00:01
For this problem, we are asked to find the ph of various solutions.
00:06
The k -a and kb values are given in the text.
00:13
In an appendix, d, if you want to look them up yourself.
00:20
And our first solution is 0 .095 molar propanoic acid, which is c2h5.
00:42
Okay, i'm just going to do each one separately.
00:47
So first, maybe i don't have the right problem here.
01:02
See which one i'm supposed to be doing.
01:04
62.
01:08
62.
01:11
I read the wrong substance.
01:18
0 .095.
01:19
That was right.
01:20
Molar hypochloric acid, which is h -c -l -o.
01:39
Okay.
01:41
And this is often written as h -o -c -l.
01:46
However you want to write it.
01:47
I'm going to write h.
02:04
And i'm given an initial concentration of 0 .095.
02:10
This is going to go down.
02:12
By this going to increase by x up by x.
02:27
Okay, so then we can put this into our equation.
02:31
Our ka, which we can look up, is 3 .0 times 10 to the minus eighth.
02:39
That will equal h plus times ocl minus over h .ocl.
02:54
And that's going to equal x squared over 0 .095 minus x.
03:03
We can see from our value here that we can ignore the part that i'm about to highlight.
03:11
We can ignore that part.
03:18
So we'll be able to solve this as follows.
03:39
And this is my h -plus concentration.
03:42
So if we take the negative log of 5 .3 times 10 to the minus 5, we get 4 .27...