Determine the concentration of the cation and anion in each aqueous solution. (Assume complete dissociation of each compound.) (a) \(0.20 \mathrm{M} \mathrm{SrSO}_{4}\) (b) \(0.15 \mathrm{M} \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (c) \(0.12 \mathrm{M} \mathrm{SrI}_{2}\)

When ionic compounds dissolve in water, they separate into individual ions, a process known as dissociation. This process occurs because water molecules are polar, with a partial positive charge near the hydrogen atoms and a partial negative charge near the oxygen atom. The charged ions are attracted to these opposites, causing the compound to split apart.

For example, strontium sulfate (\( \text{SrSO}_4 \) ) completely dissociates in water, releasing one strontium ion (\( \text{Sr}^{2+} \) ) and one sulfate ion (\( \text{SO}_4^{2-} \) ). The ratio of ions released depends on the formula of the compound. In calculations, understanding the molar ratio in which an ionic compound dissociates is crucial to accurately determine the concentration of each ion in solution.

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In aqueous solutions, stoichiometry helps to determine the concentration of dissolved ions based on the initial concentration of the ionic compound and its dissociation pattern.

Continuing with the example of (\( \text{Cr}_2(\text{SO}_4)_3 \) ), we see that for every molecule that dissociates, two chromium ions (\( \text{Cr}^{3+} \) ) and three sulfate ions (\( \text{SO}_4^{2-} \) ) are produced. Hence, the stoichiometry of 2:3 informs us that for every mole of (\( \text{Cr}_2(\text{SO}_4)_3 \) ) that dissolves, there are twice as many moles of (\( \text{Cr}^{3+} \) ) and thrice as many moles of (\( \text{SO}_4^{2-} \) ) present in the solution.

Molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution (\( M \) or mol/L). It is one of the most common ways to express solution concentration in chemistry. For instance, if we have a 0.20 M solution of strontium sulfate (\( \text{SrSO}_4 \) ), this means there are 0.20 moles of (\( \text{SrSO}_4 \) ) dissolved in one liter of solution.

Since molarity is based on the volume of the solution and not the amount of solvent, it can be affected by changes in temperature, as volume can expand or contract with temperature fluctuations. To calculate the molarity of ions after dissociation, you multiply the initial molarity of the ionic compound by the number of ions produced upon its dissociation.

The concentration of ions in a solution is directly related to the molarity of the starting ionic compound and its dissociation into constituent ions. Knowing the stoichiometry of the dissociation helps to calculate the final concentration of each ion. For each compound, consider the number of each type of ion formed per formula unit of the compound.

For example, when solving for the ion concentration in a solution of strontium iodide (\( \text{SrI}_2 \) ), recognize that for every mole of (\( \text{SrI}_2 \) ), one mole of (\( \text{Sr}^{2+} \) ) ions and two moles of iodide (\( \text{I}^- \) ) ions are produced. Hence, the concentration of (\( \text{I}^- \) ) ions will be double that of (\( \text{SrI}_2 \) )'s initial molarity. By mastering the calculation of ion concentration, students can better understand the composition and behavior of solutions.