If the solubility of lithium sodium hexafluoroaluminate, $L i_{3} N a_{3} (A l F_{6})_{2}$ is 's' $m o l L^{- 1}$ , its solubility product is equal to -

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12s3

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18s3

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2916s8

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Solution

The correct option is D
2916s8

$L i_{3} N a_{3} (A l F_{6})_{2}$ (s) $⇌$ $3 L i^{+} + 3 N a^{+} + 2 A l F_{6}^{6 -}$

Let solubility of salt is 's' $m o l L^{- 1}$

So $[L i^{+}]$ = $3 s$ = $[N a^{+}]$

$[A l F_{6}^{6 -}]$ = $2 s$

So $K_{s p}$ = $[L i^{+}]^{3} [N a^{+}]^{3} [A l F_{6}^{6 -}]^{2}$

= $(3 s)^{3} (3 s)^{3} (2 s)$

= $2916 s^{8}$

Suggest Corrections

Similar questions

{L i}_{3} {N a}_{3} {(A l F_{6})}_{2}

a

m o l

{l i t r e}^{- 1}

L i_{3} N a_{3} (A l F_{6})_{2}

^{- 1}

{L i}_{3} {N a}_{3} {(A l F_{6})}_{2}

S

m o l L^{- 1}

If the solubility of lithium sodium hexafluord aluminate, $L i_{3} N a_{3} (A l F_{6})_{2}$ is a mole/litre, the solubility product is equal to

L i_{3} N a_{3} [A l F_{6}]_{2}

m o l L^{- 1}

[A i F_{6}]^{3 -}

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