Question

calculate the concentrations of all species present at equilibrium for a .200M solution of sodium arsenate, Na3AsO4. ( Don't forget the hydrolysis reactions of the arsenate ion)

Answer 1

Concentration of species

Solution = 0.200 M Na3AsO4

Na3AsO4 ----> 3Na+ + AsO4^3-

equilibrium [Na+] = 3 x 0.200 = 0.600 M

initial [AsO4-] = 0.200 M

Hydrolysis of AsO4-

AsO4^3- + H+ <==> HASO4^2- + OH-

let x amount has hydrolyzed

Kb = Kw/Ka3 = [HAsO4^2-][OH-]/[ASO4^3-]

1 x 10^-14/3 x 10^-13 = x^2/(0.200 - x)

x^2 + 0.033x - 0.0067 = 0

x = [OH-] = 0.067 M

So equilibrium concentrations of,

[AsO4^3-] = 0.200 - 0.067 = 0.133 M

[OH-] = 0.067 M

[H+] = 1 x 10^-14/0.067 = 1.49 x 10^-13 M

[HAsO4^2-] = 0.133 M

HAsO4^2- again hydrolyzes

HAsO4^2- + H2O <==> H2AsO4- + OH-

let x amount is hydrolyzed

Kb2 = Kw/Ka2 = [H2AsO4-][OH-]/[HAsO4^2-]

1 x 10^-14/5.6 x 10^-8 = x^2/(0.133 - x)

x be a small number here can be neglected from base

x = 1.54 x 10^-4 M

So,

at equilibrium,

[H2AsO4-] concentration = 1.54 x 10^-4 M

Now hydrolysis of H2AsO4-,

H2AsO4- + H2O <==> H3AsO4 + OH-

Kb3 = Kw/Ka1 = [H3AsO4][OH-]/[H2AsO4-]

1 x 10^-14/2.5 x 10^-4 = x^2/5.4 x 10^-4

x = 4 x 10^-11 M

So,

at equilibrium,

[H3AsO4] concentration = 4 x 10^-11 M