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Chemistry · General chemistry II · Worked example

Find the molar solubility from Ksp

Silver chloride has Ksp = 1.8 × 10⁻¹⁰ at 25 °C. Find its molar solubility in water, in mol/L and in g/L.

Write the equilibrium

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). If S mol/L dissolves, [Ag⁺] = [Cl⁻] = S, so Ksp = S².

Solve for S

Take the square root. Keep a guard digit, 1.34 × 10⁻⁵ M, for the next step.

S=1.8×10−10=1.3×10−5 M

Convert to grams per liter

Multiply by the molar mass of AgCl, 107.87 + 35.45 = 143.32 g/mol. To the two significant figures of Ksp, that is 1.9 × 10⁻³ g/L.

(1.34×10−5 mol/L)⁢(143.32 g/mol)=1.92×10−3 g/L
(1.34×10−5)⁢(143.32) g/L=1.92×10−3 g/L

Result

S = 1.3 × 10⁻⁵ mol/L, about 1.9 × 10⁻³ g/L.

Your turn

Calcium fluoride has Ksp = 3.7 × 10⁻¹¹. Find its molar solubility.

Show the answer and explanation

2.1 × 10⁻⁴ M.

[Ca²⁺] = S and [F⁻] = 2S, so Ksp = S(2S)² = 4S³ and S = ∛(3.7 × 10⁻¹¹ / 4) = 2.1 × 10⁻⁴ M. Its Ksp is smaller than silver chloride’s, yet calcium fluoride is the more soluble salt.

S=3.7×10−1143=2.1×10−4

Keep exploring

In Solubility & precipitation, start with 0.10 M chloride already in the solution. The molar solubility falls to 1.8 × 10⁻⁹ M, about 7500 times less: the common-ion effect.

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