Question

The triprotic acid H3A has ionization constants of Ka1 = 1.6× 10–3, Ka2 = 2.4× 10–9, and Ka3 = 6.4× 10–11.

Calculate the following values for a 0.0800 M solution of NaH₂A.

[H⁺] = M

[H₂A^-]/[H₃A] =   

Calculate the following values for a 0.0800M solution of Na₂HA.

[H⁺] = M

[HA^2-]/[H₂A^-] =

This is all the information this problem comes with

Hint:

Consider H2A– to be the intermediate form of a diprotic acid, surrounded by H3A and HA2–. Consider HA2– to be the intermediate form of a diprotic acid, surrounded by HA– and A3–. Use the equations for [H ] associated with the intermediate form in a diprotic system.

Answer 1

Na2HA + H2O --> Na+ + NaHA- + OH-

I. . . 0.08 . . . . . . . . . . . . . . . . . .0. . . . . . .0
E. ..0.08-x. . . . . . . . . . . .. .. . . . . x. . . . . . .x

kb1 = kw/ka1 = 10^-14 / 1.6X 10^-3 = 6.25 X 10^-12

kb1 = [NaHA-][OH-]/[Na2HA]

6.25 X 10^-12 = x^2 / (0.08-x)

x << 1

0.5 X 10^-12 = x^2

x = 0.707 X 10^-6 M

so [OH-] = 0.707 X 10^-6 =

so [H+] = 10^-14 / 0.707 X 10^-6

[H+] = 1.414 X 10^-8 M

2) There will be second dissociation as

. .. . .NaHA- + H2O <----> Na+ + HA-- + OH-
I. . .0.707 X 10^-6 . . . . . . . . . . 0. . . . . . .0
E. . .0.707 X 10^-6 -x. . . . . . . . x. . . . . . .x

kb2 = kw/ka2 = 10^-14/2.4 X 10^-9 = 4.16 X 10^-6

kb2 = [A--][OH-]/[NaHA-]

4.16 X 10^-6 = x^2 / .0.707 X 10^-6 -x

x <<1

so we can ignore it in denominator

2.94 x 10^-12 = x^2

x = 1.71 X 10^-6 = [OH-] = [HA-2]

[H+] = 10^-14 / 1.71 x 10^- 6 = 5.84 X 10^-9

We can ignore third dissociaiton as the K3 is very low

So total

[H+] = 1.414 X 10^-8 M + 5.84 X 10^-9 = 1.998 x 10^8 M