The solubility product of a sparingly soluble | vedclass

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Question

(A) Step $1$: For a salt $BA$, the solubility product $K_{sp} = S^2$, where $S$ is the molar solubility.Step $2$: Calculate molar solubility $S = \sqr

Vedclass · Accepted Answer

$1.52 	imes 10^{-4}$

Vedclass · Answer

(A) Step $1$: For a salt $BA$, the solubility product $K_{sp} = S^2$, where $S$ is the molar solubility.Step $2$: Calculate molar solubility $S = \sqrt{K_{sp}} = \sqrt{6.4 	imes 10^{-13}} = \sqrt{64 	imes 10^{-14}} = 8 	imes 10^{-7} 	ext{ mol dm}^{-3}$.Step $3$: Convert molar solubility to $	ext{g dm}^{-3}$ using the molar mass $M = 190 	ext{ g mol}^{-1}$.Step $4$: Solubility in $	ext{g dm}^{-3} = S 	imes M = (8 	imes 10^{-7} 	ext{ mol dm}^{-3}) 	imes (190 	ext{ g mol}^{-1}) = 1520 	imes 10^{-7} 	ext{ g dm}^{-3} = 1.52 	imes 10^{-4} 	ext{ g dm}^{-3}$.

The solubility product of a sparingly soluble salt $BA$ is $6.4 \times 10^{-13}$. Calculate its solubility in $\text{g dm}^{-3}$. The molar mass of the salt is $190 \text{ g mol}^{-1}$.

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