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:
<em>Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻.</em>
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.
<em>a) 0.36 M AgNO₃:</em>
- The dissociation of Ag₂SO₄: Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻, will be accompanied by the ionization of AgNO₃:
<em>AgNO₃ → Ag⁺ + NO₃⁻.</em>
- 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.
<em>∴ s </em>= 1.5 x 10⁻⁵/(0.36)² =<em> 1.157 x 10⁻⁴ M.</em>
<em>b) 0.36 M Na₂SO₄:</em>
- 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.
<em>So, [SO₄²⁻] = (0.36).</em>
∴ 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.
<em>∴ s</em> = √(1.042 x 10⁻⁵) = <em>0.0032 M.</em>
