Question

A 10.0-mL sample of sulfuric acid, H2SO4, from a battery of an old car is diluted to 100.0 mL, and a 10.00-mL aliquot (portion) of the diluted acid is then titrated with 0.2500 M NaOH solution. If the concentration of H2SO4 in the original battery was 3.25 M, how many milliliters (mL) of the NaOH solution is required to titrate the sulfuric acid present in the 10.0-mL portion of dilute acid solution

Answer 1

26.0 mL

Step-by-step explanation:

The reaction that takes place is:

• H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

First we calculate the molar concentration of the diluted sulfuric acid solution, using the C₁V₁=C₂V₂ formula:

• 3.25 M * 10.0 mL = C₂ * 100 mL

• C₂ = 0.325 M

Now we calculate how many H₂SO₄ moles would there be in 10.00 mL of this diluted solution, using the definition of molarity:

• 0.325 M * 10.00 mL = 3.25 mmol H₂SO₄

Then we convert 3.25 mmoles of H₂SO₄ into mmoles of NaOH, using the stoichiometric coefficients of the balanced reaction:

• 3.25 mmol H₂SO₄ *
  `(2mmolNaOH)/(1mmolH_2SO_4)` = 6.5 mmol NaOH

Finally we calculate the required volume of the NaOH solution, using the definition of molarity:

• 6.5 mmol NaOH / 0.2500 M = 26.0 mL