00:01
First, we recognize that the hydronium ion of hcl is going to react with the acetate to produce acetic acid and water.
00:13
As long as we don't add more moles of hcl than we have of acetic acid, we'll still have, sorry, than we have of acetate, we'll still have acetate.
00:24
We would have also made acetic acid.
00:26
Having a weak acid in its conjugate base gives us a buffer solution, and we can use the henderson -hasselbalch equation.
00:35
The henderson -hasselbalch equation is the desired ph, 4 .50, will be equal to pka of the acid.
00:44
That'll be the negative log of the ka value for acetic acid.
00:49
They didn't give that to us, so we'll have to look it up.
00:52
1 .8 times 10 to the negative 5.
00:56
Then the rest of the henderson -hasselbalch equation is plus the log of the moles of acetate, still left in solution, divided by the moles of acetic acid formed.
01:09
We could also do a ratio of molarities, but it's going to be easier in this case and equivalent to just do a ratio of moles.
01:18
So the moles of acetate will be the moles we started with, 500 milliliters, which is 0 .500 liters, multiplied by its molarity.
01:30
50 millimolar is 0 .050 molar, or moles per liter...