The solubility product of AgCl is 1.44 times | vedclass

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(B) For the dissolution of $AgCl$ in water,the equilibrium is represented as: $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$Let the solubility o

Vedclass · Accepted Answer

$1.20 	imes 10^{-2}\,M$

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(B) For the dissolution of $AgCl$ in water,the equilibrium is represented as: $AgCl(s) ightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$Let the solubility of $AgCl$ be $S \, mol/L$.Then,$[Ag^{+}] = S$ and $[Cl^{-}] = S$.The solubility product expression is given by:$K_{sp} = [Ag^{+}][Cl^{-}] = S 	imes S = S^2$Given $K_{sp} = 1.44 	imes 10^{-4}$.$S^2 = 1.44 	imes 10^{-4}$$S = \sqrt{1.44 	imes 10^{-4}} = 1.20 	imes 10^{-2} \, M$.Therefore,the solubility of silver chloride is $1.20 	imes 10^{-2} \, M$.

The solubility product of $AgCl$ is $1.44 \times 10^{-4}$ at $100\,^{\circ}C$. The solubility of silver chloride in boiling water may be

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