Question

The solubility product of AgCl is $4.0\times 10^{-10}$ at 298 K. The solubility of AgCl in 0.04 M $CaC{l}_2$ will be

• A. $2.0\times 10^{-5}M$
• B. $1.0\times 10^{-4}M$
• C. $5.0\times 10^{-9}M$
• D. $2.2\times 10^{-4}M$

Answer 1

Answer: (B)

$\begin{array}{l}x=[Ag]=\left[A^{-}\right]+[HA]---(1)\\ \text{ from }\frac{[4A]}{[A]}=10\\ x=\left[A{g}^{+}\right]=\left[A^{-}\right]+10\left[A^{-}\right]=11\left[A^{-}\right]\\ x=\left[A{g}^{+}\right]=11\times \left[A^{-}\right]---(2)\end{array}$ equation (2) can be written as $\begin{array}{l}{A}^{-}+H_{2}O\rightleftharpoons HA+OH\\ kw=\left[O{H}^{-}\right]\left[H^{+}\right]\\ \left[O{H}^{-}\right]=\frac{kw}{\left[H^{+}\right]}=\frac{10^{-14}}{10^{-4}}=10^{-5}\end{array}$ At eq $\left[A^{+}\right]=\left[O{H}^{-}\right]=10^{-5}$ from eq (1) $\left[A^{+}\right]=\left[A^{-}\right]=10^{-5}$ $\begin{array}{l}\mathrm{krp}=\left[A{g}^{+}\right]\left[\left(A^{-}\right)+(HA)\right]\\ =10^{-5}\times 11\times \left[A^{-}\right]\\ =10^{-5}\times 11\times \left[10^{-5}\right]\\ =10^{-5}\times 11=\frac{11}{10}\times 10^{-10}\times 10\\ =1.1\times 10^{-9}\end{array}$