Slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2}\), is used to soften hard water by removing calcium ions from hard water through the reaction \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \rightarrow\) 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 \(9.33 \times 10^{-5}\) mol/L.

Step by step solution

01

Write the dissolution equation.

First of all, we need to write the dissolution equation for calcium carbonate in water: \[ \ce{CaCO3_{(s)} -> Ca^{2+}_{(aq)} + CO3^{2-}_{(aq)}} \]
02

Define the molar solubility.

According to the dissolution equation, when one mole of \(\mathrm{CaCO}_{3}\) dissolves, one mole of \(\mathrm{Ca}^{2+}\) ions and one mole of \(\mathrm{CO}_{3}^{2-}\) ions are produced. Let's define the molar solubility of \(\mathrm{CaCO}_{3}\) in water as "x" mol/L. This means that the concentrations of both the \(\mathrm{Ca}^{2+}\) ions and the \(\mathrm{CO}_{3}^{2-}\) ions will also be x mol/L.
03

Write the expression for Ksp.

Now we write down the expression for the Ksp of calcium carbonate. According to the dissolution equation, Ksp is equal to the product of the concentrations of the ions in the solution raised to their stoichiometric coefficients: \[ K_{sp}=[\ce{Ca^{2+}}][\ce{CO3^{2-}}] \]
04

Substitute the molar solubility into the Ksp expression.

We substitute the molar solubility value x mol/L into the solubility product constant equation: \[ 8.7 \times 10^{-9}= [x][x] = x^2 \]
05

Calculate the molar solubility of calcium carbonate.

Now, we solve the equation for x to find the molar solubility of calcium carbonate in water: \[ x = \sqrt{8.7 \times 10^{-9}} \] \[ x = \approximately 9.33 \times 10^{-5} \hspace{4pt} \text{mol/L} \] So, the molar solubility of \(\mathrm{CaCO}_{3}\) in water is approximately \(9.33 \times 10^{-5}\) mol/L.

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