# chem 2

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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=1.8x10-5

• chem 2 -

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
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.

• chem 2 -

4.57

• chem 2 -

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