[Borek]

so the difference would be 100 - 38.4 = 61.6?

That would be my approach.

[Borek]

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

(some wrong calculations not quoted)

You already wrote earlier:

At the half-titration point, the amount of HA left is equal to the amount of A- formed. At this point, the Ka = [H+].

(...)

If Ka = [H+], the pKa = pH.

Why don't you follow what you already wrote, instead of trying to reinvent the wheel?

[sallyhansen]

so for my other calculation I ended up getting 0.033%, which would then be 100 - 0.033 = 99.67%. Does that not seem to high? And also I do not understand that concept of Ka = [H+] would that mean it is 0.1 mol/L. Totally confused now.

[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.

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.

You quoted me being correct earlier but I did some mathematical error and forgot to square it, so the basis of my question is correct here right?

[sallyhansen]

I think I just had a eureka moment
Ka = [H3O+][C2H3O2-]
        HC2H3O2
Ka =

[0.1]
But since Ka = [H+] at the half way point the answer would be
Ka = [H+]
    = 2.63 x 10^-5

Calculation #2
% difference = Ka Experimental/ Ka Given * 100
           = 2.63 x 10-5/ 1.8 x 10-5 * 100
           = 1.461
100 – 1.461 = 98.54%


*Please be right*

[Borek]

I do not understand that concept of Ka = [H+] would that mean it is 0.1 mol/L. Totally confused now.

No, it wouldn't mean 0.1 mol. You are mistaking concentration of H⁺ (given by pH) with the analytical (total) concentration of the acetic acid.

since Ka = [H+] at the half way point the answer would be
Ka = [H+]
    = 2.63 x 10^-5

Yes, you got it correctly now (assuming that's the correct concentration of H⁺).

% difference = Ka Experimental/ Ka Given * 100
           = 2.63 x 10-5/ 1.8 x 10-5 * 100

OK so far.

           = 1.461

Nope. This is without multiplying by 100%.

100 – 1.461 = 98.54%

1.461 was already wrong, but here even the subtraction is not correct - both in terms of logic and in terms of the math.

[sallyhansen]

oo goodness I am making silly mistakes thanks for notifying me
Calculation #2
% difference = Ka Experimental/ Ka Given * 100
           = 2.63 x 10-5/ 1.8 x 10-5 * 100
           = 146.11% (what does this mean if it is over 100%)

100 – 146.11 = 46.11%

[sallyhansen]

So just a summary of what I did and my answers to them
1.   Write the balanced chemical equation for the neutralization reaction you observed showing state symbols. (2)
CH3COOH (aq) + NaOH (aq)   CH3COONa(aq) + H2O(l)
Acetic acid + sodium hydroxide → sodium acetate + water
2.   Plot a graph of your data, with pH on the vertical axis and volume of NaOH on the horizontal axis. Your graph should show a steep rise in pH as the volume of NaOH becomes enough to neutralize the acetic acid. Take the midpoint of this steep rise and read off the volume of NaOH. This is the volume of NaOH that was needed to neutralize all the acetic acid. Compare the volume on your graph with the volume you recorded when the phenolphthalein indicator first turned pink. (5)
The midpoint of the steep rise was 12.0 mL of NaOH that was needed to neutralize all the acetic acid.
The recorded value of phenolphthalein when it first turned pink was also at 12.0 mL of NaOH.
When an indicator is used in a titration, the color change occurs at what is called the endpoint. If the indicator has been properly selected, this point will be the same as the equivalence point. When a pH meter is used, the pH of the solution is recorded as the titrant is added. The pH versus the volume of titrant added can be plotted on what is called a titration curve. In this case the equivalence point occurs at the point where very small additions of titrant cause a very rapid rise in the pH.
 
