A course titration is performed on a 3.00 mL sample of sulfuric acid with a 0.241 M sodium hydroxide solution.

Question

A course titration is performed on a 3.00 mL sample of sulfuric acid with a 0.241 M sodium hydroxide solution. The initial and final buret readings for the titration to the endpoint were 0.35 mL and 9.60 mL, respectively. Use this data to determine how much of the sulfuric acid solution should be used for the fine titrations if you want to use approximately 40 mL of the sodium hydroxide titrant. Round your answer to a whole number.

Answer 1

H₂SO₄ + 2NaOH ==> Na₂SO₄ + 2H₂O ... balanced equation

mls NaOH used = 9.60 ml - 0.35 ml = 9.25 ml NaOH

moles NaOH used = 9.25 mls x 1 L / 1000 mls x 0.241 mols / L = 0.00223 mols NaOH used

moles H₂SO₄ present = 0.00223 mols NaOH x 1 mol H₂SO₄ / 2 mols NaOH = 0.00111 mols H₂SO₄ present

[H₂SO₄] = 0.00111 mols / 0.003 L = 0.372 M H₂SO₄

You can now use this information to determine how much of this you should use for the fine titration if you want to use about 40 mls of the NaOH titrant.

40 mls NaOH x 1 L / 1000 mls x 0.241 mol/L = 0.00964 mols NaOH

mols H₂SO₄ = 1/2 x 0.00964 = 0.00482 mols H₂SO₄ desired

Volume of H₂SO₄ that should be used = 0.00482 mols x 1 L / 0.372 mols = 0.01296 L = ~13 mls