(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

If the $pH$ of a saturated solution of $Ba(OH)_2$ is $12$,what is the value of its solubility product constant $K_{sp}$?

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The solubility product of $AgCl$ is $1 \times 10^{-6}$ at $298 \, K$. Its solubility in $mol \, L^{-1}$ would be:

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If the $K_{sp}$ of a saturated solution of $Mg(OH)_2$ is $4 \times 10^{-12}$,then the $pH$ is equal to:

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The solubility product of Lead sulphate,$PbSO_4$ is $1.3 \times 10^{-8}$. Calculate its solubility in pure water. The molecular mass of $PbSO_4 = 303 \ g \ mol^{-1}$.

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At $25\,^{\circ}C$,the solubility product of $Mg(OH)_2$ is $1.0 \times 10^{-11}$. At which $pH$ will $Mg^{2+}$ ions start precipitating in the form of $Mg(OH)_2$ from a solution of $0.001\, M\, Mg^{2+}$ ions?

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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 $NY_{3}$?

Similar Questions

If the $pH$ of a saturated solution of $Ba(OH)_2$ is $12$,what is the value of its solubility product constant $K_{sp}$?

The solubility product of $AgCl$ is $1 \times 10^{-6}$ at $298 \, K$. Its solubility in $mol \, L^{-1}$ would be:

If the $K_{sp}$ of a saturated solution of $Mg(OH)_2$ is $4 \times 10^{-12}$,then the $pH$ is equal to:

The solubility product of Lead sulphate,$PbSO_4$ is $1.3 \times 10^{-8}$. Calculate its solubility in pure water. The molecular mass of $PbSO_4 = 303 \ g \ mol^{-1}$.

At $25\,^{\circ}C$,the solubility product of $Mg(OH)_2$ is $1.0 \times 10^{-11}$. At which $pH$ will $Mg^{2+}$ ions start precipitating in the form of $Mg(OH)_2$ from a solution of $0.001\, M\, Mg^{2+}$ ions?

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