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

Open in App

Solution

$L i_{3} N a_{3} (A l F_{6})_{2} ⇌ 3 L^{+} + 3 N a^{+} + 2 A l F_{6}^{3 -}$
35 35 25
$∴ K s p = (35)^{3} \times (35)^{3} \times (25)^{2}$
$= 21163^{3}$

Suggest Corrections

Similar questions

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

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

S

m o l L^{- 1}

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

a

m o l

{l i t r e}^{- 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}

^{'} s^{'}

Join BYJU'S Learning Program