Solubility product of BaS{O₄} is 1.3 × {10^{ - 9}} . solubility of this salt in pure water will b

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6-2.Equilibrium-II (Ionic Equilibrium)

medium

The solubility product of $BaS{O_4}$ is $1.3 \times {10^{ - 9}}$. The solubility of this salt in pure water will be

A

$1.69 \times {10^{ - 9}}\,mol\,litr{e^{ - 1}}$

B

$1.69 \times {10^{ - 18}}mol\,litr{e^{ - 1}}$

C

$3.6 \times {10^{ - 18}}\,mol\,litr{e^{ - 1}}$

D

$3.6 \times {10^{ - 5}}\,mol\,litr{e^{ - 1}}$

Solution

(d) $BaS{O_4}⇌\mathop {B{a^{ 2+ }}}\limits_{(S)\,\,\,\,} + \mathop {SO_4^{ 2 – }}\limits_{(S)\,\,\,\,} $

${K_{sp}} = {S^2};\,\,S = \sqrt {{K_{sp}}} = \sqrt {1.3 \times {{10}^{ – 9}}} $

$ = 3.6 \times {10^{ – 5}}\,\,\,mol/litre$

Standard 11

Chemistry

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