Given a diprotic acid, H2A, with two ionization constants of Ka1 = 3.9× 10–2 and Ka2 = 5.9× 10–8.
Calculate the pH and molar concentrations of each protonated form for a:
(a)0.135 M solution of H2A
pH=
[H2A]=
[HA]=
[A2-]=
(b)0.135 M solution of NaHA
pH=
[H2A]=
[HA]=
[A2-]=
(c)0.135 M solution of Na2A
pH=
[H2A]=
[HA]=
[A2-]=
I showed you how to get started at your earlier post. I should point out, too, that [A^2-] is not a protonated form for H2A.
To calculate the pH and molar concentrations of each protonated form for the given solutions, we need to consider the ionization constants (Ka1 and Ka2) of the diprotic acid, H2A. We'll go through each solution one by one.
(a) 0.135 M solution of H2A:
First, we need to determine the initial concentration of each protonated form and the hydrogen ion concentration (H+). Let's set x as the concentration of [H2A] that ionizes into [HA] and [H+].
[H2A] = 0.135 M - x
[HA] = x
[A2-] = x
Now, we can write the ionization equations and balance the charges using the ionization constants:
H2A ⇌ H+ + HA
Ka1 = [H+][HA]/[H2A]
HA ⇌ H+ + A2-
Ka2 = [H+][A2-]/[HA]
Substituting the concentrations and rearranging the equations, we get:
Ka1 = (x * x) / (0.135 - x)
Ka2 = ((x * x) / (0.135 - x)) * (x / (0.135 - x))
Since Ka2 is significantly smaller than Ka1, we can assume that x is negligible compared to 0.135.
Now, we can solve for x using the approximation method. Rearranging the equation for Ka1:
x² = (Ka1)(0.135 - x)
x² = (3.9× 10^-2)(0.135) - (3.9× 10^-2)(x)
x² + (3.9× 10^-2)(x) - (3.9× 10^-2)(0.135) = 0
Now, we can solve this quadratic equation to find x, which represents [H+] and hence, the pH.
Using the quadratic formula, x ≈ 0.066 M
pH = -log[H+] = -log(0.066) ≈ 1.18
[H2A] = 0.135 M - x ≈ 0.069 M
[HA] = x ≈ 0.066 M
[A2-] = x ≈ 0.066 M
Therefore, for the given solution (a):
pH ≈ 1.18
[H2A] ≈ 0.069 M
[HA] ≈ 0.066 M
[A2-] ≈ 0.066 M
(b) 0.135 M solution of NaHA:
In this case, NaHA fully dissociates into Na+ and HA-. Since Na+ has no impact on the pH, the concentration of [H+] will solely depend on the dissociation of HA-. Therefore,
[H2A] = 0 M
[HA] = 0.135 M
[A2-] = 0 M
To find the pH, we need to calculate the concentration of [H+] using the equation:
HA ⇌ H+ + A2-
Ka2 = [H+][A2-]/[HA]
Substituting the values,
5.9×10^-8 = [H+][0]/[0.135]
[H+] = 0
Since [H+] is 0, the pH is undefined.
Therefore, for the given solution (b):
pH = undefined
[H2A] = 0 M
[HA] = 0.135 M
[A2-] = 0 M
(c) 0.135 M solution of Na2A:
Again, Na2A fully dissociates into Na+ and A2-. In this case, since there is no HA present, and A2- is the deprotonated form, the concentration of [H+] will be 0.
[H2A] = 0 M
[HA] = 0 M
[A2-] = 0.135 M
Therefore, for the given solution (c):
pH = 0 (since [H+] = 0)
[H2A] = 0 M
[HA] = 0 M
[A2-] = 0.135 M