Question

Consider \(1.0 \mathrm{~L}\) of an aqueous solution that contains \(0.10 \mathrm{M}\) sulfuric acid to which \(0.30\) mol barium nitrate is added. Assuming no change in volume of the solution, determine the \(\mathrm{pH}\), the concentration of barium ions in the final solution, and the mass of solid formed.

Short answer

The pH of the final solution is 1, the concentration of barium ions is 0.30 M, and the mass of solid formed is 23.34 g.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and barium nitrate (Ba(NO₃)₂) is: H₂SO₄ (aq) + Ba(NO₃)₂ (aq) → BaSO₄ (s) + 2 HNO₃ (aq)

Calculate the initial moles of sulfuric acid and barium nitrate

We are given a 1.0 L sulfuric acid (H₂SO₄) solution with concentration at 0.10 M, and 0.30 moles of barium nitrate (Ba(NO₃)₂). We can calculate the initial moles of sulfuric acid: moles of H₂SO₄ = (concentration) x (volume) = (0.10 M) x (1.0 L) = 0.10 moles

Determine the limiting reactant and moles of products formed

According to the balanced chemical equation, 1 mole of H₂SO₄ reacts with 1 mole of Ba(NO₃)₂. We have 0.10 moles of H₂SO₄ and 0.30 moles of Ba(NO₃)₂, so H₂SO₄ is the limiting reactant. Since the reaction goes to completion, we will have 0.10 moles of BaSO₄ formed.

Calculate the concentration of barium ions in the final solution

The concentration of Ba²⁺ ions does not change when BaSO₄ precipitates. The initial concentration of Ba²⁺ ions can be found from the mole ratio (1:1) of the reactants (H₂SO₄ and Ba(NO₃)₂): moles of Ba²⁺ ions = moles of Ba(NO₃)₂ = 0.30 moles To calculate the final concentration of Ba²⁺ ions in the solution: Concentration of Ba²⁺ ions = (moles of Ba²⁺ ions) / (volume of the solution) = (0.30 moles) / (1.0 L) = 0.30 M

Calculate the mass of solid formed

Since we formed 0.10 moles of BaSO₄, we can calculate the mass of solid formed as: mass of BaSO₄ = (moles of BaSO₄) x (molar mass of BaSO₄) = (0.10 moles) x (137.33 g/mol Ba + 32.07 g/mol S + (4 x 16.00 g/mol O)) = 0.10 x (233.43 g/mol) = 23.34 g

Calculate the concentration of excess H+ ions and find the pH

We started with 0.10 moles of H₂SO₄ and 1 mole of H₂SO₄ produces 2 moles of H⁺ ions, so we initially have 0.20 moles of H⁺ ions. In the reaction with Ba(NO₃)₂, 1 mole of H₂SO₄ consumes 1 mole of H⁺ ions. Since 0.10 moles of H₂SO₄ reacted completely, 0.10 moles of H⁺ ions remain in the solution. The concentration of H⁺ ions in the final solution is: [H⁺] = (moles of H⁺ remaining) / (volume of the solution) = (0.10 moles) / (1.0 L) = 0.10 M Finally, we can calculate the pH of the solution: pH = -log10([H⁺]) = -log10(0.10) = 1 So, the pH of the final solution is 1, the concentration of barium ions is 0.30 M, and the mass of solid formed is 23.34 g.

Key concepts

pH Calculation

The potential of hydrogen, or pH, is a measure of the acidity or alkalinity of an aqueous solution. Calculating the pH involves determining the concentration of hydrogen ions, denoted as [H⁺], in the solution. The pH is then found using the formula \( pH = -\log_{10}([H⁺]) \) which involves taking the negative logarithm, base 10, of the concentration of hydrogen ions. In the given exercise, sulfuric acid (H₂SO₄), a strong acid, dissociates completely in water to produce two hydrogen ions per molecule. To calculate the pH of the resulting solution, we determine the remaining concentration of [H⁺] after the reaction with barium nitrate (Ba(NO₃)₂) and apply the pH formula. Understanding this concept allows students to appreciate the relationship between the concentration of hydrogen ions in a solution and how it translates to pH, a scale commonly used in various sciences to express the level of acidity or basicity.

Limiting Reactant Determination

In a chemical reaction, the limiting reactant is the substance that is entirely consumed first and thus determines the amount of product that can be formed. It is akin to the ingredient in a recipe that runs out first, limiting the amount of food you can prepare. To find the limiting reactant, one needs to compare the mole ratio of the reactants used in the reaction to the ratio in the balanced chemical equation. In the example provided, the balanced equation shows a 1:1 mole ratio between H₂SO₄ and Ba(NO₃)₂. By comparing the initial moles of H₂SO₄ and Ba(NO₃)₂, we can identify H₂SO₄ as the limiting reactant because it has fewer moles than required based on the mole ratio. Learning to identify the limiting reactant is crucial as it allows students to predict quantities of substances required and formed in a chemical reaction.

Precipitation Reaction

A precipitation reaction occurs when two soluble reactants combine to form an insoluble solid, known as the precipitate. This process is a hallmark of double-displacement reactions, where the anions and cations of two different compounds swap places, occasionally resulting in a product that falls out of solution. In our exercise, barium sulfate (BaSO₄) is the precipitate formed when sulfuric acid and barium nitrate react. Identifying a precipitation reaction is straightforward if one is familiar with the solubility rules. These rules help to predict whether a compound will be soluble or insoluble in water. For example, most sulfate salts are soluble, except those of barium, strontium, and lead. Understanding precipitation reactions is essential, especially in analytical chemistry, where they are used for separation and purification techniques.

Molar Concentration

Molar concentration, noted as molarity, refers to the amount of solute present in a given volume of solution, commonly expressed in moles per liter (M). It is calculated by dividing the number of moles of solute by the volume of the solution in liters. In the context of our exercise, after Ba(NO₃)₂ is added to the sulfuric acid solution and the precipitation reaction occurs, the concentration of the barium ions (Ba²⁺) is unchanged because precipitation does not affect the dissolved ions that do not form a solid. By computing the number of moles of Ba²⁺ present initially and dividing it by the unchanged total volume of solution, we find the molarity of the barium ions. Mastering the concept of molar concentration is vital as it permits the comparison of reaction rates and yield, and informs the preparation of solutions with precise chemical concentrations in lab settings.