For the diprotic weak acid H2A, Ka1 = 2.2 × 10-5 M and Ka2 = 7.8 × 10-7 M. What is the pH of a 0.0650 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
To find the pH of a 0.0650 M solution of H2A, we need to consider the dissociation of the acid and calculate the equilibrium concentrations of H2A and A2–.
First, let's write down the dissociation reactions for H2A:
H2A ↔ H+ + HA-
HA- ↔ H+ + A2-
Next, we need to set up an ICE (Initial, Change, Equilibrium) table for each dissociation reaction to keep track of the concentrations:
For the first dissociation reaction:
[H2A] [H+] [HA-]
Initial: 0.0650 0 0
Change: -x +x +x
Equilibrium: 0.0650 - x x x
For the second dissociation reaction:
[HA-] [H+] [A2-]
Initial: x 0 0
Change: -x +x +x
Equilibrium: x - x x x
Since the equilibrium concentration of H2A is (0.0650 - x) and x represents the concentration of H+, we can assume that (0.0650 - x) ≈ 0.0650.
Now, we need to calculate the concentration of H+ using the first dissociation reaction, in which Ka1 = [H+][HA-] / [H2A].
Plugging in the equilibrium concentrations into the equation:
(2.2 × 10^-5) = (x)(x) / 0.0650
Rearranging this equation, we get:
(x)^2 = (2.2 × 10^-5)(0.0650)
Solving for x, we find:
x ≈ 0.00253
Since the concentration of H+ represents the acidity of the solution, we can use the equation pH = -log[H+] to find the pH:
pH = -log(0.00253) ≈ 2.60
Therefore, the pH of the 0.0650 M solution of H2A is approximately 2.60.
To find the equilibrium concentrations of H2A and A2-, we can substitute the value of x (0.00253) into the equilibrium expressions for each species:
[H2A] ≈ 0.0650 - x ≈ 0.0650 - 0.00253 ≈ 0.0625 M
[A2-] ≈ x ≈ 0.00253 M
So, the equilibrium concentrations of H2A and A2- in this solution are approximately 0.0625 M and 0.00253 M, respectively.