I need to calculate the Ka of the acetic acid from two titration curves involving acetic acid. Then report the % error in each how do i go about this? what values do i need to know?
Did you add a particular volume of titrant and measure the pH of the resulting solution? If not, how did you perform the experiment. The usual way of determining the Ka is to look up the pH at the halfway point to the equivalence point.
Trial 1: NaOH (titrant) and acetic acid
pH at equivalence is: 8.14
Ph at endpoint is: 7.78
Volume of Titrant: 14.6 mL
Volme of Titrant at endpointL 13.8
Halfway point to equivalence point:
112 sec/2= 56.o sec
pH at this point is 4.80
Trial 2: NH3 (titrant) and acetic acid
pH at equivalence is: 7.01
Ph at endpoint is: 8.30
Volume of Titrant: 16.6 mL
Volme of Titrant at endpoint: 36.7
Halfway point to equivalence point:
170 sec/2= 85.o sec
pH at this point is 4.79
Did you use a timer? An automatic titrator? How did you read the pH? Why is the volume of the titrant different from volume of tritrant at end point? (My guess is that those are the volume readings at those points and 36.7 - 16.6 = volume acutally used???)
My partner as in charge of writing all the points and i think she messed me up somehow. but yea we had a timer... we used a computer program that showed the titration curve ph vs time in sec. The pH was read using the curve. The volume of titrant is just how much was used in 250 sec ( that's how long the titrant was supposed to drip) and then the endpoint we just looked up on the graph and matched it up. We marked where the solution began to change color and looked at how much mL was used
OK. So you used a computer program to do all this. For the weak acid, which I will cll HA, ionizes
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
Solving this for (H^+), we get
(H^+) = Ka*(HA)/(A^-). This is the equation that determines the pH from start to the equivalence point. When the titration is exactly half way to the equivalence, then (HA) = (A^-) and (H^+) = Ka. If you take the -log of both sides you get -log(H^+) = -log Ka but that is pH = pKa. You know where the equivalence point is. (I'm a little puzzled that you make a distinction between pH at end point and pH at equivalence point.) I assume that the computer is adding titrant at a steady pace and I assume that it takes just half as long to get half way to the equivalence point as it does to get from half way there to all the way. So you look up the pH halfway to the equivalence point. That pH = pKa. You have said it ws 4.80. That converts to 1.58 x 10^-5 for Ka, which isn't half bad since Ka for acetic acid, I think, is 1.8 x 10^-5. I'm sure you have it in your set of tables. Is that the percent error you are to calculate?
%error = [(actual value - exp value)/actual value]*100 which is about 12% if I didn't goof. I used 1.8 x 10^-5 as the actual value. Check that, of course.
ok so in steps, how do i break that up to write...
so you're saying for
Trial 1: Calculate the Ka
I just uses the half way pH that i found for both and do the -log to get the ka values?
so for example Trial 1: = -log ( 4.80) = ?? im not getting the answer that you got instead i got 6.81x10^-1?
oh my sorry... i have to do the 10^(-pH). Ill post what i get post that you can check
so for trial one i got :
% error of 12.222--> 12%
ka= 1.58x10^-5
from 10^(-4.80)
trial two:
% error of 10%
ka= 1.62x10^-5 from 10^(-4.79)