00:01
Hi, so for this question we're going to have benzoic acid that i'm going to label as hb acid and we have benzoate that i'm going to label as b-.
00:15
So the first thing that we need to do is calculate the amount of mole that we have of each of them.
00:22
Why is that? if we look at our henderson -hasselbalch equation you're going to see that we have a factor that looks like this but the molarity is nothing but the amount of mole divided by the total volume and since we are dealing with a solution the total volume it is going to be the same on each of them.
00:42
So we can replace this henderson -hasselbalch equation in that factor as mb minus divided by n hb.
00:53
Let me take this guy out and write it down as an n.
00:59
Okay so now the amount that i have added of hb it is going to be equal to the volume times the molarity of my solution and the volume needs to be in recall that molarity is n divided by volume so n is volume times molarity.
01:19
So 0 .125 liters times 0 .75 molar and the amount of mole that i'm going to have of b - it is going to be 0 .2 liters times a 0 .5 molar solution.
01:41
This is going to give us 0 .1 mole and 0 .09375 mole.
01:56
Great! so we have these two amount of mole.
02:00
Next up they add 15 milliliters of a 1 molar solution of hcl.
02:10
Remember that hcl in water is going to produce h2o plus plus cl minus.
02:19
So the amount of mole of h2o plus that i'm going to have is equal to the volume again in this case 15 times 10 to the minus 3 liters times my concentration so 1 molar.
02:34
So i have 1 .5 times 10 to the minus 2 mole of h2o plus.
02:43
With this information i can set up a nice table for the reaction of the acid with the base that i have present in solution.
02:55
This is going to look like b - plus h2o plus it is going to generate hb plus water my benzoic acid...