Question

Acid/Base titrations 10. 25.0 mL of 0.100 M H2A (a weak diprotic acid) is titrated with 0.200 M NaOH. What is the pH of the solution when 0.00 mL, 10.0 mL, 12.5 mL, 20.0 mL, 25.0 mL, and 40.0 mL E.S RaWeMA 5.83 x 10 8) have been added? (Ka1= 2.46 x 10, Ka2 ANSWER: 0 mL = 2.315, 10.0 mL= 4.211 (or 4.213), 12.5 mL = 5.422, 20.0 mL 7.410, 25.0 mL = 9.966, 40.0 mL = 12.664 11. 25.0 mL of 0.100 M Az (a weak diprotic base) is titrated with 0.100 M HCl. What is the pH of the solution when 0.00 mL, 10.0 mL, 25.0 mL, 35.0 mL, 45.0 mL, 50.0 mL and 80.0 mL have been added? (Kb1 3.45 x 104, Kb2 =6.39 x 10) ANSWER: 0 mL 11.756, 10.0 mL = 10.714 (or 10.701), 25.0 mL= 8.670, 35.0 mL = 6.982, 45.0 mL = 6.203, 50.0 mL = 4.142, 80.0 mL 1.544

Answer 1

10)

concentration of H2A = 0.100 M

H2A -----------> HA- +    H+

0.100                   0           0

0.100-x                x            x

Ka1 = x^2 / 0.100 - x

2.46 x 10^-4 = x^2 / 0.100 - x

x = 4.84 x 10^-3

[H+] = 4.84 x 10^-3 M

pH = 2.315

b)

mmoles of H2A = 25 x 0.100 = 2.5

mmoles of NaOH = 10 x 0.200 = 2.0

H2A   +   NaOH    ---------> HA-   +   H2O

2.5            2.0                      0             0

0.5             0                        2.0

pH = pKa1 + log [HA- / H2A]

    = 3.61 + log [2.0 / 0.5]

pH = 4.211

c)

mmoles of NaOH = 12.5 x 0.2 = 2.5

H2A   +   NaOH    ---------> HA-   +   H2O

2.5            2.5                      0             0

0             0                        2.5

pH = pKa1 + pKa2 / 2

      = 3.609 + 7.234 / 2

pH = 5.422

d)

mmoles of NaOH = 20 x 0.2 = 4.0

H2A   +   NaOH    ---------> HA-   +   H2O

2.5           4.0                        0             0

0             1.5                        2.5

HA-   +   NaOH   -----------> A2- + H2O

2.5             1.5                       0          0

1.0              0                        1.5

pH = pKa2 + log [A2- / HA-]

    = 7.234 + log [1.5 / 1]

pH = 7.410

e)

mmoles of NaOH = 25 x 0.2 = 5

H2A   +   NaOH    ---------> HA-   +   H2O

2.5           5.0                        0             0

0             2.5                        2.5

HA-   +   NaOH   -----------> A2- + H2O

2.5             2.5                       0          0

0              0                        2.5

concentration of A2- = 2.5 / 25 + 25 = 0.05 M

A2- +   H2O   ---------> HA- + OH-

0.05                                 0           0

0.05-x                              x           x

Kb1 = x^2 / 0.05 - x

1.715 x 10^-7 = x^2 / 0.05 - x

x = 9.26 x 10^5

[OH-] = 9.26 x 10^5

pOH = 4.03

pH = 9.966

f)

mmoles of NaOH = 40 x 0.2 = 8

pH = 12.664