For a sparingly soluble salt AB_2 , the equilibrium concentrations of A^2+ ions and B^- ions are 1.2 × 10^-4 M and 0.24 × 10^-3 M , respectively. The solubility product of AB_2 is :

Correct answer: (A) 6.91 × 10^-12. AB_2(~s) leftharpoons A_(aq)^+2+2 ~B_(aq)^- K_sp= [A^+2 ] [B^- ]^2 =1.2 × 10^-4 × (2.4 × 10^-4 )^2 =6.91 × 10^-12 M^3

Q6 / 90 · JEE Main 2024 (09 Apr Shift 2) · Chemistry

For a sparingly soluble salt $\mathrm{AB}_2$, the equilibrium concentrations of $\mathrm{A}^{2+}$ ions and $B^{-}$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 :

Correct answer
A
$6.91 \times 10^{-12}$
B
$0.276 \times 10^{-12}$
C
$27.65 \times 10^{-12}$
D
$0.069 \times 10^{-12}$

Solution

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

Hence, the correct answer is option: 'A'.

For a sparingly soluble salt \(\mathrm{AB}_2\), the equilibrium concentrations of \(\mathrm{A}^{2+}\) ions and \(B^{-}\)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 :

Solution

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