We have taken a saturated solution of AgBr. K_sp of AgBr is 12x10^-14. If 10^-7 mole of AgNO₃ are added to 1 litre of this solution, find conductivity of this solution in terms of 10^-7 Sm^-1 units. Given λ^o_(Ag⁺) = 6x10^-3 Sm²mol^-1, λ^o_(Br^-) = 8x10^-3 Sm²mol^-1, λ^o_(NO3^-) = 7x10^-3 Sm²mol^-1. Neglect the contribution of solvent.

1. AgBr is sparingly soluble salt. It will dissociate in water as:
$AgBr⟶A{g}^{+}+B{r}^{-}$
$K_{sp}=\left[A{g}^{+}\right]\left[B{r}^{-}\right]=S^2$
2. So $12\times {10}^{-14}=S^2$ or Solubility $S=3.4\times {10}^{-7}$
The concentration of $A{g}^{+}$ and $B{r}^{-}$ in the solution is $3.4\times {10}^{-7}$
3. But when we add the ${10}^{-7}molperlitre$ $AgN{O}_3$ the concentration of $A{g}^{+}$ ions will increase
4. The increased concentration is equal to $3.4\times {10}^{-7}+1\times {10}^{-7}=4.4\times {10}^{-7}$ of $A{g}^{+}$
5. When the concentration of $A{g}^{+}$ has increased the concentration of Br- will reduce due to the common ion effect
To find that
$K_{sp}=\left[A{g}^{+}\right]\left[B{r}^{-}\right]$
$12\times {10}^{-14}=4.4\times {10}^{-7}\times \left[B{r}^{-}\right]$
$or\left[B{r}^{-}\right]=2.7\times {10}^{-7}$
6. So now we are having as $\left[A{g}^{+}\right]=4.4\times {10}^{-7},\left[B{r}^{-}\right]=2.7\times {10}^{-7}and\left[N{O}_3\right]=1\times {10}^{-7}$
7. The conductivity of the solution = The sum of conductivities of ions present in it in according to Kolrausch Law
8. So Conductivity$=4.4\times {10}^{-7}\times 6\times {10}^{-3}\left(forA{g}^{+}\right)+2.7\times {10}^{-7}\times 8\times {10}^{-3}+1\times {10}^{-7}7\times 7\times {10}^{-3}\left(forN{O}_3\right)$

$=26.4\times {10}^{-10}+21.6\times {10}^{-10}+7\times \times {10}^{-10}$

$=55\times {10}^{-10}S{m}^{-1}$