Question

39. Phosphoric acid (H3PO4) is a triprotic weak acid. Calculate the pH, [H3PO4],
[H2PO4
-], [HPO4
2-], and [PO4
3-] of a 4.5 M H3PO4 aqueous solution.
H3PO4 (aq) + H2O (l) ⇌ H3PO4
-
(aq) + H3O+ (aq) Ka1 = 7.5 × 10-3
H3PO4
-
(aq) + H2O (l) ⇌ HPO4
2-
(aq) + H3O+ (aq) Ka2 = 6.2 × 10-8
HPO4
2-
(aq) + H2O (l) ⇌ PO4
3-
(aq) + H3O+ (aq) Ka3 = 4.2 × 10-13

Answer 1

From KA values, we can see that we care only of the first and secon dionizations, since third ionization is too low to change H+ from first+second ionizations

so:

For pH, we need total H+

so... we need to account total H+ in equilibriums

so

intially:

H3A <--> H+ +H2A- Ka1

H2A- <--> H+ + HA-2 Ka2

HA- <--> H+ + A-3 Ka3

now...

initially

[H3A] = 4.5 M

[H2A-] = 0

[H+] = 0

in equilibrium

[H3A] = 4.5 -x

[H2A-] = 0 + x

[H+] = 0 + x

so

substitute in KA1:

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

7.5*10^-3= x*x/(4.5-x)

solve for x

x = [H+] = 0.18

now..

[H3A] = M-x = 4.5 - 0.18= 4.32 M

so...

second ionization will happen:

H2A- <--> H+ + HA-2 Ka2

initially

[H2A-] = x = 0.18

[H+] = 0.18

[HA-2] = 0

let y be the second ionization so:

in equilibrium:

[H2A-] = x - y = 0.18 - y

[H+] = 0.18 + y

[HA-2] = 0 + y

substitute in Ka2:

Ka2 = [H+][A-2]/[HA-]

6.2*10^-8 = (0.18+ y) (y) / (0.18- y)

since Ka2 is too low... we can assume safely that 0.18- y --> 0.18- 0 = 0.18

so it turns:

6.2*10^-8 = (0.18+ y) (y) / (0.18)

(6.2*10^-8)(0.18) = 0.18y + y^2

y^2 + 0.18y - 1.11*10^-8 = 0

y = 6.11*10^-8

so

[HA-] = x - y = 0.18 -  6.11*10^-8 = 0.004429

[H+] = 0.18 + y = 0.18 +6.11*10^-8 = 0.180

Clearly, there was no need to even calculate Ka2,

so

pH = -log(H) = -log(0.18) = 0.7447