Calculate the solubility product of sparingly soluble salt BA at 300 K if its solubility is $9.1 \times…

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.
  1. $9.635 \times 10^{-5}$
  2. $9.012 \times 10^{-5}$
  3. $\quad 8.281 \times 10^{-5}$
  4. $7.816 \times 10^{-5}$

Solution

$\begin{aligned} & \mathrm{BA}_{(\mathrm{s})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-} \\ & \text {Here, } x=1 \text { and } \mathrm{y}=1 \\ & \mathrm{~K}_{\mathrm{sp}}=x^x y^y \mathrm{~S}^{x+y}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2 \\ & \therefore \quad \mathrm{~K}_{\mathrm{sp}}=\left(9.1 \times 10^{-3}\right)^2 \\ & \therefore \quad \mathrm{~K}_{\mathrm{sp}}=8.281 \times 10^{-5} \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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