Q35 / 120 · JEE Advanced 2006 · Chemistry
We have taken a saturated solution of $\mathrm{AgBr}, K_{\mathrm{sp}}$ is $12 \times 10^{-14}$. If $10^{-7}$ mole of $\mathrm{AgNO}_3$ are added to 1 litre of this solution, find conductivity (specific conductance) of this solution in terms of $10^{-7} \mathrm{Sm}^{-1}$ units.
$$
\text { Given, } \begin{aligned}
\lambda_{\left(\mathrm{Ag}^{+}\right)}^{\circ} & =6 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}, \lambda_{\left(\mathrm{Br}^{-}\right)}^{\circ}=8 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}, \\
\lambda_{\left(\mathrm{NO}_3^{-}\right)} & =7 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1},
\end{aligned}
$$
Correct answer: 55
Solution
Suppose the solubility of $\mathrm{AgBr}$ in $10^{-7} \mathrm{MAgNO}_3$ is $\mathrm{mol} / \mathrm{L}$
$$
\begin{aligned}
& \underset{s \mathrm{~mol} / \mathrm{L}}{\mathrm{AgBr}} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-} \\
& \mathrm{AgNO}_3 \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{NO}_3^{-} \\
& 10^{-7} \mathrm{M} \quad 10^{-7} \mathrm{M} \times 10^{-7} \mathrm{M} \\
& \text { Total }\left[\mathrm{Ag}^{+}\right]=\left(s+10^{-7}\right) \mathrm{M} \\
& K_{\mathrm{sp}} \text { of } \mathrm{AgBr}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right] \\
& 12 \times 10^{-14}=\left(s+10^{-7}\right)(s)=s^2+10^{-7} s \\
& s^2+10^{-7} s-12 \times 10^{-14}=0 \\
&
\end{aligned}
$$
On solving $s=3 \times 10^{-7} \mathrm{M}$
Hence, $\left[\mathrm{Br}^{-}\right]=3 \times 10^{-7} \mathrm{M}=3 \times 10^{-7} \times 10^3 \mathrm{~m}^3=3 \times 10^{-4} \mathrm{~m}^3$
$$
\left[\mathrm{Ag}^{+}\right]=\left(3 \times 10^{-7}+10^{-7}\right) \mathrm{M}=4 \times 10^{-7} \times 10^3 \mathrm{~m}^3=4 \times 10^{-4} \mathrm{~m}^3
$$
$$
\left[\mathrm{NO}_3^{-}\right]=10^{-7} \mathrm{M}=10^{-7} \times 10^{+3}=1 \times 10^{-4} \mathrm{~m}^3
$$
$$
\because \quad \lambda=\frac{k}{C} \text { or } k=\lambda \times C
$$
$$
\begin{aligned}
\therefore \quad k_{\mathrm{Br}^{-}} & =3 \times 10^{-4} \times 8 \times 10^{-3} \mathrm{Sm}^{-1}=24 \times 10^{-7} \mathrm{Sm}^{-1} \\
k_{\mathrm{Ag}^{+}} & =4 \times 10^{-4} \times 6 \times 10^{-3} \mathrm{Sm}^{-1}=24 \times 10^{-7} \mathrm{Sm}^{-1} \\
k_{\mathrm{NO}_3^{-}} & =1 \times 10^{-4} \times 7 \times 10^{-3}=7 \times 10^{-7} \mathrm{Sm}^{-1} \\
k_{\text {Total }} & =k_{\mathrm{Br}^{-}}+k_{\mathrm{Ag}^{+}}+k_{\mathrm{NO}_3^{-}} \\
& =\left(24 \times 10^{-7}+24 \times 10^{-7}+7 \times 10^{-7} \mathrm{Sm}^{-1}\right. \text { ) } \\
& =55 \times 10^{-7} \mathrm{Sm}^{-1} \\
& \left.=\mathbf{5 5} \mathbf{S m}^{-1} \text { (in term of } 10^{-7} \mathrm{Sm}^{-1}\right)
\end{aligned}
$$
Hence, the correct answer is 55.