00:01
If you want to speed through it, i would not be offended.
00:03
We are asked to find the ph after each addition when we start with 10 milliliters of 0 .250 molar acetic acid.
00:11
Okay, so my first edition is before we add anything.
00:16
And this will be, i'm going to write an ice table right here.
00:28
I said i was going to abbreviate those for brevity.
00:40
Okay, and our initial change in equilibrium.
00:44
Here we have zero.
00:45
0 .10 minus x plus x and plus x x x x and 0 .10 minus x, which is approximately equivalent to 0 .10.
01:02
And my k a expression will be equal to h plus concentration times my acetate ion concentration.
01:13
I see the gaseid concentration.
01:16
So my ka is 1 .8 times 10 to the minus 5 equal x squared over 0 .100.
01:28
So x will equal second square root.
01:32
Whoops see.
01:34
Second square root 1 .8 times 10 minus 5 times point 1 .1.
01:41
This will equal 0 .0134 molar h plus.
01:47
My ph will be equal to the negative log, 0 .0134.
01:58
Did i get that right? 0134? my bad.
02:03
I was thinking that's going to be very high.
02:06
013 negative log second answer.
02:10
And i get my ph for scenario number one would be 2 .87.
02:22
Next.
02:24
My next addition is b.
02:34
We are adding 12.
02:36
0 .5 milliliters of base.
02:41
So let's figure out how many milliliters of each substance we have, or excuse me, our moles.
02:47
We have moles of our hac, and that'll equal 10 .0 millimeters times 0 .250 moles per 1 ,000 millimeters.
03:09
And that will equal 0 .00250 moles of hac.
03:23
That's my initial amount.
03:26
Then for my moles of my base, i have 12 .5 milliliters, and my concentration was 0 .1010 moles of k -o -h per 1 ,000 milliliters.
03:47
And this will equal 0 .001 .25 moles of base.
03:58
Okay.
04:00
So now let's get started on this one.
04:02
So we've got our ice table again here.
04:13
It's 10 to the minus 3.
04:15
10 to the minus 3.
04:17
Okay.
04:19
So we'll have our acid and our h plus.
04:28
I like over acid plus our oh minus.
04:35
And that will yield our.
04:38
H plus and our ac minus.
04:47
Okay, so we have, we're starting with 0 .00250 moles and 0 .00.
05:02
Then this is going to decrease by 0 .00125125.
05:09
This will decrease by 0 .00125.
05:16
And this will increase, that'll be equal to 0 .00125, 0 .00125.
05:48
And we can take our k -a expression, will be 1 .8 times 10 to the minus 5 equals 0 .00125 squared divided by 0 .00125.
06:20
I don't know how i feel about this.
06:27
So this would be one point.
06:30
I just feel like i'm doing this one wrong.
06:32
Let me think about this a minute.
06:38
That'll be to zero.
06:44
This shouldn't be here.
06:48
So i'm going to get my ac.
06:52
This should be h plus.
06:59
I don't have this over here.
07:00
This should be h2.
07:05
My bad.
07:19
And then i bet i can use henderson heselbach here instead of this.
07:29
So i'm going to use my ph.
07:31
Equals my p -k -a plus the log of my base concentration over my acid concentration.
07:41
And i can see i'm at the half -equivalence point here.
07:45
So my ph should be equal to my p -k -a.
08:06
Okay.
08:11
Okay.
08:11
And then i have the negative log of 1 .8 times 10 to the minus 5, it leaves 4 .74.
08:18
This is the half equivalence point where ph equals the pca and i don't need to worry about my volumes at this point.
08:43
Okay, my next one is c is 24 .5 mill liters of base.
08:54
And now that was 0 .10.
08:59
That'll equal 0 .00245.
09:05
Molds the base and let's continue with our acid plus our base and that'll give me my acetate plus h2o which i don't care about and my base was 0 .00250 for my acid so this will be 0 .00245 so this is going to go down 0 .00245 down 0 .0245 and up 0245.
09:58
So my equilibrium concentrations will be 0 .000...