6-2.Equilibrium-II (Ionic Equilibrium)
medium

The solubility product of a sparingly soluble salt $A{X_2}$ is $3.2 \times {10^{ - 11}}$. Its solubility (in moles / litres) is

A

$2 \times {10^{ - 4}}$

B

$4 \times {10^{ - 4}}$

C

$5.6 \times {10^{ - 6}}$

D

$3.1 \times {10^{ - 4}}$

(AIPMT-2004)

Solution

(a) $A{X_2} \to \mathop A\limits_x \;\; + \;\;\mathop {2X}\limits_{2x} $

${K_{sp}} = 4{x^3}$ ; $x = \sqrt[3]{{\frac{{3.2 \times {{10}^{ – 11}}}}{4}}}$; $x = 2 \times {10^{ – 4}}$ $mole/litre.$

Standard 11
Chemistry

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