Question

We will abbreviate the different forms of malonic acid, CH2(CO2H)2, as MH2. Find the pH concentrations of MH2, MH-, and M2- in:

a) a 0.100 M MH2 solution

b) a 0.100 M NaMH solution

c) a 0.100 M Na2M solution

Answer 1

pH calculation

a) 0.1 M MH2 solution

MH2 + H2O <==> MH- + H3O+

Ka1 = [MH-][H3O+]/[MH2]

1.15 x 10^-3 = x^2/(0.1 - x)

x^2 + 1.15 x 10^-3x - 1.15 x 10^-4 = 0

x = [H3O+] = 0.01 M

pH = -log[H3O+] = 1.99

b) 0.1 M NaMH solution

MH- + H2O <==> M^2- + H3O+

Ka2 = [M2-][H3O+]/[MH-]

2.03 x 10^-6 = x^2/0.1

x = [H3O+] = 4.50 x 10^-4 M

pH = -log[H3O+] = 3.35

c) 0.1 M Na2M solution

M^2- + H2O <==> MH- + OH-

Kb1 = [MH-][OH-]/[M^2-]

1 x 10^-14/2.03 x 10^-6 = x^2/0.1

x = [OH-] = 2.22 x 10^-5 M

pOH = -log[OH-] = 4.65

pH = 14 - pOH = 9.35