MHT CET · Chemistry · Ionic Equilibrium
Calculate solubility \(\left(\mathrm{moldm}^{-3}\right)\) of a sparingly soluble electrolyte AB at 298 K if its solubility product is \(1.6 \times 10^{-5}\) ?
- A \(1.6 \times 10^{-3}\)
- B \(2.5 \times 10^{-3}\)
- C \(4.0 \times 10^{-3}\)
- D \(8.0 \times 10^{-3}\)
Answer & Solution
Correct Answer
(C) \(4.0 \times 10^{-3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { For } \mathrm{AB}, \\ & \mathrm{AB}_{(\mathrm{s})} \rightleftharpoons \mathrm{A}_{(\mathrm{aq})}^{+}+\mathrm{B}_{(\mathrm{aq})}^{-} \\ & \text {Here }^{-} x=1, \mathrm{y}=1 \\ \therefore \quad & \mathrm{~K}_{\mathrm{sp}}=x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{x+\mathrm{y}}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2\end{aligned}\)
\(\begin{aligned} \therefore \quad \mathrm{S} & =\sqrt{\mathrm{K}_{\mathrm{sp}}} \\ & =\sqrt{1.6 \times 10^{-5}}=\sqrt{16 \times 10^{-6}} \\ & =4.0 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}\)
\(\begin{aligned} \therefore \quad \mathrm{S} & =\sqrt{\mathrm{K}_{\mathrm{sp}}} \\ & =\sqrt{1.6 \times 10^{-5}}=\sqrt{16 \times 10^{-6}} \\ & =4.0 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}\)
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