Problem 69 - Chemistry The Central Science In Si Units

Question

(a) Which will have the highest concentration of sodium ions: \(0.25 \mathrm{M} \mathrm{NaCl}, 0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3},\) or \(0.075 \mathrm{MNa}_{3} \mathrm{PO}_{4} ?(\mathbf{b})\) Which will contain the greater number of moles of sodium ion: \(20.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{NaHCO}_{3}\) or \(15.0 \mathrm{~mL}\) of \(0.04 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S} ?\)

Step-by-Step Solution

Answer

(a) Na₂CO₃ has the highest concentration of sodium ions (0.30 M). (b) NaHCO₃ contains more moles of sodium ions (0.003 mol).

Step 1: Determine Sodium Ion Concentration from Each Compound

For each compound, calculate the concentration of sodium ions based on its dissociation in water. - **NaCl:** Dissociates into Na⁺ and Cl⁻. Thus, 1 mol of NaCl yields 1 mol of Na⁺. Concentration of Na⁺ = 0.25 M - **Na₂CO₃:** Dissociates into 2 Na⁺ and CO₃²⁻. Thus, 1 mol of Na₂CO₃ yields 2 mol of Na⁺. Concentration of Na⁺ = 2 × 0.15 M = 0.30 M - **Na₃PO₄:** Dissociates into 3 Na⁺ and PO₄³⁻. Thus, 1 mol of Na₃PO₄ yields 3 mol of Na⁺. Concentration of Na⁺ = 3 × 0.075 M = 0.225 M.

Step 2: Identify the Highest Concentration of Sodium Ions

Compare the concentrations calculated in Step 1: - NaCl: 0.25 M Na⁺ - Na₂CO₃: 0.30 M Na⁺ - Na₃PO₄: 0.225 M Na⁺ The highest concentration of sodium ions is 0.30 M from Na₂CO₃.

Step 3: Calculate Moles of Sodium Ions in Each Solution

Determine the moles of sodium ions for each given solution: - **NaHCO₃:** Dissociates into Na⁺ and HCO₃⁻. Thus, 1 mol of NaHCO₃ yields 1 mol of Na⁺. Moles of Na⁺ in 20.0 mL of 0.15 M NaHCO₃ = 0.020 L × 0.15 mol/L = 0.003 mol - **Na₂S:** Dissociates into 2 Na⁺ and S²⁻. Thus, 1 mol of Na₂S yields 2 mol of Na⁺. Moles of Na⁺ in 15.0 mL of 0.04 M Na₂S = 0.015 L × 0.04 mol/L × 2 = 0.0012 mol

Step 4: Compare Moles of Sodium Ions

Compare the moles calculated in Step 3: - NaHCO₃: 0.003 mol of Na⁺ - Na₂S: 0.0012 mol of Na⁺ The solutions made from NaHCO₃ contain more moles of sodium ions.

Key Concepts

Sodium Ion Calculation

Determining the concentration of sodium ions in a solution is a key task in chemistry. Many compounds, when dissolved in water, release sodium ions (Na⁺), but the number of ions released depends on the chemical formula of the compound. For instance, NaCl will dissociate completely into one Na⁺ ion for every molecule, while Na₂CO₃ will release two Na⁺ ions. To find the concentration of Na⁺, you need to multiply the molarity (M) of the compound by the number of sodium ions it can release.

• Consider the compound's formula. For example, in Na₃PO₄, each formula unit releases three Na⁺ ions when dissociated.
• Use the formula:
  Concentration of Na⁺ = Molarity of compound × Number of Na⁺ ions released.

This is a simple method to calculate the amount of sodium ions in a solution, which is crucial for many chemical applications.

Chemical Dissociation

When a chemical compound dissolves in water, it often separates into its component ions through a process called dissociation. Understanding this process is vital for calculating ion concentrations.

• For NaCl, the formula shows it's made up of Na⁺ and Cl⁻ ions. When NaCl is dissolved, it separates into one Na⁺ and one Cl⁻ ion.
• Compounds like Na₂CO₃ dissociate differently, splitting into two Na⁺ ions and one CO₃²⁻ ion.
• A more complex example is Na₃PO₄, which dissociates to release three Na⁺ ions and one PO₄³⁻ ion.

Understanding the dissociation pattern of a compound allows you to properly calculate how many of each ion are present in a solution. Chemical dissociation is essential for predicting the outcomes of reactions and the behavior of solutions under different conditions.

Molarity

Molarity is a concentration unit in chemistry that is defined as the number of moles of solute per liter of solution. It is denoted by the symbol "M". Molarity is central to understanding solution concentrations and is one of the first steps in calculating ion concentrations.

• To calculate molarity, divide the moles of solute by the volume of the solution in liters: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
• This unit helps chemists compare the concentration of solutions and determine reaction ratios and stoichiometry.

When you know the molarity of a compound, and how it dissociates, you can calculate individual ion concentrations, as shown in sodium ion calculations.

Moles Calculation

The concept of 'the mole' is crucial for converting between atoms, molecules, or ions and measurable quantities. A mole is Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\). When calculating moles, you're converting measurable mass or volume into a countable number of particles.

• To calculate moles from volume and molarity, use the formula: \[ \text{Moles} = \text{Volume (L)} \times \text{Molarity (M)} \]
• This calculation allows for determining the exact number of sodium ions or any other species in a given volume of solution.
• For example, calculating the moles of Na⁺ ions in a 20 mL solution of 0.15 M NaHCO₃ involves converting mL to L (0.020 L) and using the equation to find 0.003 moles of Na⁺.

Mastering mole calculations enables accurate chemical analysis and is foundational for more advanced chemistry topics.