Question

A 0.500-L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a \(100.0-\mathrm{mL}\) aliquot and adding \(50.0 \mathrm{~mL}\) of \(0.213 M \mathrm{NaOH}\). After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required \(13.21 \mathrm{~mL}\) of \(0.103 \mathrm{M} \mathrm{HCl}\) 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 sample of H₂SO₄ is approximately 0.0465 M.

Step-by-step solution

Calculate the moles of OH⁻ ions from HCl neutralization

We are given that after the reaction and with the excess base, it took 13.21 mL of 0.103 M HCl to neutralize the OH⁻ ions. To find the moles of OH⁻ ions, we use the formula: Moles = Molarity (of HCl) × Volume (of HCl) Moles of OH⁻ = 0.103 mol/L × 13.21 mL × (1L/1000 mL) Moles of OH⁻ = 0.00135923 mol

Calculate the moles of OH⁻ ions reacted with H₂SO₄

We are given that 50.0 mL of 0.213 M NaOH was added, so we calculate the total moles of OH⁻ ions: Total moles of OH⁻ = Molarity (of NaOH) × Volume (of NaOH) Total moles of OH⁻ = 0.213 mol/L × 50 mL × (1L/1000 mL) Total moles of OH⁻ = 0.010650 mol Now, subtract the moles of OH⁻ ions from HCl neutralization to find the moles of OH⁻ ions that reacted with H₂SO₄: Moles of OH⁻ (reacted with H₂SO₄) = 0.010650 mol - 0.00135923 mol Moles of OH⁻ (reacted with H₂SO₄) = 0.00929077 mol

Calculate moles of H₂SO₄ and find the molarity

Since H₂SO₄ has two acidic hydrogens, it reacts with two OH⁻ ions per molecule. To find the moles of H₂SO₄ that reacted, we divide the moles of OH⁻ ions by 2: Moles of H₂SO₄ = Moles of OH⁻ (reacted with H₂SO₄) / 2 Moles of H₂SO₄ = 0.00929077 mol / 2 Moles of H₂SO₄ = 0.004645385 mol Recall that the H₂SO₄ solution was originally a 0.500-L sample, and 100 mL aliquot was taken for the reaction. We can now find the concentration (molarity) of the original H₂SO₄ solution: Molarity = Moles of H₂SO₄ / Volume of aliquot (in liters) Molarity (of H₂SO₄) = 0.004645385 mol / 0.1 L Molarity (of H₂SO₄) = 0.04645385 mol/L Therefore, the molarity of the original sample of H₂SO₄ is approximately 0.0465 M.

Key concepts

Titration Analysis

Titration analysis is a critical laboratory procedure used to determine the concentration of an unknown solution. Imagine you're on a treasure hunt, but instead of a map, you have to rely on a reaction to guide you to the 'X marks the spot'—the molarity of a sample. In the context of titration, the treasure is the point where an acid and base neutralize each other, known as the equivalence point.

To reach this, a known concentration of titrant (in this case, NaOH and then HCl) is gradually added to the analyte (H2SO4 solution). Observing the volume of titrant used until neutralization indicates how much of the analyte is present. But, finding out the molarity doesn't end here. What's crucial is the stoichiometry of the reaction—understanding how the acid and base interact on a mole-to-mole basis. Improving your grasp of this concept will make titration a powerful tool in your chemistry toolkit.

Stoichiometry

Stoichiometry bridges the gap between the molecular world and measurable quantities. It is like a baking recipe that tells you how much of each ingredient you need to mix together. In chemistry, stoichiometry uses balanced chemical equations to relate the amounts of reactants and products. It focuses on mole ratios, which are the coefficients of the chemicals involved in a reaction.

For example, in the neutralization of H2SO4 by NaOH, we must consider that sulfuric acid can donate two H+ ions, meaning it reacts with two OH- ions, a detail that's pivotal when computing the exact moles of acid in the original sample. If we miss this, we would incorrectly estimate the concentration of our 'recipe'. Dissecting the stoichiometric relationships in the provided problem is fundamental for students. Breaking down complex recipes into simpler, digestible parts will ensure you don't just memorize; you understand and apply.

Acid-Base Neutralization

Acid-base neutralization is a reaction where an acid like H2SO4 interacts with a base like NaOH, to produce a salt and water. It's a chemical dance where the acid's H+ ions pair with the base's OH- ions to create H2O. The beauty of this reaction is its predictability. With acids releasing H+ and bases releasing OH-, when they meet, they always aim to form water, reaching a balance point, or neutrality.

To improve the understanding of neutralization, visualize it as a seesaw where both acid and base are trying to balance each other out. Too much of one, and we'd need to add the other to level things. This is essentially what's happening when excess hydroxide ions from NaOH are neutralized by HCl. The exercise showcases how much acid is needed to bring the seesaw to a level position after introducing a known amount of base, highlighting the symmetry inherent in acid-base reactions. Always remember, the goal is to achieve that perfect harmony, also known as the equivalence point.