The conductivity of a saturated solution of $BaSO_4$ is $3.06 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and its molar conductivity is $1.53 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is the $K_{sp}$ of $BaSO_4$?
- A$4 \times 10^{-12}$
- B$2.5 \times 10^{-9}$
- C$2.5 \times 10^{-13}$
- D$4 \times 10^{-6}$
Explore More
Similar Questions
Formula Solubility product $(K_{sp})$ $AgBrO_3$ $2.5 \times 10^{-10}$ $MgSO_4$ $2.7 \times 10^{-27}$ $CaCO_3$ $2.7 \times 10^{-18}$ $FeC_2O_4$ $2.4 \times 10^{-15}$
Which of the following salts has the maximum solubility?
| Formula | Solubility product $(K_{sp})$ |
|---|---|
| $AgBrO_3$ | $2.5 \times 10^{-10}$ |
| $MgSO_4$ | $2.7 \times 10^{-27}$ |
| $CaCO_3$ | $2.7 \times 10^{-18}$ |
| $FeC_2O_4$ | $2.4 \times 10^{-15}$ |
Medium
View SolutionSolubility product is
Easy
View SolutionThe solubility of $AgBr_{(s)}$,having solubility product $5 \times 10^{-10}$ in $0.2 \ M$ $NaBr$ solution,equals
$1 \ dm^{3}$ solution containing $10^{-5} \ mol$ each of $Cl^{-}$ ions and $CrO_{4}^{2-}$ ions is treated with $10^{-4} \ mol$ of silver nitrate. Which one of the following observations is made?
$[K_{sp} \ Ag_{2}CrO_{4} = 4 \times 10^{-12}]$
$[K_{sp} \ AgCl = 1 \times 10^{-10}]$
$[K_{sp} \ Ag_{2}CrO_{4} = 4 \times 10^{-12}]$
$[K_{sp} \ AgCl = 1 \times 10^{-10}]$
MediumKCET 2010
View SolutionIf the solubility product $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$, the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is:
DifficultAP EAMCET 2014
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