The molar solubility of $PbI_2$ in $0.2 \ M \ Pb(NO_3)_2$ solution in terms of solubility product of $PbI_2$ is

  • A
    $\sqrt{K_{sp}/0.2}$
  • B
    $\sqrt{K_{sp}/0.4}$
  • C
    $\sqrt[3]{K_{sp}/0.8}$
  • D
    $\sqrt{K_{sp}/0.8}$

Explore More

Similar Questions

The solubility of a sparingly soluble salt $A_x B_y$ in water at $25\,^{\circ}C$ is $1.4 \times 10^{-4} \ M$. The solubility product is $1.1 \times 10^{-11}$. The values of $x$ and $y$ may be

Difficult
View Solution

Solubility of a binary sparingly soluble salt is $1.12 \times 10^{-4} \ g \ dm^{-3}$. Calculate its solubility product $(K_{sp})$ if the molar mass of the salt is $112 \ g \ mol^{-1}$.

How many grams of $CaC_2O_4$ will dissolve in distilled water to make one litre of saturated solution (in $g$)? (Solubility product of $CaC_2O_4$ is $2.5 \times 10^{-9} \ mol^2 \ L^{-2}$ and its molecular weight is $128$)

In the reaction $Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+}_{(aq)} + 3OH^{-}_{(aq)}$, if the concentration of $OH^{-}$ ions is decreased by $\frac{1}{4}$ times, then the equilibrium concentration of $Fe^{3+}$ will increase by: (in $\times$)

The $K_{sp}$ of an electrolyte $AB$ is $1 \times 10^{-10}$. If $[A^{+}] = 10^{-5} \ M$,which concentration of $B^{-}$ will not result in the precipitation of $AB$?

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