Question

A solution is made by equilibrating the two solids barium carbonate (K_sp = 8.1×10^-9) and barium sulfite (K_sp = 8.0×10^-7) with water. What are the concentrations of the three ions at equilibrium, if some of each of the solids remain?

[Ba²⁺] = _________M
[CO₃^2-] = _________M
[SO₃^2-] = ____________ M

Answer 1

Sol :-

Solubility product (K_sp) is the product of the molar concentration of products raised to the power of stoichiometric coefficient at equilibrium stage of the reaction.

Let solubility of BaCO₃ = S mol/L

Now,

Partial dissociation of BaCO₃ is

BaCO₃ (s) <---------------------> Ba²⁺ (aq) ..................+....................CO₃^2- (aq)

....................................................S mol/L..............................................S mol/L

Expression of K_sp is :

K_sp = [Ba²⁺].[CO₃^2-]

S² = (8.1 x 10^-9)

S = (8.1 x 10^-9)^1/2

S = 9.0 x 10^-5 mol/L

So,

[CO₃^2-] = 9.0 x 10^-5 M

again

Let solubility of BaSO₃ in water = S mol/L

Partial dissociation of BaSO₃ :

BaSO₃ (s) <---------------------> Ba²⁺ (aq) ........................+......................CO₃^2- (aq)

...................................................S mol/L........................................................S mol/L

Expression of K_sp is :

K_sp = [Ba²⁺].[CO₃^2-]

S² = (8.0 x 10^-7)

S = (8.0 x 10^-7)^1/2

S = 8.94 x 10^-4

So,

[SO₃^2-] = 8.94 x 10^-4 M

Total concentration of Ba²⁺ = [Ba²⁺] = 8.94 x 10^-4 M + 9.0 x 10^-5 M = 9.84 x 10^-4 M

So,

[Ba²⁺] = 9.84 x 10^-4 M