Question

Solid $AgN{O}_3$ is gradually added to a solution containing $C{l}^{-}$ and $I^{-}$. If $K_{sp}$ values of $AgCl$ and $AgI$ are respectively $1.7\times {10}^{-10}$ and $1.5\times {10}^{-16}$, which one will precipitate first? Also, find the relative concentration of $\left[I^{-}\right]$ to $\left[C{l}^{-}\right]$ just before the precipitation of $AgCl$.

Answer 1

$Ksp\left(AgCl\right)=1.7\times {10}^{-10}$

$Ksp\left(AgI\right)=1.5\times {10}^{-16}$

That ion will precipitate first which has lower

$Ksp$ value of $Agl$ is smaller so $AgI$ will ppt first.

For ppt of $AgCl$ to start

$Q_{AgCl}=Ks{p}_{AgCl}$

$\left[A{g}^{+}\right]\left[C{l}^{-}\right]=1.7\times {10}^{-10}$

$\left[C{l}^{-}\right]=\frac{1.7\times {10}^{-10}}{\left[A{g}^{+}\right]}$

Also $\left[I^{-}\right]$ at this instant $=\frac{Ksp\left(AgI\right)}{\left[A{g}^{+}\right]}=\frac{1.5\times {10}^{-16}}{\left[A{g}^{+}\right]}$

$\therefore \frac{\left[I^{-}\right]}{C{l}^{-}}=\frac{1.5\times {10}^{-16}}{\left[A{g}^{+}\right]}\times \frac{\left[A{g}^{+}\right]}{1.7\times {10}^{-10}}$

$\frac{\left[I^{-}\right]}{\left[C{l}^{-}\right]}=\frac{1.5}{1.7}\times {10}^{-6}\approx {10}^{-6}$