Solubility (in molarity) of sparingly soluble salts $MX, MX_2$ and $MX_3$ in water are same. The order of $K_{sp}$ of $MX, MX_2$ and $MX_3$ is

  • A
    $K_{sp}(MX) = K_{sp}(MX_2) = K_{sp}(MX_3)$
  • B
    $K_{sp}(MX) < K_{sp}(MX_2) < K_{sp}(MX_3)$
  • C
    $K_{sp}(MX) = \frac{1}{2}K_{sp}(MX_2) = \frac{1}{3}K_{sp}(MX_3)$
  • D
    $K_{sp}(MX) > K_{sp}(MX_2) > K_{sp}(MX_3)$

Explore More

Similar Questions

At $T(K)$,the solubility product of $AgBr$ is $4 \times 10^{-13}$. What is its solubility in $0.1 \ M$ $KBr$ solution?

For a sparingly soluble strong electrolyte $AgIO_3$ (molar mass = $283 \, g/mol$),the equilibrium in a saturated solution is given by $AgIO_3(s) \rightleftharpoons Ag^+(aq) + IO_3^-(aq)$. If the solubility product constant $K_{sp}$ of $AgIO_3$ at a given temperature is $1.0 \times 10^{-8}$,how many grams of $AgIO_3$ are contained in $100 \, mL$ of its saturated solution?

Difficult
View Solution

$A$ solution is $0.1 \ M$ in $Cl^{-}$ and $0.001 \ M$ in $CrO_{4}^{2-}$. Solid $AgNO_{3}$ is gradually added to it. Assuming that the addition does not change in volume and $K_{sp}(AgCl) = 1.7 \times 10^{-10} \ M^{2}$ and $K_{sp}(Ag_{2}CrO_{4}) = 1.9 \times 10^{-12} \ M^{3}$. Select the correct statement from the following:

If the solubility product of $CaSO_4$ is $2.5 \times 10^{-5}$,then its solubility will be .......

The solubility product $({K_{sp}})$ of $BaCO_3$ is $1.5 \times 10^{-9}$. At what concentration of $Ba^{2+}$ ions will precipitation begin when solid $Ba(NO_3)_2$ is added to a $10^{-4} \ M$ solution of $Na_2CO_3$?

Difficult
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