00:01
We have 150 milliliter buffer, and it's 1 .1 molar.
00:05
Its page is 8 .3, and its p .a is 8 .3.
00:09
And we add 3 milliliters of one molar and a new h to it.
00:12
And we want to find the ph after this edition.
00:14
And so one really helpful thing here is our ph is equal to our pk.
00:20
And what that means for buffers is our weak acid form is equal to our weak base form concentrations.
00:31
So when this is 0 .1 molar at this ph, what this is really is 0 .05 molar of its acid form and 0 .05 mole of its base form.
00:41
And together, that is where we get our 0 .1 molar.
00:46
So when we add an oh, it's only going to react with our h .a.
00:51
It's not going to affect, well, it's not going to react with a -mise.
00:54
It's going to react with h .a.
00:55
So the actual concentration of reacting with is going to be this lower 0 .05 and not this higher point.
01:01
And so the reaction is h -a plus oh minus from that n -a is just a spectator ion.
01:12
And that's going to make a -minus in water.
01:14
And we don't really care about the water here.
01:16
And so if we can find the concentrations of both of these, and then we can just kind of find out what the final concentration is over the species is from there.
01:26
And so we can either convert everything to moles, and then find everything in moles, and then convert back into concentration, or we can just give everything in concentration.
01:34
Keep everything in concentration.
01:35
But if you want to do moles, you can totally do everything in moles as well.
01:40
When doing things in concentrations, you do have to account for dilution.
01:44
So we're going from 150 millimeters to 153 millimeters, right? we're diluting by 3 millimeters.
01:50
And so we have to adjust the concentrations.
01:52
And so when you do dilution, you can use this dilution formula here.
01:56
With concentration of 1 times following 1 equals concentrated 2 times volume 2.
02:00
And we know what the volume 2 is...