Question

The solubility product of silver chloride is $1.8 \times 10^{-10}$ at 298 K. The solubility of AgCl in 0.01 M HCl solution in $mol/dm^3$ is

• A. $2.4 \times 10^{-9}$
• B. $3.6 \times 10^{-8}$
• C. $0.9 \times10^{-10}$
• D. $1.8 \times 10^{-8}$

Answer 1

Answer: (D)

Let solubility of $AgCl$ in $0.01\,M\,HCl=x\,mol\,L^{-1}$
$\therefore $ $[Ag^{+}]=x\,mol\,L^{-1}$
$[Cl^{-}]=[HCl]=0.01\,M=10^{-2}\,M$
$[Cl^{-}]$ from $AgCl$ can be neglcctcd
$K_{sp}=[Ag^{+}] [ Cl^{-}]$
$1.8 \times 10^{-10}=x \times 10^{-2}$
$\Rightarrow x=1.8\times10^{-8}$