Consider a 0.10 M solution of a weak polyprotic acid (H2A) with the possible values of Ka1 and Ka2 given below, calculate the contributions to [H3O+] from each ionization step.
Part 1: Ka1=1.0*10^-4 Ka2=1.0*10^-5
Part 2: Ka1=1.0*10^-4 Ka2=1.0*10^-6
I don't know how accurate these are to be but as a first approximation they are as follows:
Note that
........H2A ==> H^+ + HA^- and
........HA^- ==> H^+ + A^2-
If you calculate H^+ from k1 (both acids) you obtain about 3.16E-4. That comes from k1 = (x)(x)/(0.1-x).
Since (H^+) = (HA^-), then those cancel each other when you substitute into k2 so (A^2- ) = k2 for each and that is the contribution of H^+ from the second k. You can see that it is almost negligible for #1 but has a greater impact on #2.
To calculate the contributions to [H3O+] from each ionization step, you will need to use the given values of Ka1 and Ka2 and the concentration of the weak polyprotic acid (H2A) solution.
Part 1: Ka1 = 1.0 * 10^-4, Ka2 = 1.0 * 10^-5
1. Calculate the initial concentration of H2A:
[H2A]initial = 0.10 M
2. Calculate the concentration of H3O+ from the first ionization step:
[H3O+]1 = √(Ka1 * [H2A]initial)
3. Calculate the concentration of A- from the first ionization step:
[A-]1 = [H3O+]1
4. Calculate the concentration of HA- from the first ionization step:
[HA-]1 = [H2A]initial - [H3O+]1
5. Calculate the concentration of H3O+ from the second ionization step:
[H3O+]2 = √(Ka2 * [HA-]1)
6. Calculate the total concentration of [H3O+] as the sum of [H3O+] from each ionization step:
[H3O+]total = [H3O+]1 + [H3O+]2
Part 2: Ka1 = 1.0 * 10^-4, Ka2 = 1.0 * 10^-6
Follow the same steps as in Part 1 using the given values of Ka1 and Ka2.
Note: The concentration of H3O+ from each ionization step will depend on the values of Ka1 and Ka2, as well as the initial concentration of the weak polyprotic acid (H2A) solution.
To calculate the contributions to [H3O+] from each ionization step, we need to use the equations for dissociation and the equilibrium expressions for weak acids.
The dissociation reactions for the weak polyprotic acid (H2A) can be written as follows:
Part 1:
H2A ⇌ H+ + HA- (First ionization step)
HA- ⇌ H+ + A2- (Second ionization step)
Part 2:
H2A ⇌ H+ + HA- (First ionization step)
HA- ⇌ H+ + A2- (Second ionization step)
We are given the values of Ka1 and Ka2 for each part of the question. The equilibrium expressions for the ionization steps can be written as follows:
Ka1 = [H+][HA-]/[H2A]
Ka2 = [H+][A2-]/[HA-]
Now let's calculate the contributions to [H3O+] from each ionization step for each part of the question.
Part 1:
For the first ionization step, the initial concentration of H2A is 0.10 M.
So, [H2A] = 0.10 M
We can assume that the concentration of H+ formed in the first ionization step is negligible compared to 0.10 M. Therefore, we can use the approximation that [H+], [HA-], and [A2-] are approximately equal to the initial concentration of H2A, which is 0.10 M.
[H2A] ≈ [H+]
[HA-] ≈ [H2A] = 0.10 M
[A2-] ≈ 0
Using the equilibrium expression for Ka1:
Ka1 = [H+][HA-]/[H2A]
Replacing the approximate values:
1.0*10^-4 = [H2A] × [H2A] / [H2A]
Simplifying the equation:
[H+] ≈ √(Ka1 × [H2A])
Substituting the values:
[H+] ≈ √(1.0*10^-4 × 0.10) ≈ 3.16 × 10^-3 M
Therefore, the contribution to [H3O+] from the first ionization step is approximately 3.16 × 10^-3 M.
For the second ionization step, the initial concentration of HA- is `[H2A] - [H+]`.
[H+] ≈ 3.16 × 10^-3 M
[HA-] ≈ [H2A] - [H+] = 0.10 M - 3.16 × 10^-3 M
Using the equilibrium expression for Ka2:
Ka2 = [H+][A2-]/[HA-]
Replacing the approximate values:
1.0*10^-5 = [H+] × 0 / [HA-]
Simplifying the equation:
[H+] ≈ √(Ka2 × [HA-])
Substituting the values:
[H+] ≈ √(1.0*10^-5 × (0.10 M - 3.16 × 10^-3 M)) ≈ 9.0 × 10^-4 M
Therefore, the contribution to [H3O+] from the second ionization step is approximately 9.0 × 10^-4 M.
Part 2:
The calculations for Part 2 are similar to Part 1.
For the first ionization step, the contribution to [H3O+] is approximately 3.16 × 10^-3 M.
For the second ionization step, [HA-] ≈ [H2A] - [H+] = 0.10 M - 3.16 × 10^-3 M
Using the equilibrium expression for Ka2, we can calculate that the contribution from the second ionization step is approximately 2.68 × 10^-4 M.
Therefore, the contributions to [H3O+] from each ionization step in Part 2 are approximately 3.16 × 10^-3 M for the first ionization step and 2.68 × 10^-4 M for the second ionization step.
Please note that these calculations are based on approximations, assuming negligible changes in concentration during each ionization step. The actual values may vary slightly.