Question

What is the molarity of a saturated solution of \(\mathrm{CaCO}_{3} ?\left(K_{s p}=2.8 \times 10^{-9}\right)\) (a) \(2.6 \times 10^{-5}\) (b) \(2.8 \times 10^{-9}\) (c) \(5.2 \times 10^{-5}\) (d) \(5.6 \times 10^{-9}\)

Short answer

\(5.29 \times 10^{-5}\) M or \(5.2 \times 10^{-5}\) M when rounded to two significant figures, so the correct answer is (c) \(5.2 \times 10^{-5}\).

Step-by-step solution

Write the dissolution equation of \(\mathrm{CaCO}_3\)

The dissolution of calcium carbonate can be represented by the following chemical equation: \(\mathrm{CaCO}_3(s) \rightleftharpoons \mathrm{Ca^{2+}}(aq) + \mathrm{CO_3^{2-}}(aq)\). In a saturated solution, the solid is in equilibrium with its ions.

Express the solubility product constant (Ksp)

The solubility product constant for the dissolution is represented by the expression \(K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{CO_3^{2-}}]\), where the concentrations are those at equilibrium in a saturated solution.

Calculate the molarity using the Ksp value

Assuming that the solubility of \(\mathrm{CaCO}_3\) is \(s\) molar, the dissolution leads to \(s\) moles of \(\mathrm{Ca^{2+}}\) and \(s\) moles of \(\mathrm{CO_3^{2-}}\) in the solution. Therefore, the \(K_{sp}\) expression can be written as \(K_{sp} = s^2\). Solving for \(s\), we get \(s = \sqrt{K_{sp}} =\sqrt{2.8 \times 10^{-9}}\).

Calculate the square root of the Ksp

To find the value of \(s\), take the square root of \(K_{sp}\), which is \(\sqrt{2.8 \times 10^{-9}} \approx 5.29 \times 10^{-5}\). This value corresponds to the molarity of the saturated solution for both \(\mathrm{Ca^{2+}}\) and \(\mathrm{CO_3^{2-}}\) ions.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is essential in solving problems related to the solubility of compounds. Equilibrium occurs when a reversible reaction balances out, meaning that the rate of the forward reaction equals the rate of the reverse reaction, leading to a constant concentration of reactants and products.

This state does not mean that the reactants and products are in equal concentrations, but that their ratios remain steady over time. When referring to solubility, a saturated solution represents a dynamic equilibrium between the undissolved solute and the dissolved ions.

In the context of the exercise, \(\mathrm{CaCO}_3(s) \rightleftharpoons \mathrm{Ca^{2+}}(aq) + \mathrm{CO_3^{2-}}(aq)\), the solid calcium carbonate is in equilibrium with its dissociated ions in water. This is represented by the solubility product constant, \(K_{sp}\), which gives us a quantifiable value for the extent of solubility at a given temperature.

Molarity Calculation

Molarity is a way of expressing the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. It's a crucial concept in chemistry, especially when discussing reactions in solution.

To find the molarity of a saturated solution, \(s\), from the \(K_{sp}\) value, we use the formula \(s = \sqrt{K_{sp}}\), as demonstrated in the exercise. This is possible because, at saturation, the concentration of \(\mathrm{Ca^{2+}}\) and \(\mathrm{CO_3^{2-}}\) ions in the solution equals the solubility \(s\) of \(\mathrm{CaCO}_3\). Hence, the molarity calculation is a straightforward way of determining the concentration of ions in a saturated solution.

Ionic Product of Salts

The ionic product is pivotal in discussing the solubility of ionic compounds. It represents the product of the concentrations of the ions raised to the power of their stoichiometric coefficients in the balanced dissolution equation.

In our case, \(K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{CO_3^{2-}}]\), this simplifies to \(s^2\) because, for every mole of \(\mathrm{CaCO}_3\) dissolved, one mole of \(\mathrm{Ca^{2+}}\) and one mole of \(\mathrm{CO_3^{2-}}\) ions are formed. If the ionic product exceeds the \(K_{sp}\), it indicates the solution is supersaturated and precipitation occurs. However, if it is less, the solution can dissolve more solute.

Physical Chemistry JEE

Physical Chemistry forms a crucial part of the Joint Entrance Examination (JEE) syllabus, which is a standardized test for admission to engineering programs in India. Key concepts such as solubility product, chemical equilibrium, and molarity are frequently tested.

These concepts are interlinked, and aspirants must have a solid grasp of them to solve questions effectively. Students must be comfortable with applying mathematical formulas to chemical problems, as evidenced by the exercise for calculating the molarity of \(\mathrm{CaCO}_3\) from the \(K_{sp}\) value and understanding its implication in terms of solubility and chemical equilibrium. A well-rounded understanding of these topics is indispensable not only for JEE but for higher studies in chemistry.