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.
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?
2 answers
Got it! Thank you very much!
1 answer