Electrochemistry 2 Question 36
35. We have taken a saturated solution of $\mathrm{AgBr}, K_{\mathrm{sp}}$ is $12 \times 10^{-14}$. If $10^{-7} \mathrm{M}$ of $\mathrm{AgNO}_{3}$ are added to $1 \mathrm{~L}$ of this solution, find conductivity (specific conductance) of this solution in terms of $10^{-7} \mathrm{Sm}^{-1}$ units.
$(2006,6 \mathrm{M})$ Given,
$$ \begin{aligned} & \lambda_{\left(\mathrm{Ag}^{+}\right)}^{\circ}=6 \times 10^{-3} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}, \ & \lambda_{\left(\mathrm{Br}^{-}\right)}^{\circ}=8 \times 10^{-3} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}, \ & \lambda_{\left(\mathrm{NO}_{3}^{-}\right)}^{\circ}=7 \times 10^{-3} \mathrm{Sm}^{2} \mathrm{~mol}^{-1} . \end{aligned} $$
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Solution:
- The solubility of $\mathrm{AgBr}$ in $10^{-7} \mathrm{M} \mathrm{AgNO}_{3}$ solution is determined as
$$ \begin{aligned} & \mathrm{AgBr} \rightleftharpoons \underset{S+10^{-7}}{\mathrm{Ag}^{+}}+\mathrm{Br}^{-} \ & \mathrm{AgNO}{3} \longrightarrow \underset{S+10^{-7}}{\mathrm{Ag}^{+}}+\underset{10^{-7}}{\mathrm{NO}{3}^{-}} \ & K_{\mathrm{sp}}=14 \times 10^{-14}=S\left(S+10^{-7}\right) \end{aligned} $$
Solving for $S$ gives : $\quad S=3 \times 10^{-7} \mathrm{M}$
$$ \begin{array}{ll} \Rightarrow \quad & {\left[\mathrm{Br}^{-}\right]=3 \times 10^{-7} \mathrm{M},} \ & {\left[\mathrm{Ag}^{+}\right]=4 \times 10^{-7} \mathrm{M},} \ & {\left[\mathrm{NO}{3}^{-}\right]=10^{-7} \mathrm{M}} \ \Rightarrow \kappa(\text { sp. conductance })=\kappa{\mathrm{Br}^{-}}+\kappa_{\mathrm{Ag}^{+}}+\kappa_{\mathrm{NO}_{3}^{-}} \ = & {\left[8 \times 10^{-3} \times 3 \times 10^{-7}+6 \times 10^{-3} \times 4 \times 10^{-7}\right.} \ & \left.+7 \times 10^{-3} \times 10^{-7}\right] 1000 \ & =24 \times 10^{-7}+24 \times 10^{-7}+7 \times 10^{-7} \ & =55 \times 10^{-7} \mathrm{~S} \mathrm{~m}^{-1} \ & =55\left(\text { in terms of } 10^{-7} \mathrm{~S} \mathrm{~m}^{-1}\right) \end{array} $$
