Question

$AgBr\left(s\right)+2S_2{O}_3^{2-}\left(aq\right)\rightleftharpoons Ag\left(S_2{O}_3{)}_2^{3-}\right(aq)+B{r}^{-}(aq)$
Given $K_{sp\left(AgBr\right)}=5\times {10}^{-13}$, $K_{f\left(Ag\right(S_2{O}_3{)}_2{)}^{3-}}=5\times {10}^{13}$
What is the molar solubility of $AgBr$ in $0.1{\text{M Na}}_2{\text{S}}_2{\text{O}}_3$?

• A. 0.5 M
• B. 0.45 M
• C. 0.045 M
• D. None of these

Answer 1

The correct option is C 0.045 M

$AgBr\left(s\right)\leftrightharpoons A{g}^{+}+B{r}^{-}$ $K_1=K_{sp}$
$A{g}^{+}+2S_2{O}_3^{2-}\leftrightharpoons Ag(S_2{O}_3{)}_2^{3-}$ $K_2=K_f$
adding both equation we will get
$AgBr\left(s\right)+2S_2{O}_3^{2-}\leftrightharpoons Ag(S_2{O}_3{)}_2^{3-}+B{r}^{-}$ so $K_{eq}=K_f\times K_{sp}=(5\times {10}^{13})(5\times {10}^{-13})=25$
Assume that the solubility of $AgBr$ at equilibrium is `x`
$\frac{\left[Ag\right(S_2{O}_3{)}_2^{3-}]\times [B{r}^{-}]}{[S_2{O}_3^{2-}{]}^2}=\frac{x^2}{(0.1-2x{)}^2}=25$
$\frac{x}{0.1-2x}=5$
$\therefore x=0.0454M$