(B) For $MY$: $K_{SP} = s^2$,so $s = \sqrt{K_{SP}} = \sqrt{6.2 \times 10^{-13}} \approx 7.87 \times 10^{-7} \ M$. For $NY_{3}$: $K_{SP} = 27s^4$,so $s

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$NY_{3}$ is more soluble than $MY$

(B) For $MY$: $K_{SP} = s^2$,so $s = \sqrt{K_{SP}} = \sqrt{6.2 \times 10^{-13}} \approx 7.87 \times 10^{-7} \ M$. For $NY_{3}$: $K_{SP} = 27s^4$,so $s = (K_{SP} / 27)^{1/4} = (6.2 \times 10^{-13} / 27)^{1/4} \approx 1.18 \times 10^{-4} \ M$. Comparing the two,the molar solubility of $NY_{3}$ is greater than that of $MY$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

Medium

Two insoluble salts $MY$ and $NY_{3}$ have the same $K_{SP} = 6.2 \times 10^{-13}$ at room temperature. Which of the following statements is true regarding $MY$ and $NY_{3}$?

A
$MY$ is more soluble than $NY_{3}$
B
$NY_{3}$ is more soluble than $MY$
C
Both have the same solubility
D
The solubility of $MY$ is $1/3$ of $NY_{3}$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product

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Two insoluble salts $MY$ and $NY_{3}$ have the same $K_{SP} = 6.2 \times 10^{-13}$ at room temperature. Which of the following statements is true regarding $MY$ and $NY_{3}$?

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If the solubility of $AgCl$ (molar mass $= 143 \, g/mol$) in water at $25 \, ^\circ C$ is $1.43 \times 10^{-4} \, g/100 \, mL$ of solution,then the value of $K_{sp}$ is?

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The solubility of $CdSO_{4}$ in water is $8.0 \times 10^{-4} \ mol \ L^{-1}$. Its solubility in $0.01 \ M \ H_{2}SO_{4}$ solution is $..... \ \times 10^{-6} \ mol \ L^{-1}$.
(Round off to the Nearest integer) (Assume that solubility is much less than $0.01 \ M$)