Posted by Reen on Wednesday, March 3, 2010 at 10:55pm.
9.00mL of 1.10M CH3COOH and 1.00mL of 0.900M CH3COO- what is the pH?
- buffers - DrBob222, Wednesday, March 3, 2010 at 10:56pm
Use the Henderson-Hasselbalch equation.
- buffers - Reen, Wednesday, March 3, 2010 at 11:03pm
we havn't learnt that equation.
- buffers - DrBob222, Wednesday, March 3, 2010 at 11:12pm
pH = pKa + log(base/acid)
pKa is pKa for CH3COOH
base is (CH3COO^-)
acid is (CH3COOH)
- buffers - Reen, Wednesday, March 3, 2010 at 11:18pm
thanks. so for this question is the pH 3.70?
- buffers - DrBob222, Wednesday, March 3, 2010 at 11:27pm
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.
- buffers - DrBob222, Wednesday, March 3, 2010 at 11:30pm
I calculate 3.698 pH which rounds to 3.70. Yes. Very good.
- buffers - Reen, Thursday, March 4, 2010 at 12:03am
thanks a lot :D
- buffers - DrBob222, Thursday, March 4, 2010 at 12:17am
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