solubility product

Question

The solubility product of AgCl at 298 K is 1.0 × 10⁻¹⁰. The solubility of AgCl in water is:

RankGuru · Accepted Answer

Silver chloride ($AgCl$) is a sparingly soluble salt.
$AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$
If the solubility of $AgCl$ is '$s$' mol/L, then:
- $[Ag^+] = s$
- $[Cl^-] = s$
The **Solubility Product Constant ($K_{sp}$)** is:
$K_{sp} = [Ag^+][Cl^-] = s 	imes s = s^2$
Given $K_{sp} = 1.0 	imes 10^{-10}$:
$s^2 = 10^{-10} \implies s = \sqrt{10^{-10}} = 10^{-5} 	ext{ M}$

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