Chemistry - Ionic Equilibrium Question with Solution | TestHub

${\left[{\text{I}}^{-}\right]}_{{\text{solution}}_{\text{I}}}=10^{-3}\text{M}$

${\left[{\text{Cl}}^{-}\right]}_{{\text{solution}}_{\text{I}}}=10^{-3}\text{M}$

${\left[{\text{Ag}}^{+}\right]}_{\text{solution }\left(\text{II}\right)}=$$\sqrt{{\text{K}}_{\text{sp}}}=10^{-8}\text{M}$

${\left[{\text{I}}^{-}\right]}_{\text{solution }\left(\text{II}\right)}=$$\sqrt{{\text{K}}_{\text{sp}}}=10^{-8}\text{M}$. on mixing the equal volume of two solutions.

$\left[{\text{I}}^{-}\right]=\frac{10^{-3}+10^{-8}}{2}=\frac{10^{-3}\left[1+10^{-5}\right]}{2}$$\approx 5\times {10}^{-4} \text{M}$

$\left[{\text{Cl}}^{-}\right]=\frac{10^{-3}}{2}\text{M}$

$\left[{\text{Ag}}^{+}\right]=$$\frac{{10}^{-16}}{5\times {10}^{-4}} =2\times {10}^{-13}$

${\left[{\text{K}}_{\text{IP}}\right]}_{\text{AgI}}=\frac{10^{-8}}{2}\times \frac{10^{-3}\left[1+10^{-5}\right]}{2}\approx \frac{10^{-11}}{4}>{\text{K}}_{\text{sp}}\text{AgI}$

${\left[{\text{K}}_{\text{IP}}\right]}_{\text{AgCl}}=\frac{10^{-8}}{2}\times \frac{10^{-3}}{2}\approx \frac{10^{-11}}{4}<{\text{K}}_{\text{sp}}\text{AgCl}$

Hence$\text{AgI}$will ppt, but$\text{AgCl}$will not ppt.