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Solubility Product

The solubilit...

Question

The solubility of ${L i}_{3} {N a}_{3} {(A l F_{6})}_{2}$ is $0.075 g$ per $100 m L$ at $25^{o} C$ . $K_{s p}$ of salt at $25^{o} C$ is $X \times 10^{- 19}$ . The value of $X$ is:

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Solution

First convert the solubility from g/100mL to M.
$S = \frac{0.075 g}{100 m L} = \frac{0.75 g}{1 L} = \frac{0.75 g / L}{371.7 g / m o l} = 2 \times 10^{- 3} M$ .
The expression for the solubility product is
$K_{s p} = [L i^{+}]^{3} [N a^{+}]^{3} [A l F_{6}^{3 -}]^{2} = (3 S)^{3} (3 S)^{3} (2 S)^{2}$
$K_{s p} = 2916 S^{8} = 2916 (2 \times 10^{- 3})^{8} = 8 \times 10^{- 19} = x \times 10^{- 19}$ .
Thus $X = 8$ .

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