Question

A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0-mL aliquot and adding \(50.0 \mathrm{mL}\) of \(0.213 \mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required \(13.21 \mathrm{mL}\) of \(0.103 M\) HCI for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Sulfuric acid has two acidic hydrogens.

Short answer

The molarity of the original H₂SO₄ solution is 0.00929 M.

Step-by-step solution

Find moles of OH⁻ ions neutralized by excess HCl

We know that: Volume of HCl = 13.21 mL Molarity of HCl = 0.103 M Let's convert the volume to liters and then use the molarity formula to calculate moles of OH⁻ ions in the excess base. \(Volume_{HCl} (L) = \frac{13.21~mL}{1000} = 0.01321 L\) Moles of OH⁻ ions = Molarity × Volume Moles of OH⁻ ions = 0.103 M × 0.01321 L Moles of OH⁻ ions = 0.00136 mol

Find total moles of OH⁻ ions added to H₂SO₄

We know that: Volume of NaOH = 50.0 mL Molarity of NaOH = 0.213 M Convert the volume to liters and then use the molarity formula to calculate moles of OH⁻ ions added to H₂SO₄. \(Volume_{NaOH} (L) = \frac{50.0~mL}{1000} = 0.05 L\) Moles of OH⁻ ions = Molarity × Volume Moles of OH⁻ ions = 0.213 M × 0.05 L Moles of OH⁻ ions = 0.01065 mol

Calculate moles of H⁺ ions in H₂SO₄

First, let's find the moles of OH⁻ ions that reacted with H₂SO₄ by subtracting the moles of OH⁻ ions that reacted with the excess acid from the total moles of OH⁻ ions added. Moles of OH⁻ ions reacted with H₂SO₄ = Total moles of OH⁻ ions - Moles of OH⁻ ions in excess base Moles of OH⁻ ions reacted with H₂SO₄ = 0.01065 mol - 0.00136 mol Moles of OH⁻ ions reacted with H₂SO₄ = 0.00929 mol Since H₂SO₄ has two acidic hydrogens, it reacts with two moles of OH⁻ ions for every mole of H₂SO₄. Therefore, the moles of H₂SO₄ can be calculated as: Moles of H₂SO₄ = Moles of OH⁻ ions reacted with H₂SO₄ ÷ 2 Moles of H₂SO₄ = 0.00929 mol ÷ 2 Moles of H₂SO₄ = 0.004645 mol

Calculate the molarity of the original H₂SO₄ solution

We are given that the volume of the original H₂SO₄ solution is 500 mL. Let's calculate the molarity of the solution using the formula: Molarity = moles of solute ÷ volume of solution (in L) First, convert the volume to liters: \(Volume_{H_2SO_4} (L) = \frac{500~mL}{1000} = 0.5 L\) Now, calculate the molarity of the solution: Molarity of H₂SO₄ = 0.004645 mol ÷ 0.5 L Molarity of H₂SO₄ = 0.00929 M The molarity of the original H₂SO₄ solution is 0.00929 M.

Key concepts

Acid-Base Titration

Acid-base titration is a fundamental process in chemistry used to determine the concentration of an unknown acid or base by neutralizing it with a base or acid of known concentration, respectively. The endpoint of the titration is reached when the number of moles of hydrogen ions, H⁺, in the acid equals the number of moles of hydroxide ions, OH⁻, from the base, signaling complete neutralization.

Here's the basic premise: A carefully measured volume of a standard solution (the titrant with known molarity) is added to a known volume of the solution with the unknown concentration until the reaction reaches the endpoint. This is indicated by a color change from an indicator or by reaching a specified pH value using a pH meter. In our exercise, we can infer the molarity of the sulfuric acid solution by neutralizing it with a known quantity of sodium hydroxide (NaOH) and then back-titrating the excess NaOH with hydrochloric acid (HCl).

Molarity

Molarity, symbolized as M, is a unit of concentration within chemistry that measures the number of moles of a solute per liter of solution. To calculate molarity, you use the formula: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). For our problem, this concept allows us to find the molarity of sulfuric acid by determining the moles of H₂SO₄ and dividing it by the total volume of the solution in which H₂SO₄ has dissolved. This calculation helps to understand the strength of the acid solution and plays a crucial role in various applications, including chemical reactions and titration exercises like the one we explored.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It allows us to predict the amounts of substances consumed and produced in a given reaction. In acid-base reactions, stoichiometry helps us to understand the mole-to-mole ratios between acid and base. For sulfuric acid, which has a stoichiometric factor of two due to its two acidic hydrogens, it will react with two moles of hydroxide ions for each mole of H₂SO₄.

By using stoichiometry, we can convert the volume and concentration (molarity) of NaOH added to moles, determine the moles of OH⁻ involved in the reaction, and hence calculate the moles of H₂SO₄ in the sample. Understanding stoichiometry is pivotal for accurate chemical calculations and is a cornerstone of high-quality chemistry education.

Sulfuric Acid (H₂SO₄)

Sulfuric acid is a strong acid widely used in industry and laboratories. It's characterized by its chemical formula H₂SO₄ and is known for having two acidic protons, which means it can donate two H⁺ ions per molecule in reactions. In the context of our titration exercise, recognizing that sulfuric acid can react with two moles of base per mole of acid is essential to accurately perform calculations.

Given its nature, when it dissociates in water, it contributes to the acid's molarity based on the relation that 1 M of sulfuric acid provides 2 M of H⁺ ions, demonstrating why stoichiometry is a critical factor in getting the correct molarity. When dealing with sulfuric acid in calculations and experimental setups, always consider its dual acidic nature for precise results.