Calculate the solubility of sparingly soluble salt $\mathrm{BA}^{\prime}$ in $\mathrm{mol} \mathrm{dm}^{-3}$…

Calculate the solubility of sparingly soluble salt $\mathrm{BA}^{\prime}$ in $\mathrm{mol} \mathrm{dm}^{-3}$ at 300 K if its solubility product is $4.9 \times 10^{-9}$ at same temperature.
  1. $5.72 \times 10^{-5}$
  2. $6.40 \times 10^{-5}$
  3. $7.00 \times 10^{-5}$
  4. $7.81 \times 10^{-5}$

Solution

$\begin{aligned} & \mathrm{BA}_{(\mathrm{s})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-} \\ & x=1, \mathrm{y}=1 \\ & \mathrm{~K}_{\mathrm{sp}}=x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{x+y}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2 \\ \therefore \quad & \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}=\sqrt{4.9 \times 10^{-9}}\end{aligned}$ $\begin{aligned} & =\sqrt{49 \times 10^{-10}} \\ & =7.00 \times 10^{-5} \mathrm{M}\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 2)

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