MHT CET · Chemistry · Ionic Equilibrium
Solubility of a salt \(A_2 B_3\) is \(1 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3}\). What is the value of its solubility product?
- A \(1.08 \times 10^{-13}\)
- B \(8.1 \times 10^{-15}\)
- C \(2.7 \times 10^{-15}\)
- D \(2.0 \times 10^{-13}\)
Answer & Solution
Correct Answer
(A) \(1.08 \times 10^{-13}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{A}_2 \mathrm{~B}_{3(\mathrm{~s})} \rightleftharpoons 2 \mathrm{~A}_{(\text {(aq) }}^{3+}+3 \mathrm{~B}_{(\text {aq })}^{2-}\)
\(\begin{aligned} & \text { Here, } x=2, \mathrm{y}=3 \\ & \begin{aligned} \mathrm{K}_{\text {sp }} & =x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{x+y} \\ & =(2)^2(3)^3 \mathrm{~S}^{2+3} \\ & =108 \mathrm{~S}^5\end{aligned} \\ & \begin{aligned} \mathrm{K}_{\text {sp }} & =108 \times\left(1 \times 10^{-3}\right)^5 \\ & =108 \times 10^{-15} \\ & =1.08 \times 10^{-13}\end{aligned}\end{aligned}\)
\(\begin{aligned} & \text { Here, } x=2, \mathrm{y}=3 \\ & \begin{aligned} \mathrm{K}_{\text {sp }} & =x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{x+y} \\ & =(2)^2(3)^3 \mathrm{~S}^{2+3} \\ & =108 \mathrm{~S}^5\end{aligned} \\ & \begin{aligned} \mathrm{K}_{\text {sp }} & =108 \times\left(1 \times 10^{-3}\right)^5 \\ & =108 \times 10^{-15} \\ & =1.08 \times 10^{-13}\end{aligned}\end{aligned}\)
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