MHT CET · Chemistry · Ionic Equilibrium
The solubility of sparingly soluble salt \(\mathrm{AB}_2\) is \(1.0 \times 10^{-4} \mathrm{~mol}\) \(\mathrm{dm}^{-3}\). What is its solubility product?
- A \(2 \times 10^{-12}\)
- B \(4 \times 10^{-8}\)
- C \(4 \times 10^{-12}\)
- D \(2 \times 10^{-8}\)
Answer & Solution
Correct Answer
(C) \(4 \times 10^{-12}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{AB}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{A}_{(\mathrm{aq})}^{2+}+2 \mathrm{~B}_{(\mathrm{aq})}^{-}\)
At equilibrium, \(-\mathrm{S} \quad 2 \mathrm{~S}\)
\(\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{2+}\right]\left[\mathrm{B}^{-}\right]^2 \\
& =\mathrm{S} .(2 \mathrm{~S})^2=4 \mathrm{~S}^3 \\
& =4 \times\left(1 \times 10^{-4}\right)^3 \\
& \mathrm{~K}_{\mathrm{sp}}=4 \times 10^{-12}
\end{aligned}\)
At equilibrium, \(-\mathrm{S} \quad 2 \mathrm{~S}\)
\(\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{2+}\right]\left[\mathrm{B}^{-}\right]^2 \\
& =\mathrm{S} .(2 \mathrm{~S})^2=4 \mathrm{~S}^3 \\
& =4 \times\left(1 \times 10^{-4}\right)^3 \\
& \mathrm{~K}_{\mathrm{sp}}=4 \times 10^{-12}
\end{aligned}\)
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