Question

The solubility (in $\mathrm{mol}{L}^{-1}$) of $AgCl(K_{sp}=1.0\times {10}^{-10})$ in a $0.1MKCl$ solution will be:

• (A) $1.0\times {10}^{-9}$
• (B) $1.0\times {10}^{-10}$
• (C) $1.0\times {10}^{-5}$
• (D) $1.0\times {10}^{-11}$

Answer 1

The Correct Option is A

Explanation:
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. The solubility product constant describes the equilibrium between a solid and its constituent ions in a solution,Let solubility of $AgCl=x$ mole/$L$$AgCl\rightleftharpoons {Ag}^{+}+{Cl}^{-}$i.e., $K_{sp}(AgCl)=x^2$$KCl\to K^{+}+{Cl}^{-}$ $[{Cl}^{-}]$ from $KCl=0.1m$Total $[{Cl}^{-}]$ in solution $=x+0.1$$K_{sp}(AgCl)=\left[A{g}^{+}\right][C{l}^{-}]=x(x+0.1)$$1.0\times {10}^{-10}=x(x+0.1)$$1.0\times {10}^{-10}=0.1x($ as $x^2<<1)$$x=1.0\times {10}^{-9}mol/L$Hence, the correct option is (A).