titration problem

Question

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1954 M0.1954 M standard sodium hydroxide solution. He takes a 25.00 mL25.00 mL sample of the original acid solution and dilutes it to 250.0 mL.250.0 mL. Then, he takes a 10.00 mL10.00 mL sample of the dilute acid solution and titrates it with the standard solution. The endpoint was reached after the addition of 13.02 mL13.02 mL of the standard solution. What is the concentration of the original sulfuric acid solution?

Answer 1

Use the diluted sample of sulfuric acid (H2SO4), and calculate moles of acid present:

H2SO4 + 2NaOH ==> Na2SO4 + 2H2O .. balanced equation for titration with NaOH

mols NaOH used = 13.02 mls x 1 L / 1000 mls x 0.1954 mol / L = 0.002544 mols NaOH

mols H2SO4 present = 0.002544 mols NaOH x 1 mol H2SO4 / 2 mol NaOH = 0.001272 mols H2SO4

Concentration of H2SO4 in 10 ml sample = 0.001272 mol / 10 ml x 1000 ml / L = 0.1272 M

This is the amount in the 10.00 ml of the diluted sample, but this sample has been diluted from the original sample by a factor of 10 (25.00 mls diluted to 250.0 ml is a 10 fold dilution).

Concentration of original H2SO4 solution = 0.1272 M x10 = 1.272 M