Ionic EquilibriumHard

Question

The solubility of sparingly soluble salt A3B2 (molar mass = ‘M’ g/mol) in water is ‘x’ g/L. The ratio of molar concentration of B3– to the solubility product of the salt is

Options

A.$\frac{108x^{5}}{M^{5}}$
B.$\frac{x^{4}}{108M^{4}}$
C.$\frac{x^{4}}{54M^{4}}$
D.$\frac{M^{4}}{54x^{4}}$

Solution

$A_{3}B_{2}(s) \rightleftharpoons \underset{\frac{3x}{M}}{3A^{2 +}} + \underset{\frac{3x}{M}}{2B^{3 -}}\left( \text{Solubility =}\frac{x}{M}\text{ mol/l} \right)$

$K_{sp} = \left\lbrack A^{2 +} \right\rbrack^{3}\left\lbrack B^{3 -} \right\rbrack^{2} = \left( \frac{3x}{M} \right)^{3}.\left( \frac{2x}{M} \right)^{2} = \frac{108x^{5}}{M^{5}} $$$\therefore\frac{\left\lbrack B^{3 -} \right\rbrack}{K_{sp}} = \frac{2x/M}{108x^{5}/M^{5}} = \frac{M^{4}}{54x^{4}}$$

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