[numberonenacho]

To 10.0 mL of 0.100 M acetic acid (Ka= 1.8  10-5) is added 25 mL of deionized water and this solution is titrated with 0.100 M NaOH. Calculate the pH after the addition of the following volumes of 0.100 M NaOH. Round your pH values to one decimal place.

they want me to calculate the pH after the addition of various volumes, if someone could walk me through how to do it for 1.00 mL of 0.100 M NaOH, i should be able to do the rest. Please
Whats confusing me is the deionized water!

Thanks! 

[enahs]

What is the pH of 10.0mL of 0.100 M acetic acid? Now, what is it when you dilute it with 25mL of the water?

Now, what is the balanced equation for the reaction? Does one mol of NaOH react with 1 mol of Acetic Acid, 2, 58, etc?

So after 1 mL of NaOH is added and neutralized the acetic acid; how much acetic acid is left and what is the new concentration? If you know it's concentration you know it's pH. Remember, when you add 1mL of NaOH the volume changes, so do not forget to factor that in.

Now, you try it and somebody will tell you if you are correct!

[numberonenacho]

Thanks! this is how I'm doing so far..

For finding the initial pH this is what i've done
CH₃COOH + H₂O >> CH₃COO + H₃O

I found the pH to be 2.88 for initial. I think this is wrong because i need to know the initial pH after adding the 25mL of  deionized water, How does that affect the pH?

Then for adding the 1mL..

OH^- + HC₂H₃O₂ >> C₂H₃O₂ + H₂O

1.00 mL x 0.10 M = .1mmol for the OH-
10.0ml x 0.10 M = 1.00 mmol for the HC₂H₃O₂

so the .1 must be completely used up and subtracted from both sides, leaving .9mmol
Then I try to find the pH

Ka = [H][C₂H₃O₂]/]HC₂H₃O₂]

Initial Concentration:                     Equilibrium:
[HC2H3O2] : .9/11                         .9/11 - X
[C2H3O2]:    .1/11                         .1/11 - X

then
1.8 x 10^-5 = Ka = .1/.9 x (1.8 x 10^-5) = 2 x 10^=6

And then taking the -log of that gives me a pH of 5.69

Is this right?
Please Correct me =)
Thanks.   

[Borek]

General remarks:

http://www.chembuddy.com/?left=pH-calculation&right=titration-curves-calculation

I found the pH to be 2.88 for initial. I think this is wrong because i need to know the initial pH after adding the 25mL of  deionized water, How does that affect the pH?

What is the initial (after dilution) concentration of acetic acid?

Then for adding the 1mL..

OH^- + HC₂H₃O₂ >> C₂H₃O₂ + H₂O

1.00 mL x 0.10 M = .1mmol for the OH-
10.0ml x 0.10 M = 1.00 mmol for the HC₂H₃O₂

so the .1 must be completely used up and subtracted from both sides, leaving .9mmol
Then I try to find the pH

Ka = [H][C₂H₃O₂]/]HC₂H₃O₂]

So far so good. At this moment you don't have to use ICE table - you know Ka, you know concentration of acid, you know concentration of conjugated base - the only unknown is [H⁺]. For acids with pKa in the 3-11 range you may safely assume nothing else happens in the solution.

Do you know what Henderson-Hasselbalch equation is?

[numberonenacho]

General remarks:

What is the initial (after dilution) concentration of acetic acid?

That is where I'm having the trouble, How does the water affect the M? if its 10 ml of .100 M Acetic acid , and you add 25ml. 10ml/35 = .285, and then do you multiply that by .100 and get .0285 M Acetic acid after? I used 35, because that would be the total amount after adding the 25ml..

So far so good. At this moment you don't have to use ICE table - you know Ka, you know concentration of acid, you know concentration of conjugated base - the only unknown is [H⁺]. For acids with pKa in the 3-11 range you may safely assume nothing else happens in the solution.

Do you know what Henderson-Hasselbalch equation is?

Could you explain how it would be used in this problem?
I was really hoping for a step by step process, where in I could figure out how its done. and try to do the rest of the volumes of NaOH from there.
Thanks.

[Borek]

What is the initial (after dilution) concentration of acetic acid?

That is where I'm having the trouble, How does the water affect the M?

Geez, how is it possible you are asked to calculate pH during titration if you were not taught how to calculate concentration of diluted solution? Change school or teacher ASAP.

Check this page:

http://www.chembuddy.com/?left=concentration&right=dilution-mixing

Do you know what Henderson-Hasselbalch equation is?

Could you explain how it would be used in this problem?
I was really hoping for a step by step process, where in I could figure out how its done. and try to do the rest of the volumes of NaOH from there.

pH = pKa + log([CH₃COO^-]/[CH₃COOH])

Simply assume amount of neutralized acid equals amount of added base and amount of acid is that that was left after part was neutralized. Then calculate concentrations of both and plug the numbers into Henderson-Hasselbalch equation.