Chemistry help

panksea06

So normally chem comes quite naturally to me but I am finding this bit somewhat difficult and would like a guide as to how do to it.

We are titrating weak acid with a strong base.

Take 10mls .100M acetic acid (Ka= 1.8 10-5) and add 25mls of water.
Find the pH at this point and after you add 1 ml of .100 NaOH.
Then at 2.00 mls NaOH total then 3.00 etc etc. up to 15mls NaOH.

I have the first few answers in this but am getting confused and somewhat lost as I think I am at the point now were I need to account for the self ionization of water.
0 ml NaOH= 3.1 pH
1mls = 3.8 pH
2mls = 4.2
3mls =4.3
and that is as far as I got, and the last only really by guessing.

So PA can you help me out?

Edit: Also it should be noted since at 10mls you will have equal amounts of the acid and base in there in some form that this will cause the equations to model the situation to change afterword.

Unknown

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panksea06

electricitylikesme wrote:

Also, since you're using a strong base you can assume in fact that the conjugate acid is just the concentration of NaOH in the solution.

I dont quite follow this.

HA (representing acetic acid) will eventually have its component in solution, A- act as a conjugate base and I will have to account for that I know, but there is no conjugate acid of any significance (2 figures of accuracy in my case) with NaOH... Right?

Mostly I am looking for the steps in the process to solve for the pH at say 5mls, since I have forgotten the proccess when its a weak acid that is that dillute (ie water contributes heavily).

Rialcel

You can find the pH by using the Henderson-Hasslebach equation, pH=pKa+log([A-]/[HA]), pKa is a constant, and for acetic acid it is 4.74. To find the pH of the solution when 4ml of NaOH is added, the equation would look like this.

pH=4.74+log((.4/(35+4))/(.6/(35+4)))=4.6

Hope this helps.

corcorigan

Assume acetic (ethanoic?) acid is completely unionised, and the sodium hydroxide is completely ionised?

Should be pretty easy then.

What level are you at? I know a-level gave no benefits to doing all the complex not-entirely-ionised stuff.

panksea06

Rialcel wrote:

You can find the pH by using the Henderson-Hasslebach equation, pH=pKa+log([A-]/[HA]), pKa is a constant, and for acetic acid it is 4.74. To find the pH of the solution when 4ml of NaOH is added, the equation would look like this.

pH=4.74+log((.4/(35+4))/(.6/(35+4)))=4.6

Hope this helps.

Yes I think that does solve what I am looking for. Thanks.

corcorigan: I am at the 2nd quarter of the gen chem series at univeristy (chem 152) and just was blanking on how to do these.

I think this is solved now so feel free to lock.

Thanks again!