Question

The solubility product of silver chloride is 1.8 × 10^-10 at 298 k. The solubility of AgCl in 0.01 M KCl aqueous solution in mol dm^-3 is

• (A) 9.0 × 10^-9
• (B) 1.8 × 10^-8
• (C) 3.6 × 10^-8
• (D) 2.4 × 10^-9

Answer 1

The Correct Option is B

Explanation:
Let solubility of AgCl in water = s$$\begin{array}{l}{\mathrm{AgCI}}_{(\mathrm{s})}\frac{\text{ inwater }}{⟶}{\mathrm{Ag}}_{(\mathrm{aq})}^{+}+{\mathrm{C}}_{(\mathrm{aq})}^{-}\\ {\mathrm{k}}_{\mathrm{sp}}\mathrm{AgCl}={\mathrm{s}}^2\end{array}$$$s^2=1.8\times {10}^{-10}$Now in $0.01\mathrm{M}\mathrm{KCl}$ aqueous solution Complete hydrolysis of $\mathrm{KCl}$ will take place and there will be a concentration of $\left[{\mathrm{Cl}}^{-}\right]=0.01\mathrm{M}$$$0.01\mathrm{MKCl}⟶0.01\mathrm{M}{\mathrm{K}}^{+}+0.01\mathrm{M}{\mathrm{Cl}}^{-}$$Now if solubility of AgCl in KCI solution is s' Then$$\begin{array}{l}{\mathrm{AgCI}}_{(s)}\frac{\mathrm{inKC}1}{⟶}{\mathrm{Ag}}_{(\mathbf{a}\mathbf{q})}^{+}+{\mathrm{CI}}_{(\mathbf{a}\mathbf{q})}^{-}\\ {\mathrm{s}}^{\prime }0.01+{\mathrm{s}}^{\prime }\\ {\mathrm{k}}_{\mathrm{sp}}\mathrm{AgCl}={\mathrm{s}}^{\prime }(0.01+{\mathrm{s}}^{\prime })\end{array}$$${\mathrm{s}}^{\prime }$ is small compare to 0.01 so neglecting it with respect to 0.01 ${\mathrm{k}}_{\mathrm{sp}}\mathrm{AgCl}={\mathrm{s}}^{\prime }(0.01)$$1.8\times {10}^{-10}=s^{\prime }\times {10}^{-2}$$s^{\prime }=1.8\times {10}^{-8}\mathrm{mol}{\mathrm{L}}^{-1}$$$=1.8\times {10}^{-8}\mathrm{mol}{\mathrm{dm}}^{-3}(1{\mathrm{dm}}^3=1\mathrm{Liter})$$