Question

A sample of $AgCl$ was treated with $5.00mL$ of $1.5MN{a}_{2}C{O}_3$ solution to give $A{g}_{2}C{O}_3$ . The remaining solution contained $0.0026g$ of $C{l}^{-}$ ions per litre. Calculate the solubility product of $AgCl[K_{sp}(A{g}_{2}C{O}_3)=8.2\times 10^{-12}]$

Answer 1

$2AgCl(g)+C{O}_3^{2-}\rightleftharpoons A{g}_{2}C{O}_3(g)+2C{l}^{-}$
$K=\frac{[C{l}^{-}{]}^2}{[C{O}_3^{2-}]}=\frac{[C{l}^{-}{]}^2}{[C{O}_3^{2-}]}\times \frac{[A{g}^{+}{]}^2}{[A{g}^{+}{]}^2}=\frac{[K_{sp}(AgCl){]}^2}{k_{sp}(A{g}_{2}C{O}_3)}$
$[C{l}^{-}]=\frac{0.0026}{35.5}M=7.3\times 10^{-5}$ M
The above concentration of $C{l}^{-}$ indicates that $[C{O}_3^{2-}$ ] remains almost unchanged.
$\frac{7.3\times 10^{-5}}{1.5}=\frac{[K_{sp}(AgCl){]}^2}{8.2\times 10^{-12}}$
$K_{sp}=(AgCl)=2\times 10^{-8}$