Slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2},\) is used to soften hard water by removing calcium ions from hard water through the reaction $$ \begin{array}{r}{\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \rightarrow} \\ {2 \mathrm{CaCO}_{3}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$ Although \(\mathrm{CaCO}_{3}(s)\) is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of \(\mathrm{CaCO}_{3}\) in water $\left(K_{\mathrm{sp}}=8.7 \times 10^{-9}\right) .$

Short Answer

Expert verified
The molar solubility of \(\mathrm{CaCO}_{3}\) in water is approximately \(2.95 \times 10^{-5}\) M, using the given \(K_{\mathrm{sp}} = 8.7 \times 10^{-9}\) and solving for \(s\).

Step by step solution

01

Write the dissociation reaction and \(K_{\mathrm{sp}}\) expression for \(\mathrm{CaCO}_{3}\).

Dissolving \(\mathrm{CaCO}_{3}\) in water, we have: \[ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca^{2+}}(aq) + \mathrm{CO}_{3}^{2-}(aq) \] Let \(s\) be the molar solubility of \(\mathrm{CaCO}_{3}\) in water. Then, the concentration of \(\mathrm{Ca^{2+}}\) and \(\mathrm{CO}_{3}^{2-}\) will both be \(s\). The \(K_{\mathrm{sp}}\) expression for the dissociation is: \[K_{\mathrm{sp}} = [\mathrm{Ca^{2+}}][\mathrm{CO}_{3}^{2-}] = (s)(s) = s^2\] Now, we are given \(K_{\mathrm{sp}} = 8.7 \times 10^{-9}\). We will use this information to solve for \(s\).
02

Solve for the molar solubility (\(s\)).

To find the molar solubility, we set up the following equation: \[ K_{\mathrm{sp}} = s^2 \] Substitute the given value of \(K_{\mathrm{sp}}\): \[ 8.7 \times 10^{-9} = s^2 \] Now, solve for \(s\): \[ s = \sqrt{8.7 \times 10^{-9}} \] Calculating this value gives: \[ s \approx 2.95 \times 10^{-5} \] Hence, the molar solubility of \(\mathrm{CaCO}_{3}\) in water is approximately \(2.95 \times 10^{-5}\) M.

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