JEE MAIN - Chemistry (2023 - 1st February Evening Shift - No. 17)
$$1 \times 10^{-5} ~\mathrm{M} ~\mathrm{AgNO}_{3}$$ is added to $$1 \mathrm{~L}$$ of saturated solution of $$\mathrm{AgBr}$$. The conductivity of this solution at $$298 \mathrm{~K}$$ is _____________ $$\times 10^{-8} \mathrm{~S} \mathrm{~m}^{-1}$$.
[Given : $$\mathrm{K}_{\mathrm{SP}}(\mathrm{AgBr})=4.9 \times 10^{-13}$$ at $$298 \mathrm{~K}$$
$$ \begin{aligned} & \lambda_{\mathrm{Ag}^{+}}^{0}=6 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{Br}^{-}}^{0}=8 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \left.\lambda_{\mathrm{NO}_{3}^{-}}^{0}=7 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\right] \end{aligned} $$
Answer
13039
Explanation
$$
\begin{aligned}
& \operatorname{AgBr}(\mathrm{S}) \rightleftharpoons \underset{\left(10^{-5}+\mathrm{x}\right)}{\rightleftharpoons \mathrm{Ag}^{+}}(\mathrm{aq})+\mathrm{Br}_{\mathrm{x}}^{-}(\mathrm{aq}) \\\\
& x\left(x+10^{-5}\right)=4.9 \times 10^{-13} \\\\
& x \simeq 4.9 \times 10^{-8} \mathrm{M} \\\\
& \lambda_{\mathrm{Ag}^{+}}^{\circ}=6 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\
& \lambda_{\mathrm{Br}^{-}}^0=8 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\\\
& \mathrm{~K}_{\text {solution }}=\mathrm{K}_{\mathrm{Ag}^{+}}+\mathrm{K}_{\mathrm{Br}^{-}}+\mathrm{K}_{\mathrm{NO}_3^{-}} \\\\
& =6 \times 10^{-3} \times 10^{-5} \times 10^3+8 \times 10^{-3} \times 4.9 \times 10^{-8} \times 10^3 \\\\
& +7 \times 10^{-3} \times 10^{-5} \times 10^3 \\\\
& =(6000+39.2+7000) \times 10^{-8} \\\\
& =13039.2 \times 10^{-8} ~\mathrm{Sm}^{-1}
\end{aligned}
$$
Comments (0)
