Question

A sample of $AgCl$ was treated with $5mL$ of $1.5M$ ${Na}_2{CO}_3$ solution to give ${Ag}_2{CO}_3$. The remaining solution contained $0.0026g{Cl}^{-}$ per litre. Calculate the solubility product of $AgCl$.
$\left(K_{sp}\right({Ag}_2{CO}_3)=8.2\times {10}^{-12})$

Answer 1

$A{g}_{2}C{O}_3\rightleftharpoons \underset{2S}{2A{g}^{+}}+\underset{S}{C{O}_3^{2-}}$

$\therefore K_{sp}=\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}]=(2S{)}^{2}S=4S^3$

$\therefore 4S^3=8.2\times {10}^{-12}$

$\Rightarrow S=1.27\times {10}^{-4}$

$\therefore \left[A{g}^{+}\right]=1.27\times {10}^{-4}M$

$\left[C{l}^{-}\right]=\frac{0.0026}{35.5}=7.32\times {10}^{-5}M$

$\therefore K_{sp}$ of $AgCl=\left[A{g}^{+}\right]\times \left[C{l}^{-}\right]$

$=1.27\times {10}^{-4}\times 7.32\times {10}^{-5}$

$=9.3\times {10}^{-9}$