Question

We will abbreviate malonic acid, CH₂(CO₂H)₂, as H2M. Find the pH and concentrations of H₂M, HM⁺, and M²⁺ in (a) 0.100 M H₂M; (b) 0.100 M NaHM; (c) 0.100 M Na₂M.

 

 

Answer 1

Ka1 = 1.48 x 10-3 ; Ka2 = 2.04 x 10-6 2.83, 5.69

 

a. 0.1 M H2M

Ka1 = [H+][HM-]/[H2M]

1.48 x 10^-3 = x^2/0.1

x = [H+] = 0.012 M

pH = 1.91

[H2M] = 0.1 - 0.012 = 0.088 M

Ka2 = [H+][M2-]/[HM-]

2.04 x 10-6 = x2/0.088

x = [M2-] = 4.23 x 10-4 M

[HM-] = 0.087 M

 

b. 0.1 M NaHM

Kb = Kw/Ka1 = [H2M][OH-]/[HM-]

1 x 10-14/1.48 x 10-3 = x2/0.1

x = [OH-] = 8.22 x 10-7 M

[H+] = 1 x 10-14/8.22 x 10^-7 = 1.21 x 10-8 M

pH = -log[H+] = 7.91

[H2M] = 1.21 x 10-8 M

Ka2 = [H+][M2-]/[HM-]

2.04 x 10-6 = x2/0.1

x = [M2-] = 4.51 x 10-4 M

[HM-] = 0.099 M

 

c. 0.1 M Na2M

Kb2 = Kw/Ka2 = [HM-][OH-]/[M2-]

1 x 10-14/2.04 x 10-6 = x2/0.1

x = [OH-] = 2.21 x 10-5 M

[H+] = 1 x 10-14/2.21 x 10-5 = 4.51 x 10-10 M

pH = -log[H+] = 9.34

[M2-] = 0.1 M

Kb1 = 1 x 10-14/1.48 x 10-3 = x2/0.1

x = [H2M] = 8.22 x 10-7 M

 [HM-] = 0.1 M

Ka1 = 1.48 x 10-3 ; Ka2 = 2.04 x 10-6 2.83, 5.69

 

a. 0.1 M H2M

Ka1 = [H+][HM-]/[H2M]

1.48 x 10^-3 = x^2/0.1

x = [H+] = 0.012 M

pH = 1.91