Question

For the diprotic weak acid H2A, Ka1 = 2.8

Answer 1

Since Ka1 >> Ka2, then nearly all of the H3O+ produced comes from the first hydrolysis reaction.

Molarity . . . . .H2A + H2O <==> H3O+ + HA-
Initial . . . . . .0.0550 . . . . . . . . . . . . .0 . . . . . .0
Change . . . . . .-x . . . . . . . . . . . . . . .x . . . . . .x
Equilibrium . .0.0550-x . . . . . . . . . . . x . . . . . .x
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PART-A

Ka1 = [H3O+][HA-] / [H2A]

Ka1= (x)(x) / (0.0550-x)

Ka1= 2.8 x 10^-6

x^2 / (0.0550-x) = 2.8 x 10^-6

. . .because Ka1 is small compared to [H2A]o, we can delete the -x term.

x^2 / 0.0550 = 2.8 x 10^-6

x = 3.924x 10^-4 M = [H3O+]

pH = -log [H3O+] =

pH-log 3.924 x 10^-4

pH= 3.406

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PART-B

[H2A] = 0.0550 - x

x = 3.924x 10^-4 M

[H2A] =0.0550-3.924*10^-4

[H2A] = 0.0546 M

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PART- C

For the second reaction,

Ka2 = [A(^2-)]*[H+]/[HA-]

Ka2= [A(^2-)]*x/x

Ka2=[A(^2-]

[A^2]- = 7.7*10^-9 M

ANSWERS

1) pH= 3.406

2)[H2A] = 0.0546 M

3)[A^2]- = 7.7*10^-9 M