(D) Let the solubility of all salts be $s$. For $M_2X$: $M_2X \rightleftharpoons 2M^+ + X^{2-}$. $K_{sp} = (2s)^2(s) = 4s^3$. For $QY_2$: $QY_2 \right

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$K_{sp}(M_2X) = K_{sp}(QY_2) = K_{sp}(PZ_2)$

(D) Let the solubility of all salts be $s$. For $M_2X$: $M_2X \rightleftharpoons 2M^+ + X^{2-}$. $K_{sp} = (2s)^2(s) = 4s^3$. For $QY_2$: $QY_2 \rightleftharpoons Q^{2+} + 2Y^-$. $K_{sp} = (s)(2s)^2 = 4s^3$. For $PZ_2$: $PZ_2 \rightleftharpoons P^{2+} + 2Z^-$. $K_{sp} = (s)(2s)^2 = 4s^3$. Since all salts have the same solubility $s$,their $K_{sp}$ values are equal: $K_{sp}(M_2X) = K_{sp}(QY_2) = K_{sp}(PZ_2)$.

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

Medium

If the solubility of salts $M_2X$,$QY_2$,and $PZ_2$ is the same,what is the relationship between their $K_{sp}$ values?

A
$K_{sp}(M_2X) > K_{sp}(QY_2) > K_{sp}(PZ_2)$
B
$K_{sp}(M_2X) = K_{sp}(QY_2) < K_{sp}(PZ_2)$
C
$K_{sp}(M_2X) > K_{sp}(QY_2) = K_{sp}(PZ_2)$
D
$K_{sp}(M_2X) = K_{sp}(QY_2) = K_{sp}(PZ_2)$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

6-2.Equilibrium-II (Ionic Equilibrium)

Subtopic

Solubility product

Express the relationship between $K_{sp}$ and solubility $(S)$ for the salt $Ca_3(PO_4)_2$.

Easy

View Solution

The solubility of $AgCl$ in $0.1 \text{ M } NaCl$ is $S \text{ mol/L}$. If the solubility product of $AgCl$ is $1.8 \times 10^{-10}$, then $S$ is approximately:

DifficultMHT CET 2026

View Solution

What is the concentration of $Ba^{2+}$ when $BaF_2$ $(K_{sp} = 1.0 \times 10^{-6})$ begins to precipitate from a solution that is $0.30 \, M \ F^{-}$?

Medium

View Solution

The solubility product of a binary weak electrolyte is $4 \times 10^{-10}$ at $298 \ K$. Its solubility in $\text{mol} \ dm^{-3}$ at the same temperature is

Medium

View Solution

$25.0 \ mL$ of $0.050 \ M \ Ba(NO_3)_2$ is mixed with $25.0 \ mL$ of $0.020 \ M \ NaF$. The $K_{sp}$ of $BaF_2$ is $0.5 \times 10^{-6}$ at $298 \ K$. The ratio of $[Ba^{2+}][F^{-}]^2$ to $K_{sp}$ is $.......$. (Nearest integer)

DifficultJEE MAIN 2023

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If the solubility of salts $M_2X$,$QY_2$,and $PZ_2$ is the same,what is the relationship between their $K_{sp}$ values?

Similar Questions

Express the relationship between $K_{sp}$ and solubility $(S)$ for the salt $Ca_3(PO_4)_2$.

The solubility of $AgCl$ in $0.1 \text{ M } NaCl$ is $S \text{ mol/L}$. If the solubility product of $AgCl$ is $1.8 \times 10^{-10}$, then $S$ is approximately:

What is the concentration of $Ba^{2+}$ when $BaF_2$ $(K_{sp} = 1.0 \times 10^{-6})$ begins to precipitate from a solution that is $0.30 \, M \ F^{-}$?

The solubility product of a binary weak electrolyte is $4 \times 10^{-10}$ at $298 \ K$. Its solubility in $\text{mol} \ dm^{-3}$ at the same temperature is

$25.0 \ mL$ of $0.050 \ M \ Ba(NO_3)_2$ is mixed with $25.0 \ mL$ of $0.020 \ M \ NaF$. The $K_{sp}$ of $BaF_2$ is $0.5 \times 10^{-6}$ at $298 \ K$. The ratio of $[Ba^{2+}][F^{-}]^2$ to $K_{sp}$ is $.......$. (Nearest integer)

Vedclass Test Series

Exam Paper Generator

Online Exam Module