Problem 77 - Chemistry An Atoms First Approach

Question

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 \(\left(K_{\mathrm{sp}}=8.7 \times 10^{-9}\right)\).

Step-by-Step Solution

Answer

The molar solubility of \(\mathrm{CaCO}_3\) in water is \(9.3 \times 10^{-5}\, \mathrm{M}\).

Step 1: Write the balanced dissolution reaction of \(\mathrm{CaCO}_{3}\)

The balanced dissolution reaction for \(\mathrm{CaCO}_{3}\) can be written as: \[\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\]

Step 2: Write the \(K_{sp}\) expression

The expression for the solubility product constant, \(K_{sp}\), can be written as: \[K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{CO_{3}^{2-}}]\]

Step 3: Define the molar solubility

Let the molar solubility of \(\mathrm{CaCO}_{3}\) in water be \(S\). Then, the equilibrium concentrations of the ions are as follows: \[[\mathrm{Ca}^{2+}] = S\] \[[\mathrm{CO}_{3}^{2-}] = S\]

Step 4: Substitute the concentrations into the \(K_{sp}\) expression

Now substitute the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\) into the \(K_{sp}\) expression: \[K_{sp} = S \times S\]

Step 5: Solve for molar solubility (\(S\))

Given that \(K_{sp} = 8.7 \times 10^{-9}\), we have: \[8.7 \times 10^{-9} = S^{2}\] Solving for \(S\), we get: \[S = \sqrt{8.7 \times 10^{-9}} = 9.3 \times 10^{-5}\, \text{M}\] Thus, the molar solubility of \(\mathrm{CaCO}_3\) in water is \(9.3 \times 10^{-5}\, \mathrm{M}\).

Key Concepts

Solubility Product Constant

The solubility product constant, commonly represented as \(K_{sp}\), is a vital concept in understanding how compounds dissolve in solution. It is specific to sparingly soluble salts, like calcium carbonate \(\mathrm{CaCO}_{3}\).
\(K_{sp}\) is the equilibrium constant for the dissolution of a solid substance into its ions in a saturated solution. For \(\mathrm{CaCO}_{3}\), the dissolution reaction is written as:
\[\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\] The \(K_{sp}\) value indicates the level at which a compound dissolves in water and reaches saturation. The expression is defined as:

• \(K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{CO_{3}^{2-}}]\)

In this expression, \([\mathrm{Ca^{2+}}]\) and \([\mathrm{CO}_{3}^{2-}]\) represent the concentrations of the dissolved ions at equilibrium. Because\( K_{sp}\) is a constant at a given temperature, it is critical for predicting whether the precipitate will form under certain conditions.

Dissolution Reaction

A dissolution reaction describes the process by which an ionic compound, such as calcium carbonate \(\mathrm{CaCO}_{3}\), dissolves in a solvent to form a solution. In the context of sparingly soluble compounds, dissolution reaches an equilibrium between the solid and its ions in solution. For calcium carbonate, the equation is written as:

• \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\)

At equilibrium, the rate of the solid dissolving equals the rate of ion recombination to form the solid.
This balance determines the molar solubility, \(S\), which is the number of moles of \(\mathrm{CaCO}_{3}\) that dissolve to reach this equilibrium. Using the solubility product constant \(K_{sp}\), the equilibrium concentrations can be determined by substituting \([\mathrm{Ca}^{2+}] = S\) and \([\mathrm{CO}_{3}^{2-}] = S\) into the \(K_{sp}\) expression:
\[K_{sp} = S \times S = S^2\] Thus, solving for \(S\) gives the molar solubility of the compound in water.

Hard Water Softening

Hard water contains high concentrations of calcium (\(\mathrm{Ca^{2+}}\)) and magnesium ions, which can cause scale buildup in pipes and affect soap efficacy. Softening methods aim to remove these ions to improve water quality. An effective way to remove calcium ions is through the addition of slaked lime, \(\mathrm{Ca(OH)_{2}}\).
During the hard water softening process, \(\mathrm{Ca(OH)_{2}}\) reacts with bicarbonate ions \(\mathrm{HCO_{3}^{-}}\) in water. This reaction produces calcium carbonate \(\mathrm{CaCO}_{3}\), which is insoluble and precipitates out:

• \(\mathrm{Ca(OH)_{2}(aq)} + \mathrm{Ca}^{2+}(aq) + 2\mathrm{HCO}_{3}^{-}(aq) \rightarrow 2 \mathrm{CaCO}_{3}(s) + 2 \mathrm{H}_{2} \mathrm{O}(l)\)

This method clarifies water by removing hardness ions. As some calcium carbonate dissolves back into the solution, understanding the dissolution equilibrium and \(K_{sp}\) helps in determining the extent of water softening and predicting leftover hardness in the water.