00:01
Okay, we're told that we have a student who has 100 milliliters of 0 .50 molar dimethylamine.
00:08
We're given the kb of dimethylamine as 1 .3 times 10 to the minus third.
00:13
We're asked to find the mass of the bromine salt for dimethylamine that needs to be dissolved for the buffer to give us a ph of 10 .67.
00:26
And we're asked to round to two sig figs and assume no volume change.
00:33
Okay, let's begin.
00:35
So to start here, i'm going to start by doing getting a ka and that'll equal my k or 1 times 10 to the minus 14th, 3 times 10 to the minus 3.
00:48
Let me see what this equals.
00:51
1 times 10 to the minus 14th minus 1 .3 times 10 to the minus third.
00:57
I get, oh, i subtracted.
01:03
My bad.
01:04
That didn't make sense.
01:05
1 times 10 to the minus 14th divide 1 .3 times 10 to the minus 3.
01:11
And that's 7 .69.
01:16
And i'll go 2 times 10 to the minus 12th.
01:20
Now we're going to use henderson -hasselbalch to get our concentrations and this will be our salt and our dimethylamine.
01:45
Okay, so let's go ahead and set this up.
01:48
My ph was given as 10 .67 and my pka will equal the negative log of 7 .69 equals, let me do this.
02:12
This will be fast because it's negative log second answer.
02:15
It's 11 .114.
02:26
My dimethylamine was 0 .50 molar, 0 .50 molar over, okay, so let's do this.
03:04
And this will be negative 0 .444 equals the log of 0 .5 over x.
03:33
So 10 to the minus 0 .444 equals 0 .5 over x.
03:47
Am i doing this right? i want to make sure i got my x in the right spot here.
03:56
So this will be 0 .359749.
04:02
X will equal 0 .5 divided by second answer equals...