Question

Calculate the concentrations of all species in a 0.610 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

A. [HSO3-]

B. [H2SO3]

C. [OH-]

D. [H+]

Answer 1

solution:

From given values of Ka1 and Ka2

Kb₁ for SO₃^-2 = Kw / Ka₂= 10^-14 / 6.3 X 10^-8

Kb₁ = 1.587 X 10^-7

Kb₂ for SO₃^-2 = Kw / Ka1 = 10^-14 / 1.4 X 10^-2

Kb₂ = 7.143 X 10^-13

Hydrolysis equation is

                        SO₃^-2 + H2O ---> HSO₃^- + OH^-

Initial 0.61 0            0

Change             -x                           +x           +x

Equilibrium 0.61 - x x              x

Kb = x .x / 0.61 - x = 1.587 X 10^-9

We can ignores x in denominator as Kb is very low

x² / 0.61 = 1.587 X 10^-7

x = 3.11 X 10^-4

Therefore,

[SO₃^-] = [OH-] = x = 3.11 X 10^-4

[SO₃^-2] = 0.61 - 3.11 X 10^-4 = 0.609 M

Now dissociation of HSO3^-

                      HSO₃^- + H2O ---> H₂SO₃ + OH^-

Initial         3.11 X 10^-4 0         3.11 X 10^-4

Change             -x +x          3.11 X 10^-4 + x

Equilibriu 3.11 X 10^-4- x x        3.11 X 10^-4 + x

Kb = x (3.11 X 10^-4 + x) / (3.11 X 10^-4) = 7.143 X 10^-13

On solving x = 7.143 X 10^-13

Therefore,

[SO3^-2] = 0.609 M

[Na+] = 2 X [Na₂SO₃] = 2 X 0.610 = 1.21 M

[HSO3-] = 3.11 X 10^-4- x = 3.11 X 10^-4- 7.143 X 10^-13 = 3.11 X 10^-4 M

[OH-] =4.12 X 10^-4 + x = 4.12 X 10^-4 + 7.143 X 10^-13 = 3.11 X 10^-4 M

[H+] = 10^-14 / [OH-] = 10^-14 /3.11 X 10^-4 = 3.21 X 10^-11 M

[H2SO3] = 7.143 X 10^-13 M