Indicate the concentration of each ion or molecule present in the following solutions: (a) 0.25\(M\) NaNO \(_{3}\) , (b) \(1.3 \times 10^{-2} M \mathrm{MgSO}_{4},(\mathbf{c}) 0.0150 M \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},(\mathbf{d})\) a mixture of 45.0 \(\mathrm{mL}\) of 0.272 \(\mathrm{M} \mathrm{NaCl}\) and 65.0 \(\mathrm{mL}\) of 0.0247 \(\mathrm{M}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) Assume that the volumes are additive.

Molarity, symbolized by the letter 'M', is a measure of concentration that reveals the number of moles of a solute in one liter of solution. Specifically, molarity is calculated using the formula

\[\begin{equation}M = \frac{\text{moles of solute}}{\text{liters of solution}}\end{equation}\]

To clarify, a mole is a unit in chemistry that represents a specific quantity of particles, such as atoms or molecules—6.022 \(\times\) 10^23 particles to be precise. When you're working with a solution's molarity, it's like knowing the 'strength' of that solution—the higher the molarity, the more concentrated the solution is. For instance, in the exercise, a 0.25M NaNO3 solution indicates that there are 0.25 moles of sodium nitrate dissolved per liter of water.

Understanding molarity is crucial in chemistry as it helps you to precisely prepare solutions with desired properties, and it's fundamental when it comes to reactions in solutions, referencing their stoichiometry, and calculating the resulting concentrations after chemical processes.

When ionic compounds dissolve in water, they undergo a process called dissociation, where the compound separates into its constituent ions. This is an important concept because the degree of dissociation impacts the concentration of ions in a solution and, in turn, the solution's reactivity and properties.

Ionic compounds, like NaNO3 and MgSO4 featured in our exercise, dissociate to release positively charged cations (like Na+, Mg2+) and negatively charged anions (like NO3-, SO4 2-). The dissociation can be represented by simple equations, for example:

\[\begin{equation}NaNO3 (s) \rightarrow Na^+ (aq) + NO3^- (aq)\end{equation}\]
\[\begin{equation}MgSO4 (s) \rightarrow Mg^{2+} (aq) + SO4^{2-} (aq)\end{equation}\]

The (s) symbol denotes the solid ionic compound, and (aq) represents an aqueous, or dissolved, ion. In the provided solutions, each ionic compound has a 1:1 stoichiometry, meaning that for every mole of compound that dissolves, one mole of each ion is produced. Understanding dissociation is key to predicting the behavior of ions in a solution and calculating their concentrations for various applications.

Stoichiometry is essentially the 'math of chemistry', offering a way to describe the quantitative relationships between reactants and products in chemical reactions. In the context of ion concentration, stoichiometry enables us to understand how many ions will be present in a solution, once a compound dissociates.

The stoichiometry of an ionic compound provides a ratio of the number of cations to anions released upon dissociation. For example, the stoichiometry of NaNO3 and MgSO4 is 1:1, meaning each formula unit that dissolves produces one Na+ and one NO3-, or one Mg2+ and one SO4 2-, respectively. Differently, for the compound (NH4)2CO3, which has a stoichiometry of 2:1 with respect to NH4+ and CO3 2-, two ammonium ions are formed for every one carbonate ion when it dissociates.

To calculate ion concentration from stoichiometry, you simply multiply the molarity of the original compound by the number of each ion produced. For instance, the exercise starts with a 0.0247 M (NH4)2CO3 solution—since the stoichiometry is 2:1, the concentration of NH4+ ions will be 2 times 0.0247 M, while the CO3 2- ion concentration will match that of the original compound.

Mixing solutions is a common practice in chemistry and requires careful consideration of both concentration and volume to determine the resulting solution's properties. The combined mixture will only have the same concentration as the individual solutions if their volumes and concentrations are equal, which is rarely the case.

In the exercise, we mix two solutions with different volumes and concentrations. To comprehend their combined effect, one must first calculate the moles of each ion in the individual solutions and then find the total volume of the mixture. The molarity of each ion in the mixture is then the total moles of that ion divided by the total volume of the mixture. As mentioned in the exercise, it's assumed that the volumes are additive, which simplifies the math, but in real scenarios, due to intermolecular forces, this might not strictly be the case.

Understanding the principles involved in mixing solutions is paramount since it applies to a myriad of situations, from titration in analytical chemistry to diluting stock solutions in molecular biology. Remember, accurate calculations can make the difference between success and failure in any chemical experiment.