Question

125.0-mL aliquot of 0.124 M diprotic acid H2A (pK1 = 4.03, pK2 = 8.06) was titrated with 1.24 M NaOH. Find the pH at the following volumes of base added: Vb = 0.00, 1.80, 6.25, 11.50, 12.50, 13.50, 18.75, 24.00, 25.00, and 28.00 mL. (Assume Kw = 1.01 ✕ 10−14.)

Answer 1

1) 0.00 mL base added

C = 0.124 M

pH = 1/2 [pK1 -log C]

pH = 1/2 [4.02 - log 0.124]

pH = 2.46

2) 1.80 mL added

millimoles of acid = 125 x 0.124 = 15.5

millimoles of base = 1.80 x 1.24 = 2.32

H2A    +   B -------------------> HA-

15.5      2.32                       0

13.27      0                          2.32

pH = pK1 + log [HA- / H2A]

pH = 4.03 + log (2.32 / 13.27)

pH = 3.27

3) 6.25 mL added

millimoles of acid = 125 x 0.124 = 15.5

millimoles of base = 6.25 x 1.24 = 7.75

H2A    +   B -------------------> HA-

15.5      7.75                       0

7.75     0                          7.75

pH = pK1 + log [HA- / H2A]

pH = 4.03 + log (7.75 / 7.75)

pH = 4.03

4 ) 11.70 mL

millimoles of total base = 11.50 x 1.24 = 14.26

actual base after half equilvanece = 14.26 - 7.75 = 6.51

HA- + B ---------------> A-2

7.75    6.51                   0

1.24     0                      6.51

pH = pK2 + log [A-2 / HA-]

pH = 8.06 + log (6.51 / 1.24)

pH = 8.78