3.   Determine the amount of NaOH added (in mols).(1)
0.1 mol/L of NaOH and 15 mL of NaOH  0.015L
(0.015L)(0.1mol/L)
=0.0015 moles
4.   Use the ratio in which the acid and base react, determined from the chemical equation. Calculate the molar concentration of the acetic acid. (2)
CH3COOH (aq) + NaOH (aq)   CH3COONa(aq) + H2O(l)

From the equation, there was 1 mole of NaOH neutralized with 1 mole of acetic acid; so therefore, there are 0.0015 moles of acid.

n=CV
0.0015= C(0.015)
C= 0.1 mol/L
5.   Write the expression for Ka for the ionization of acetic acid in water. (2)
HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2-(aq)
 Ka =       [H3O+(aq)][ C2H3O2-(aq)]_____       
           [HC2H3O2(aq)]

6.   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 x 10-3 mol/L for both [H3O+] and [CH3COO]-
7.   Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the acid. Calculate Ka for the acid. (2)
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 10-3]2
        [0.1]
Ka = 6.916 x 10-6
8.   Refer to the volume of NaOH on your graph from question 2. Calculate half this volume and on your graph, find the pH when the solution was half neutralized. (1)
Original volume of NaOH = 12
Half the volume of NaOH = 6
pH for 12 mL of NaOH = 10.66
pH for 6 mL of NaOH = 4.58
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]
But since Ka = [H+] at the half way point the answer would be
Ka = [H+]
    = 2.63 x 10-5
10.   Calculate the percent difference between your value for Ka (from the calculations and the graph) and the accepted value. (3)
Calculation #1
% difference = Ka Experimental/ Ka Given * 100
           = 6.916 x 10-6/ 1.8 x 10-5 * 100
           = 38.4%
100 – 38.4 = 61.6%
Calculation #2
% difference = Ka Experimental/ Ka Given * 100
           = 2.63 x 10-5/ 1.8 x 10-5 * 100
           = 146.11%
100 – 146.11 = 46.11%
11.   Do the values you calculated for [H3O+] and [CH3COOH] prove that CH3COOH is a weak acid? Explain. (2)
The values calculated for [H3O+] and [CH3COOH], which is 2.63 x 10-3 mol/L and 2.63 x 10-5 mol/L for when it was half-neutralized, show that it has not been completely ionized. So the stronger the acid the larger their Ka value is, but the weaker the acid the smaller their Ka value is. A strong acid is a better proton donor, resulting in more products. Since the concentration of the products is in the numerator of the Ka expression, the stronger the acid, the larger the Ka. The two values for the acetic acid are Ka = 6.916 x 10-6 and Ka = 2.63 x 10-5. The exponents show that the value contains numerous zeros before it not having a large number for the Ka shows that acetic acid is a weak acid.

AND there is an attachment of how my graph looks like

[Borek]

The midpoint of the steep rise was 12.0 mL of NaOH

The recorded value of phenolphthalein when it first turned pink was also at 12.0 mL of NaOH.

0.1 mol/L of NaOH and 15 mL of NaOH looks like

So 12, or 15?

[sallyhansen]

The total amount is 15 but at the midpoint it was 12 so half of that would be 6

[Borek]

So why do you use 15 mL to calculate amount of acid?

And 12 mL was not a midpoint, but endpoint. Midpoint was 6 mL.

[sallyhansen]

So why do you use 15 mL to calculate amount of acid?

And 12 mL was not a midpoint, but endpoint. Midpoint was 6 mL.

okay if the midpoint is 6 then that is when it was neutralized meaning that the half neutralized is 3? And I use 15 mL because as you can see on the graph that is where the volume and pH correlation ends ... ?

[Borek]

Midpoint is the point in the middle, half way between the start and equivalence point - that's where the acid was half neutralized. So if the equivalence point was at 12 mL, midpoint was at 6 mL.

To calculate amount of titrant (and stoichiometric equivalent of titrated substance) you should use volume required to arrive at the end point (so 12 mL), not some random point after that.

