Question

Calculate the molar solubility of ag2so4 in each solution below. the ksp of silver sulfate is 1.5 × 10−5. (a) 0.36 m agno3: (b) 0.36 m na2so4:

Answer 1

Answer:

a) 1.157 x 10⁻⁴ M.

b) 0.0032 M.

Explanation:

• Ag₂SO₄ is sparingly soluble salt in water which is dissociate according to:

Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻.

Ksp = [Ag⁺][SO₄²⁻] = (2s)²(s) = 1.5 x 10⁻⁵.

4s³ = 1.5 x 10⁻⁵.

s³ = 1.5 x 10⁻⁵/4 = 3.75 x 10⁻⁶.

∴ s = ∛(3.75 x 10⁻⁶) = 0.0155 M.

a) 0.36 M AgNO₃:

• The dissociation of Ag₂SO₄: Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻, will be accompanied by the ionization of AgNO₃:

AgNO₃ → Ag⁺ + NO₃⁻.

• So, the concentration of Ag⁺ will be that of AgNO₃  and Ag₂SO₄:

[Ag⁺] = (2s + 0.36)², s is neglected with respect to 0.36 M the concentration resulted from AgNO₃.

So, [Ag⁺] = (0.36)².

∴ Ksp = [Ag⁺][SO₄²⁻] = (0.36)²(s) = 1.5 x 10⁻⁵.

∴ s = 1.5 x 10⁻⁵/(0.36)² = 1.157 x 10⁻⁴ M.

b) 0.36 M Na₂SO₄:

• The dissociation of Ag₂SO₄: Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻, will be accompanied by the ionization of Na₂SO₄:

Na₂SO₄ → 2Na⁺ + SO₄²⁻.

• So, the concentration of SO₄²⁻ will be that of Na₂SO₄  and Ag₂SO₄:

[SO₄²⁻] = (s + 0.36), s is neglected with respect to 0.36 M the concentration resulted from Na₂SO₄.

So, [SO₄²⁻] = (0.36).

∴ Ksp = [Ag⁺][SO₄²⁻] = (2s)²(0.36) = 1.5 x 10⁻⁵.

∴ 4s² = 1.5 x 10⁻⁵/(0.36) = 4.166 x 10⁻⁵.

∴ s² = 4.166 x 10⁻⁵/4 = 1.042 x 10⁻⁵.

∴ s = √(1.042 x 10⁻⁵) = 0.0032 M.