Question

We have taken a saturated solution of $AgBr,K_{sp}$ is $12 \times 10^{-14}.$ if $10^{-7}\, M$ of $AgNO_3$ are added to $1 \,L$ of this solution, find conductivity (specific conductance) of this solution in terms of $10^{-7}\, Sm^{-1}$ units.
Given, $ \lambda^\circ_{(Ag^+)}=6 \times 10^{-3} Sm^2\, mol^{-1},$
$\lambda^\circ_{(Br^-)}=8 \times 10^{-3} Sm^2\, mol^{-1},$
$\lambda^\circ_{(NO^-_3)}=7 \times 10^{-3} Sm^2\, mol^{-1}.$

Answer 1

The solubility of $AgBr$ in $10^{-7} M\, AgNO _{3}$ solution is determined as

Solving for $S$ gives: $S=3 \times 10^{-7} M$
$\Rightarrow \quad\left[ Br ^{-}\right]=3 \times 10^{-7} M ,\left[ Ag ^{+}\right]=4 \times 10^{-7} M $
$\left.\qquad NO _{3}^{-}\right]=10^{-7} M$
$\Rightarrow \kappa \text { (sp. conductance) }=\kappa_{ Br ^{-}}+\kappa_{ Ag _{+}}+\kappa_{ NO _{3}^{-}} $
$=\left[8 \times 10^{-3} \times 3 \times 10^{-7}+6 \times 10^{-3} \times 4 \times 10^{-7}\right. \left.+7 \times 10^{-3} \times 10^{-7}\right] 1000$
$=24 \times 10^{-7}+24 \times 10^{-7}+7 \times 10^{-7} $
$=55 \times 10^{-7} S m ^{-1} $
$=55\left(\text { in terms of } 10^{-7} S m ^{-1}\right)$