00:01
Because we've added half of the amount of moles sodium hydroxide to the amount of benzoic acid, we have created a buffer solution where the benzoic acid, when reacting with the hydroxide from sodium hydroxide, produce the conjugate base to benzoic acid.
00:25
Maybe i'll do it this way.
00:27
We'll represent benzoic acid as ha, producing its conjugate base.
00:34
And water.
00:36
So every mole of hydroxide we add will decrease the benzoic acid by the same amount and increase its conjugate base by the same amount.
00:47
So to calculate the ph, that'll be equal to the pca of benzoic acid, which is the negative log of its ka value 6 .5 times 10 to the negative 5, plus the log of the moles of base over the moles of acid or the concentration base over the concentration acid.
01:09
It's going to be just as easy to use moles because then we don't need to worry about the volume.
01:15
So the moles of the base will be the moles of strong base added.
01:20
Every mole of strong base will make the same moles of weak conjugate base...
