Question

The solubilities of the following salts are at a certain temperature. Ag₂SO₄ 0.03m, PbF₂ 1.0x10^-3m, and BaC₂O₄ 10^-4m. Calculate the concentrations of all ions present in each solution. Identify v, v_{_}, and v₊. Find m+/- for each solution.

Answer 1

1)

Solubility of Ag₂SO₄ = 0.03m (given)

The chemical equation is

Ag₂SO₄(s) ----------> 2 Ag⁺(aq) + SO₄^2-(aq)

The ratio of cations to anions 2:1

The dissociation equation shows that, for every mole of Ag₂SO₄ that dissociates, 2 mol of Ag⁺ and 1 mol of SO₄²⁻ are produced. Therefore, at equilibrium, the concentrations of the ions are

[Ag^-]= 2 x 0.03M = 0.06M

[SO₄^2-] = 0.03M

2) The chemical equation is

PbF₂(s) --------> Pb²⁺(aq) +2 F^-(aq)

Solubility of PbF₂ = 1 x 10^-3 M (given)

The ratio of cations to anions 1:2

The dissociation equation shows that, for every mole of PbF₂ that dissociates, 1 mol of Pb²⁺ and 2 mol of F− are produced. Therefore, at equilibrium, the concentrations of the ions are:

[Pb²⁺] = 1 × 10^−3 M

  [F⁻] = 2 × 1 × 10^−3 = 2 × 10^−3 M

3)

BaC₂O₄ x 10^-4M

The chemical equation is-

BaC₂O₄ ------> Ba²⁺(aq) +C₂O₄^2- (aq)

The ratio of cations to anions is 1:1

The dissociation equation shows that, for every mole of BaC₂O₄ that dissociates, 1 mol of Ba²⁺ and 1 mol of C₂O₄²⁻ are produced. Therefore, at equilibrium, the concentrations of the ions are

[Ba²⁺] = 10^-4 M

[C₂O₄^2-] = 10^-4 M