Question

59. The solubility product of Baso, is 4 x 10-10. The
solubility of BaSO4 in presence of 0.02 N H2SO4
will be
(1) 4 x 10-8 M
(2) 2 x 10-8 M
(3) 2 x 10-5 m (4) 2 x 10-4 M​

Answer 1

Given:

• The solubility product of BaSO₄ (Ksp) = 4*10⁻¹⁰
• The normality of H₂SO₄ = 0.02 N

To find:

The solubility of BaSO₄ (S) in presence of 0.02 N H₂SO₄.

Solution:

• Molarity of H₂SO₄ = normality/2 = 0.01 M (since, n-factor = 2)
• One mole of BaSO₄ dissociates into one mole of Ba²⁺ ion and one mole of SO₄²⁻ ions.
• [Ba²⁺] = S , [SO₄²⁻] = 0.01 + S ( contribution from both BaSO₄ and H₂SO₄) .
• Ksp = [Ba²⁺].[SO₄²⁻] = S*(S+0.01) ≅ 0.01*S (since, S << 0.01)
• S = Ksp/0.01 = 4*10⁻⁸ M

Answer:

The solubility of BaSO₄ (S) in presence of 0.02 N H₂SO₄ = 4*10⁻⁸ M