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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2024 · 9 April · Evening Shift · Chemistry · Ionic Equilibrium

For a sparingly soluble salt (AB_2), the equilibrium concentrations...

For a sparingly soluble salt (\mathrm{AB}_2), the equilibrium concentrations of (\mathrm{A}^{2+}) ions and B−B^{-}ions are (1.2 \times 10^{-4} \mathrm{M}) and (0.24 \times 10^{-3} \mathrm{M}), respectively. The solubility product of (\mathrm{AB}_2) is :

Options

  1. A(6.91 \times 10^{-12})
  2. B(0.276 \times 10^{-12})
  3. C(27.65 \times 10^{-12})
  4. D(0.069 \times 10^{-12})

Correct answer

A. (6.91 \times 10^{-12})

Step-by-step solution

AB2( s)⇌A(aq)+2+2 B(aq)−Ksp=[A+2][B−]2=1.2×10−4×(2.4×10−4)2=6.91×10−12M3\begin{aligned} & \mathrm{AB}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{A}_{(\mathrm{aq})}^{+2}+2 \mathrm{~B}_{(\mathrm{aq})}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+2}\right]\left[\mathrm{B}^{-}\right]^2 \\ &=1.2 \times 10^{-4} \times\left(2.4 \times 10^{-4}\right)^2 \\ &=6.91 \times 10^{-12} \mathrm{M}^3\end{aligned}
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