Question

Calculate the pH at each point listed for the titration of 100.00mL of 0.100M Diprotic acid (H2A) with 1.00M Sodium hydroxide:

0.0, 9, 10, 15, 20, and 22mL. The pKa's for the diprotic acid are: pKa1=4.94 and pKa2=9.82.

Please show any work, a photo of a solution would be preferred but not necessary...Thanks for any help!

Answer 1

H2A millimoles = 100 x 0.1 = 10

. pKa1=4.94 and pKa2=9.82.

a) 0.0 mL NaOH added :

pH = 1/2(pKa1 -log C)

      = 1/2 (4.94 - log 0.1)

     = 2.97

b) 9.0 mL NaOH added :

millimoles of NaOH added = 9 x 1 = 9

H2A + NaOH ----------------> NaHA   +   H2O

10             9                                 0               0

1                0                               9                9

here acid and salt remained . so it forms acidic buffer

pH = pKa1 + log[salt/acid]

     = 4.94 + log[9/1]

    = 5.89

c) 10 ml NaOH added :

millimoles of NaOH = 10 x 1 = 10

it is first half equivalence point. so here

pH = pKa1 + pKa2 / 2

      = 4.94 + 9.82 / 2

     = 7.38

d) 15 mL NaOH added :
millimoles of NaOH = 15 x 1 = 15

it is second half equivalence point . here

pH = pKa2

     = 9.82