Question

A 1.0 M solution of a compound with 2 ionizable groups (pKa's = 6.2 and 9.5; 100 mL total) has a pH of 6.8. What are the concentrations of the relevant acid and conjugate base?

Answer 1

pKa1 = 6.2

pKa1 = -logKa1

Ka1 = 10^-pKa1

Ka1 = 6.31 x 10^-7

pH = 6.8

[H+] = 10^-6.8

[H+] = 1.58 x 10^-7 M

H2A -------------------------> H+ + HA-

1                                     0           0 --------------------> initial

1-x                                 x             x ------------------------> equilibrium

Ka1 = [H+][HA-]/[H2A]

6.31 x 10^-7 = x^2 / 1-x

x^2 + 6.31 x 10^-7 x - 6.31 x 10^-7= 0

x = 7.94 x 10^-4

x = [H+] = 7.94 x 10^-4 M

   [HA-] = 7.94 x 10^-4 M

[H2A] = 1 - 7.94 x 10^-4

           = 0.999 M

HA-         ---------------------> H+    +            A-2

7.94 x 10^-4               7.94 x 10^-4             0         ----------------------> initial

7.94 x 10^-4 -y           7.94 x 10^-4+y        y --------------------------> equilibrium

Ka2 = (7.94 x 10^-4+y )(y)/(7.94 x 10^-4-y )

3.16 x 10^-10 = (7.94 x 10^-4+y )(y)/(7.94 x 10^-4-y )

3.16 x 10^-10 = (7.94 x 10^-4 y + y^2 / 7.94 x 10^-4-y

2.5 x 10^-13 - 3.16 x 10^-10 y = 7.94 x 10^-4 y + y^2

y^2 + 7.94 x 10^-4 y - 2.5 x 10^-13 =0

y = 3.14 x 10^-10

[A-2] = y = 3.14 x 10^-10 M