Question

The solubility product of $ AgCl $ is $ 4.0\times {{10}^{-10}} $ at $ 298K $ . The solubility of $ AgCl $ in $ 0.04\,\,M\,\,CaC{{l}_{2}} $ will be:

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

Answer 1

Answer: (A)

In $CaCl _{2},\left[ Cl ^{-}\right]=2 \times 0.004=0.008\, mol L ^{-1}$
$\left[ Cl ^{-}\right]$from $AgCl$ is too small and is neglected
$K_{s p} =\left[ Ag ^{+}\right]\left[ Cl ^{-}\right] $
$4 \times 10^{-10} =\left[ Ag ^{+}\right] \times 0.08$
$\left[ Ag ^{+}\right] =\frac{4 \times 10^{-10}}{0.08} $
$=\frac{4}{8} \times 10^{-8}$
$=5.0 \times 10^{-9} M$