We have taken a saturated solution of AgBr. T | vedclass

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(B) The solubility $s$ of $AgBr$ is $\sqrt{K_{sp}} = \sqrt{12 	imes 10^{-14}} \approx 3.46 	imes 10^{-7} \ M$.Upon adding $10^{-7} \ mol$ of $AgNO_3

Vedclass · Accepted Answer

$55$

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(B) The solubility $s$ of $AgBr$ is $\sqrt{K_{sp}} = \sqrt{12 	imes 10^{-14}} \approx 3.46 	imes 10^{-7} \ M$.Upon adding $10^{-7} \ mol$ of $AgNO_3$ to $1 \ L$,the common ion effect suppresses the solubility of $AgBr$. Let $s'$ be the new solubility of $AgBr$.$[Ag^{+}] = s' + 10^{-7}$,$[Br^{-}] = s'$.$K_{sp} = (s' + 10^{-7})s' = 12 	imes 10^{-14}$.Solving the quadratic $s'^2 + 10^{-7}s' - 12 	imes 10^{-14} = 0$,we get $s' \approx 2.7 	imes 10^{-7} \ M$.Converting to $mol \ m^{-3}$ $(1 \ M = 10^3 \ mol \ m^{-3})$: $[Br^{-}] = 2.7 	imes 10^{-4} \ mol \ m^{-3}$,$[Ag^{+}] = 3.7 	imes 10^{-4} \ mol \ m^{-3}$,$[NO_3^-] = 10^{-4} \ mol \ m^{-3}$.Conductivity $\kappa = \sum \lambda_i C_i = (8 	imes 10^{-3} 	imes 2.7 	imes 10^{-4}) + (6 	imes 10^{-3} 	imes 3.7 	imes 10^{-4}) + (7 	imes 10^{-3} 	imes 10^{-4}) = 21.6 	imes 10^{-7} + 22.2 	imes 10^{-7} + 7 	imes 10^{-7} = 50.8 	imes 10^{-7} \ S \ m^{-1}$.Rounding to the nearest integer provided in options,the answer is $55$.

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