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6-2.Equilibrium-II (Ionic Equilibrium)
hard
If the solubility products of $AgCl$ and $AgBr$ are $1.0 \times {10^{ - 8}}\,M$ and $3.5 \times {10^{ - 13}}$ respectively, then the relation between the solubilities (denoted by the symbol$'S'$) of these salts can correctly be represented as
A
$S$of $AgBr$ is less than that of $AgCl$
B
$S$ of $AgBr$ is greater than that of $AgCl$
C
${10^{ - 11}}M$ of $AgBr$ is equal to that of $AgCl$
D
$S$ of $AgBr$ is ${10^6}$ times greater than that of $AgCl$
Solution
(a) $AgCl$ ${K_{sp}} = 1.2 \times {10^{ – 10}}$
$S = \sqrt {1.2 \times {{10}^{ – 10}}} $; $S = 1.09 \times {10^{ – 5}}$
$AgBr $ ${K_{sp}} = 3.5 \times {10^{ – 13}}$
$S = \sqrt {3.5 \times {{10}^{ – 13}}} $$ = 5.91 \times {10^{ – 6}}$
So that $S $ of $AgBr $ is less than that of $ AgCl.$
Standard 11
Chemistry
