Question

the dibasic compound b (pka1=10.55 & pka2=12.75) was titrated with 0.500 M HCl. the initial solution of B was 0.200 M and had a volume of 100.0 mL. find the pH after 40 mL, 50 mL, and 60 mL of acid has been added... i found that the first equivalence point is at 40 mL and the second equivalence is 80 mL... i am not sure this is right...

Answer 1

pKa1 = 10.55 , pKb1 = 1.25

pKa2 = 12.75 , pKb2 = 3.45

a) at 40.0 mL

mmoles of base = 0.200 x 100 = 20

mmoles of HCl = 40 x 0.500 = 20

this is first equivalence point : so

pH = pKa1 + pKa2 / 2

    = 10.55 + 12.75 / 2

pH = 11.65

b)

mmoles of acid = 50 x 0.500 = 25

B   +    HCl    --------------> BH+  

20       25                             0

0          5                               20

BH+   +    H+    - ------------> BH2+

20             5                              0

15             0                               5

pOH = pKb2 + log [BH2+ / BH+]

        = 3.45 + log [5 / 15]

         = 2.97

pH = 11.03

c)

mmoles of acid = 60 x 0.5 = 30

this is second half equivalence point .

pOH = pKb2

        = 3.45

pH = 10.55

second equivalence point at 80.0 mL .

yes you are correct