Chemistry - Ionic Equilibrium Question with Solution | TestHub

Let solubility of $\mathrm{AgCl}=x\mathrm{mole}/\mathrm{L}$ $$\begin{array}{rl}& \mathrm{AgCl}\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}\\ & \text{i.e., }{K}_{\mathrm{sp}(\mathrm{AgCl})}=x\times x\\ & \mathrm{KCl}⟶{\mathrm{K}}^{+}+{\mathrm{Cl}}^{-}\\ & \left[{\mathrm{Cl}}^{-}\right]\text{from }\mathrm{KCl}=0.1\mathrm{m}\\ & \text{ Total }\left[\mathrm{Cl}{}^{-}\right]\text{in solution }=x+0.1\\ & K_{sp}(\mathrm{AgCl})=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]=x(x+0.1)\\ & 1.0\times 10^{-10}=x(x+0.1)\\ & 1.0\times 10^{-10}=x^2+0.1x\\ & 1.0\times 10^{-10}=0.1x\left(\text{ as }{x}^2<<1\right)\\ & x=1.0\times 10^{-9}\mathrm{mol}/\mathrm{L}\end{array}$$