Problem 115 - General Chemistry

Question

A barium mineral was dissolved in hydrochloric acid to give a solution of barium ion. An excess of potassium sulfate was added to \(50.0 \mathrm{~mL}\) of the solution, and \(1.128 \mathrm{~g}\) of barium sulfate precipitate formed. Assume that the original solution was barium chloride. What was the molarity of \(\mathrm{BaCl}_{2}\) in this solution?

Step-by-Step Solution

Answer

The molarity of \( \text{BaCl}_2 \) in the solution is 0.0966 M.

Step 1: Understanding the Reaction

The barium ion ({Ba^{2+}}) from barium chloride ({BaCl_2}) reacts with sulfate ions (SO_4^{2-}) from potassium sulfate ({K_2SO_4}) to form barium sulfate precipitate ({BaSO_4}). The reaction is: \[ \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \rightarrow \text{BaSO}_4 (s) \].

Step 2: Calculate Moles of Barium Sulfate

First, determine the number of moles of barium sulfate formed. The molar mass of barium sulfate ({BaSO_4}) is calculated as follows: \[ \text{Ba: } 137.33 + \text{S: } 32.07 + \text{O: } 4\times16.00 = 233.40 \text{ g/mol} \].The moles of BaSO_4 are \[ \frac{1.128 \text{ g}}{233.40 \text{ g/mol}} \approx 0.00483 \text{ moles} \].

Step 3: Relate to Moles of Barium Chloride

The reaction shows a 1:1 ratio between {Ba^{2+}} and {BaSO_4}. Therefore, the number of moles of {Ba^{2+}} ions from {BaCl_2} is also approximately 0.00483 moles.

Step 4: Calculate Molarity of Original Solution

Molarity is calculated using the formula: \[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \].Here, the volume is 50.0 mL, which is \(0.0500\) L.Thus, the molarity of {BaCl_2} is \[ \frac{0.00483 \text{ moles}}{0.0500 \text{ L}} = 0.0966 \text{ M} \].

Key Concepts

Barium Chloride

Barium chloride, represented by the chemical formula \( \text{BaCl}_2 \), is an ionic compound consisting of barium ions (\( \text{Ba}^{2+} \)) and chloride ions (\( \text{Cl}^- \)). This compound is soluble in water. When dissolved, it releases barium ions into the solution.
Barium chloride is often used in laboratories to demonstrate precipitation reactions and analyze sulfate ions because of its ability to form insoluble barium sulfate when mixed with solutions containing sulfate ions.
This property is crucial when determining the concentration of barium in a solution, as shown in the exercise where barium chloride was completely dissolved and reacted to form a barium sulfate precipitate.

Barium Sulfate Precipitation

A precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid—called a precipitate. In this scenario, barium sulfate is the precipitate. When a solution containing barium ions from barium chloride is mixed with a solution containing sulfate ions from potassium sulfate, barium sulfate (\( \text{BaSO}_4 \)) forms as a solid.
The insolubility of barium sulfate is a key factor in many chemical analyses, where it helps to fully extract barium ions from an aqueous solution. In the given exercise, 1.128 grams of barium sulfate were formed, making it possible to back-calculate the concentration of barium ions originally present in the barium chloride solution.

Stoichiometry

Stoichiometry is an area of chemistry that deals with the quantitative relationships or ratios of reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction.
In this exercise, stoichiometry is used to determine the relationship between barium ions and barium sulfate. The balanced chemical equation \( \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \) highlights a 1:1 molar ratio between barium ions and barium sulfate.
This means that every mole of barium ions reacts with a mole of sulfate ions to produce a mole of barium sulfate, which plays a critical role in calculating the original concentration of the barium chloride solution.

Moles Calculation

Calculating the number of moles is crucial for determining concentrations and involves using the molar mass of a substance. Moles are a base unit in chemistry that measure the quantity of a substance. This exercise relies on converting the mass of barium sulfate precipitate into moles using its molar mass.
The molar mass of barium sulfate (\( \text{BaSO}_4 \)) is calculated by summing the atomic masses of barium (137.33 g/mol), sulfur (32.07 g/mol), and oxygen (64.00 g/mol from 4 oxygens) to get 233.40 g/mol.
With 1.128 grams of \( \text{BaSO}_4 \) formed, calculate moles by dividing mass by molar mass: \( \frac{1.128 \text{ g}}{233.40 \text{ g/mol}} \approx 0.00483 \text{ moles} \).
These moles directly correspond to the amount of barium ions, aiding the calculation of the initial molarity of barium chloride in the solution.