Electrochemistry Result Question 35
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 L of this solution, find conductivity (specific conductance) of this solution in terms of $10^{-7} \mathrm{Sm}^{-1}$ units.
(2006, 6M)
Given,
$ \begin{aligned} \lambda_{\left(\mathrm{AA}^{+}\right)}^{\circ} & =6 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}, \\ \lambda_{\left(\mathrm{Br}^{-}\right)}^0 & =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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Answer:
Correct Answer: 35. $( 55 )$
Solution:
- The solubility of AgBr in $10^{-7} $ $\mathrm{M}$ $ \mathrm{AgNO}_3$ solution is determined as
$ \begin{aligned} & \mathrm{AgBr} \rightleftharpoons \underset{S+10^{-7}}{\mathrm{Ag}^{+}}+\underset{S}{\mathrm{Br}^{-}} \\ & \mathrm{AgNO}_3 \longrightarrow \underset{S+10^{-7}}{\mathrm{Ag}^{+}}+\underset{10^{-7}}{\mathrm{NO}3^{-}} \\ & K{\text {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{aligned} & \Rightarrow {\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} \\ &=\left.55 \text { (in terms of } 10^{-7} \mathrm{~S} \mathrm{~m}^{-1}\right) \end{aligned} $
BETA
