For the diprotic weak acid h2a, ka1 = 2.2 × 10-6 m and ka2 = 8.2 × 10-9 m. what is the ph of a 0.0500 m soluti? for the diprotic weak acid h2a, ka1 = 2.2 × 10-6 m and ka2 = 8.2 × 10-9 m. what is the ph of a 0.0500 m solution of h2a? what are the equilibrium concentrations of h2a and a2– in this solution?

At equilibrium, the pH of the solution is 3.48; [H2A] = 0.0496 and [A2-] = 8.2 × 10^-9 M

H2A <> H+ + HA- ;Ka1 = 2.2 × 10^-6

Ka1 = 2.2x10^-6 = [H+][HA-]/[H2A]

At equilibrium:

Substituting in the equation of Ka1

2.2 × 10^-6 = X^2/0.050 - X

X^2 = 1.1 × 10^-7

X = 3.31 × 10^-4

Since Ka1 <<< Ka2 and [H+] = X

[H+] = 3.31 × 10^-4

pH = - log 3.31 × 10^-4

pH = 3.48

[H2A] = 0.05 - 3.31 × 10^-4

[H2A] = 0.0496

HA- <> H+ + A- Ka2 = 8.2 × 10^-9

Ka2 = 8.2 × 10^-9 = [H+][A2-]/[HA-]

HA- > H+ + A2-

3.31 × 10^-4 3.31 × 10^-4 0 Initial

-y +y +y Change

{3.31 ×10^-4 -y} {3.31 x10^-4+y} {+y} Equilibrium

Assuming y is very small:

Substituting in the equation of Ka2

8.2 × 10^-9 = (3.31x10^-4) × y /3.31 ×10^-4

y = 8.2 × 10^-9

Since [A2-] = y

[A2-] = 8.2 × 10^-9 M

Therefore, at equilibrium, the pH of the solution is 3.48; [H2A] = 0.0496 and [A2-] = 8.2 × 10^-9 M

Learn more about pH and equilibrium concentration of acid-base conjugates of weak acid at: link

20 drops hope this

dynamic equilibrium is the answer.