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

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

Step by step solution

01

Write the balanced chemical equation for the reaction

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
02

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
03

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.
04

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
05

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.

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Most popular questions from this chapter

What mass of \(\mathrm{NaOH}\) is contained in \(250.0 \mathrm{~mL}\) of a \(0.400 \mathrm{M}\) sodium hydroxide solution?

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