Determine the concentration of the cation and anion in each aqueous solution. (Assume complete dissociation of each compound.) (a) \(0.12 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (b) \(0.25 \mathrm{M} \mathrm{K}_{2} \mathrm{CO}_{3}\)

The process of a solid ionic compound breaking down into its individual ions when dissolved in water is known as dissociation. This is crucial in the world of chemistry because it determines the number of available ions that can participate in reactions. For instance, the ionic compound sodium sulfate \( \mathrm{Na}_2\mathrm{SO}_4 \) dissociates into two sodium ions \( \mathrm{Na}^+ \) and one sulfate ion \( \mathrm{SO}_4^{2-} \). Knowing how a compound dissociates helps us calculate the concentration of ions in a solution.

To further enhance understanding, one should visualize the process as the compound dissolving and then separating into its ions. A helpful tip when considering dissociation is to look at the compound's formula and determine the ratio of ions formed upon dissociation. Two sodium ions for every unit of sodium sulfate is a key takeaway here.

In chemistry, molarity is a measure of concentration, expressed as moles of a solute per liter of solution (\( \mathrm{M} = \text{moles of solute} / \text{liters of solution} \)). It provides a way to relate the volume of the solution to the amount of substance dissolved in it. A 0.12 M solution of sodium sulfate means it contains 0.12 moles of sodium sulfate per liter of solution.

To get a better grasp of molarity, it's helpful to imagine it as the number of 'pieces' (moles) of the substance in a defined volume of liquid. Recognizing how molarity relates to the dissolved substance and its resulting ions is essential. It is also the starting point for calculating the concentration of individual ions, as shown in the exercises for sodium sulfate and potassium carbonate.

Once the dissociation of an ionic compound is understood, we can calculate the concentration of each ion in the solution. The concentration of ions depends on the stoichiometry of the dissociation: for every 1 mole of \( \mathrm{Na}_2\mathrm{SO}_4 \) that dissociates, we get 2 moles of \( \mathrm{Na}^+ \) and 1 mole of \( \mathrm{SO}_4^{2-}\). Therefore, if you have a 0.12 M solution of sodium sulfate, the concentration of sodium ions \( \mathrm{Na}^+ \) will be 0.24 M, since each unit produces two sodium ions.

A vital tip for students is to always multiply the molarity of the original compound by the number of ions produced upon dissociation to find the resulting ionic concentration. Be sure to account for compounds that produce more than one mole of ions per mole of compound, such as potassium carbonate which provides two moles of \( \mathrm{K}^+ \) ions for each mole of \( \mathrm{K}_2\mathrm{CO}_3 \) dissolved.