The molar solubility (in $mol \cdot L^{-1}$) of a sparingly soluble salt $MX_4$ is $S$. The corresponding solubility product is $K_{sp}$. $S$ 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$S = (256 K_{sp})^{1/5}$
- D$S = (128 K_{sp})^{1/4}$
Explore More
Similar Questions
The solubility of $BaSO_4$ in pure water (in $g \, L^{-1}$) is closest to
[Given: $K_{sp}$ for $BaSO_4$ is $1.0 \times 10^{-10}$ at $25^{\circ} C$. Molecular weight of $BaSO_4$ is $233 \, g \, mol^{-1}$]
[Given: $K_{sp}$ for $BaSO_4$ is $1.0 \times 10^{-10}$ at $25^{\circ} C$. Molecular weight of $BaSO_4$ is $233 \, g \, mol^{-1}$]
DifficultKVPY 2018
View SolutionThe solubility of a sparingly soluble salt $AB_2$ in water is $1.0 \times 10^{-5} \ mol \ L^{-1}$. Its solubility product is .......?
Medium
View SolutionThe solubility product of Lead sulphate,$PbSO_4$ is $1.3 \times 10^{-8}$. Calculate its solubility in pure water. The molecular mass of $PbSO_4 = 303 \ g \ mol^{-1}$.
Medium
View SolutionAt $310 \, K$,the solubility of $CaF_{2}$ in water is $2.34 \times 10^{-3} \, g / 100 \, mL$. The solubility product of $CaF_{2}$ is $x \times 10^{-8} \, (mol / L)^{3}$. Find the value of $x$. (Given molar mass: $CaF_{2} = 78 \, g \, mol^{-1}$)
The solubility of $Al(OH)_3$ in a buffer solution at $pH = 8.0$ is [$K_{sp}$ of $Al(OH)_3 = 2.7 \times 10^{-27}$]
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo