Question

Chalk is \(\mathrm{CaCO}_{3},\) and at \(25^{\circ} \mathrm{C}\) its \(K_{\mathrm{sp}}=3.4 \times 10^{-9}\) What is the molar solubility of \(\mathrm{CaCO}_{3}\) ? How many grams of \(\mathrm{CaCO}_{3}\) dissolve in \(0.100 \mathrm{~L}\) of aqueous solution? (Ignore the reaction of \(\mathrm{CO}_{3}^{2-}\) with water.)

Short answer

The molar solubility \(s\) of \(\mathrm{CaCO}_{3}\) is \(\sqrt{3.4 \times 10^{-9}}\) mol/L, and \(0.100\) L of solution can dissolve approximately \(5.83 \times 10^{-3}\) grams of \(\mathrm{CaCO}_{3}\).

Step-by-step solution

Write the Dissolution Equation

Chalk, or \(\mathrm{CaCO}_{3}\), dissolves into its ions in water according to the equation: \(\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\). Since it is a 1:1 electrolyte, one mole of \(\mathrm{CaCO}_{3}\) will produce one mole of \(\mathrm{Ca}^{2+}\) ions and one mole of \(\mathrm{CO}_{3}^{2-}\) ions.

Set up the Expression for Solubility Product Constant (\(K_{sp}\))

The expression for the solubility product constant is based on the equilibrium concentrations of the ions in the solution: \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{CO}_{3}^{2-}]\). Considering the stoichiometry of the dissolution, if \(s\) moles per liter of \(\mathrm{CaCO}_{3}\) dissolve, then the concentration of each ion in the solution will also be \(s\), since they dissolve in a 1:1 ratio. The expression simplifies to \(K_{sp} = s^2\).

Calculate the Molar Solubility (\(s\))

Using the given solubility product constant \(K_{sp} = 3.4 \times 10^{-9}\), the molar solubility can be calculated. \(s^2 = K_{sp} \therefore s = \sqrt{K_{sp}} = \sqrt{3.4 \times 10^{-9}}\).

Find the Grams of \(\mathrm{CaCO}_{3}\) that Dissolve

First find the molar mass of \(\mathrm{CaCO}_{3}\): it's around 100 g/mol. Using the molar solubility, calculate the grams of \(\mathrm{CaCO}_{3}\) that can dissolve in 0.100 L of solution: grams = molar solubility (mol/L) \(\times\) molar mass (g/mol) \(\times\) volume (L).

Key concepts

Solubility Product Constant

Understanding the solubility product constant (\(K_{sp}\)) is fundamental when assessing the degree to which a compound can dissolve in water. It is an equilibrium constant that applies to the dissolution of solids into their ionic constituents in a saturated solution. In simpler terms, \(K_{sp}\) reflects the maximum amount of a substance that can be dissolved in water at a given temperature before the solution becomes saturated.

For chalk, or calcium carbonate (\(\mathrm{CaCO}_{3}\)), the \(K_{sp}\) value is given as \(3.4 \times 10^{-9}\) at \(25^\circ\mathrm{C}\). This low value indicates that \(\mathrm{CaCO}_{3}\) is not very soluble in water, which is consistent with its prominent use in materials like classroom chalk that do not easily dissolve upon contact with moisture.

Dissolution Equation

The dissolution equation exemplifies the stoichiometry of a solute breaking down into its component ions in a solution. For \(\mathrm{CaCO}_{3}\), the equation is written as: \(\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\). This process demonstrates how a solid compound splits into its ions when it dissolves. Here, each molecule of \(\mathrm{CaCO}_{3}\) results in the formation of one \(\mathrm{Ca}^{2+}\) ion and one \(\mathrm{CO}_{3}^{2-}\) ion - a 1:1 relationship that is the cornerstone of understanding its solubility behavior and calculating its equilibrium concentration in solution.

Moreover, the dissolution equation forms the basis for setting up the \(K_{sp}\) expression, essential for calculating the molar solubility, which dictates how much of the compound can dissolve in a given volume of solvent at equilibrium.

Stoichiometry in Solutions

In solutions, stoichiometry involves the quantitative relationship between the amounts of reactants and products during dissolution. When applying stoichiometry to the solubility of \(\mathrm{CaCO}_{3}\), we consider the molar ratio of ions produced upon dissolution in solution. Because the dissolution of calcium carbonate is a one-to-one-to-one (1:1:1) ratio, for every mole of \(\mathrm{CaCO}_{3}\) that dissolves, one mole of \(\mathrm{Ca}^{2+}\) ions and one mole of \(\mathrm{CO}_{3}^{2-}\) ions are produced.

Therefore, if the molar solubility of \(\mathrm{CaCO}_{3}\) is designated as \(s\) moles per liter (mol/L), the resulting concentrations of both \(\mathrm{Ca}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\) will also be \(s\). This straightforward stoichiometric relationship allows for the solubility product to be represented by \(K_{sp} = s^2\), which is then used to calculate molar solubility by taking the square root of \(K_{sp}\).

Equilibrium Concentration

The equilibrium concentration is the concentration of each ionic species when the solution has reached a dynamic equilibrium between the dissolved ions and the undissolved solid. At this state, the rate of dissolving solid equals the rate at which ions combine to form the solid, creating a constant concentration of ions in solution.

To find the equilibrium concentration for \(\mathrm{CaCO}_{3}\), it is necessary to invoke the stoichiometric and \(K_{sp}\) relationships determined prior, recognizing that the equilibrium concentrations of both \(\mathrm{Ca}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\) will be equal to the molar solubility \(s\). This is due to their stoichiometric ratio in the dissolution equation of \(\mathrm{CaCO}_{3}\). Understanding equilibrium concentration is vital when predicting the solubility of salts in water and serves as the key to calculating how much of the substance will dissolve, as shown in the chalk solubility example.