Question

For the diprotic weak acid H2A, Ka1 = 3.3 × 10-6 and Ka2 = 8.2 × 10-9.

What is the pH of a 0.0400 M solution of H2A?

What are the equilibrium concentrations of H2A and A2– in this solution?

pH =

[H2A] =

[A2-]=

Answer 1

First, assume the acid:

to be H2A, for simplicity, so it will ionize as follows:

H2A <-> H+ + HA-

HA- <-> H+ + A-2

where, H+ is the proton and HA- and A-2 are the conjugate bases, H2A is molecular acid

Ka1 = [H+][HA-]/[H2A]; by definition

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

initially

[H+] = 0

[HA-] = 0

[H2A] = M;

the change

initially

[H+] = + x

[HA-] = + x

[HA2] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

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

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.04 M; then

x^2 + (3.3*10^-6)x - 0.04 *(3.3*10^-6) = 0

solve for x

x =3.6167*10^-4

substitute

[H+] = 0 + 3.6167*10^-4= 3.6167*10^-4 M

[HA-] = 0 + 3.6167*10^-4= 3.6167*10^-4 M

[H2A] = M - x = 0.04-3.6167*10^-4=

[H2A] = 0.039638 M

Now...

Second ionization

HA- <-> H+ + A-2

initially

[H+] = x

[A-2] = 0

[HA-] = M-x

the change

[H+] = +y

[A-2] = + y

[HA-] = - y

in equilbrirum

[H+] = x+y

[A-2] = 0 + y

[HA-] = M - x - y

substitute in Ka

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

Ka2 = (x+y)(y) / (M - x - y)

assume M-x >> M-x-y so (due to Ka2)

Ka2 = (x+y)(y) / (M - x)

y^2+ xy = Ka2(M-x)

y^2+ xy - Ka2(M-x) = 0

y^2+ ( 3.6167*10^-4) *y - (8.2*10^-9)*(0.039638 ) = 0

y = 8.96*10^-7

and we konw

[A-2] = 8.96*10^-7

ph = -log(x+y) = -log(3.6167*10^-4+8.96*10^-7) = 3.44061