Question

You have a 500ml solution of sulfuric acid but you don't know what the concentration is. You titrate the solution with 25mL of 2.00 M of NaOH but overshoot the equivalence point. You then back titrate (add acid to go back to the equivalence point) with 3 mL of 0.1 M HCl. What is the concentration of the sulfuric acid solution?

Answer 1

balanced eqations

H2SO4    + 2 NaOH -----------------------> Na2SO4 + H2O

NaOH + HCl ---------------------> NaCl + H2O

moles of HCl =3 x 0.1 / 1000 = 3 x 10^-4

moles of NaOH = 25 x 2 / 1000 = 0.05

moles of actual NaOH reacted with H2SO4 = 0.05 - (3 x 10^-4) = 0.0497

1 mol H2SO4 -----------------------> 2 mol NaOH

x mol H2SO4 -----------------------> 0.0497 mol NaOH

x = 1 x 0.0497 / 2

x = 0.02485

moles of H2SO4 = 0.02485

volume = 500 mL = 0.5 L

molarity = moles / volume

              = 0.02485 / 0.5

               = 0.0497 M

concentration of the sulfuric acid solution= 0.0497 M