Question

When equal volumes of the following solutions are mixed, precipitation of $AgCl(K_{sp}=1.8\times {10}^{-10})$ will occur only with

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

Answer 1

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

For a precipitation to occur
Solubility product < Ionic product
Given, $K_{sp}=1.8\times {10}^{-10}$
Calculating ionic products in each
$\because \text{Equal volumes are mixed,}so\left[A{g}^{+}\right]=\frac{{10}^{-4}}{2}\text{and}\left[C{l}^{-}\right]=\frac{{10}^{-4}}{2}$
Ionic product $=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=\frac{{10}^{-4}}{2}\times \frac{{10}^{-4}}{2}$
$=2.5\times {10}^{-9}$.
which is greater than $K_{sp}(1.8\times {10}^{-10}).$ Thus precipitaton will not occur in case of (a) only.
In all other cases the ionic product of silver and chloride ions are less than the solubility product of AgCl and hence the precipitaton will not occur.