Question

What is the hydronium ion concentration in a 0.0294 M aqueous solution of sodium arsenate (Na₃AsO₄)?

Answer 1

Na3AsO4 will dissociate as follows

Na3AsO4 --> 3Na+ + AsO4-3

and we know arsenate ion is a strong weak conjugte base so

AsO4-3 + H2O <--< HAsO4-2 + OH- Kb3

and follows:

HAsO4-2 + H2O <-> H2AsO4- + OH - Kb2

finally

H2AsO4- + H2O <-> H3AsO4 + OH- Kb1

note that we only have Ka values so calculate Kb values via Kw = Ka*Kb --> Kb = Kw/Ka

Ka1 = 3*10^-12; Ka2 = 9.3*10^-8; Ka3 = 5*10^-3

Kb3 = Kw/Ka3 = (10^-14)/(3*10^-12) = 3.3333*10^-3

Kb2 = Kw/Ka2 = (10^-14)/(9.3*10^-8) = 1.07526*10^-7

Kb1 = Kw/Ka1 = (10^-14)/(5*10^-3) = 2*10^-12

now...

from first ionization:

AsO4-3 + H2O <--< HAsO4-2 + OH- Kb3

Kb3 = [HAsO4-2][OH-]/[AsO4-3]

3.3333*10^-3 = x*x/(0.0294-x)

solving for x

x = 0.0083686

[HAsO4-2] = 0.0083686

[OH-] = 0.0083686

[AsO4-3] = M-x = 0.0294-0.0083686 = 0.0210314

for 2nd ionization:

HAsO4-2 + H2O <-> H2AsO4- + OH - Kb2

Kb2 = [H2AsO4-][OH-]/[HAsO4-2]

in equilibrium:

[H2AsO4-] = y

[OH-] = x+y = 0.0083686 + y

[HAsO4-2] = 0.0210314-y

substitute in kb2

1.07526*10^-7 = y*(x+y)/(0.0210314-y)

since Y is too small, we can assume:

x+y = x, and 0.0210314-y = 0.0210314

1.07526*10^-7 = y*(0.0210314)/(0.0210314)

y = (1.07526*10^-7 ) / ((0.0210314)/(0.0210314)) = 1.07*10^-7

clearly, there is no need to continue for 3rd ionization, since value will be pretty low

so

[OH-] = 0.0210314-y = 0.0210314-1.07*10^-7 = 0.0210

so

[H3O+] = (10^-14)/(OH-)= (10^-14)/(0.0210) = 4.76*10^-13