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Chapter 20: Problem 77

Slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2}\), is used to soften hard water by removing calcium ions from hard water through the reaction \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \rightarrow\) \(2 \mathrm{CaCO}_{3}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) 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 \(\left(K_{\text {sp }}=8.7 \times 10^{-9}\right)\).

Short Answer

Expert verified
The molar solubility of \(\mathrm{CaCO_3}\) in water is \(9.3 \times 10^{-5}\) mol/L.

Step by step solution

01

Write the dissolution reaction and the solubility product expression

First, we need to write the dissolution reaction for calcium carbonate in water. The balanced reaction is: \[ \mathrm{CaCO_3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_3^{2-}(aq) \] Now, we can write the solubility product expression for this reaction as: \[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{CO}_3^{2-}] \]
02

Define the molar solubility and determine the equilibrium concentrations of the ions

The molar solubility (S) of \(\mathrm{CaCO_3}\) is the number of moles of \(\mathrm{CaCO_3}\) that would dissolve in 1 L of water at equilibrium. As one mole of \(\mathrm{CaCO_3}\) dissolves, it produces one mole of \(\mathrm{Ca}^{2+}\) and one mole of \(\mathrm{CO}_3^{2-}\). Therefore, at equilibrium, we have the following concentrations: - \(\left[ \mathrm{Ca}^{2+} \right] = \mathrm{S}\) - \(\left[ \mathrm{CO}_3^{2-} \right] = \mathrm{S}\)
03

Substitute the equilibrium concentrations into the solubility product expression and solve for S

Now, we can substitute the equilibrium concentrations into the solubility product expression: \[ K_{sp} = \left( [\mathrm{Ca}^{2+}] \right) \left( [\mathrm{CO}_3^{2-}] \right) \] \[ 8.7 \times 10^{-9} = (S)(S) \] Now, we solve for S by taking the square root of both sides of the equation: \[ S = \sqrt{8.7 \times 10^{-9}} \]
04

Calculate the molar solubility of CaCO3

To obtain the molar solubility, we now compute the value of S, which gives: \[ S = \sqrt{8.7 \times 10^{-9}} = 9.3 \times 10^{-5} \, \mathrm{mol/L} \] Therefore, the molar solubility of \(\mathrm{CaCO_3}\) in water is \(9.3 \times 10^{-5}\) mol/L.

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