00:01
Long but sort of fun problem.
00:01
We're asked to find the ph, and there will be a total of eight calculations, for the titration of 20 millimeters of 0 .000 molar tri -methylamine.
00:11
And we're given the kb.
00:14
And we're titrating with 0 .1 -0 -00 molar hydrochloric acid.
00:21
Okay, so we're going to go ahead on this one and do, find the ph at zero.
00:57
So my kb is going to equal x squared over 0 .1 .000.
01:10
So x will equal the square root of the kb times 0 .1 .000.
01:22
X will equal 5 .2 times 10 to the minus 4 times 0 .10 .000.
01:31
X will equal, let me do this, second square root of 5 .2 times 10 .2.
01:44
And that's 0 .007211 molar.
01:52
And the p .o .h will be the negative log of 0 .007211, which will equal negative log, is 2 .142 all around later.
02:16
And my ph will be 14 minus 2 .14 is 11 .86.
02:30
And i'm going to go, i've got four significant figures here, four here, only two here.
02:36
So i'm going to go to two decimal places.
02:38
My ph will be 11 .86.
02:46
Okay.
02:48
Next, we're going to be at, and we're adding our strong base.
02:54
How much b is 10 .00 milliliters of 0 .10 .10 molar hcl added.
03:17
So here my ph will equal my p -k -a plus the log of n for my tri -ethylene and my hcl.
03:44
So this will be the negative log of kw divided by kb that'll be my pk a and that'll equal 10 .72 so my ph is going to equal 10 .72 plus the log of and i have 0 .02 liters of 0 .1 molar and i'm going to subtract from that my 0 .01 times my 0 .01 times point one.
04:44
I need to make a little bit more room for this so you know what i'm doing.
04:54
Okay, so my methylomey, this will be 0 .02 -0000 liters times 0 .1 0 .00 mules per liter.
05:12
We're going to subtract from that 0 .01 0 .00 liters times i'm just going to go 0 .01 1 .01 moles per liter.
05:23
I'm going to divide by that my moles of hcl.
05:34
This will equal 10 .72.
05:38
I guess i can just do this.
05:40
Ph will equal 10 .72.
05:45
Next, we're going to add a little bit more.
06:02
So what are we adding next? i've got 15 mils.
06:15
And this time it's going to be the same calculation.
06:17
What's going to change is this and this.
06:22
I should have put this down here different.
06:24
Hang on a moment.
06:47
And that will be moles per liter.
07:02
Next, we will have 10 .72 plus the log.
07:14
Again, we'll have 0 .02 liters.
07:18
I'm going to leave sig figs up.
07:19
Well, i better not.
07:41
Bowls per liter, divided by.
07:57
So my ph will equal, doing the map on this one, 10 .24.
08:02
C, d is 19.
08:33
And again, it's going to be the same thing.
09:09
There i got at that time, divided by 0 .0190 liters, holes per liter.
09:35
And here my ph will be 9 .44.
09:42
Then e is 19 .95 milliliters.
09:56
And i'm just going to substitute my 19 .95 into that same spot.
10:05
This time i'm just going to leave out my units for brevity.
10:29
995 times 0 .1 .000, divided by 0 .095 times 0 .195 times 0 .1905 times 0 .100.
10:42
So here my ph will be down to an 8 .12.
10:51
I'm guessing that my next one will be at my half equivalence point.
11:05
F is 20 milliliters, which will be my equivalence point.
11:21
10 should have been like my half equivalence point.
11:25
It's just been my pca.
11:28
Okay.
11:34
So i'm going to have volume total will equal 40 millimeters, which is 0 .0 4000000 liters.
12:04
All of my weak base is converted to the weak acid...