The solubility of $\mathrm{CaCO}_3$ is $7 \times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3}$ at $25^{\circ}…

The solubility of $\mathrm{CaCO}_3$ is $7 \times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3}$ at $25^{\circ} \mathrm{C}$. What is its solubility product at same temperature?
  1. $6.7 \times 10^{-9}$
  2. $9.0 \times 10^{-9}$
  3. $1.12 \times 10^{-9}$
  4. $4.9 \times 10^{-9}$

Solution

$\begin{array}{ll} & \mathrm{CaCO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{Ca}_{(\mathrm{aq})}^{2+}+\mathrm{CO}_{3(\text { aq })}^{2-} \\ & x=1, \mathrm{y}=1 \\ & \mathrm{~K}_{\mathrm{sp}}=x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{(x+\mathrm{y})}=(1)^1(1)^1 \mathrm{~S}^{1+1}=\mathrm{S}^2 \\ \therefore \quad & \mathrm{~K}_{\mathrm{sp}}=\left(7 \times 10^{-5}\right)^2=49 \times 10^{-10}=4.9 \times 10^{-9}\end{array}$

Asked in: MHT CET 2024 (11 May Shift 2)

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