Question

Exercise 18.77 18.0 mL of 0.117 M sulfurous acid (H2SO3) was titrated with 0.1013 M KOH.

Part A At what added volume of base does the first equivalence point occur? V = _______mL Part B At what added volume of base does the second equivalence point occur? V = _________mL

Answer 1

Part A.

18.0mL of 0.117M H₂SO₃ titrated with 0.1013 M KOH.

First equivalent point

Calculate the moles of H+ in 18ml of H₂SO₃

  1000ml of solution, contains 0.117 moles of H₂SO₃

Then, 18ml of solution contain 0.117 * 18/1000 = 0.002106 = 2.106 x 10^-3

2.106 x 10^-3 moles of H₂SO₃ neutralizes 2.106 x 10^-3 moles of KOH

The volume of KOH used at 1st equivalence = moles of KOH/concentration of KOH

=> 2.016*10^-3/0.1013 = 0.00199012 L =19.9 ml

Part B.

The second equivalence point will be identical to the first equivalence point because the number of moles of H₂SO₃ will determine the number of moles of HSO₃^-

So, at second equivalence point volume of KOH = 19.9 x 2 = 39.8 ml