For the diprotic weak acid H2A, Ka1 = 2.8 × 10-6 and Ka2 = 5.8 × 10-9.
What is the pH of a 0.0700 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
To find the pH of a 0.0700 M solution of H2A, we need to determine the concentrations of H2A and A2– at equilibrium.
First, let's write the chemical equation for the dissociation of H2A:
H2A ⇌ H+ + HA–
Since H2A is a diprotic acid, it will have two equilibrium reactions:
1) H2A ⇌ H+ + HA– (with Ka1 = 2.8 × 10^(-6))
2) HA– ⇌ H+ + A2– (with Ka2 = 5.8 × 10^(-9))
Let's assume that at equilibrium, x is the concentration of H+ and HA– formed from the first dissociation, and y is the concentration of H+ and A2– formed from the second dissociation.
From the equilibrium constant expression for the first dissociation:
Ka1 = [H+][HA–] / [H2A]
Substituting the given values:
2.8 × 10^(-6) = x * x / (0.0700 - x)
Since the initial concentration of H2A (0.0700 M) is much greater than x (assumed to be small), we can ignore the change in [H2A] and approximate it as 0.0700 M.
Therefore,
2.8 × 10^(-6) = x² / 0.0700
Now we solve this equation for x:
x² = 0.0700 * 2.8 × 10^(-6)
x ≈ 5.884 × 10^(-4)
Now, using the value of x, we can find the concentration of HA–:
[HA–] = x ≈ 5.884 × 10^(-4) M
To find the concentration of A2–, we consider the second dissociation reaction:
Ka2 = [H+][A2–] / [HA–]
Substituting the given values:
5.8 × 10^(-9) = y * y / [HA–]
5.8 × 10^(-9) = y² / 5.884 × 10^(-4)
Now we solve this equation for y:
y² = 5.884 × 10^(-4) * 5.8 × 10^(-9)
y ≈ 1.346 × 10^(-6)
Now that we have the concentrations of H2A (0.0700 M), HA– (5.884 × 10^(-4) M), and A2– (1.346 × 10^(-6) M), we can calculate the pH of the solution.
The pH is given by the equation:
pH = -log[H+]
Substituting the value of x, which is [H+], we get:
pH ≈ -log(5.884 × 10^(-4))
pH ≈ 3.23
Therefore, the pH of the 0.0700 M solution of H2A is approximately 3.23. The equilibrium concentrations of H2A and A2– are approximately 0.0700 M and 1.346 × 10^(-6) M, respectively.