[sallyhansen]

Okay so changing everything else up since the pH calculation would have changed since the half neutralized point would now be 3 and the pH for it is 4.17 these are my answers again

1.   Write the balanced chemical equation for the neutralization reaction you observed showing state symbols. (2)
CH3COOH (aq) + NaOH (aq)   CH3COONa(aq) + H2O(l)
Acetic acid + sodium hydroxide → sodium acetate + water
2.   Plot a graph of your data, with pH on the vertical axis and volume of NaOH on the horizontal axis. Your graph should show a steep rise in pH as the volume of NaOH becomes enough to neutralize the acetic acid. Take the midpoint of this steep rise and read off the volume of NaOH. This is the volume of NaOH that was needed to neutralize all the acetic acid. Compare the volume on your graph with the volume you recorded when the phenolphthalein indicator first turned pink. (5)
The midpoint of the steep rise was 6.0 mL of NaOH that was needed to neutralize all the acetic acid.
The recorded value of phenolphthalein when it first turned pink was at 12.0 mL of NaOH.
When an indicator is used in a titration, the color change occurs at what is called the endpoint. If the indicator has been properly selected, this point will be the same as the equivalence point. When a pH meter is used, the pH of the solution is recorded as the titrant is added. The pH versus the volume of titrant added can be plotted on what is called a titration curve. In this case the equivalence point occurs at the point where very small additions of titrant cause a very rapid rise in the pH.
 
3.   Determine the amount of NaOH added (in mols).(1)
0.1 mol/L of NaOH and 12 mL of NaOH  0.012L
(0.012L)(0.1mol/L)
=0.0012 moles
4.   Use the ratio in which the acid and base react, determined from the chemical equation. Calculate the molar concentration of the acetic acid. (2)
CH3COOH (aq) + NaOH (aq)   CH3COONa(aq) + H2O(l)

From the equation, there was 1 mole of NaOH neutralized with 1 mole of acetic acid; so therefore, there are 0.0012 moles of acid.

n=CV
0.0012= C(0.012)
C= 0.1 mol/L
5.   Write the expression for Ka for the ionization of acetic acid in water. (2)
HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2-(aq)
 Ka =       [H3O+(aq)][ C2H3O2-(aq)]_____       
           [HC2H3O2(aq)]

6.   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 x 10-3 mol/L for both [H3O+] and [CH3COO]-
7.   Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the acid. Calculate Ka for the acid. (2)
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 10-3]2
        [0.1]
Ka = 6.916 x 10-6
8.   Refer to the volume of NaOH on your graph from question 2. Calculate half this volume and on your graph, find the pH when the solution was half neutralized. (1)
Original volume of NaOH = 6
Half the volume of NaOH = 3
pH for 6 mL of NaOH = 4.58
pH for 3 mL of NaOH = 4.12
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 3 mL = 4.12
pH = -log[H+]=4.12
     = [H+] = 10-4.12
      = 7.6 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]
But since Ka = [H+] at the half way point the answer would be
Ka = [H+]
    = 7.6 x 10-5
10.   Calculate the percent difference between your value for Ka (from the calculations and the graph) and the accepted value. (3)
Calculation #1
% difference = Ka Experimental/ Ka Given * 100
           = 6.916 x 10-6/ 1.8 x 10-5 * 100
           = 38.4%
100 – 38.4 = 61.6%
Calculation #2
% difference = Ka Experimental/ Ka Given * 100
           = 7.6 x 10-5 / 1.8 x 10-5 * 100
           = 422.22%
100 – 422.22= 322.22%
11.   Do the values you calculated for [H3O+] and [CH3COOH] prove that CH3COOH is a weak acid? Explain. (2)
The values calculated for [H3O+] and [CH3COOH], which is 2.63 x 10-3 mol/L and 7.6 x 10-5 mol/L for when it was half-neutralized, show that it has not been completely ionized. So the stronger the acid the larger their Ka value is, but the weaker the acid the smaller their Ka value is. A strong acid is a better proton donor, resulting in more products. Since the concentration of the products is in the numerator of the Ka expression, the stronger the acid, the larger the Ka. The two values for the acetic acid are Ka = 6.916 x 10-6 and Ka = 7.6 x 10-5. The exponents show that the value contains numerous zeros before it not having a large number for the Ka shows that acetic acid is a weak acid.

For the part that I bolded the percent difference is a very large number, how may that be?

[Borek]

Okay so changing everything else up since the pH calculation would have changed since the half neutralized point would now be 3

No, it won't be. Midpoint and the half neutralization is the same, as explained in my previous post.