Question

Calculate the number of moles of each ion present in each of the following solutions. a. \(1.25 \mathrm{~L}\) of \(0.250 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution b. \(3.5 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution c. \(25 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{AlCl}_{3}\) solution d. 1.50 L of \(1.25 M\) BaCl \(_{2}\) solution

Short answer

a. Moles of Na\(^+\) = \(0.9375\ mol\), Moles of PO\(_{4}$$^{3-}\) = \(0.3125\ mol\) b. Moles of H\(^+\) = \(0.042\ mol\), Moles of SO\(_{4}$$^{2-}\) = \(0.021\ mol\) c. Moles of Al\(^{3+}\) = \(0.00375\ mol\), Moles of Cl\(^-\) = \(0.01125\ mol\) d. Moles of Ba\(^{2+}\) = \(1.875\ mol\), Moles of Cl\(^-\) = \(3.75\ mol\)

Step-by-step solution

a. Calculating moles of ions in Na\(_{3}\)PO\(_{4}\) solution

To calculate the number of moles of ions in the \(1.25 L\) of \(0.250 M\) Na\(_{3}\)PO\(_{4}\) solution, we first need to find the total number of moles of Na\)_{3}\(PO\)_{4}\) using the formula: Moles of solute = Molarity × Volume Moles of Na\(_{3}\)PO\(_{4}\) = \(0.250 M × 1.25 L = 0.3125\ mol\) Now, considering the stoichiometry of the compound Na\(_{3}\)PO\(_{4}\), we have 3 moles of Na\(^+\) and 1 mole of PO\(_{4}$$^{3-}\) ions per mole. Moles of Na\(^+\) = \(3 × 0.3125 \ mol = 0.9375\ mol\) Moles of PO\(_{4}$$^{3-}\) = \(1 × 0.3125 \ mol = 0.3125\ mol\).

b. Calculating moles of ions in H\(_{2}\)SO\(_{4}\) solution

First, we convert the volume from milliliters to liters: \(3.5\ ml = 0.0035\ L\) Next, we find the total number of moles of H\(_{2}\)SO\(_{4}\) using the formula: Moles of H\(_{2}\)SO\(_{4}\) = \(6.0 M × 0.0035 L = 0.021\ mol\) Considering the stoichiometry of the compound H\(_{2}\)SO\(_{4}\), we have 2 moles of H\(^+\) and 1 mole of SO\(_{4}$$^{2-}\) ions per mole. Moles of H\(^+\) = \(2 × 0.021\ mol = 0.042\ mol\) Moles of SO\(_{4}$$^{2-}\) = \(1 × 0.021\ mol = 0.021\ mol\)

c. Calculating moles of ions in AlCl\(_{3}\) solution

First, we convert the volume from milliliters to liters: \(25\ ml = 0.025\ L\) Next, we find the total number of moles of AlCl\(_{3}\) using the formula: Moles of AlCl\(_{3}\) = \(0.15 M × 0.025 L = 0.00375\ mol\) Considering the stoichiometry of the compound AlCl\(_{3}\), we have 1 mole of Al\(^{3+}\) and 3 moles of Cl\(^-\) ions per mole. Moles of Al\(^{3+}\) = \(1 × 0.00375\ mol = 0.00375\ mol\) Moles of Cl\(^-\) = \(3 × 0.00375\ mol = 0.01125\ mol\)

d. Calculating moles of ions in BaCl\(_{2}\) solution

We find the total number of moles of BaCl\(_{2}\) using the formula: Moles of BaCl\(_{2}\) = \(1.25 M × 1.50 L = 1.875\ mol\) Considering the stoichiometry of BaCl\(_{2}\), we have 1 mole of Ba\(^{2+}\) and 2 moles of Cl\(^-\) ions per mole. Moles of Ba\(^{2+}\) = \(1 × 1.875\ mol = 1.875\ mol\) Moles of Cl\(^-\) = \(2 × 1.875\ mol = 3.75\ mol\)

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the relationships between reactants and products in chemical reactions. It tells us how much of one substance will react with a given amount of another substance. Think of it like a recipe in cooking; if you know how much flour you have, stoichiometry tells how much sugar to use. To solve stoichiometry problems, we use balanced chemical equations and the concept of the mole, which represents a specific number of molecules or atoms.
In the exercise, stoichiometry helps us to determine how many ions are produced when a certain amount of substance dissolves in water. Understanding the stoichiometry of compounds is crucial because it allows us to predict the proportions of different ions. For example, in Na\(_{3}\)PO\(_{4}\), the stoichiometry indicates that dissolving 1 mole of Na\(_{3}\)PO\(_{4}\) produces 3 moles of Na\(^+\) ions and 1 mole of PO\(_{4}^{3-}\) ions.

• Use the balanced chemical formula to understand the ratio of ions produced.
• Convert volumes of solutions to moles using molarity.
• Use stoichiometry ratios to find moles of specific ions.

Molarity

Molarity (M) is a common way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. This concept is crucial in solution chemistry because it helps us determine how concentrated a solution is, much like how sugar concentration can vary between lightly sweetened tea and a syrup.
When given a molarity and a volume, you can find out how many moles you have of a substance by multiplying them. For instance, if a solution has a molarity of 0.25 M and the volume is 1.25 L, the number of moles of solute is calculated as \( 0.25 \times 1.25 = 0.3125 \ mol \).

• Molarity gives the concentration and is measured in moles per liter.
• To find the moles from molarity, multiply the molarity by the solution's volume in liters.
• Understand how the change in either molarity or volume affects the number of moles.

Solution Chemistry

Solution chemistry is the study of solutions, which are mixtures where one substance (the solute) is dissolved in another (the solvent). Understanding solution chemistry is essential for recognizing how various substances interact when dissolved and how they form ions.
In our exercise, we deal with salts like Na\(_{3}\)PO\(_{4}\), which dissociate into their respective ions when added to water. The process of dissociation increases the number of particles in the solution, which can affect properties like conductivity and reactivity.
Key aspects to consider in solution chemistry include:

• Solutes like Na\(_{3}\)PO\(_{4}\) and AlCl\(_{3}\) dissociate to produce multiple ions.
• The type of ions produced and their ratio determines the properties of the solution.
• Solutions are described by concentration (molarity), which determines how many solute molecules are present in a given volume of solvent.

Using solution chemistry principles, one can accurately predict the behavior of solutions in different chemical processes. This is fundamental in fields ranging from biology to industrial applications.