00:01
Okay, so we're looking for the ph of a solution when a, we have ch3cooh with a concentration of 0 .1 molar and 20 ml of it being titrated with sodium hydroxide and 0 .100 molar of that and 20 ml.
00:24
Okay, so really i guess this one would give us sodium acetate, c2h3o2, plus water.
00:35
I guess that's the balanced chemical equation.
00:38
Now underneath that equation, write down what you're given.
00:41
So the 0 .1 molar of acetic acid and the 20 ml and the 0 .1 molar of sodium hydroxide and the 20 ml.
00:49
What you want to do now is multiply these quantities together because remember molarity is moles per volume.
00:55
So i will get an answer in millimoles if i don't convert the milliliters to liters.
01:00
So doing that, i get 2 millimoles of this guy and when i go over here on the right, i get 2 millimoles on that side.
01:20
So what you do now is you subtract the smallest mole.
01:23
So 2 millimoles on the left and we have 2 millimoles on the right.
01:30
So both of these are 0.
01:36
So i guess that means that if you're subtracting 2 millimoles over here, you better add it over here.
01:41
So that's plus 2 millimoles of sodium acetate right there.
01:46
And then i guess we better rewrite our equation.
01:49
So we get nac2h3o2 and we have 2 millimoles of that over the total volume of our solution which is 40 ml.
02:02
And that's going to go to naoh plus c2h3o2.
02:17
And then maybe there's a, you know, technically i guess it goes to that.
02:21
Okay.
02:23
So this i guess would be our x and this would be our x.
02:26
Now we need to write down what we call the kb expression.
02:31
So that's going to be products over reactants.
02:33
So in this case, it's x squared over 2 over 40.
02:37
And again, why is this kb? because you're creating base.
02:43
But in the problem, we were given a ka.
02:46
Ka is 1 .76 times 10 to the negative 5th.
02:50
You have to remember that kw, the water concentration, 1 times 10 to the negative 14 equals ka times kb.
02:59
So we're going to set this equal to 1 times 10 to the negative 14 over 1 .76 times 10 to the negative 5th.
03:09
Okay.
03:10
I'm now going to solve for x.
03:13
So here we go.
03:28
I get x to be about 5 .33 times 10 to the negative 6.
03:35
Because this is a kb and this is a base problem, i can find the poh by the negative log of x.
03:46
And that gives me about 5 .27.
03:50
But the question wanted ph.
03:51
And i know that ph plus poh equals 14...