Question

The precipitation of $AgCl(K_{sp}=1.8\times {10}^{-10})$ will occur when which of the following two solutions are mixed?

• A. ${10}^{-4}M\left(A{g}^{+}\right)$ and ${10}^{-4}M\left(C{l}^{-}\right)$
• B. ${10}^{-5}M\left(A{g}^{+}\right)$ and ${10}^{-5}M\left(C{l}^{-}\right)$
• C. ${10}^{-5}M\left(A{g}^{+}\right)$ and ${10}^{-6}M\left(C{l}^{-}\right)$
• D. ${10}^{-10}M\left(A{g}^{+}\right)$ and ${10}^{-10}M\left(C{l}^{-}\right)$

Answer 1

The correct option is A ${10}^{-4}M\left(A{g}^{+}\right)$ and ${10}^{-4}M\left(C{l}^{-}\right)$

A precipitate will be formed when, the ionic product exceeds solubility product $(Q>K_{sp})$.
$Q=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$Q={10}^{-4}\times {10}^{-4}={10}^{-8}$
$K_{sp}=1.8\times {10}^{-10}$
$Q>K_{sp}$
Hence, the precipitation of $AgCl(K_{sp}=1.8\times {10}^{-10})$ will occur when ${10}^{-4}M\left(A{g}^{+}\right)$ and ${10}^{-4}M\left(C{l}^{-}\right)$ are mixed.