Question

For the diprotic weak acid H2A, Ka1 = 3.8 × 10-6 and Ka2 = 8.2 × 10-9. What is the pH of a 0.0750 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

1) pH

2) [H2A]

3) A^2-

Please explain all steps and equations, please...

Answer 1

For a diprotic weak acid H2A the equations are

H₂A HA^- + H⁺

Ka1 = 3.8 x 10^-8

To find the concentrations of the individual species we can use as ICE table

	H2A	HA-	H+
Initial	0.0750	0	0
Change	-X	+X	+X
Equilibrium	0.0750-X	X	X

Ka1 = 3.8 x 10^-8 = [HA^-][H⁺]/[H₂A]

3.8 x 10^-8 = x²/0.0750-x since it is a weak acid amount of ionization will be small and so x in the denominator can be ignored

x² = 3.8 x 10^-8 x 0.0750

x = 5.338 x 10^-5

so the H+ ion as well as the HA^- ion concentration is 5.338 x 10^-5M

Now

HA^- A^2- + H⁺

Ka2 = 8.2 × 10^-9

We will set up another ICE table

	HA-	A2-	H+
Initial	5.338 x 10-5M	0	0
Change	-X	+X	+X
Equilibrium	5.338 x 10-5-x	x	x

Ka2 = 8.2 × 10^-9 =  [A^2-][H⁺]/[HA^-]

8.2 × 10^-9 = x²/5.338 x 10^-5-x again we will ignore x in the denominator

x² = 8.2 × 10^-9 x 5.338 x 10^-5

x = 6.61 x 10^-7 This is the concentration of A^2- and H⁺ from the second equilibrium

so the total concentration of H⁺ is 5.338 x 10^-5M + 6.61 x 10^-7 = 5.404 x 10^-5M

pH = -log[H⁺] = -log (5.404 x 10^-5)

pH is 4.267

concentration of H₂A is 0.0750-x (where x = 5.338 x 10^-5M)

concentration of H₂A is 0.0749 M

concentration of A^2- is 6.61 x 10^-7M