00:01
Okay, so for this question we're told to determine the ph during the titration of 23 .1 ml of a 0 .415 molar nitrous acid that has a ka of 4 .5 x 10ā»ā“ by 0 .357 molar sodium hydroxide at the following points.
00:23
So a is based on before any sodium hydroxide was added.
00:29
So that means that we're looking at nitrous acid which is a weak acid.
00:33
We're putting that into equilibrium.
00:45
Okay, so now we have initially then a 0 .415 molar solution.
00:52
So that means initially we're 0 and 0.
00:55
We're going to go minus x plus x plus x.
00:59
0 .415 molar is a pretty decent concentration.
01:04
Our magnitude is 10ā»ā“.
01:09
So we're going to go ahead and use the approximation of 0 .415 assuming that x is negligible compared to 0 .415 molar.
01:21
And then we're going to write our ka expression which would be the concentration of h3o plus times the concentration of no2 minus divided by the concentration of hno2 which would give us x times x divided by 0 .415 equals 4 .5 x 10ā»ā“.
01:46
So the x value then is going to be equal to 4 .5 x 10ā»ā“ x 0 .415.
02:01
And then we're going to take the square root of that value and x is equal to 0 .0130137.
02:13
I'm just comparing that to 0 .415 and we do get a 5 percent so we're at a 3 percent.
02:24
So we're okay with the approximation.
02:27
So that would then equal the concentration of h3o plus.
02:33
And now the ph of that solution is going to be the negative log of that concentration.
02:42
So our ph before anything is added is 1 .482.
02:56
Okay so now going back to the question, the second part says after the addition of 6 .5 milliliters of naoh.
03:08
So let me rewrite that equation.
03:11
So we have hno2 plus oh minus to give us h2o plus no2 minus.
03:25
So we started with 23 .1 milliliters of a 0 .415 molar solution.
03:37
And now we're going to be adding 6 .50 milliliters of the 0 .357 molar solution.
03:45
So now what's key here is that we're going to have to break those down into moles.
03:50
So we're going to look at the hno2.
03:54
So we're going to take 0 .0231 liters times 0 .415 molar.
04:04
And that's going to give us a total of 0 .0231 times 0 .415 gives us 0 .00959 moles.
04:19
And then for the base we're going to look at 0 .00650 liters times 0 .357 molar gives me 0 .0065 times 0 .357 gives me 0 .00232 moles of the hydroxide.
04:46
So that's what we have initially.
04:49
So i'm going to put zero here.
04:51
So that would be our initial value.
04:53
And if you notice the hydroxide is limiting here.
04:57
So we're going to in terms of what's being consumed, we're going to consume 0 .00232 moles.
05:06
0 .00232 moles.
05:10
But then we're going to be making 0 .00232 moles of the nitrite.
05:19
So that at the end we look at our number of moles is going to be the 0 .00959 minus 0 .00232.
05:33
Gives me 0 .00727 moles.
05:39
0 and 0 .00232 moles.
05:44
So now what we have to do is we have to find the concentration of those two ion, of those two substances, the hno2, which is the acid and the no2 minus, which is the conjugate base.
05:57
So we had 23 .1 milliliters and we added to it 6 .5 milliliters.
06:06
So our total volume then is 29 .6 milliliters.
06:12
So in liters that's going to be 0 .0296 liters.
06:20
So that's going to give me a concentration then of 0 .246 molar.
06:32
And for the nitrite ion it's going to be 0 .0784 molar.
06:47
So now what we're going to do is we're going to use the henderson -hasselbalch equation because what i have present in the solution is the acid and its conjugate base.
06:57
So we're going to use the fact that the ph is going to be equal to the pka plus the log of the concentration of the nitrite ion, which is the base, divided by the concentration of the nitrous acid.
07:21
So we're going to put on the side here what the pka is because we're going to use that.
07:26
So the pka is the negative log of the ka and the ka was 4 .5 times 10 to the negative fourth.
07:38
So the pka is a negative log of 4 .5 times 10 to the negative fourth.
07:47
So the pka is equal to 3 .347.
08:00
So now the ph then is going to be equal to 3 .347 plus the log of 0 .0784 divided by 0 .246.
08:15
So the ph at this point is going to be 0 .0784 divided by 0 .246, 2 .850.
08:31
So that was the first point, well, the second part of that question.
08:37
Part c is asks for at the half -equivalence point.
08:42
Now i'm just going to go through the...actually, i'm going to rewrite it so i can just kind of show you what happens.
08:50
So if we look at hno2 plus oh minus gives us h2o plus no2 minus.
09:01
So i'm going to just change the concentrations for a second because it really doesn't matter.
09:06
So what happens at the half -equivalence point? so if this was 1 molar, then what happens at the half -equivalence point, that means that i'm going to consume half of what's present.
09:19
So that means that it would go down to 0 .5 molar.
09:22
Well, if this goes down by 0 .5, this has to go up by 0 .5 molar.
09:28
So what happens in the henderson -hasselbalch equation is the ph is equal to the pka plus the log of 0 .5 over 0 .5.
09:40
So that becomes the log of 1, which is 0.
09:44
So what happens is, is no matter what the concentrations are at the half -equivalence point, the ph is equal to the pka...