Question

For the titration of 50.00 mL of
a 0.1000 M solution of compound H₂SO₄ with a 0.2000
M solution of compound NaOH in the following table.
For each titration, calculate the pH after the addition
of 0.00, 20.00, 25.00, 37.50,

50.00, and 60.00 mL of com-
pound NaOH.

Please explain all the steps.

Answer 1

millimoles of H2SO4 = 2 x 50 x 0.1 = 10

1) 0.00 ml NaOH added

H2SO4 molarity = 0.1 x 2 = 0.2

pH = -log [H+] = -log 0.2 = 0.699

pH = 0.699

2) 20 ml NaOH added

millimoles of NaOH = 20 x 0.2 = 4

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 4) / (50 +20)

      = 0.0857 M

pH = 1.067

3) 25 ml NaOH added

millimoles of NaOH = 25 x 0.2 = 5

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 5) / (50 +25)

      = 0.067 M

pH = 1.176

4) 37.5 ml NaOH added

millimoles of NaOH = 37.5 x 0.2 = 7.5

millimoles of H2SO4 > millimoles of NaOH

so acid dominent

[H+] = (millimoles of H2SO4 - millimoles of NaOH) / total volume

       = (10 - 7.5) / (50 +37.5)

      = 0.02857 M

pH = 1.544

5) 50 ml NaOH added

millimoles of NaOH = 50 x 0.2 = 10

millimoles of acid = millimoles of base

so it is equivalece point . strong acid + strong base here

pH= 7

6) 60 ml NaOH added

millimoles of NaOH = 60 x 0.2 = 12

millimoles of H2SO4 < millimoles of NaOH

so acid dominent

[OH] = (millimoles of NaOH - millimoles of H2SO4) / total volume

       = (12 - 10) / (50 +60)

      = 0.0182 M

pOH = -log [OH-]

pOH = 1.74

pH + pOH = 14

pH = 12.26