The Ka values of H3PO4 are shown below.

Ka1 7.5 multiplied by 10-3
Ka2 6.2 multiplied by 10-8
Ka3 4.8 multiplied by 10-13

What is the pH of a 0.14 M solution of Na2HPO4?

thanks!

To find the pH of a solution of Na2HPO4, we need to consider the dissociation of the HPO4^2- ion in water. The dissociation reaction can be written as follows:

Na2HPO4 ⇌ 2Na+ + HPO4^2-

Since HPO4^2- is a weak base, it will partially dissociate in water and produce OH- ions. However, to determine the pH, we need to find the concentration of H3O+ ions, which is related to the concentration of OH- ions.

First, let's write the expression for the dissociation constant (Kb) for HPO4^2-:

Kb = [OH-][HPO4^2-] / [Na2HPO4]

Since we only have the Ka values, we need to convert them to Kb values using the relationship: Ka x Kb = Kw (water's autoionization constant)

Kb1 = Kw / Ka1 = (1.0 x 10^-14) / (7.5 x 10^-3) = 1.33 x 10^-12
Kb2 = Kw / Ka2 = (1.0 x 10^-14) / (6.2 x 10^-8) = 1.61 x 10^-7
Kb3 = Kw / Ka3 = (1.0 x 10^-14) / (4.8 x 10^-13) = 2.08 x 10^-2

Now, let's calculate the concentration of OH- ions using the Kb value for HPO4^2-. We'll assume that x is the concentration of OH- ions formed when HPO4^2- dissociates:

Kb1 = [OH-][HPO4^2-] / [Na2HPO4]
1.33 x 10^-12 = x^2 / 0.14

Simplifying the equation:
x^2 = 1.33 x 10^-12 * 0.14
x^2 = 1.862 x 10^-13
x ≈ 4.32 x 10^-7 M

Since HPO4^2- dissociates to produce the same amount of OH- ions, the concentration of OH- in the solution is 4.32 x 10^-7 M.

To determine the concentration of H3O+ ions, we can use the following equation based on the water autoionization constant (Kw):

Kw = [H3O+][OH-]
1.0 x 10^-14 = [H3O+](4.32 x 10^-7)
[H3O+] ≈ 2.32 x 10^-8 M

Finally, to find the pH, we can use the equation: pH = -log10([H3O+]):

pH = -log10(2.32 x 10^-8)
pH ≈ 7.64

Therefore, the pH of a 0.14 M solution of Na2HPO4 is approximately 7.64.

(H^+) = sqrt(k2k3)

pH = -log (H^+)