Calculate the pH and concentrations of H2A, HA− , and A 2− , at equilibrium for a 0.236 M solution of Na2A. The acid dissociation constants for H2A are Ka1=7.68×10−5 and Ka2=6.19×10−9

[H₂A] = 5.0409x10⁻⁷M

[HA⁻] = 0.001951M

[A²⁻] = 0.234

11.29 = pH

Explanation:

When Na₂A = A²⁻is in equilibrium with water, the reactions that occurs are:

A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + OH⁻(aq)

By definition of Ka of reaction, you can find Kb (the inverse basic reaction), thus:

Kb1 = KwₓKa2 =

Where Kw is K of equilibrium of water, 1x10⁻¹⁴

1x10⁻¹⁴/ 6.19x10⁻⁹ =

1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]  

And HA⁻ will be in equilibrium as follows:

HA⁻(aq) + H₂O(l) ⇄ H₂A(aq) + OH⁻(aq)

Kb2 = KwₓKa1 = 1x10⁻¹⁴/ 7.68x10⁻⁵ =

1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

Thus, the Na₂A has 2 equilibriums, for the first reaction, concentrations wil be:

[HA⁻] = X

[OH⁻] = X

[A²⁻] = 0.236M - X

X as how much will react, Reaction coordinate

Replacing in Kb1:

1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]

1.6155x10⁻⁶ = [X] [X] / [0.236-X]

3.8126x10⁻⁶ - 1.6155x10⁻⁶X = X²

3.8126x10⁻⁶ - 1.6155x10⁻⁶X - X² = 0

Solving for X :

X = -0.00195 → False solution. There is no negative concentrations

X = 0.001952.

Replacing, concentrations for the first equilibrium are:

Now, in the second equilibrium:

[HA⁻] = 0.001952 - X

[OH⁻] = X

[H₂A] = X

Replacing in Kb1:

1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

1.3021x10⁻¹⁰ = [X] [X] / [0.001952 - X]

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X = X²

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X - X² = 0

Solving for X :

X = -5.04x10⁻⁷ → False solution. There is no negative concentrations

X = 5.0409x10⁻⁷

Replacing, concentrations for the second equilibrium are:

Thus, you have concentrations of H2A, HA−, and A2−

Now, for pH, you need [OH⁻] concentration that is:

[OH⁻] = 0.0019525

pOH = -log[OH⁻] = 2.709

As 14 = pH+ pOH

14-2.709=pH

-be a crystalline solid

- be soluble in water

-conduct electricity

hope this : )