buffers
posted by Reen on .
9.00mL of 1.10M CH3COOH and 1.00mL of 0.900M CH3COO what is the pH?

Use the HendersonHasselbalch equation.

we havn't learnt that equation.

pH = pKa + log(base/acid)
pKa is pKa for CH3COOH
base is (CH3COO^)
acid is (CH3COOH) 
thanks. so for this question is the pH 3.70?

Or you can do it another way.
Write ionization of CH3COOH.
CH3COOH ==> H^+ + CH3COO^
Now write the Ka expression.
Ka = (H^+)(CH3COO^)/(CH3COOH)
Now plug in (CH3COO^) from the problem.
Substitute (CH3COOH) from the problem.
Substitute Ka for CH3COOH. Try to find Ka in your text or notes but it's very close to 1.8 x 10^5. Calculate (H^+) and convert that to pH. 
I calculate 3.698 pH which rounds to 3.70. Yes. Very good.

thanks a lot :D

You're welcome.