00:01
Very long calculation coming up.
00:03
I'm just going to get started.
00:04
We have three to do.
00:13
First equation is pretty easy.
00:17
I've got it written down here.
00:18
I just got to get to it.
00:22
Sorry about this.
00:23
I should have been a little bit more prepared.
00:30
We know that nacn dissociates into na plus and cn minus.
00:39
We'll be doing similar calculations for all three of these.
00:45
And we know that cn, this is the equation that we're starting, going to be starting, with the, i guess, matters, if you will.
01:15
And i will be using this.
01:16
I'm going to go ahead and do that right now.
01:23
Our initial concentration is 0 .10 and minus x.
01:27
So 0 .10 minus x.
01:31
0 on 0.
01:38
We can find the ka for hcn.
01:43
So the ka for this reaction, if i can find it here, is 4 .9 times 10 to the minus 10th.
02:04
Just find my kb, i'm going to take 1 times 10 to the minus 14th, divided by 4 .9 times 10 to the minus 10th, which is 2 .04 times 10 to the minus 5th.
02:22
Now we can set up our kb expression using this and this.
02:30
Our kb expression will be these two concentrations divided by this concentration.
02:46
So that will be 2 .04 times 10 to the minus fifth equals x squared over 0 .10 minus x.
03:02
We can go ahead and solve for this.
03:05
Um, x is negligible.
03:09
So solving for x here, x is going to equal the square root.
03:13
Of and that's going to be 1 .43 times 10 to the minus third molar and that's the oh h concentration for this one.
03:37
Now lastly we have to find our ph and that will be calculated as follows.
03:46
Negative log of 1 .43 times 10 to the minus 3 equals 11 .15 so there are both my answers for part a.
04:07
Part b.
04:14
We have 0 .080 molar sodium carbonate.
04:25
And it's one there.
04:26
Carbonate will dissociate as follows.
04:40
That's what that's going to do.
04:47
And then my next equation will be.
04:56
Make sure i write this down all correctly.
05:27
And by initial concentration 0 .080.
05:33
0 and 0, i believe i've got that correct here...