Question

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

Answer 1

To calculate the pH and concentrations of H2A, HA−, and A2− at equilibrium for a 0.236 M solution of Na2A, we need to consider the acid dissociation constants (Ka) for H2A. Given that the values of Ka1 = 7.68 × 10−5 and Ka2 = 6.19 × 10−9, we can use the Henderson-Hasselbalch equation to calculate the pH.

Step-by-step explanation:

To calculate the pH and concentrations of H2A, HA−, and A2− at equilibrium for a 0.236 M solution of Na2A, we need to consider the acid dissociation constants (Ka) for H2A. Given that the values of Ka1 = 7.68 × 10−5 and Ka2 = 6.19 × 10−9, we can use the Henderson-Hasselbalch equation to calculate the pH.

First, we need to determine the concentrations of H2A, HA−, and A2− at equilibrium. We know that Na2A is a salt of the weak acid H2A, so it will dissociate into Na+ and A2− ions. Since the concentration of Na2A is 0.236 M, the concentration of A2− will also be 0.236 M. The concentration of H2A can be calculated using the equation: [H2A] = [A2−] * Ka2 / [HA−]. Using the given values, we can substitute them into the equation to find the concentration of H2A.

Finally, we can use the Henderson-Hasselbalch equation, which is pH = pKa + log([A2−] / [H2A]), to calculate the pH. Substitute the calculated concentrations of A2− and H2A into the equation along with the values of the pKa's, and solve for the pH.

Answer 2

[H₂A] = 5.0409x10⁻⁷M

[HA⁻] = 0.001951M

[A²⁻] = 0.234

11.29 = pH

Step-by-step 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:

[HA⁻] = 0.001952

[OH⁻] = 0.001952

[A²⁻] = 0.234

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:

[HA⁻] = 0.001951M

[OH⁻] = 5.0409x10⁻⁷M

[H₂A] = 5.0409x10⁻⁷M

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

11.29 = pH