Question

A \(10.00-\mathrm{mL}\) sample of sulfuric acid from an automobile battery requires \(35.08 \mathrm{~mL}\) of \(2.12 M\) sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid? Sulfuric acid contains two acidic hydrogens.

Short answer

The molarity of the sulfuric acid solution is \(3.72 M\).

Step-by-step solution

Write the balanced chemical equation for the reaction

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

Calculate the moles of NaOH used in the reaction

moles of NaOH = (volume in L) × molarity moles of NaOH = (35.08 mL × (1L / 1000 mL)) × 2.12 M = 0.0743696 mol

Determine the stoichiometry of the reaction between H₂SO₄ and NaOH

From the balanced chemical reaction: 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

Calculate the moles of H₂SO₄ in the reaction

moles of H₂SO₄ = (moles of NaOH) / 2 moles of H₂SO₄ = 0.0743696 mol / 2 = 0.0371848 mol

Calculate the molarity of the H₂SO₄ solution

molarity = (moles of H₂SO₄) / (volume in L) molarity = 0.0371848 mol / (10.00 mL × (1L / 1000 mL)) = 3.72 M The molarity of the sulfuric acid solution is 3.72 M.

Key concepts

Sulfuric Acid

Sulfuric acid, chemically known as \( \text{H}_2\text{SO}_4 \), is a strong acid known for its ability to donate two protons or hydrogen ions \( (\text{H}^+) \). This characteristic makes it diprotic. Due to having two acidic hydrogens, sulfuric acid can engage in reactions where each mole of the acid reacts with two moles of a base. It's commonly found in batteries and is a vital industrial chemical. In the discussed problem, sulfuric acid is neutralized using sodium hydroxide, allowing us to determine its molarity based on the reacted quantities.

Sodium Hydroxide

Sodium hydroxide \( (\text{NaOH}) \) is a strong base commonly used in laboratories and industry due to its reactivity. In a chemical reaction, sodium hydroxide will readily dissociate in water to produce \( \text{Na}^+ \) and \( \text{OH}^- \) ions.
Given its strong basic nature, sodium hydroxide can neutralize strong acids like sulfuric acid through a process called neutralization. In the exercise, the exact amount of \( 2.12 \text{ M} \) sodium hydroxide solution was calculated to find how many moles of \( \text{NaOH} \) were required to completely react with the sulfuric acid.

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. This equation is a typical example:
\( \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \).
In this reaction, sulfuric acid \( (\text{H}_2\text{SO}_4) \) reacts with sodium hydroxide \( (\text{NaOH}) \), leading to the formation of sodium sulfate \( (\text{Na}_2\text{SO}_4) \) and water \( (\text{H}_2\text{O}) \). Neutralization is a fundamental concept in chemistry, demonstrating how acids and bases interact to form compounds with a pH closer to 7.

Stoichiometry

Stoichiometry is the area of chemistry that involves quantifying the amounts of substances consumed and produced in chemical reactions. In our problem, stoichiometry helps us determine how much sulfuric acid reacts based on the amount of sodium hydroxide used.
According to the balanced equation, one mole of sulfuric acid reacts with two moles of sodium hydroxide. Understanding this mole ratio is key to determining how much of each reactant is needed for complete neutralization. By calculating the moles of \( \text{NaOH} \) used, we can back-calculate to find the moles of \( \text{H}_2\text{SO}_4 \). Finally, using the initial volume of sulfuric acid, we can determine its molarity, revealing the concentration of the acid in the solution.