Calculate the solubility product of sparingly soluble salt $BA$ at $300 \ K$ if its solubility is $9.1 \times 10^{-3} \ mol \ dm^{-3}$ at same temperature.

  • A
    $9.635 \times 10^{-5}$
  • B
    $9.012 \times 10^{-5}$
  • C
    $8.281 \times 10^{-5}$
  • D
    $7.816 \times 10^{-5}$

Explore More

Similar Questions

The solubility product of $Mg(OH)_2$ is $1.0 \times 10^{-12}$. Concentrated aqueous $NaOH$ solution is added to a $0.01 \ M$ aqueous solution of $MgCl_2$. The $pH$ at which precipitation occurs is

The solubility product of $AgCl$ is $10^{-10} \, M^2$. The minimum volume (in $m^3$) of water required to dissolve $14.35 \, mg$ of $AgCl$ is approximately

The solubility of a sparingly soluble salt $AB_2$ is $1 \times 10^{-6} \text{ mol/dm}^3$. Calculate its solubility product.

Which from the following equations represents the relation between solubility ($S$ in $mol \ L^{-1}$) and solubility product $(K_{sp})$ for a salt $B_3A_2$?

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:

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