Question

A 100.0 mL solution containing 0.750 g of maleic acid (MW=116.072 g/mol) is titrated with 0.264 M KOH. Calculate the pH of the solution after the addition of 49.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27.

pH=

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 as H2M, HM−, and M2−, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.

[M2−]=

[HM−]=

[H2M]=

Answer 1

H2M moles = mass / molar mass

                     = 0.750 / 116.072

                    = 6.46 x 10^-3

KOH moles = 0.264 x 49 / 1000

                   = 0.01294

H2M + 2 KOH ------------------> K2M + 2H2O

moles of KOH neede = 6.46 x 10^-3 x 2

                                   = 0.01292

it is equivalnece point second

M2- formed = 6.46 x 10^-3

total volume = 149 mL = 0.149 L

[M2-] = 6.46 x 10^-3 / 0.149 = 0.0434 M

pKb1 = 14 - pKa2

          = 14- 6.27

           = 7.73

Kb1 = 10^-7.73 = 1.86 x 10^-8

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

0.0434                             0           0

0.0434 -x                         x             x

Kb1 = x^2 / 0.0434-x = 1.86 x 10^-8

x = 2.84 x 10^-5

pOH = -log (2.84 x 10^-5) = 4.55

pH + pOH = 14

pH = 9.45

[H2M] = pKb2 = 14 - 1.92 = 12.08

Kb2 = 8.32 x 10^-13

[H2M] = 8.32 x 10^-13 M

[HM-] = 2.84 x 10^-5

[M2-] = 0.0434 M

pH = 9.45