Calculate the molar solubility of \(\mathrm{Ca}\left(\mathrm{IO}_{3}\right)_{2}\) in (a) \(0.060 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) and \((\mathrm{b}) 0.060 \mathrm{M} \mathrm{NaIO}_{3} .\) (See Appendix C.)

When working out the molar solubility of a compound, understanding its dissociation equation is crucial. For \(\text{Ca(IO}_3)_2\), it dissociates in water to form ions. This process is represented by the following equation:

\[ \text{Ca(IO}_3)_2 \rightleftharpoons \text{Ca}^{2+} + 2 \text{IO}_3^- \]
Here, one formula unit of \(\text{Ca(IO}_3)_2\) produces one calcium ion (\(\text{Ca}^{2+}\)) and two iodate ions (\(\text{IO}_3^-\)). This relationship is important as it helps us determine how the solute breaks down in a solution and impacts the molar solubility calculations.

The balanced dissociation equation tells us that for every one mole of \(\text{Ca(IO}_3)_2\) that dissolves, we get one mole of \(\text{Ca}^{2+}\) and two moles of \(\text{IO}_3^-\). Knowing this stoichiometry is essential when plugging in values into the solubility product expression.

The solubility product constant, often denoted as \(K_{sp}\), is a key factor in solubility calculations. It represents the product of the concentrations of the ions, each raised to the power of their coefficients from the dissociation equation. For \( \text{Ca(IO}_3)_2 \), we have:

\[ K_{sp} = [\text{Ca}^{2+}][\text{IO}_3^-]^2 \]

The \(K_{sp}\) value for a given compound is constant at a given temperature. In this exercise, the \(K_{sp}\) for \( \text{Ca(IO}_3)_2 \) is given as \(7.1 \times 10^{-7}\). This constant helps us relate the concentrations of the ions at equilibrium in a saturated solution.

To determine the molar solubility, we need to find the concentrations of the ions that would satisfy this \(K_{sp}\) expression. The dissociation equation provides the ratio in which these ions appear, guiding us in solving for the unknown concentrations.

The common ion effect refers to the decrease in solubility of a salt when another compound, providing a common ion, is present in the solution. This is illustrated in the exercise by using \(0.060 \text{ M} \text{Ca(NO}_3)_2\) and \(0.060 \text{M} \text{NaIO}_3\) as the solvents.

In \(0.060 \text{M} \text{Ca(NO}_3)_2\), the common ion \(\text{Ca}^{2+}\) from the \(\text{Ca(NO}_3)_2\) reduces the solubility of \(\text{Ca(IO}_3)_2\). This happens because the increased concentration of \(\text{Ca}^{2+}\) shifts the equilibrium to favor the formation of the solid salt over its dissociation.

Similarly, in \(\text{0.060 M NaIO}_3\), the presence of the common ion \(\text{IO}_3^-\) lowers the solubility of \(\text{Ca(IO}_3)_2\). The excess iodate ions shift the equilibrium towards the formation of the undissolved compound.

Understanding the common ion effect helps in accurately calculating the molar solubility of a compound in solutions containing common ions.

When solving solubility problems, especially those involving common ions, an approximation method can simplify the calculations. This method makes assumptions that reduce the complexity of solving equations involving \(K_{sp}\).

In the given exercise, for \(0.060 \text{ M Ca(NO}_3)_2\), we assume the solubility \(s\) is much smaller than the concentration of the common ion. Thus, \( \text{Ca}^{2+} = 0.060 + s \) simplifies to \( \text{Ca}^{2+} \approx0.060 \text{M}\). This lets us solve the \(K_{sp}\) expression more easily.

Similarly, for \(0.060 \text{M NaIO}_3\), the assumption \(2s \ll 0.060 \) makes \(\text{IO}_3^- = 0.060 + 2s \approx 0.060 \text{M}\). These approximations significantly reduce algebraic complexity while keeping the solutions reasonably accurate.

This method is helpful when the contribution of the solubility term to the total concentration is negligible, allowing for simpler and quicker calculations in solubility problems.