Question

Suppose \(250.0 \mathrm{~mL}\) of a \(0.600 \mathrm{M}\) solution of barium nitrate is combined with enough \(0.500 \mathrm{M}\) sodium sulfate solution to precipitate the maximum amount of barium sulfate. What volume of the sodium sulfate solution is required?

Short answer

A volume of 300.0 mL of 0.500 M sodium sulfate solution is required to precipitate the maximum amount of barium sulfate when combined with 250.0 mL of 0.600 M barium nitrate solution.

Step-by-step solution

1. Write down the balanced chemical equation.

The balanced chemical equation for this reaction is: \[Ba(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaNO_3(aq).\]

2. Calculate the moles of barium nitrate.

The volume of the barium nitrate solution is given as 250.0 mL, and its concentration is 0.600 M. We can convert the volume to liters and calculate the mol of the given substance: \[moles \: of \: Ba(NO_3)_2 = Volume \times Concentration.\] Moles of \(Ba(NO_3)_2 = 0.250 \mathrm{~L} \times 0.600 \mathrm{~M} = 0.150 \mathrm{~mol}.\]

3. Determine the moles of sodium sulfate required.

From the balanced chemical equation, we see that 1 mole of \(Ba(NO_3)_2\) reacts with 1 mole of \(Na_2SO_4\). Therefore, moles of \(Na_2SO_4\) required for the reaction is equal to the moles of \(Ba(NO_3)_2\): Moles of \(Na_2SO_4 = 0.150 \: \mathrm{mol}.\)

4. Calculate the volume of sodium sulfate solution required.

We are given the concentration of sodium sulfate solution as 0.500 M, now we will use this to find the volume required: \[Volume \: of \: Na_2SO_4 = \frac{Moles \: of \: Na_2SO_4}{Concentration}\] \(Volume\: of\: Na_2SO_4 = \frac{0.150\: \mathrm{mol}}{0.500\: \mathrm{M}} = 0.300\: \mathrm{L}\) or 300.0 mL.

Summary

A volume of 300.0 mL of 0.500 M sodium sulfate solution is required to precipitate the maximum amount of barium sulfate when combined with 250.0 mL of 0.600 M barium nitrate solution.

Key concepts

Chemical Equation Balancing

Understanding the basics of chemical equation balancing is essential in stoichiometry. It's akin to ensuring you have the right amount of ingredients when cooking a recipe. To balance a chemical equation like
\[Ba(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaNO_3(aq)\],
we must make sure that the quantity of each type of atom on the reactant side equals that on the product side. In our case, we have one barium atom, one sulfate ion, and two nitrate ions on both sides, indicating that our equation is balanced. When equations are unbalanced, coefficients must be used to equalize the atom counts, emphasizing the concept of conservation of matter.

Molarity Calculations

Molarity calculations provide a way to measure the concentration of a solution, allowing chemists to know precisely how much of a substance is present in a given volume of liquid. It's defined as moles of solute per liter of solution (mol/L). For example, to find the moles of barium nitrate (\(Ba(NO_3)_2\)) in a 250.0 mL (0.250 L) solution with a molarity of 0.600 M, we use the formula:
\[moles\:of\:Ba(NO_3)_2 = 0.250\:L \times 0.600\:M\].
This results in 0.150 mol of \(Ba(NO_3)_2\). Understanding this concept is crucial for progressing through more advanced stoichiometric calculations.

Limiting Reactant Determination

In chemical reactions, the limiting reactant is the substance that is completely used up first, thus dictating the maximum amount of product that can be formed. This concept is much like running out of eggs when baking; you cannot make more cake than the eggs allow. In our exercise, because each mole of \(Ba(NO_3)_2\) reacts with an equal mole of \(Na_2SO_4\), they are used in a 1:1 molar ratio. Since we start with 0.150 mol of \(Ba(NO_3)_2\), we'll need the same amount of \(Na_2SO_4\) to react with it. No reactant is in excess in this perfectly proportioned scenario, demonstrating a stoichiometrically balanced reaction.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form one or more insoluble products. In the context of our problem, when a solution of barium nitrate is mixed with sodium sulfate, an insoluble solid called a precipitate forms, in this case, barium sulfate (\(BaSO_4\)). The ability to predict and calculate the outcomes of precipitation reactions is a vital skill in chemistry. It relies upon understanding solubility rules, balancing chemical equations, and calculating molarities, as demonstrated in solving our given problem to find the volume of sodium sulfate required to precipitate the maximum amount of barium sulfate.