Calculate the solubility of $AgCl$ in a $0.1 \ M \ NaCl$ solution. (Given: $K_{sp}$ of $AgCl = 1.6 \times 10^{-10}$)

(N/A) The solubility product expression for $AgCl$ is $K_{sp} = [Ag^+][Cl^-]$.
In a $0.1 \ M \ NaCl$ solution,the concentration of $Cl^-$ ions is $0.1 \ M$ due to the common ion effect.
Let $s$ be the solubility of $AgCl$ in the presence of $NaCl$.
Then,$[Ag^+] = s$ and $[Cl^-] = 0.1 + s \approx 0.1 \ M$ (since $s$ is very small).
Substituting these values into the $K_{sp}$ expression:
$1.6 \times 10^{-10} = s \times 0.1$
$s = \frac{1.6 \times 10^{-10}}{0.1} = 1.6 \times 10^{-9} \ M$.