MHT CET · Chemistry · Ionic Equilibrium
Calculate the solubility product of sparingly soluble salt BA at 300 K if its solubility is \(9.1 \times 10^{-3} \mathrm{moldm}^{-3}\) at same temperature.
- A \(9.635 \times 10^{-5}\)
- B \(9.012 \times 10^{-5}\)
- C \(8.281 \times 10^{-5}\)
- D \(7.816 \times 10^{-5}\)
Answer & Solution
Correct Answer
(C) \(8.281 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{BA}_{(\mathrm{s})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-} \\
& \text {Here, } x=1 \text { and } \mathrm{y}=1 \\
& \mathrm{~K}_{\mathrm{sp}}=x^x y^y \mathrm{~S}^{x+y}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2 \\
& \therefore \quad \mathrm{~K}_{\mathrm{sp}}=\left(9.1 \times 10^{-3}\right)^2 \\
& \therefore \quad \mathrm{~K}_{\mathrm{sp}}=8.281 \times 10^{-5}
\end{aligned}\)
& \mathrm{BA}_{(\mathrm{s})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-} \\
& \text {Here, } x=1 \text { and } \mathrm{y}=1 \\
& \mathrm{~K}_{\mathrm{sp}}=x^x y^y \mathrm{~S}^{x+y}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2 \\
& \therefore \quad \mathrm{~K}_{\mathrm{sp}}=\left(9.1 \times 10^{-3}\right)^2 \\
& \therefore \quad \mathrm{~K}_{\mathrm{sp}}=8.281 \times 10^{-5}
\end{aligned}\)
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