Electrochemistry Question 13
Question 13 - 01 February - Shift 2
$1 \times 10^{-5} M AgNO_3$ is added to $1 L$ of saturated
solution of $AgBr$. The conductivity of this solution at $298 K$ is $\times 10^{-8} S m^{-1}$.
[Given : $K _{sp}(AgBr)=4.9 \times 10^{-13}$ at $298 K$
$\lambda _{Ag^{+}}^{0}=6 \times 10^{-3} Sm^{2} mol^{-1}$
$\lambda _{Br^{-}}^{0}=8 \times 10^{-3} Sm^{2} mol^{-1}$
$.\lambda _{NO_2^{-}}^{0}=7 \times 10^{-3} Sm^{2} mol^{-1}]$
Show Answer
Answer: (14)
Solution:
$ \begin{aligned} & {[Ag^{\top}]=10^{-3}} \\ & \lfloorNO_3^{-}\rfloor=10^{-5} \end{aligned} $
$[Br^{-}]=\frac{Ksp}{[Ag^{+}]}=4.9 \times 10^{-8}$
$ \Lambda_m^{ath}=\frac{k}{1000 \times M} $
For $Ag^{+}$
$ 6 \times 10^{-3}=\frac{K _{Ag^{+}}}{1000 \times 10^{-5}} $
$K _{Ag+}^{atho}=6 \times 10^{-5}$
$\Rightarrow 6000 \times 10^{-8}$
for $Br^{-}$
$8 \times 10^{-3}=\frac{K _{Br^{-}} \text{mathong }}{1000 \times 4.9 \times 10^{-8}}$
$K _{Br-}=39.2 \times 10^{-8}$
for $NO_3^{-}$
$7 \times 10^{-3}=\frac{K _{NO_3^{-}}}{1000 \times 10^{-5}}$[^0]
Conductivity of solution
$\Rightarrow(6000+7000+39.2) \times 10^{-8}$
$\Rightarrow 13039.2 \times 10^{-8} S m^{-1}$
