Question

Determine the value of Keq for CaSO3 (Ksp = 1.0 x 10^-4) dissolving in acid solution, given that Ka2 = 6.3 x 10 ^-8 for H2SO3.

Answer 1

Reaction 1 : CaSO₃ (s) Ca²⁺ (aq) + SO₃^2- (aq) : K₁ = K_sp = 1.0 x 10^-4

Reaction 2 : HSO₃^- (aq) H⁺ (aq) + SO₃^2- (aq) : K₂ = K_a2 = 6.3 x 10^-8

If reverse reaction 2, we get

Reaction 3 : H⁺ (aq) + SO₃^2- (aq) HSO₃^- (aq) : K₃ = 1/K_a2 = 1/(6.3 x 10^-8) = 1.6 x 10⁷

Now add Reaction 1 and Reaction 3

CaSO₃ (s) + H⁺ (aq) + SO₃^2- (aq) Ca²⁺ (aq) + SO₃^2- (aq) + HSO₃^- (aq)

Cancelling common terms

CaSO₃ (s) + H⁺ (aq) Ca²⁺ (aq) + HSO₃^- (aq)

K_eq = (K₁) * (K₃)

K_eq = (1.0 x 10^-4) * (1.6 x 10⁷)

K_eq = 1.6 x 10³