1. 1.6×10⁻⁹ M 2. 1.6×10⁻¹¹ M 3. zero 4. 1.26×10⁻⁵ M

Answer: 1. 1.6×10⁻⁹ M Key idea: the common ion Cl^- from NaCl suppresses the solubility of AgCl. Let solubility be s . [Ag^+]=s , [Cl^-]≈ 0.1 M (s is negligible). K_sp=s(0.1)⇒ s=frac1.6×10⁻¹⁰0.1=1.6×10⁻⁹\ M Option 4 is the solubility in pure water ( sqrtK_sp ); it ignores the common ion.

Topic: Solubility product· Equilibrium

The solubility of AgCl (s) with solubility product 1.6×10^-10 in 0.1 M NaCl solution would be

1. $1.6\times10^{-9}\,M$

2. $1.6\times10^{-11}\,M$

3. zero

4. $1.26\times10^{-5}\,M$

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Answer: 1. $1.6\times10^{-9}\,M$

Key idea: the common ion $Cl^-$ from NaCl suppresses the solubility of AgCl.

Let solubility be $s$. $[Ag^+]=s$, $[Cl^-]\approx 0.1$ M (s is negligible).

$$K_{sp}=s(0.1)\Rightarrow s=\frac{1.6\times10^{-10}}{0.1}=1.6\times10^{-9}\ \mathrm{M}$$

Option 4 is the solubility in pure water ($\sqrt{K_{sp}}$); it ignores the common ion.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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The solubility of AgCl (s) with solubility product 1.6×10^-10 in 0.1 M NaCl solution would be

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