Question

$50\text{mL}$ of a solution containing ${10}^{-3}$ moles of $A{g}^{+}$ is mixed with $50\text{mL}$ 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 $2.5\times {10}^{-9}$

$\text{Concentration of}\left[A{g}^{+}\right]=\frac{{10}^{-3}}{50+50}\times 1000=0.01\text{M}$
$\text{Concentration of}\left[C{l}^{-}\right]=\frac{0.1\times 50}{50+50}=0.05\text{M}$
$\text{So, the reaction quotient is}Q=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=5\times {10}^{-4}\text{M}$
$\because Q>K_{sp},\text{so precipitate forms}$
$\begin{array}{cccc}& A{g}^{+}(aq.)& +& C{l}^{-}(aq.)\rightleftharpoons AgCl\left(s\right)\\ \text{At given condition}& 0.01& & 0.05\\ \text{After precipitation}& x& & \approx 0.04\end{array}$

$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]\Rightarrow {10}^{-10}=x\times 0.04$
$\Rightarrow x=\left[A{g}^{+}\right]=2.5\times {10}^{-9}$