X (X+Y) / (0.04971- X)

1.86 x 10^-8 = X^2 / 0.04971 - X

X = 3.04 x 10^-5 M

Kb1 = 8.32 x 10^-13 = Y (X+Y) / (X-Y) = Y

=> Y = 8.32 x 10^-13 M

[M2-] = 0.0497 - X = 0.0497 M (approx.)

[HM-] = X - Y = 3.04 x 10^-5 M

[H2M] = Y = 8.32 x 10^-13 M

[OH-] = X + Y = 3.04 x 10^-5

pOH = 4.52

pH = 14 - 4.52

pH = 9.48