Ionic EquilibriumMedium
Question
The solubility of Li3Na3(AlF6)2 is 0.0744 g per 100 ml at 298 K. Calculate the solubility product of the salt (Atomic masses: Li = 7, Na = 23, Al = 27, F = 19).
Options
A.2.56 × 10−22
B.2 × 10−3
C.7.46 × 10−19
D.3.46 × 10−12
Solution
Solubility, $S = \frac{0.00744/372}{100/1000} = 2 \times 10^{- 3}\text{ M}$
$Li_{3}Nal3\left( AlF_{6} \right)_{2}(s) \rightleftharpoons \underset{3S}{3Li^{+}} + \underset{3S}{3Na^{+}} + \underset{2S}{2AlF_{6}^{3 -}}$
$K_{sp} = \left\lbrack Li^{+} \right\rbrack^{3}\left\lbrack Na^{+} \right\rbrack^{3}\left\lbrack AlF_{6}^{3 -} \right\rbrack^{2} = (3S)^{3}.(3S)^{3}.(2S)^{2} = 7.46496 \times 10^{- 19}$
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