Question

The fixing of photographic plate film involves dissolution of unexposed silver bromide in hypo solution according to the equation that follows.
$AgBr\left(s\right)+2S_2{O}_3^{2-}\left(eq\right)\rightleftharpoons Ag\left(S_2{O}_3{)}_2^{3-}\right(aq)+B{r}^{-}(aq)$. If the molar solubility of $AgBr$in a 0.1 M hypo solution assuming its complete dissociation, is $4.3\times {10}^{-x}$. Then find the value of $x$.
$K_{sp}$ of $AgBr=4.5\times {10}^{-13}$
$K_f$ of $Ag(S_2{O}_3{)}_2^{3-}=2.0\times {10}^{13}$

Answer 1

$AgBr\left(s\right)\leftrightharpoons A{g}^{+}+B{r}^{-}$ $K_1=K_{sp}$
$A{g}^{+}+2S_2{O}_3^{2-}\leftrightharpoons Ag(S_2{O}_3{)}_2^{3-}$ $K_2=K_f$
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 $K_{eq}=K_f\times K_{sp}$
$K_{eq}=9$ lets say m is solubility of $AgBr$ then at eqlbm
$\frac{\left[Ag\right(S_2{O}_3{)}_2^{3-}]\times [B{r}^{-}]}{[S_2{O}_3^{2-}{]}^2}=\frac{m^2}{(0.1-2m{)}^2}=9$
$\frac{m}{0.1-2m}=3$
$\therefore m=0.043=4.3\times {10}^{-x}$
So, the value of `x` will be 2