The solubility of AgCl is 1 times 10^{-5} mo | vedclass

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Question

(C) The solubility product constant $(K_{sp})$ of $AgCl$ is calculated as: $K_{sp} = (	ext{solubility})^2 = (1 	imes 10^{-5})^2 = 1 	imes 10^{-10}$

Vedclass · Accepted Answer

$1 	imes 10^{-9}$

Vedclass · Answer

(C) The solubility product constant $(K_{sp})$ of $AgCl$ is calculated as: $K_{sp} = (	ext{solubility})^2 = (1 	imes 10^{-5})^2 = 1 	imes 10^{-10}$. Let the solubility of $AgCl$ in $0.1 \ M \ NaCl$ be $x \ mol/L$. In the presence of $0.1 \ M \ NaCl$,the concentration of chloride ions is $[Cl^-] = (x + 0.1) \ M$. Since $x$ is very small compared to $0.1$,we can approximate $[Cl^-] \approx 0.1 \ M$. The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$. Substituting the values: $1 	imes 10^{-10} = x 	imes 0.1$. Solving for $x$: $x = \frac{1 	imes 10^{-10}}{0.1} = 1 	imes 10^{-9} \ mol/L$.

The solubility of $AgCl$ is $1 \times 10^{-5} \ mol/L$. Its solubility in $0.1 \ M$ sodium chloride solution is

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