化學科 - Solubility Product Calculations?
From Past Exam Paper without answer
回答 (1)
(b)
Volume of final solution = (50 + 50 + 100) cm³ = 200 cm³
Consider only dilution effect just after mixing of the three solution:
[Ca²⁺] = (0.10 M) × (50/200) = 0.025 M
[Ag⁺] = (0.10 M) × (50/200) = 0.025 M
Ksp(CaSO₄) = [Ca²⁺][SO₄²⁻] = (0.025)[SO₄²⁻] = 2.4 × 10⁻⁵
[SO₄²⁻] needed to precipitate CaSO₄ = (2.4 × 10⁻⁵)/(0.025) = 9.6 × 10⁻⁴ M
Ksp(Ag₂SO₄) = [Ag⁺]²[SO₄²⁻] = (0.025)²[SO₄²⁻] = 1/2 × 10⁻⁵
[SO₄²⁻] needed to precipitate CaSO₄ = (1.2 × 10⁻⁵)/(0.025)² = 0.0192 M
Only one of the cations precipitates (i.e. Ca²⁺). Hence,
9.6 × 10⁻⁴ M ≤ [SO₄²⁻] < 0.0192 M
Since [SO₄²⁻] = 2[H⁺],
9.6 × 10⁻⁴ M ≤ 2[H⁺] < 0.0192 M
4.8 × 10⁻⁴ M ≤ [H⁺] < 9.6 × 10⁻³ M
-log(4.8 × 10⁻⁴) ≥ -log[H⁺] > -log(9.6 × 10⁻³)
2.0 < pH ≤ 3.3
收錄日期: 2021-04-24 08:00:39
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