00:02
Buffer and the buffer is acetic acid and sodium acetate we're given the initial concentrations of the buffer and we're given the acid uh the ka at 1 .8 times 10 to the minus fifth for the first question we're asked to find the initial ph okay so the initial ph is going to be simple um let's write our equation here i'm going to write it because it's just a little little quicker for me.
00:49
Ch3cooh come to equilibrium with the acetate ion and my ka expression will be equal to my ch3coo times h plus over ch3cooh.
01:20
This will give me my x.
01:22
Go ahead and substitute here here, 1 .8 times 10 to the minus 5.
01:30
This will equal x squared.
01:43
What do i have here? i have 0 .560, 0 .250.
02:15
Actually, i don't want to do this one like this.
02:17
I'm going to do henderson here.
02:18
So i'm going to do my ph is equal to my pka plus the log of my conjugate base concentration over my acid concentration.
02:33
And the negative log of 1 .8 times 10 to the minus 5 equals negative log 1 .8 times 10 to the minus 5.
02:47
4 .744, 745.
02:52
So i have my ph 4 .745 plus the log of 0 .560, 0 .250, and my ph will be 4 .745 plus the log 0 .56 divided by 0 .25.
03:22
Close, enter.
03:24
And i get a ph of 5 .095.
03:31
5 .095.
03:32
Let me check what i'm given here with my ka.
03:41
I believe i can go to three sigs.
03:43
5 .010.
03:51
Okay, then we're asked to find the ph when 0 .0060 moles of hcl is added.
04:14
Okay, so my initial amount was of acid is 0 .300 liters times 0 .250 molar acid and that'll be 0 .25 times 0 .3 would be 0 .0750 moles of acid.
04:49
That was what was originally present...