[idankor]

So,

we have 40 ml of NaOH 0.05 N
and
200 ml of CH3COOH 0.01 N (Ka = 1.8 x 10^-5)

Obviously 0.002 mol of the acid and 0.002 mol of the base, so I thought the pH would be 7, but according to the answers it's not true.

The NaOH goes to Na+ and OH- , 0.002 mole each.
The CH3COOH reacts completely with the OH- to give H2O and CH3COO- , so we have 0.002 mole of CH3COO-. I am a bit confused. What reacts with what and how much ? 

Thanks to all! Great forum.

[Mr Peanut]

Yes, you are at the equivalence point. But, because CH3COOH is a weak acid, the CH3COO- now reacts with water to establish its solubility equilibrium in a process called "hydrolysis":

CH3COO- + HOH == CH3COOH + OH- (that's right!, the solution is actually above pH 7)

Write the equilibrium expression for the above and see that it is a combination of two simultaneous equilibria: the dissociation of water and the reverse of the disociation of CH3COOH. (Hint: multiply the numerator and denominator by [H]).

Once you have your reaction written down and you have your equilibrium expression set-up, you'll see that the expression is equal to the appropriate combination of the equilibrium constants for each of the two simultaneous equilibria. (I won't give the details away yet.)

You should see a nasty looking quadratic equation that is usually simplified with an approximation.

Let us know after you noodle with it for a while if you are still stuck.

[idankor]

Thanks a lot, you really helped me. My main problem was that I didn't know how to get the Kb out of the Ka and friend of mine told me it's 10^-14 divide the Ka, so then I just used the qdr(Kb + Cb) and solved it. Can you explain me the reason why Kb = (10^-14) / Ka. I know that [HO-]*[H3O+] = 10^-14, but how do you get from there to the Kb = (10^-14) / Ka equation ? Thanks again.

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=bronsted-lowry-theory

[Mr Peanut]

Yes, I think that website does a good job. Myself, I like to see one more step in the explanation.

There is a three part secret to all of these equilibrium problems. It applies to all equilibrium problems, not just acid-base reactions. Follow it and you will have success.

1) Don't get anxious about the problem.
2) Think out, then write out, all of the balanced reaction equations associated with the problem.
3) Write the equilibrium expressions for the reactions. (product of the conc of the products raised to the power of their coefficients divided by the product of the conc of the reactants raised to the power of their coefficients)

Again, for the hydrolysis reaction, we have:

OAc- + HOH == HOAc + OH-

The equilibrium expression:

[HOAc][OH]
[OAc][HOH]

The trick here is we will multiply by 1: [H]/[H]=1 and get:


[HOAc][OH][H]
[H][OAc][HOH]

=1/Ka*Kw