[Borek]

Of course, you have to do a complete titration first, then redo the titration but add only have as much NaOH to reach the half titration point.

Not exactly - it is enough that you do titration once, registering whole titration curve. Half titration point is already there then, you just have to find it.

[sallyhansen]

Okay so for the part of the first question I find confusing is that from question #1 you are given pH and from there you can calculate [H+] and [CH3COO-] which are both X.
 
Then, notice that in question #2 you're not given the Ka, but expected to calculate it with the X from question #1.
So the first wuestion I had was this
1.   Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2)
Initial pH = 2.58
pH = -log[H+]=2.58
   = [H+] = 10^-2.58
   = 2.63*10^-3 m

so What I would do is (2.63 x 10^-3)/0.1 = 0.0263 or 2.63 x 10^-2

Would that be the correct answer then?

[Borek]

Okay so for the part of the first question I find confusing is that from question #1 you are given pH and from there you can calculate [H+] and [CH3COO-] which are both X.
 
Then, notice that in question #2 you're not given the Ka, but expected to calculate it with the X from question #1.

What does the fact that something is marked with X has to do with the calculations? Something is an unknown, once the problem is solved and you know its value, it becomes a known, no matter what symbol is/was/will be used.

So the first wuestion I had was this
1.   Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2)
Initial pH = 2.58
pH = -log[H+]=2.58
   = [H+] = 10^-2.58
   = 2.63*10^-3 m

so What I would do is (2.63 x 10^-3)/0.1 = 0.0263 or 2.63 x 10^-2

Would that be the correct answer then?

Close, but you should check your math. You were right up to 2.63×10^-3 M.

[sallyhansen]

whoops forgot to square it the answer is 6.0 x 10-9

[sallyhansen]

okay another question I would like to attempt
10.   Calculate the percent difference between your value for Ka (from the calculations and the graph) and the accepted value.

Would I use the initial Ka value which was 6.916 x 10-6 or the half neutralized one which was 6.0 x 10-9?
And I am not sure what it means by the calculation of the graph. Also is there a formula that is needed to calculate it? 

[Borek]

Would I use the initial Ka value which was 6.916 x 10-6 or the half neutralized one which was 6.0 x 10-9?

Both, each separately.

And I am not sure what it means by the calculation of the graph.

Are we reading the same question? Not "calculations of the graph" but "calculations AND the graph". First refers to the number you just calculated, the other to the number you are expected to read from the titration curve.

Also is there a formula that is needed to calculate it?

(Kaexp/Katables-1)*100%

[sallyhansen]

so the ka expected is the 6.0 x 10-9 and the other one and they are each done separately and the Ka table is the value I would get on the table which would be 1.8 x 10^-5?

[Borek]

so the ka expected is the 6.0 x 10-9

Please pay attention to what you write, you make so many small mistakes it is pretty difficult o know what you are talking about.

exp=experimental

Basically it is about calculating ratio of the value you got and the value taken from tables. Trick is, what you will get is something like 120% (for example). You are asked about the difference, so it is 20%. That was already reflected by the formula I have listed.

[sallyhansen]

Also you wrote Katable-1 would I subtract 1.8 x 10^-5 by 1?

[sallyhansen]

But basically this is what I got

% difference = Ka Experimental/ Ka Given * 100
           = 6.916 x 10^-6/ 1.8 x 10^-5 * 100
           = 38.4%

[sallyhansen]

And for the second calculation I got

Calculation #2
% difference = Ka Experimental/ Ka Given * 100
           = 6.0 x 10-9/ 1.8 x 10-5 * 100
           = 0.033%

[Borek]

Where did you got 6.0×10^-9 from.

[Borek]

But basically this is what I got

% difference = Ka Experimental/ Ka Given * 100
           = 6.916 x 10^-6/ 1.8 x 10^-5 * 100
           = 38.4%

If one is 38.4% of the other, difference is NOT 38.4%.

[sallyhansen]

Where did you got 6.0×10^-9 from.

This was the second calculation I got when acetic acid is half neutralized
9.   Calculate Ka when the acetic acid was half-neutralized. How does this value compare with your Ka value for acetic acid? (2)
pH for 6 mL = 4.58
pH = -log[H+]=4.58
     = [H+] = 10^-4.58
      = 2.63 x 10^-5 mol/L

HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2-(aq)
   HC2H3O2   H3O+   C2H3O2-
Initial   0.1   0   0
Change   -x   +x   +x
Equilibrium   0.1-x (the x is relatively small compared to 0.1)   x   x
Ka = [H3O+][C2H3O2-]
        HC2H3O2
Ka =

[0.1]
Ka= [2.63 x 10^-5]^2
       [0.1]
Ka = 6.0 x 10^-9

[sallyhansen]

But basically this is what I got

% difference = Ka Experimental/ Ka Given * 100
           = 6.916 x 10^-6/ 1.8 x 10^-5 * 100
           = 38.4%

If one is 38.4% of the other, difference is NOT 38.4%.

so the difference would be 100 - 38.4 = 61.6?