Problem 15 - Chemistry The Science in Context

Question

Calculate the molarity of each of the following ions: a. \(0.33 \mathrm{g} \mathrm{Na}^{+}\) in \(100.0 \mathrm{mL}\) of solution b. \(0.38 \mathrm{g} \mathrm{Cl}^{-}\) in \(100.0 \mathrm{mL}\) of solution c. \(0.46 \mathrm{g} \mathrm{SO}_{4}^{2-}\) in \(50.0 \mathrm{mL}\) of solution d. \(0.40 \mathrm{g} \mathrm{Ca}^{2+}\) in \(50.0 \mathrm{mL}\) of solution

Step-by-Step Solution

Answer

Answer: The molarities of the ions are as follows: 𝑁𝑎⁺ is 0.143 M, 𝐶𝑙⁻ is 0.107 M, 𝑆𝑂₄²⁻ is 0.0958 M, and 𝐶𝑎²⁺ is 0.199 M.

Step 1: Calculate the moles of each ion

First, we need to divide the mass of each ion by its molar mass. For \(\mathrm{Na}^{+}\), molar mass is \(22.99 \mathrm{g/mol}\); For \(\mathrm{Cl}^{-}\), molar mass is \(35.45 \mathrm{g/mol}\); For \(\mathrm{SO}_{4}^{2-}\), molar mass is \(32.07 + 4(16.00) = 96.07 \mathrm{g/mol}\) (Sulfur and 4 Oxygen atoms); For \(\mathrm{Ca}^{2+}\), molar mass is \(40.08 \mathrm{g/mol}\).

Step 2: Convert mass to moles of each ion

Divide the mass of each ion by their respective molar mass: a. Moles of \(\mathrm{Na}^{+} = \cfrac{0.33 \mathrm{g}}{22.99 \mathrm{g/mol}} \approx 0.0143 \mathrm{mol}\) b. Moles of \(\mathrm{Cl}^{-} = \cfrac{0.38 \mathrm{g}}{35.45 \mathrm{g/mol}} \approx 0.0107 \mathrm{mol}\) c. Moles of \(\mathrm{SO}_{4}^{2-} = \cfrac{0.46 \mathrm{g}}{96.07 \mathrm{g/mol}} \approx 0.00479 \mathrm{mol}\) d. Moles of \(\mathrm{Ca}^{2+} = \cfrac{0.40 \mathrm{g}}{40.08 \mathrm{g/mol}} \approx 0.00998 \mathrm{mol}\)

Step 3: Convert volume to liters

Divide the volume of each solution by 1000 to convert it from mL to L: a. Volume of solution = \(\cfrac{100.0 \mathrm{mL}}{1000} = 0.100 \mathrm{L}\) b. Volume of solution = \(\cfrac{100.0 \mathrm{mL}}{1000} = 0.100 \mathrm{L}\) c. Volume of solution = \(\cfrac{50.0 \mathrm{mL}}{1000} = 0.0500 \mathrm{L}\) d. Volume of solution = \(\cfrac{50.0 \mathrm{mL}}{1000} = 0.0500 \mathrm{L}\)

Step 4: Calculate the molarity of each ion

Divide the moles of each ion by the volume of the solution in L: a. Molarity of \(\mathrm{Na}^{+} = \cfrac{0.0143 \mathrm{mol}}{0.100 \mathrm{L}} = 0.143 \mathrm{M}\) b. Molarity of \(\mathrm{Cl}^{-} = \cfrac{0.0107 \mathrm{mol}}{0.100 \mathrm{L}} = 0.107 \mathrm{M}\) c. Molarity of \(\mathrm{SO}_{4}^{2-} = \cfrac{0.00479 \mathrm{mol}}{0.0500 \mathrm{L}} = 0.0958 \mathrm{M}\) d. Molarity of \(\mathrm{Ca}^{2+} = \cfrac{0.00998 \mathrm{mol}}{0.0500 \mathrm{L}} = 0.199 \mathrm{M}\)

Key Concepts

Ions

Ions are tiny particles that carry an electrical charge, and they play a crucial role in chemistry. They form when atoms gain or lose electrons, resulting in a positive or negative charge. Ions can be:

• Cations: Positively charged ions, formed when an atom loses electrons. For instance, sodium (Na) becomes Na⁺ by losing one electron.
• Anions: Negatively charged ions, formed when an atom gains electrons. For example, chlorine (Cl) becomes Cl^- after gaining one electron.

Ions are vital in many chemical reactions, particularly in determining a solution's properties, like conductivity. Understanding ions is essential for calculating properties like molarity, as seen in the given exercise.

Moles

The mole is a fundamental concept in chemistry that allows chemists to count particles, like atoms or ions, in a given mass. Instead of counting individual particles, which would be impractical due to their tiny size, we use moles. One mole equates to Avogadro's number, which is approximately 6.022 x 10²³ particles.
To find the number of moles, we need the mass of the substance and its molar mass (which we'll cover next). The formula to convert mass to moles is:

• Moles = \( \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \)

For example, in the exercise, the moles of Na⁺ are calculated using its molar mass of 22.99 g/mol for a given mass of 0.33 g. This concept allows you to convert weighable amounts into a count of entities—a critical step in many calculations, such as determining concentration.

Molar Mass

Molar mass is the weight of one mole of a given substance, typically expressed in grams per mole (g/mol). It's crucial for converting between mass and moles. To calculate the molar mass of a compound, sum up the molar masses of all the atoms within its molecular formula.
For individual elements, their molar mass is the atomic mass found on the periodic table. However, compounds require adding atomic masses of all constituent atoms:

• Example: For sulfate (SO₄^2-), calculate by adding the molar mass of sulfur (32.07 g/mol) to four times the molar mass of oxygen (16.00 g/mol each), resulting in a total of 96.07 g/mol.

In the exercise, having the molar mass allows the conversion of ion mass to moles, which is necessary for calculating solution concentrations.

Solution Concentration

Solution concentration describes how much solute is present in a given volume of solvent. The most common unit for solution concentration in chemistry is molarity (M), representing moles of solute per liter of solution.
Calculating molarity involves two main steps:

• First, determine the moles of the solute using the formula: moles = mass/molar mass.
• Second, divide this number by the volume of the solution in liters: Molarity (M) = moles of solute/volume of solution (L).

In the provided exercise, the volume of each solution is converted from milliliters to liters to apply this formula correctly. Solution concentration is a critical concept because it affects how chemical reactions proceed, influencing factors like reaction speed and equilibrium.