Question

2. Consider the triprotic acid, H₃A, with the following acid dissociation contants: K_a1 = 1.5 x 10^-5 , K_a2= 5.0 x 10^-9 , and K_a3 = 5.0 x 10^-12

a.) What is the pH of an aqueous solution of 0.200 M H₃A?

b.)   What is the pH of an aqueous solution of 0.200 M NaH₂A?

c.)   What is the pH of an aqueous solution of 0.200 M Na₂HA?

d.)   What is the pH of an aqueous solution of 0.200 M Na₃A?

Answer 1

a) pH for 0.2 M H3A

H3A <==> H2A- + H+

let x amount has dissociated

Ka1 = [H2A-][H+]/[H3A]

1.5 x 10^-5 = x^2/0.2

x = [H+] = 1.73 x 10^-3 M

pH = -log[H+] = 2.76

b) pH for 0.2 M NaH2A

H2A- <==> HA^2- + H+

let x amount has dissociated

Ka2 = [HA^2-][H+]/[H2A-]

5 x 10^-9 = x^2/0.2

x = [H+] = 3.16 x 10^-5 M

pH = 4.5

c) pH of 0.2 M Na2HA

HA^2- + H2O <==> H2A- + OH-

let x amount has reacted

Kb2 = Kw/Ka2 = x^2/0.2

1 x 10^-14/5 x 10^-9 = x^2/0.2

x = [OH-] = 6.32 x 10^-4 M

pOH = -log[OH-] = 3.20

pH = 14 - pOH = 10.8

d) pH of Na3A

A^3- + H2O <==> HA^2- + OH-

let x amount has reacted

Kb3 = Kw/Ka3 = x^2/0.2

1 x 10^-14/5 x 10^-12 = x^2/0.2

x = [OH-] = 0.02 M

pOH = 1.70

pH = 12.3