The solubility of AgCl in 0.1 text{ M } NaCl | vedclass

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Question

(A) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.Let the solubility of $AgCl$ be $S 	ext{ mol/L}$.In $0.1

Vedclass · Accepted Answer

$1.8 	imes 10^{-9} 	ext{ M}$

Vedclass · Answer

(A) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.Let the solubility of $AgCl$ be $S 	ext{ mol/L}$.In $0.1 	ext{ M } NaCl$, the concentration of $Cl^-$ ions is $0.1 	ext{ M}$ from $NaCl$ and $S 	ext{ M}$ from $AgCl$. Since $S$ is very small, $[Cl^-] \approx 0.1 	ext{ M}$.The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.Substituting the values: $1.8 	imes 10^{-10} = (S)(0.1)$.Solving for $S$: $S = \frac{1.8 	imes 10^{-10}}{0.1} = 1.8 	imes 10^{-9} 	ext{ M}$.

The solubility of $AgCl$ in $0.1 \text{ M } NaCl$ is $S \text{ mol/L}$. If the solubility product of $AgCl$ is $1.8 \times 10^{-10}$, then $S$ is approximately:

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