This is regarding the Biochemistry question postef by Grace..
I'm wasn't able to mention this in the subject section due to the large no.of characters! I apologize for any inconveniences made!
In this case,to calculate pH of the solution, we should find the ka value of HAc.
Is this how we do that?
HAc<===>H^+ + Ac^ --->(1)
NaAc--->Na^+ Ac^- --->(2)
[HAc]=0.01 M
[NaAc]=0.03 M
As NaAc dissociates 100℅ we can take the [Ac^-]=0.03
And as usual,from 1,can we take [Ac^-]=[H^+] ,when finding ka for HAc?
So ka=[Ac^-] ^2/[HAc]
As HAc is a weak acid,we can take its equilibrium concentration =initial concentration
Which gives,
Ka=[ (9*10^-4)/(1*10^-2)] M
Ka=9*10^-2 M
Is this possible for HAc to have this much of a bigger value as its Ka???
Where did I made a mistake?
It is not possible to calculate Ka HAc from the data posted by Grace. The standard Ka value for HAc is used (e.g., from tables) and pH is calculated from that. As for errors, your first assumption is not right. (Ac^-) is not = to (H^+). It IS when you have a solution of HAc but the instant you add NaAc, the increased Ac^- forces the HAc (your equation 1) to the left and H^+ is decreased. From your equation it would look like this.
.....HAc ==> H^+ + Ac^-
I....0.01....0.....0
C....-x......x.....x
E..0.01-x....x.....x
So Ka (nothing else), then
Ka = (H^+)(Ac^-)/(HAc). Now we add NaAc of 0.03 and have this, as you correctly note.
......NaAc ==> Na^+ + Ac-
I.....0.03.....0.......0
C....-0.03...0.03....0.03
E.....0......0.03....0.03
Back to the Ka expression, here is what we enter.
Ka = (H^+)(Ac^-)/(HAc)
(H^+) = x
(Ac^-) = x + 0.03 (that's x from the HAc and 0.03 from the NaAc and that shows that your first assumption is in error).
(HAc) = (0.01-x)
Thus, if we know Ka we can calculate (H^+) and pH or knowing (H^+) or pH we can calculate Ka. There are two unknowns (Ka and H^+) and only one equation In fact, starting with your equation 1 the Henderson-Hasselbalch equation can be derived.
Got it! Thank you very much!
To calculate the pH of a solution containing a weak acid such as acetic acid (HAc), you need to first find the Ka value. Here's a step-by-step explanation of the process:
1. Write the balanced chemical equation for the dissociation of the weak acid:
HAc ⇌ H+ + Ac-
2. Determine the initial and equilibrium concentrations of the species involved:
[HAc] = 0.01 M (initial concentration)
[H+] = unknown
[Ac-] = unknown
3. Since NaAc completely dissociates, you can assume that the concentration of the acetate ion (Ac-) provided by NaAc is equal to the concentration of Ac- in the solution:
[Ac-] = 0.03 M
4. Under equilibrium conditions, the concentration of Ac- is equal to the concentration of H+ (by the principles of stoichiometry in the reaction), so [Ac-] = [H+].
5. Now, you can substitute the concentrations into the Ka expression for HAc:
Ka = [H+][Ac-] / [HAc]
Since [Ac-] = [H+] = 0.03 M and [HAc] = 0.01 M:
Ka = (0.03)*(0.03) / 0.01 = 0.09 M
Therefore, the calculated Ka value for acetic acid is 0.09 M.
The Ka value represents the extent of dissociation for the weak acid. A larger Ka value indicates a stronger acid, while a smaller Ka value corresponds to a weaker acid. In the case of acetic acid, a Ka value of 0.09 M is within the typical range for weak acids, indicating that it is a moderately weak acid.
So, it seems like you have calculated the Ka value correctly, and there doesn't appear to be any mistakes in your approach. Acetic acid does have a relatively small Ka value compared to strong acids, but 0.09 M is a reasonable value for a weak acid like acetic acid.