The Ka for a weak acid, HA, is 1.6x10-6. What are the (a) pH and (b) degree of ionization of the acid in a 10-3 M solution? (c) Calculate pKa and pKb
Why don't you scroll down and read some of the chemistry (this one was under biochemistry). I have worked this one before. The url is here.
https://www.jiskha.com/questions/1840757/ka-for-a-weak-acid-ha-is-1-6x10-6-what-are-the-a-ph-and-b-degree-of-ionization-of
(a) pH:
Well, well, well, looks like we've got ourselves a weak acid party! To find the pH of a weak acid solution, we're gonna need to know the degree of ionization. So, let's put that on hold for a second and move to the next question.
(b) Degree of Ionization:
Now we're talking! To calculate the degree of ionization, we divide the concentration of ionized HA by the initial concentration of HA. In this case, the initial concentration of HA is 10-3 M, and since it's a weak acid, only a fraction of it will ionize. So, let's say "x" is our degree of ionization.
The equation would look like this:
HA ⇌ H+ + A-
Initial Concentration: 10-3 M 0 M 0 M
Concentration at Equilibrium: 10-3 - x x x
Now, remember, the Ka is 1.6x10-6, which means it's gonna be pretty unimpressive compared to strong acids. But hey, you can't please everyone, right?
So, now we set up the expression for the Ka:
Ka = [H+][A-] / [HA]
Plug in the values we have:
1.6x10-6 = x^2 / (10-3 - x)
Now, since x is small compared to 10-3, we can assume that x is negligible compared to the initial concentration. That means we can simplify the equation to:
1.6x10-6 = x^2 / (10-3)
Alright, time for some math magic! By solving this equation, we can find x, which represents the degree of ionization of the weak acid. Once we find it, we can finally go back and calculate the pH.
(c) pKa and pKb:
Hang on tight, pun lovers! To find the pKa and pKb, we need to know the value of Ka or Kb and then convert it to logarithmic form.
In this case, we have Ka, which is 1.6x10-6. So, for pKa, we use the formula pKa = -log10(Ka). Plug in the values, crunch some numbers, and voila!
As for pKb, it's calculated for bases, not acids. And since we have an acid, we're out of luck on that one. But hey, there's always the next question!
Good luck with your calculations, and may the weak acid gods be with you!
To solve this problem, we can use the formula for the degree of ionization (α) of a weak acid:
α = √(Ka * C)
where Ka is the acid dissociation constant and C is the initial concentration of the acid.
(a) pH Calculation:
To calculate the pH, we need to find the concentration of H+ ions.
In a weak acid, the concentration of H+ ions is equal to the degree of ionization (α) multiplied by the initial concentration of the acid (C). Let's substitute the values into the formula:
α = √(Ka * C)
α = √(1.6x10^(-6) * 10^(-3))
α ≈ √(1.6x10^(-9)) (1)
Since the concentration of H+ ions is equal to the degree of ionization:
[H+] = α * C
Plugging in the values, we get:
[H+] = √(1.6x10^(-9)) * 10^(-3)
[H+] ≈ 4x10^(-6) M (2)
To find the pH, we use the equation:
pH = -log[H+]
Substituting the value from (2):
pH = -log(4x10^(-6))
pH ≈ 5.4
Therefore, the pH of the 10^(-3) M solution is approximately 5.4.
(b) Degree of ionization Calculation:
From equation (1), we found that α ≈ √(1.6x10^(-9)). Therefore, the degree of ionization of the acid in a 10^(-3) M solution is approximately √(1.6x10^(-9)).
(c) pKa and pKb Calculation:
pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka). Since we have Ka = 1.6x10^(-6), we can calculate pKa as follows:
pKa = -log(Ka)
pKa = -log(1.6x10^(-6))
pKa ≈ 5.8
Similarly, pKb can be calculated using the relationship:
pKb = 14 - pKa
pKb = 14 - 5.8
pKb ≈ 8.2
Therefore, the pKa of the acid is approximately 5.8, and the pKb is approximately 8.2.
To find the answers to these questions, we need to use the equations related to the ionization of weak acids.
(a) pH of a weak acid solution:
The pH of a solution can be calculated using the equation: pH = -log[H+].
In the case of a weak acid, ionization is important. So we need to determine the concentration of H+ ions produced by the ionization of HA.
For a weak acid HA, the ionization reaction can be represented as:
HA ⇌ H+ + A-
Here, HA represents the undissociated weak acid, H+ represents the hydrogen ion, and A- represents the conjugate base.
Since HA ionizes to H+ and A-, the concentration of H+ ions formed can be assumed to be equal to the degree of ionization (α) multiplied by the initial concentration of the acid (HA) in the solution.
Therefore, [H+] = α x [HA].
Given that the concentration of HA is 10^(-3) M, we have [HA] = 10^(-3) M.
Now, let's calculate the degree of ionization (α):
Since Ka is the equilibrium constant for the ionization reaction, Ka = [H+][A-] / [HA].
Using the given Ka value of 1.6x10^(-6) and letting x be the degree of ionization (α), we have:
(α * 10^(-3))^2 / (10^(-3) - α * 10^(-3)) ≈ 1.6x10^(-6)
Simplifying the equation:
α^2 / (1 - α) ≈ 1.6x10^(-6)
α^2 ≈ 1.6x10^(-6) * (1 - α)
α^2 + 1.6x10^(-6)α - 1.6x10^(-6) ≈ 0
Since α is small, we can approximate the equation:
1.6x10^(-6)α ≈ 1.6x10^(-6)
α ≈ 1
Therefore, the degree of ionization (α) is approximately 1. This means that the weak acid nearly completely ionizes in the given concentration range.
Since the concentration of H+ ions formed ([H+]) is equivalent to the degree of ionization (α), we have:
[H+] = α x [HA] = 1 * 10^(-3) = 10^(-3) M
Now, let's calculate the pH:
pH = -log[H+]
pH = -log(10^(-3))
pH = -(-3) = 3
So, the pH of the 10^(-3) M solution of the weak acid HA is 3.
(b) Degree of Ionization:
As calculated earlier, the degree of ionization (α) is approximately 1, indicating that the acid nearly completely ionizes.
(c) Calculation of pKa and pKb:
The pKa and pKb values are related to the acid-base equilibrium. For a weak acid with Ka, we can find the pKa using the equation: pKa = -log(Ka).
Using the given Ka value of 1.6x10^(-6), we calculate pKa as follows:
pKa = -log(1.6x10^(-6))
pKa ≈ 5.8
Since the weak acid is given, we cannot directly calculate the pKb value without having the Kb information for its conjugate base.
In summary:
(a) The pH of the 10^(-3) M solution of the weak acid HA is 3.
(b) The degree of ionization (α) is approximately 1, indicating near-complete ionization.
(c) The pKa value is approximately 5.8.