# Homework Help: Titration and Calculating pH

1. Feb 13, 2010

### erok81

1. The problem statement, all variables and given/known data

100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide . An appropriate
indicator is used. Ka for acetic acid is 1.7 x 10 -5Calculate the pH in the flask at the following points in the titration.
a. when no NaOH has been added.
b. after 25.0 ml of NaOH is added
c. after 50.0 ml of NaOH is added
d. after 75.0 ml of NaOH is added
e. after 100.0 ml of NaOH is added

2. Relevant equations

n/a

3. The attempt at a solution

Here is what I have so far.

To calculate "a" I think I can use the equation ka=[CH3COON]/[CH3COOH][NaOH] (which uses CH3COOH +NaOH -> CH3COON + H20) and use a table to figure out the equilibrium values. Then plug those values into the formula to get the values of the acid, then figure the pH from there.

The part I am not sure on (well two parts) is do I have the equation right, I don't think I do since there isn't an H+ to calculate. The second thing is...how do I factor those different volumes in? And I am sure the 100mL in the beginning matters, but I can't remember how to factor those in.

2. Feb 13, 2010

### erok81

Ok..forget the part I said about A. That uses a different equation. I figured the pH of part a to be 2.89.

The rest of them I am not sure.

Do I just set up the table (intial, change, equilibrium) with each of the different measurements for NaOH?

3. Feb 13, 2010

### Staff: Mentor

Calculating titration curves - scroll down the page - third paragraph from the end summarizes how to calculate whole titration curve of a weak acid.

Last edited by a moderator: Aug 13, 2013
4. Feb 13, 2010

### erok81

I think I've got it figured out until I got to 100mL of NaOH.

Using that formula, I get a divide by zero. Unless I am still doing it wrong?

Here is what I have so far.

b. 4.28
c. 4.76
c. 5.23
d. I end up with no CH3COOH after reacting and can't divide by that zero.

5. Feb 13, 2010

### Staff: Mentor

At the end use different approach - assume you have just a solution of salt. That means solution of a weak base.

Last edited by a moderator: Aug 13, 2013
6. Feb 13, 2010

### erok81

Got it. Does 8.38 sound about right?

7. Feb 13, 2010

### Staff: Mentor

Close, but wrong by about 0.3 unit, that's too much of an error for such a simple calculation.

Last edited by a moderator: Aug 13, 2013
8. Feb 13, 2010

### erok81

I think my value for Kb is wrong. I used 5.88x10^-10.

This is where the problem lies I am pretty sure.

I found that by using the ka value for acetic acid then using kw/ka=kb.

Is that wrong?

9. Feb 13, 2010

### Staff: Mentor

Nope, Kb is OK.

Last edited by a moderator: Aug 13, 2013
10. Feb 13, 2010

### erok81

Ok...tried it again. I had the wrong value in for NaOH.

Does 8.85 sound a bit better?

Here is how I came up with that figure.

NaOH -> Na+ + OH-
Initial 0.100 0 0
Change -x x x
Equil 0.100-x x x

kb=[Na][OH]/[NaOH]

Multiplying everything out...

5.88x10^-11 - 5.88x10^-10x - x^2

Then for x I get 7.66x10^-16

And finally for the pH

-log(7.66x10^-16) = 5.115

For the base pH 14-5.115 = 8.89.

Last edited: Feb 13, 2010
11. Feb 14, 2010

### Staff: Mentor

Still wrong. It is not NaOH that is a weak base responsible for pH, but acetate. Kb for acetate is not that for NaOH, beides, NaOH is a STRONG base, with large Kb, not a small one.

Have you taken dilution into account?

Last edited by a moderator: Aug 13, 2013