Question

A sample of $AgCl$ was treated with $500mL$ of $1.5M$ $N{a}_{2}C{O}_3$ solution to give $A{g}_{2}C{O}_3$. The remaining solution contained $0.0026g$ of $C{l}^{-}$ per litre. The solubility product of $AgCl$ is if $K_{sp}\left(A{g}_{2}C{O}_3\right)=8.2\times {10}^{-12}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The double displacement reaction is:

$2AgCl+N{a}_{2}C{O}_3$$\to A{g}_{2}C{O}_3+2NaCl$

$K_{sp}\left(A{g}_{2}C{O}_3\right)$ = $8.2\times {10}^{-12}\left(given\right)$ = $\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}]$

Sodium Carbonate will completely dissociate into $N{a}^{+}$ and $C{O}_3^{2-}$ ions. $\left[C{O}_3^{2-}\right]$ = [$N{a}_{2}C{O}_3$] = $1.5M$

From this , we can arrive at $\left[A{g}^{+}\right]$ = $(\frac{K_{sp}\left(A{g}_{2}C{O}_3\right)}{\left[C{O}_3^{2-}\right]}{)}^{0.5}$ = $2.34\times {10}^{-6}M$

Amount of $C{l}^{-}$ remaining = $00026g$ = $\frac{0.0026}{35.5}mol{L}^{-1}$ = $7.33\times {10}^{-5}M$

We need to figure out $K_{sp}\left(AgCl\right)$ = $\left[A{g}^{+}\right]\left[C{l}^{-}\right]$ = $2.34\times {10}^{-6}M\times 7.33\times {10}^{-5}M$ = $1.71\times {10}^{-10}$