Question

$50\text{mL}$ of a solution containing ${10}^{–3}mol$ of $A{g}^{+}$ is mixed with $50\text{mL}$ of a $0.1\text{M}HCl$ solution. How much $\left[A{g}^{+}\right]$ remains in the solution? (Given: $K_{sp}$ of $AgCl={10}^{–10}$)

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

Answer 1

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

$[A{g}^{+}{]}_{initial}\text{just after mixing}=\frac{{10}^{-3}}{50+50}\times 1000=0.01M$
$[C{l}^{-}{]}_{initial}\text{just after mixing}=\frac{0.1\times 50}{50+50}=0.05M$
$I.P.=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=5\times {10}^{-4}$ (where I.P. = ionic product)
$\because I.P.>K_{sp}$
$\begin{array}{ccccc}A{g}^{+}\left(aq\right)& +& C{l}^{-}\left(aq\right)& \to & AgCl\left(s\right)\\ 0.01M& & 0.05& & \text{Initial conc.}\\ X& & 0.04M& & \text{conc. after precipitation}\end{array}$

$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
${10}^{-10}=\left[A{g}^{+}\right]\left(0.04\right)$
$\therefore \left[A{g}^{+}\right]=2.5\times {10}^{-9}M$