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

a^{8}

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12 a^{3}

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18 a^{3}

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2916 a^{8}

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Solution

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

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If the solubility of lithium sodium hexafluord aluminate, Li3Na3(AlF6)2 is a mole/litre, the solubility product is equal to

The correct option is D 2916a8Li3Na3(AlF6)2→3Li++3Na++2AlF3–6 A 3a 3a 2aKsp=(3a)3 (3a)3 (2a)2=2916a8

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

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