Problem 80 - Chemistry: The Central Science in SI Units, Global Edition

Question

You want to analyze a silver nitrate solution. What mass of \(\mathrm{NaCl}\) is needed to precipitate \(\mathrm{Ag}^{+}\) ions from \(45.0 \mathrm{~mL}\) of \(0.2500 \mathrm{MAgNO}_{3}\) solution?

Step-by-Step Solution

Answer

To precipitate all the Ag⁺ ions from the 45.0 mL of 0.2500 M AgNO₃ solution, approximately 0.6574 grams of NaCl are needed.

Step 1: Write the chemical equation

The chemical reaction between silver nitrate (AgNO₃) and sodium chloride (NaCl) produces a precipitate of silver chloride (AgCl) and sodium nitrate (NaNO₃). The balanced chemical equation is: AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)

Step 2: Determine the moles of Ag⁺ ions in the solution

Using the given volume and concentration of the silver nitrate (AgNO₃) solution, we can calculate the moles of Ag⁺ ions present: Moles of Ag⁺ = Volume * Concentration Moles of Ag⁺ = 45.0 mL * 0.2500 M Keep in mind that we need to convert the volume from mL to L before multiplying with the concentration: Moles of Ag⁺ = 0.045 L * 0.2500 M Moles of Ag⁺ = 0.01125 mol

Step 3: Determine the moles of NaCl needed using stoichiometry

Using the balanced chemical equation, we can determine the moles of NaCl needed to precipitate all the Ag⁺ ions: 1 mol AgNO₃ : 1 mol NaCl Given that there are 0.01125 mol of Ag⁺ ions, the required moles of NaCl will be equal: Moles of NaCl = 0.01125 mol

Step 4: Calculate the mass of NaCl required

Using the molar mass of NaCl, we can calculate the mass of NaCl needed to precipitate all the Ag⁺ ions: Molar mass of NaCl = 58.44 g/mol Mass of NaCl = Moles of NaCl * Molar mass of NaCl Mass of NaCl = 0.01125 mol * 58.44 g/mol Mass of NaCl = 0.6574 g Therefore, approximately 0.6574 grams of NaCl are needed to precipitate all the Ag⁺ ions from the 45.0 mL of 0.2500 M AgNO₃ solution.

Key Concepts

Chemical Equations

Chemical equations are representations of chemical reactions where reactants transform into products. In this exercise, the chemical reaction involves silver nitrate

• (AgNO₃)
• and sodium chloride (NaCl).

Both are reactants that participate in the reaction.
When they mix in solution, they form a precipitate of silver chloride

• (AgCl)
• and sodium nitrate (NaNO₃).

This is represented by the balanced chemical equation:

\[\text{AgNO}_3 \text{ (aq)} + \text{NaCl (aq)} \rightarrow \text{AgCl (s)} + \text{NaNO}_3 \text{ (aq)}\]
This equation indicates that one molecule of

• AgNO₃ reacts with one molecule of NaCl
• to produce one molecule of AgCl
• and one molecule of NaNO₃.

Writing chemical equations helps us understand the quantity and relationship of substances involved in the reactions.

Molar Mass

Molar mass is a crucial concept in stoichiometry as it connects the mass of a substance to the amount in moles. It is the mass of one mole of a given substance, often expressed in grams per mole (g/mol).
In solving this exercise, you need to know the molar mass of sodium chloride (NaCl) to find out how much mass is needed.
The molar mass of NaCl is the sum of the molar masses of its constituent atoms:

• The molar mass of sodium (Na) is approximately 22.99 g/mol,
• and that of chlorine (Cl) is approximately 35.45 g/mol.

Adding these gives the molar mass of NaCl as 58.44 g/mol.
Knowing the molar mass allows us to convert from moles to grams, helping us find that approximately 0.6574 grams of NaCl are needed.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions mix and an insoluble solid forms. This solid is called a precipitate. In this problem, the precipitation reaction occurs between the silver nitrate (AgNO₃) and sodium chloride (NaCl) solutions.
When AgNO₃ is added to NaCl, silver chloride (AgCl) precipitates as a solid. This happens because Ag⁺ ions in silver nitrate combine with Cl⁻ ions in sodium chloride.

• The result is the formation of a white solid, AgCl, which is insoluble in water.
• The remaining sodium ions (Na⁺) and nitrate ions (NO₃⁻) stay dissolved in the solution.

Precipitation reactions help in removing ions from solutions, useful in various chemical processes especially analytical chemistry to identify specific ions.

Solution Concentration

Solution concentration describes how much solute is present in a given volume of solvent, often expressed in moles per liter (Molarity, M). In this exercise, we start with a silver nitrate solution of 0.2500 M concentration in a 45.0 mL volume.
Understanding molarity is key to determining how much

• NaCl is needed to react with the AgNO₃ solution.

To calculate the amount of Ag⁺ ions present, convert the volume from mL to L, and then multiply by the concentration:

• Volume = 0.045 L (since 1 L = 1000 mL)
• Moles = Volume × Concentration = 0.045 L × 0.2500 M = 0.01125 mol of Ag⁺

The stoichiometry of the reaction tells us equal moles of NaCl are needed to completely precipitate the Ag⁺ ions, leading us to calculate the mass of NaCl required based on the number of moles.