(c) If 55.8 mL of a BaCl₂ solution is needed to precipitate all the sulfate ion in a 752-mg sample of Na₂SO₄, what is the molarity of the BaCl₂ solution?

Verified step by step guidance

1. Identify the chemical reaction: BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl.
2. Calculate the moles of Na_2SO_4 using its molar mass: \text{moles of Na}_2\text{SO}_4 = \frac{\text{mass of Na}_2\text{SO}_4}{\text{molar mass of Na}_2\text{SO}_4}.
3. Use the stoichiometry of the balanced equation to find the moles of BaCl_2 needed: \text{moles of BaCl}_2 = \text{moles of Na}_2\text{SO}_4.
4. Convert the volume of BaCl_2 solution from mL to L: \text{volume in L} = \frac{55.8 \text{ mL}}{1000}.
5. Calculate the molarity of the BaCl_2 solution: \text{Molarity} = \frac{\text{moles of BaCl}_2}{\text{volume of solution in L}}.

Key Concepts

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using balanced chemical equations to determine the relationships between the amounts of substances consumed and produced. In this question, stoichiometry is essential to relate the moles of BaCl2 used to the moles of sulfate ions from Na2SO4 that are being precipitated.

Molarity

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in units of moles per liter (mol/L). To find the molarity of the BaCl2 solution in this problem, one must calculate the moles of BaCl2 that reacted and divide that by the volume of the solution in liters.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react to form an insoluble product, known as a precipitate. In this case, BaCl2 reacts with Na2SO4 to form barium sulfate (BaSO4), which precipitates out of solution. Understanding this concept is crucial for determining how much BaCl2 is needed to completely react with the sulfate ions present in the sample.