Question

$50mL$ of a solution containing ${10}^{-3}$ mole of $A{g}^{+}$ is mixed with $50mL$ of a $0.1MHCl$ solution how much $A{g}^{+}$ remains in solution? $(K_{sp}ofAgCl=1.0\times {10}^{-10})$

• A. $2.5\times {10}^{-9}$
• B. $2.5\times {10}^{-7}$
• C. $2.5\times {10}^{-8}$
• D. $2.5\times {10}^{-10}$

Answer 1

The correct option is A $2.5\times {10}^{-9}$

$\left[A{g}^{+}\right]=\frac{{10}^{-3}}{50+50}\times 1000=0.01\left[C{l}^{-}\right]=\frac{0.1\times 50}{50+50}=0.05Q=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=5\times {10}^{-4}\because Q>K_{sp}A{g}^{+}(aq.)+C{l}^{-}(aq.)\to AgCl\left(s\right)$
$$\begin{array}{ccc}\text{at given condition}& 0.01& 0.05\\ \text{after precipitation}& x& \approx 0.04\end{array}$$
$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]{10}^{-10}\approx x\times 0.04x=\left[A{g}^{+}\right]=2.5\times {10}^{-9}$