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

plants need sunlight, water, and carbon dioxide to make food. plants use the raw materials of water, nutrients they absorb from soil, and carbon dioxide to make sugars the plant can use. energy from the sun and chlorophyll are also needed.

jjj89t'ji; it hink the answer is your mom is not si