Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The solubility product of a sparingly soluble salt $\mathrm{AX}_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles/litre) is:
ChemistryIonic EquilibriumNEETNEET 2004
Options:
  • A $5.6 \times 10^{-6}$
  • B $3.1 \times 10^{-4}$
  • C $2 \times 10^{-4}$
  • D $4 \times 10^{-4}$
Solution:
2082 Upvotes Verified Answer
The correct answer is: $2 \times 10^{-4}$
$$
\begin{aligned}
& \mathrm{K}_{s p}= 3.2 \times 10^{-11} \\
& \mathrm{AX}_2 \rightleftharpoons \mathrm{A}^{2+}+2 \mathrm{X}^{-} \\
& \mathrm{S} \quad 2 \mathrm{~S} \\
& \mathrm{~K}_{s p}= \mathrm{S} \times(2 \mathrm{~S})^2=4 \mathrm{~S}^3 ; \\
&= 3.2 \times 10^{-11}=4 \mathrm{~S}^3
\end{aligned}
$$
or, $\quad \mathrm{S}^3=0.8 \times 10^{-11}=8 \times 10^{-12}$
$$
\therefore \quad \mathrm{S}=2 \times 10^{-4} \text {. }
$$


Related Theory
The solubility of a substance in a solvent is the total amount of the solute that can be dissolved in the solvent at equilibrium. On the other hand, the solubility product constant is an equilibrium constant that provides insight into the equilibrium between the solid solute and its constituent ions that are dissociated across the solution.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.