Question

You are given a 1.50-g mixture of sodium nitrate and sodium chloride. You dissolve this mixture into 100 mL of water and then add an excess of 0.500 M silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of 0.641 g. a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the mass percent of sodium chloride in the original unknown mixture.

Short answer

The species present in the solution include Na⁺ (aq), NO₃⁻ (aq), Cl⁻ (aq), and Ag⁺ (aq). The balanced net ionic equation for the reaction is \(Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)\). The mass percent of sodium chloride in the original unknown mixture is 17.4%.

Step-by-step solution

Identify the species present in the solution (Question a)

When the sodium nitrate (NaNO₃) and sodium chloride (NaCl) mixture is dissolved in water, we would observe the following species, including their charges: 1. Na⁺ (aq) - sodium ions 2. NO₃⁻ (aq) - nitrate ions 3. Cl⁻ (aq) - chloride ions Additionally, once we add the silver nitrate (AgNO₃) solution, we would also observe: 4. Ag⁺ (aq) - silver ions

Write the balanced net ionic equation (Question b)

The reaction that takes place is between silver ions (Ag⁺) and chloride ions (Cl⁻) to form an insoluble precipitate, silver chloride (AgCl). The balanced net ionic equation, including phases and charges, is: \( \newline Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s) \newline \)

Calculate the mass percent of sodium chloride (Question c)

First, we need to calculate the number of moles of silver chloride formed using its mass: Mass of AgCl = 0.641 g Molar mass of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol Number of moles of AgCl = (mass of AgCl)/(molar mass of AgCl) Number of moles of AgCl = (0.641 g)/(143.32 g/mol) = 0.00447 mol Since one mole of silver chloride is formed per mole of chloride ion, the number of moles of sodium chloride is equal to that of silver chloride: Number of moles of NaCl = 0.00447 mol Now we can calculate the mass of sodium chloride originally present in the mixture: Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol Mass of NaCl = (number of moles of NaCl) × (molar mass of NaCl) Mass of NaCl = (0.00447 mol) × (58.44 g/mol) = 0.261 g Finally, we can find the mass percent of sodium chloride in the original mixture: Mass percent of NaCl = (mass of NaCl)/(total mass of the mixture) × 100 Mass percent of NaCl = (0.261 g)/(1.50 g) × 100 = 17.4 % So, the mass percent of sodium chloride in the original unknown mixture is 17.4%.

Key concepts

Net Ionic Equation

In chemistry, net ionic equations help us focus on the components that undergo change during a reaction. When solutions mix and a reaction occurs, not all ions participate. Some remain unchanged and are called spectator ions. Removing these gives us the net ionic equation.

In the given reaction, we mix sodium chloride (NaCl) and silver nitrate (AgNO₃). While both dissociate into their respective ions, only silver ions (Ag⁺) and chloride ions (Cl⁻) react.

They form silver chloride (AgCl), the white precipitate. The net ionic equation, focusing on these ions, is:

• Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

This equation shows the formation of the solid precipitate, omitting spectator ions like Na⁺ and NO₃⁻.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products. In our example, stoichiometry helps us track how much sodium chloride reacts to form silver chloride.

We know from the balanced equation that the reaction occurs in a 1:1 mole ratio, so every mole of NaCl gives exactly one mole of AgCl. This simple ratio means we can directly convert moles of AgCl to moles of NaCl, simplifying our calculations.

Understanding this relationship allows us to determine quantities needed or produced, ensuring reactions are efficient and predictable.

Molar Mass Calculation

Molar mass is crucial for converting between mass and moles, making it a key step in stoichiometry. In this problem, we use molar mass to find how much sodium chloride contributes to our white precipitate.

The molar mass of a compound is the sum of the atomic masses of its elements.

• For AgCl, it is 107.87 g/mol (Ag) + 35.45 g/mol (Cl), resulting in 143.32 g/mol.
• For NaCl, it's 22.99 g/mol (Na) + 35.45 g/mol (Cl), equating to 58.44 g/mol.

By dividing the mass of our AgCl precipitate by its molar mass, we find the moles of AgCl, which directly gives us the moles of NaCl due to the 1:1 ratio. This conversion allows calculating the mass of NaCl in the mixture.

Precipitation Reaction

Precipitation reactions occur when two soluble salts react to form an insoluble solid, known as a precipitate. In this experiment, silver chloride (AgCl) is our precipitate.

These reactions are often used in analytical chemistry to isolate or detect specific ions. When silver nitrate is added to a solution containing chloride ions, they quickly form solid silver chloride, which appears as a white precipitate.

Understanding precipitation reactions is vital for predicting the outcomes of various chemical reactions and for practical applications, such as water purification and salt formation. Identifying the insoluble product in these reactions is crucial for successful analysis and application.