weak acid, HA, is 0.1 % ionized in a 0.2 M solution.
(a) What is the equilibrium constant for the dissociation of the acid (Ka)?
(b) What is the pH of the solution?
..................HA ==> H^+ + A^-
I...............0.2M.........0.........0
C................-x............x..........x
E..............0.2-x...........x.........x
HA = 0.2 M and 0.1% ionized; therefore,
(H^+) = (A^-) = 0.2 M x 0.001 = 2E-4 and (HA) = 0.2-2E-4
Ka = (H^+)(A^-)/(HA)
Substitute the numbers into the Ka expression and solve for Ka.
b part. pH = -log(H^+). You know (H^+) from above, calculate pH. Post your work if you get stuck.
(a) Well, if the weak acid (HA) is only 0.1% ionized, it's looking a little shy! To find the equilibrium constant (Ka), we need to divide the concentration of the ionized acid ([A-]) by the concentration of the non-ionized acid ([HA]). So if the concentration of HA is 0.2 M, and it's only 0.1% ionized, then the concentration of [A-] is like a tiny fraction of that. It's probably easier if I just give you the answer: Ka is approximately 0.001 M.
(b) Now, talking about the pH of the solution... well, that's a delicate matter. Since we know the concentration of the weak acid, we can use the equilibrium constant (Ka) to find the concentration of H+ ions. And with that, we can calculate the logarithm of the H+ concentration and get the pH. Oh wait! You didn't give me the pKa value of the weak acid or its ionization equation, so I can't really do any calculations here. Looks like the pH of the solution wants to remain a secret! Don't worry, we'll find a way to squeeze it out of them next time!
To find the answers, we need to use the given information and some equations related to acid dissociation and pH calculations.
(a) The equilibrium constant for the dissociation of the weak acid (Ka) can be calculated using the expression:
Ka = ([H+][A-])/[HA]
Given that the weak acid (HA) is 0.1% ionized, we can infer that the concentration of the undissociated acid [HA] will be (100 - 0.1)% = 99.9%.
So, [HA] = 0.2 M * 99.9% = 0.1998 M (rounded to 0.2 M)
Since the weak acid (HA) is monoprotic, we will assume that the concentration of [H+] will be the same as [A-].
Let's represent the concentration of [H+] as x.
Now, we can substitute the values into the Ka expression:
Ka = (x * x) / (0.2 - x)
Since the percent ionization is 0.1%, we can consider x as very small compared to 0.2. Therefore, we approximate 0.2 - x as 0.2.
The expression simplifies to:
Ka = (x * x) / 0.2
Now, we can solve the equation to find the value of Ka.
(b) To calculate the pH of the solution, we need to determine the concentration of [H+]. Since the weak acid is 0.1% ionized, we can assume the concentration of [H+] and [A-] to be the same, denoted by x.
To find x, we will use the expression:
Ka = (x * x) / (0.2 - x)
We will substitute the value of Ka from part (a), and solve the equation to find the value of x.
Once we have the value of [H+], we can calculate the pH using the equation:
pH = -log[H+]
Let's perform the calculations for both parts.
(a) Substitute the values into the Ka expression:
Ka = (x * x) / 0.2
(b) Follow these steps to calculate the pH:
1. Substitute the value of Ka into the equation:
(x * x) / (0.2 - x) = Ka
2. Solve this equation for x.
3. Once you have the value of x, calculate the pH using the equation:
pH = -log[H+]
Please note that the calculations will depend on the specific value of Ka, which needs to be provided in the question.
To find the answers to these questions, we need to apply the concepts of percent ionization and equilibrium constants for acids. Let's go step by step:
(a) To determine the equilibrium constant for the dissociation of the acid (Ka), we can use the percent ionization:
Percent Ionization (α) = (Ionized concentration / Initial concentration) x 100
In this case, the percent ionization is given as 0.1%. Therefore, α = 0.1/100 = 0.001.
The ionized concentration is equal to [H+], which can be assumed to be the same as [A-] since it's a monoprotic acid.
Let's assume the initial concentration of the weak acid (HA) is [HA]0 = 0.2 M.
Then, the ionized concentration [H+] (or [A-]) can be calculated as follows:
[H+] = α x [HA]0 = 0.001 x 0.2 = 0.0002 M
The equilibrium expression for the dissociation of HA is:
HA ⇌ H+ + A-
The equilibrium constant Ka can be expressed as:
Ka = [H+][A-] / [HA]
Since [H+] = [A-] = 0.0002 M and [HA] = 0.2 M, we can substitute these values into the equation:
Ka = (0.0002)(0.0002) / 0.2
Ka = 4 x 10^-7
Therefore, the equilibrium constant for the dissociation of the weak acid is 4 x 10^-7.
(b) Once we have the equilibrium constant (Ka), we can use it to find the pH of the solution. The pH is defined as the negative logarithm (base 10) of the concentration of H+ ions.
pH = -log[H+]
In this case, the concentration of H+ ions is 0.0002 M.
pH = -log(0.0002)
Using a calculator, we find that pH ≈ 3.70.
Thus, the pH of the solution is approximately 3.70.