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