00:01
If we add hcl, which i'll represent as hydronium ion, to a solution that contains sodium acetate, which i'll represent as just a minus, it will create acetic acid, represented as ha and water, in a one -to -one stoichiometric relationship.
00:21
So to determine the ph of the buffer that is created after the addition of hcl, we use the henderson -hasselbalch equation, where ph is going to be equal to pka for acetic acid, which they gave us at 4 .76, and then we add to that the log of, we can use the molarity of acetate over the molarity of acetic acid, or moles acetate over moles acid.
00:53
It's easier to use moles rather than molarity, and the calculation is equivalent.
01:00
So the moles of acetate will be the moles we started with, 2 .5 milliliters, which is 0 .0025 liters, at a concentration of 2 .00 moles per liter.
01:15
We'll then subtract off the moles of hcl added, because every mole of hcl we add consumes a mole of acetate.
01:25
So we're adding 3 .3 milliliters, which is 0 .0033 liters, at a concentration of 0 .500 moles per liter...