If the molar solubility (in $mol \cdot L^{-1}$) of a sparingly soluble salt $AB_4$ is $S$ and the corresponding solubility product is $K_{sp}$,then $S$ in terms of $K_{sp}$ is given by the relation:

  • A
    $S = \left(\frac{K_{sp}}{128}\right)^{1/4}$
  • B
    $S = \left(\frac{K_{sp}}{256}\right)^{1/5}$
  • C
    $S = \left(256 K_{sp}\right)^{1/5}$
  • D
    $S = \left(128 K_{sp}\right)^{1/4}$

Explore More

Similar Questions

The conductivity of a saturated solution of $BaSO_4$ is $3.06 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and its equivalent conductance is $1.53 \ \Omega^{-1} \ cm^{2} \ equivalent^{-1}$. The $K_{sp}$ of the $BaSO_4$ will be

Assign $A, B, C, D$ from the given type of reaction.
$CaCl_2 + Na_2C_2O_4 \longrightarrow CaC_2O_4\downarrow + 2NaCl$

The values of $K_{sp}$ of $CaCO_{3}$ and $CaC_{2}O_{4}$ are $4.7 \times 10^{-9}$ and $1.3 \times 10^{-9}$ respectively at $25 \, ^oC$. If the mixture of these two is washed with water,what is the concentration of $Ca^{2+}$ ions in water $\dots \times 10^{-5} \, M$?

If the solubility product of $CuS$ is $9 \times 10^{-16}$,then what will be the maximum molarity of $CuS$ in an aqueous solution?

When solid $Pb(NO_3)_2$ is added to $1 \, L$ of $H_2SO_4$ $(1 \times 10^{-3} \, M)$ to make the concentration of $Pb(NO_3)_2$ equal to $0.002 \, M$,what will happen? $(K_{sp} \text{ of } PbSO_4 = 1.3 \times 10^{-8})$

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