Titration in Chemistry

Question

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. She has a 0.1323 M standard sodium hydroxide solution. She takes a 25.00 mL sample of the original acid solution and dilutes it to 250.0 mL. Then, she takes a 10.00 mL sample of the dilute acid solution and titrates it with the standard solution in the buret. The initial buret reading was 3.55 mL, and after the endpoint was reached, the final buret reading was 23.27 mL. What is the concentration of the original sulfuric acid solution?

  

Answer 1

Looking at the balanced equation for the titration, we have...

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

volume of NaOH used = 23.27 ml - 3.55 ml = 19.72 ml

moles NaOH used = 19.72 ml x 1 L / 1000 ml x 0.1323 mol/L = 0.002609 mols NaOH used

moles H₂SO₄ present = 0.002609 mol NaOH x 1 mol H₂SO₄/2 mol NaOH = 0.001304 mols H₂SO₄

This is the amount of H₂SO₄ present in the 10.00 mls of diluted sample being titrated.

[H₂SO₄] = 0.001304 mols / 10 ml x 1000 ml / L = 0.1304 M

To find the concentration of the undiluted original sample, we must take into account the dilution that was made, i.e. 25 ml to 250 ml = 10x

Thus, the [H₂SO₄] in the original sample is 10x the concentration in the diluted sample.

[H₂SO₄] in original sample = 10 x 0.1304 M = 1.304 M