The solubility product of $BaSO_4$ is $1.3 \times 10^{-9}$. The solubility of this salt in pure water will be

  • A
    $1.69 \times 10^{-9} \ mol \ L^{-1}$
  • B
    $1.69 \times 10^{-18} \ mol \ L^{-1}$
  • C
    $3.6 \times 10^{-18} \ mol \ L^{-1}$
  • D
    $3.6 \times 10^{-5} \ mol \ L^{-1}$

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 $1.53 \ \Omega^{-1} \ cm^2 \ eq^{-1}$. Then the $K_{sp}$ for $BaSO_4$ will be:

Solubility of a salt $M_2X_3$ is $y \ mol \ dm^{-3}$. The solubility product of the salt will be

The best explanation for the solubility of $MnS$ in dil. $HCl$ is that

If the solubility product of $Zr_3(PO_4)_4$ is denoted by $K_{SP}$ and its molar solubility is denoted by $S$,then which of the following relations between $S$ and $K_{SP}$ is correct?

The moles of $Ag^{+}$ which must be added to decrease the concentration of $Cl^{-}$ from $4 \times 10^{-5} \ M$ to $10^{-5} \ M$ in $100 \ mL$ solution,if $K_{sp}$ for $AgCl$ is $10^{-10} \ M^2$ at $25 \ ^oC$.

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