00:01
Consider that we have 25 milliliter solution of 0 .10 .10 .molar acetic acid.
00:05
And this is titrated with 0 .200 molar potassium hydroxide.
00:09
And we're asked to calculate the ph after the following additions.
00:13
Well, first we're getting our initial ph.
00:15
So i'm assuming i've got 1 .8 times 10 to the minus 5.
00:21
Let me go verify that.
00:32
My k .a.
00:35
For acetic acid is 1 .8 times 10 to the minus 5.
00:39
The minus five.
00:41
And my dissociation is going to be, i'm going to call my acetic acid just for brevity.
00:49
I'm going to call it hac, and i'm going to cheat a little bit and not put my hydronium ion in.
00:59
So for my first one, i'm not adding anything.
01:06
So if we look at our ice, which will be approximately equal to 0 .10, according to 5 % rule.
01:25
And my expression will be, and that's my acetate ion.
01:37
Again, i just did that for brevity.
01:42
So my k -a, i'll have 1 .8 times 10 to the minus 5 will equal x squared over 0 .10 .0.
01:54
So x will equal, let me stick this in my calculator, second square root, 1 .8 times 10 to the negative 5 times 0 .1, enter, and i'll get my x is 0 .00 -10 .1 .1 .1 .0 .0 .1 .0 .1 .1.
02:11
Three, four, two.
02:14
My ph will equal the negative logarithm of that number.
02:23
So my ph, my initial ph, negative log, second answer will be 2 .87.
02:34
That's my answer for a.
02:38
B will be five.
02:45
Let's see what i think things look like, 5 .0 milliliters of my 0 .0 .0.
02:56
0 .20 molar.
03:01
It was a k -o -h -k -o -h.
03:04
Okay.
03:08
That one will be too bad here.
03:10
So my, how much did i have in my original? 25 milliliters.
03:16
For this one, my new volume total will now be 20, 25 plus 5.
03:30
This will equal 30 millimeters total.
03:40
And i have everything else.
03:43
Okay.
03:46
This is also going to be, just in case i use it, 0 .0 330 liters.
03:54
And we're going to figure out our moles of acid, and that's going to be 0 .0 -250 liters times 0 .1010 moles per liter will equal 0 .00250 moles per liter will equal 0 .00250 moles per liter will equal 0 .00250 moles per liter will equal.
04:17
And my moles of base will be equal to 0 .050 liters times 0 .200 moles per liter.
04:38
And that will equal 0 .0010 moles a base.
04:48
Once i've got that information, we can see that we have excess acid, and we know how much we have 0 .00 -250 -0 minus 0 .00 -10 moles will give me 0 .00 -150 moles of acid.
05:23
And then if i'm going to divide that, my concentration of acid will be 0 .00 -150 moles divided by 0 .030 -0 -30 .000 -3 -0 -130 1 -2 -3 -1 -1 -3.
05:45
Yep, liters, and that equals 0 .05 -0 molar.
05:54
Okay.
05:59
And then that's my concentration of my acid.
06:04
My concentration of my acetate is going to be 0 .0010 moles per 0 .030310 liters.
06:26
And that will be 0 .0333 liters.
06:31
Now let's use our henderson.
06:34
Hessel.
06:34
Which i'll write down once.
06:43
And this will be my base concentration or my acid concentration.
06:50
And my ph is going to equal my pca, which i believe is 4 .7.
06:57
Oh, let me see.
07:00
Negative log 1 .8 times 10 to the minus 5, 4 .744, plus the log of 0 .0333 molar divided by 0 .0503 .000.
07:23
Molar and my ph will equal 4 .57.
07:30
Feel free to check my math.
07:33
My second edition, good heaven i've got five.
07:38
So next we have 10 milliliters.
07:50
And then we're going to go ahead and take that time 0 .200 moles per liter.
07:59
And i believe that'll be two.
08:16
Now i've got my new.
08:22
No, i'll need that.
08:23
This will give me my moles.
08:24
And this will be 0 .0020 moles of base were added.
08:38
And i was going to have what i calculated in my last problem.
08:43
I've got 0 .00 -250 moles of acid minus 0 .0020 moles of base.
08:58
Give me 0 .005 moles of acid.
09:13
Then where am i here? so i'm going to take 0 .005 moles divided by 0 .0035.
09:33
So i had 0 .25 liters.
09:42
So i had 0 .25 liters...