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?
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To find the pH of a solution of a diprotic weak acid H2A, we need to consider the dissociation of the acid into its individual ions and the equilibrium concentrations of these ions.
Let's denote the initial concentration of H2A as [H2A]0. In this case, it is 0.0650 M.
The dissociation of the acid occurs in two steps:
H2A ⇌ H+ + HA-
HA- ⇌ H+ + A^2-
Based on the given equilibrium constant (Ka1), we can establish the expression for the first step of dissociation:
Ka1 = [H+][HA-] / [H2A]
Similarly, using the second equilibrium constant (Ka2), we can establish the expression for the second step of dissociation:
Ka2 = [H+][A^2-] / [HA-]
We can assume that at equilibrium, the concentrations of H+ and HA- are equal because H2A initially dissociates only partially. Therefore, [H+]eq = [HA-]eq.
Let's denote the concentration of H+ at equilibrium as [H+].
Now, we can derive a quadratic equation using the given values:
Ka1 = [H+][HA-] / [H2A]
Ka1 = [H+]^2 / [H2A] (since [HA-]eq = [H+]eq)
[H+]^2 = Ka1[H2A]
[H+]^2 = (2.2 × 10^-5 M)(0.0650 M)
[H+]^2 = 1.43 × 10^-6 M^2
[H+] = √(1.43 × 10^-6) M
[H+] = 1.20 × 10^-3 M
Since [H+] is the same as [HA-] at equilibrium, the concentration of HA- is also 1.20 × 10^-3 M.
Now, let's calculate the concentration of A^2- using the second equilibrium constant (Ka2):
Ka2 = [H+][A^2-] / [HA-]
Solving for [A^2-]:
[A^2-] = (Ka2[HA-]) / [H+]
[A^2-] = (7.8 × 10^-7 M)(1.20 × 10^-3 M) / (1.20 × 10^-3 M)
[A^2-] = 7.8 × 10^-10 M
Now, to find the pH of the solution, we can use the equation:
pH = -log[H+]
pH = -log(1.20 × 10^-3)
pH ≈ 2.92
So, the pH of the 0.0650 M solution of H2A is approximately 2.92.
The equilibrium concentrations of H2A and A^2- in this solution are approximately 1.20 × 10^-3 M and 7.8 × 10^-10 M, respectively.