Question

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

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

Answer 1

Answer: (B)

$1.0\times {10}^{-4}M$

Let solubility of $AgA$ be $x$ moles/liter

$x=\left[Ag\right]=\left[A^{-}\right]+\left[HA\right]---\left(1\right)$

from $\frac{\left[4A\right]}{\left[A\right]}=10$

$x=\left[A{g}^{+}\right]=\left[A^{-}\right]+10\left[A^{-}\right]=11\left[A^{-}\right]$

$x=\left[A{g}^{+}\right]=11\times \left[A^{-}\right]---\left(2\right)$

equation (2) can be written as

$A^{-}+H_{2}O\rightleftharpoons HA+OH$

$kw=\left[O{H}^{-}\right]\left[H^{+}\right]$

$\left[O{H}^{-}\right]=\frac{kw}{\left[H^{+}\right]}=\frac{{10}^{-14}}{{10}^{-4}}={10}^{-5}$

At eq $\left[A^{+}\right]=\left[O{H}^{-}\right]={10}^{-5}$

from eq (1) $\left[A^{+}\right]=\left[A^{-}\right]={10}^{-5}$

$krp=\left[A{g}^{+}\right]\left[\right(A^{-})+(HA\left)\right]$

$={10}^{-5}\times 11\times \left[A^{-}\right]$

$={10}^{-5}\times 11\times \left[{10}^{-5}\right]$

$={10}^{-5}\times 11=\frac{11}{10}\times {10}^{-10}\times 10$

$=1.1\times {10}^{-9}$