Question

We have taken a saturated solution of $AgBr,K_{sp}$ is $12\times 10^{-14}.$ if $10^{-7}M$ of $AgN{O}_3$ are added to $1L$ of this solution, find conductivity (specific conductance) of this solution in terms of $10^{-7}S{m}^{-1}$ units.
Given, ${\lambda }_{(A{g}^{+})}^{\circ }=6\times 10^{-3}S{m}^{2}mo{l}^{-1},$
${\lambda }_{(B{r}^{-})}^{\circ }=8\times 10^{-3}S{m}^{2}mo{l}^{-1},$
${\lambda }_{(N{O}_3^{-})}^{\circ }=7\times 10^{-3}S{m}^{2}mo{l}^{-1}.$

Answer 1

The solubility of $AgBr$ in $10^{-7}MAgN{O}_3$ solution is determined as

Solving for $S$ gives: $S=3\times 10^{-7}M$
$\Rightarrow \left[B{r}^{-}\right]=3\times 10^{-7}M,\left[A{g}^{+}\right]=4\times 10^{-7}M$
$N{O}_3^{-}]=10^{-7}M$
$\Rightarrow \kappa \text{ (sp. conductance) }={\kappa }_{B{r}^{-}}+{\kappa }_{A{g}_{+}}+{\kappa }_{N{O}_3^{-}}$
$=[8\times 10^{-3}\times 3\times 10^{-7}+6\times 10^{-3}\times 4\times 10^{-7}+7\times 10^{-3}\times 10^{-7}]1000$
$=24\times 10^{-7}+24\times 10^{-7}+7\times 10^{-7}$
$=55\times 10^{-7}S{m}^{-1}$
$=55\left(\text{ in terms of }10^{-7}S{m}^{-1}\right)$