If the solubility of lithium sodium hexafluoro aluminate, ${L i}_{3} {N a}_{3} {(A l F_{6})}_{2}$ is $S$ $m o l L^{- 1}$ , 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 C $2916 S^{8}$
The given aluminate will dissociate into the following ions in water.
${L i}_{3} {N a}_{3} {(A l F_{6})}_{2} ⇌ 3 {L i}^{+} + 3 {N a}^{+} + 2 {(A l F_{6})}^{3 -} S 3 S 3 S 2 S$
The solubility product is calculated as $= {(3 S)}^{3} \times {(3 S)}^{3} \times {(2 S)}^{2} = 2916 S^{8}$

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If the solubility of lithium sodium hexafluoro aluminate, Li3Na3(AlF6)2 is S molL−1, its solubility product is equal to:

The correct option is C 2916S8The given aluminate will dissociate into the following ions in water.Li3Na3(AlF6)2⇌3Li++3Na++2(AlF6)3−S3S3S2SThe solubility product is calculated as =(3S)3×(3S)3×(2S)2=2916S8

If the solubility of lithium sodium hexafluoro aluminate, ${L i}_{3} {N a}_{3} {(A l F_{6})}_{2}$ is $S$ $m o l L^{- 1}$ , its solubility product is equal to: