General ChemistryHard

Solubility and solubility product

K_sp of AgCl is 1.8×10⁻¹⁰. Calculate molar solubility of AgCl in water and in 0.010 M NaCl.
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).

Given data

K<sub>sp</sub>(AgCl) = 1.8×10⁻¹⁰

Steps

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In water: s = [Ag⁺] = [Cl⁻]. K_sp = s². s = √(K_sp).
Write your working — the numerical result must appear in the text M
In 0.010 M NaCl: [Cl⁻]₀ = 0.010 M. Approx s << 0.010. s = K_sp/[Cl⁻].

Full worked solution

In water: s = [Ag⁺] = [Cl⁻]. K_sp = s². s = √(K_sp).
$\sqrt{1.8\times10^{-10}}$
s = √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ M.
In 0.010 M NaCl: [Cl⁻]₀ = 0.010 M. Approx s << 0.010. s = K_sp/[Cl⁻].
$1.8\times10^{-10} / 0.010$
s = 1.8×10⁻¹⁰/0.010 = 1.8×10⁻⁸ M.

Result:Solubility in water: 1.34×10⁻⁵ M — in 0.010 M NaCl: 1.8×10⁻⁸ M (common ion effect).

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