The solubility product of $CaCO_3$ is $5 \times 10^{-9}$. Its solubility is equal to .......

  • A
    $2.2 \times 10^{-9}$
  • B
    $7 \times 10^{-5}$
  • C
    $2.5 \times 10^{-4}$
  • D
    $2.5 \times 10^{-5}$

Explore More

Similar Questions

Calculate the minimum concentration of sulphate ion required to precipitate $BaSO_4$ in a solution containing $10^{-4} \ mol/L$ of $Ba^{2+}$. $(K_{sp} \ BaSO_4 = 4 \times 10^{-10})$

The $K_{sp}$ for bismuth sulphide $(Bi_{2}S_{3})$ is $1.08 \times 10^{-73}$. The solubility of $Bi_{2}S_{3}$ in $mol \ L^{-1}$ at $298 \ K$ is ......

The solubility product constants $(K_{sp})$ of $CuS$,$Ag_2S$,and $HgS$ are $10^{-37}$,$10^{-44}$,and $10^{-54}$ respectively. The order of their solubility is:

The solubility product constants of $Ag_2CrO_4$ and $AgBr$ are $32x$ and $4y$ respectively at $298 \text{ K}$. The value of $(\frac{\text{molarity of } Ag_2CrO_4}{\text{molarity of } AgBr})$ can be expressed as :

The solubility product of $BaCl_2$ is $4 \times 10^{-9}$. Its solubility in moles/litre would be:

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