If the solubility of lithium sodium hexafluoridoaluminate, $L i_{3} N a_{3} [A l F_{6}]_{2}$ is 's' $m o l L^{- 1}$ , its solubility product is:
(Assume no ionisation of $[A i F_{6}]^{3 -}$ )

18 s^{3}

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186624 s^{8}

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1458 s^{8}

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2916 s^{8}

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Solution

The correct option is D $2916 s^{8}$
$L i_{3} N a_{3} (A l F_{6})_{2} ⇌ 3 L i^{+} + 3 N a^{+} + 2 A l F_{6}^{3 -}$
$\Rightarrow K_{s p} = [L i^{+}]^{3} [N a^{+}]^{3} [A l F_{6}^{3 -}]^{2}$
$= (3 s)^{3} (3 s)^{3} (2 s)^{2}$
$= 2916 s^{8}$ .

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If the solubility of lithium sodium hexafluoridoaluminate, Li3Na3[AlF6]2 is 's' mol L−1, its solubility product is: (Assume no ionisation of [AiF6]3−)

The correct option is D 2916 s8Li3Na3(AlF6)2⇌3Li++3Na++2AlF3−6⇒Ksp=[Li+]3[Na+]3[AlF3−6]2 =(3s)3(3s)3(2s)2 =2916s8 .

If the solubility of lithium sodium hexafluoridoaluminate, $L i_{3} N a_{3} [A l F_{6}]_{2}$ is 's' $m o l L^{- 1}$ , its solubility product is:
(Assume no ionisation of $[A i F_{6}]^{3 -}$ )

The correct option is D $2916 s^{8}$
$L i_{3} N a_{3} (A l F_{6})_{2} ⇌ 3 L i^{+} + 3 N a^{+} + 2 A l F_{6}^{3 -}$
$\Rightarrow K_{s p} = [L i^{+}]^{3} [N a^{+}]^{3} [A l F_{6}^{3 -}]^{2}$
$= (3 s)^{3} (3 s)^{3} (2 s)^{2}$
$= 2916 s^{8}$ .