If the solubility of lithium sodium hexafluoro aluminate ${L i}_{3} {N a}_{3} {(A l F_{6})}_{2}$ is 'S' moles/litre. Its solubility product is equal to:

S^{8}

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12 S^{3}

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18 S^{3}

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2916 S^{8}

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Solution

The correct option is D $2916 S^{8}$
$L i_{3} N a_{3} (A l F_{6})_{2} \to 3 L i^{+} + 3 N a^{+} + 2 (A l F_{6})^{3 -}$
3s 3s 2s

Solubility product= $(3 s)^{3} . (3 s)^{3} . (2 s)^{2}$
= $2916 s^{8}$

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