chem 2
posted by tina .
A buffer is prepared by adding 300.0mL of 2.0M NaOH to 500.0mL of 2.0M CH3COOH. What is the pH of this buffer? ka=1.8x105

This one is a little different than your posts under umpteen other screen names. Here you are MAKING the buffer by neutralizing an excess of the acid.
First the mols are:
millimols NasOH = mL x M = 300 x 2.0 = 600
mmols CH3COOH = 500 x 2.0M = 1000
Now make an ICE chart.
.......CH3COOH + NaOH ==> CH3COONa + H2O
initial..1000......0........0........0
add..............600................
change...600....600.....+600.....+600
equil....400......0........600......600
Now substitute the millimoles in the ICE chart (the equilibrium line) into the HH equation and solve for pH.
Note: Technically, base/acid in the HH equation goes in in molarity (millimoles/mL)and one should convert 400 mmoles CH3COOH and 600 mmoles CH3COONa to M by 400/800 and 600/800 respectively; however, since the volume of 800 mL is the same for both, the volume cancels, and one can take a short cut and use mmoles alone. Chemically it isn't exactly right but mathematically it is.
Post your work if you get stuck. 
4.57

Amanda is wrong. It is 4.92.
Do the ICE tablee (kinda like DrBob's) then plug it into pH=pKa+log(600/400)
pka=log(ka)=4.74472
pH=4.92