Question

You are given a \(1.50-\mathrm{g}\) mixture of sodium nitrate and sodium chloride. You dissolve this mixture into \(100 \mathrm{~mL}\) of water and then add an excess of \(0.500 \mathrm{M}\) silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of \(0.641 \mathrm{~g}\). 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 percent sodium chloride in the original unknown mixture.

Short answer

The species in the solution are Na⁺, NO₃⁻, Cl⁻, and Ag⁺. The balanced net ionic equation for the reaction producing the solid is Cl⁻ (aq) + Ag⁺ (aq) → AgCl (s). The percent of sodium chloride in the original unknown mixture is approximately 17.4%.

Step-by-step solution

Identify dissolved species in water

After dissolving the mixture of sodium nitrate (NaNO₃) and sodium chloride (NaCl) in water, the species present are sodium ions (Na⁺), nitrate ions (NO₃⁻), and chloride ions (Cl⁻). When the silver nitrate (AgNO₃) solution is added, we also have silver ions (Ag⁺) in the solution. So, the species we would see in the solution are Na⁺, NO₃⁻, Cl⁻, and Ag⁺. b. Write the balanced net ionic equation

Write complete ionic equation

In the complete ionic equation, all soluble ionic compounds are shown as individual ions. In our case, the equation is: Na⁺ (aq) + NO₃⁻ (aq) + Na⁺ (aq) + Cl⁻ (aq) + Ag⁺ (aq) + NO₃⁻ (aq) → Na⁺ (aq) + NO₃⁻ (aq) + AgCl (s) + Na⁺ (aq) + NO₃⁻ (aq)

Cross out the spectator ions

Spectator ions are the ions that do not participate in the reaction. In this case, Na⁺ and NO₃⁻ are the spectator ions. Crossing them out, we get: Cl⁻ (aq) + Ag⁺ (aq) → AgCl (s) b. Balanced net ionic equation:

Balanced net ionic equation for the reaction

The balanced net ionic equation for the reaction that produces the solid is: Cl⁻ (aq) + Ag⁺ (aq) → AgCl (s) c. Calculate the percent sodium chloride in the original mixture

Find the mass of silver chloride formed

The mass of the white solid (silver chloride) formed is 0.641 g.

Use stoichiometry to find the mass of sodium chloride

Using the stoichiometry of the balanced net ionic equation, we can find the mass of sodium chloride that reacted to form silver chloride. The molar mass of AgCl = 107.87 (Ag) + 35.45 (Cl) = 143.32 g/mol The molar mass of NaCl = 22.99 (Na) + 35.45 (Cl) = 58.44 g/mol 1 mol of NaCl reacts with 1 mol of AgCl. So, mass of NaCl = (mass of AgCl / molar mass of AgCl) * molar mass of NaCl = (0.641 g / 143.32 g/mol) * 58.44 g/mol ≈ 0.261 g

Calculate the percent sodium chloride in the original mixture

The total mass of the original mixture is given as 1.50 g. The mass of sodium chloride we found is 0.261 g. Percent NaCl = (mass of NaCl / total mass of the mixture) * 100 Percent NaCl = (0.261 g / 1.50 g) * 100 ≈ 17.4 % c. The percent sodium chloride in the original unknown mixture is approximately 17.4 %.