1.4 moles of a 3.0M NaOH solution neutralizes .25L of H2SO4. What is the molarity of H2SO4?

Question

Balanced equation: 2NaOH + H2SO4 → Na2SO4 + 2H2O

Answer 1

The concentration of the NaOH solution is extraneous information- you can ignore it. What is important is that 1.4 MOLES of NaOH react completely with H₂SO₄. According to the balanced equation, the mole ratio of NaOH to H₂SO₄ is 2/1 = 2. This means that two NaOH "molecules" react with every one molecule of H₂SO₄.

(I put quotation marks around "molecules" because NaOH is not a molecule- it is an ionic compound. A single Na⁺ ion bonded to an OH^- ion is called a "formula unit.")

Anyway, this means there are half as many H₂SO₄ moles in solution as the number of moles of NaOH that were added to neutralize it. Half of 1.4 moles = 0.70 moles. Now we know that there are 0.70 moles of H₂SO₄ in the solution.

Molarity = (moles of solute)/(liters of solution). We were given the denominator, and we just solved for the numerator, so now we can calculate molarity:

0.70 mol H₂SO₄ / 0.25L solution = 2.8 molar H₂SO₄