Question

Calculate the concentrations of all species in a 1.49 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8. [Na+]= [SO3^2-]= [HSO3^-]= [H2SO3]= [OH^-]= [H^+]=

Answer 1

Sodium sulfite dissociates completely.

2Na⁺= 2.98

SO₃^2-= 1.49

Na₂SO₃(aq) + H₂O(l) 2Na⁺ + SO₃^2-

When Na₂SO₃(aq) dissociates it forms 2 moles of Na⁺ and sulfite ions i.e. weak base.

When weak base reacts with H₂O it forms OH^- ions and HSO₃^-

Kw= [H⁺] [OH^-] = 1.0 x 10^-14

                    SO₃^2-(aq) ←→ HSO₃^- + OH^-

I:                 1.49                     0              10^-7

C:               -x                      +x               +x

E:             1.49 – x                  x           10^-7+x

Since it is base so we have to found Kb first.

Ka x Kb = Kw = [H⁺] [OH^-]= 1.0 x 10^-14 -----------------------(1)

We have Ka₂ = 6.3 × 10^–8

So, Kb = 1.0 x 10^-14 / 6.3 x 10^-8

For above ICE chart we got

Kb = [HSO₃^-] [OH^-] / [SO₃^2-]

1 x 10^-14 / 6.3 x 10^-8 = [x][x] / [1.49 - x]

x² = (1.49)(1.6 x 10^-7) ;              1.49 – x ~ 1.49; drop down x

x² = 2.4 x 10^-7

x = 4.9 x 10^-4

So, x = 4.9 x 10^-4 = [HSO₃^-] = [OH^-]

A small amount of the hydrogen sulfite ion (HSO3–) will also react with water:

                HSO₃^-(aq) ←→ H₂SO₃(aq) + OH^-(aq)

I .            0.00049                   0                 0.00049

C:              -y                        +y                   +y

E:          0.00049-y                 y             0.00049+y

Ka x Kb = Kw = [H⁺] [OH^-]= 1.0 x 10^-14-----------------------(1)

We have Ka₁ = 1.4 × 10^–2

So, Kb₂ = 1.0 x 10^-14 / 1.4 x 10^-2

For above ICE chart we got

Kb = [H₂SO₃] [OH^-] / [HSO₃^-]

1 x 10^-14 / 1.4 x 10^-2 = [y][0.00049 + y] / [0.00049 - y]

7.1 x 10^-13 = [y] [0.00049 + y] / [0.00049] ;

(Dropping down y from the denominator since y << 0.00049)

3.5 x 10^-16 = y² + 0.00049y

y² + 0.00049y – 3.5 x 10^-16 =0

Solving the quadratic equation, we get;

y = 7.14 x 10^-13

y = 7.14 x 10^-13= [H₂SO₃]

Now, from equation (1)

[H⁺] [OH^-]= 1.0 x 10^-14

[H⁺] = 1.0 x 10^-14 / [OH^-] = (1.0 x 10^-14) / (4.9 x 10^-4) = 2.0 x 10^-11