00:01
This question is asking us about titration of phosphoric acid, which has three protons on it.
00:07
The question says it's diprotic.
00:08
It's actually tripodic.
00:09
You can see three right there.
00:11
But there's only two that appear in the equation or the question.
00:19
So we only need to worry about two.
00:20
And so that's why it's okay to say it's diprotic.
00:22
At least it's diproduct in the cases we care about.
00:25
But we're told the pca of the first proton loss is 1 .3.
00:29
The pca of the second proton loss, is 6 .7.
00:32
So each p .k is telling us, or classifying essentially how to the, when these different steps happen, b.
00:48
04, this is minus, and this is 2 minus.
00:52
So when we shoot off in h plus, and when we shoot off our second h pluses, these happen at pk1 and pka2.
01:01
And so that's what they're really telling us is, is when these start, these protons start to fall off.
01:06
We're told we have 1 .5 molar h3 opium the of our phosphoric acid as well as 1 .5 molar k -oh that we're titrating this reaction with and the titration curve is going to look something like this and so if you were to plot ph over these additions against the volume of k -o -h added your curve will look something like this this should be increasing slightly.
01:43
I'm not the best drawer, so i'll make this slightly better.
01:50
So basically, it's going to stare step, where you have an initial increase, it's going to hit what's called a buffer region, hit c sharp increase, hit a second buffer region, and then c sharp increase.
02:10
And since there is a third proton, that would continue one more step.
02:12
But again, that's beyond the scale of this question.
02:15
So these buff regions happen when you have similar concentrations or relatively similar concentrations of the acid form and the conjugate base form at the relevant species.
02:34
So when we have 50 % of our compound is hzopo4 and 50 % is h2po4, we're at, buffer 1.
02:51
Conversely, this looks like 58%, that should be 50.
02:54
Conversely, if this was 50%, and this was 50%, we'd be at buffer 2.
03:08
So if buffer regions happen when you have appraisal amounts of each of these, when you're at 100%, let's say i'm right to have colors here, 100 % h3p04, you're down here.
03:23
If you're out, oh, denote that's one.
03:27
If you're at 100 % h2po4, so we've converted everything to h2po4, there's nothing else, that's when you're in this region.
03:35
And when you're around 100 % hpo42 minus, that's when you're in this region here.
03:43
Nope.
03:45
And so this is where each of these regions, the chemistry happening behind each of the regions as you titrate with k -o -h.
03:56
Remember it's k -o -h is what's pulling off these protons.
03:59
These protons are joining oh to make water and neutralizing the acid as you continue to add.
04:06
And so to find the ph at each of these, we need to do five calculations, but two of the calculations you don't are pretty straightforward.
04:17
So for point, what's volume zero, volume 0, 25, 50, and so on.
04:25
So for volume 0, we've added 0 k -oh.
04:27
The only thing in solution is going to be h3p -o -4.
04:33
So we need to find the k -a of the species that matters, which should be the first proton, or the first p -k, right? because that's the extent to which that disassociates the first one.
04:45
And so we need to take 10 to the negative p -k -a -1, or 10 to the negative 1 .3.
04:51
That is the inverse on the side note.
04:55
Here that is the inverse of negative log k a so we essentially need to solve undo this negative log to get to the k a buried in there and that's what this is doing here and so 10 negative 1 .3 is equal to 0 .0501 so this is k -a -1 and so what k -a -1 is describing extent that h3 -p -o -4 disassociates or interacts with water alone to make an acid.
05:26
And since that's all we have, solution h3 -p -o -4 plus water, this is the expression that's actually going to be driving our ph here.
05:34
And so the k -a expression is equal to the concentration of h -3 -p -o -4 times h -2 -p -o -4 minus and h -plus.
05:50
Well, we don't actually have any h -2 -p -o -4, we don't know what concentration h2p4 is, and at this point, because we haven't added any qh.
05:58
So we can do a mathematical trick, where if we're thinking about it, every time we make h2p -o -4, we're making one h -plus, these are always going to be the same number.
06:07
We can do a mathematical trick and simplify this down to h -plus squared over the concentration of h -3 -p -o -4.
06:14
These are always going to be the same number.
06:17
We can just go ahead and square that value and eliminate one of those variables.
06:21
And so now we know what h3po4 concentration is.
06:24
It's 1 .5 molar.
06:25
We know what the k is.
06:26
We just solved it here.
06:27
We can solve for h plus.
06:30
So if we take ka times h3po4 concentration and square root it, we'd be left with h plus concentration.
06:41
So let's go ahead do that.
06:42
I'll plug in the numbers.
06:44
We have 0 .0501 times our 1 .5 .5 .1 times our 1 .5 .4.
06:51
And we take the square root of that, that is equal to 0 .274 molar h plus.
07:04
So now we can take our ph equation, which is just negative log or h plus concentration, and plug our h plus concentration.
07:14
So ph is equal to negative log 0 .274 molar, and we find it is equal to 0 .561.
07:23
Or 6 -2.
07:26
So this is the ph at 0 milliliters k -o -h added.
07:39
And box that in.
07:40
This is going to be lots of numbers on the screen.
07:44
Okay.
07:48
So volume 25.
07:51
Now we're, i'm going to write the chemical reaction that's happening in here.
07:56
We added 25 milliliters of oh minus, and that is reacting to form h2, p -o -o -4.
08:06
Plus water.
08:08
We don't really care about water.
08:09
It doesn't affect our ph.
08:10
So i'm going to go ahead, ignore the water throughout the rest of the expression.
08:15
So h3 -04 reacts with oh minus, and we get h2po4 out of it.
08:19
And so we have this at 1 .5 molar.
08:27
We have these at 1 .5 molar each.
08:32
So we can do some quick calculation here or 1 .5 milliliters your 1 .5 molar times 50 milliliters over 75 milliliters so this is the amount of base or the amount of original had out 50 milliliters this is 50 plus the 25 we added on base and so what we're doing is finding the new concentration of h3 p o4 after we add 25 millimeters of koh .h.
09:08
And so you can see we delete it down to one molar.
09:16
We can do the same thing with k -o -h.
09:18
This is h3 p -04.
09:22
This is k -oh -h here.
09:25
And so that is our 25 millimeters of base added to the total volume of our solution, which is so component over total volume.
09:38
And so we just, we just find we have 0 .5 molar k -o -h.
09:46
Alright, and we start with 0 -h2 -p -o -4.
09:49
So this reaction progresses forward.
09:51
We see oh is limiting, right? it's one -to -one, and oh has fewer, or a smaller concentration, so minus 0 .05, plus 0 .5, 0 .5, and 0 .5, and 0 .5.
10:05
So here, i'll start it, because this is an important point, is we have equal concentrations of h2po4 and h3po4.
10:15
And so this right here is our buffer.
10:21
Anytime these are the only two concentrations of note within your solution left over, you have a buffer.
10:28
And so that point is right here in the buffer.
10:31
It's right in the middle of the buffer region.
10:33
And so when we're considering buffers, the equation to consider is the henderson -hoss.
10:41
Hendersen ossebock equation has one purpose and that's to describe the behavior or the ph of a buffer and so both of these species are contributing to ph and so we need this equation to consider them in unison and so we have the pk a we know the pkk for this is our 1 .3 1 .3 yeah there we go it's further up and we have the concentrations of our acid and our base in this case.
11:21
So this is acid and its conjugate base.
11:23
It's always an acid -based pair.
11:25
So we are not considering the oh concentration here.
11:27
We're considering h2 -p -o -4 here.
11:30
And so we have these concentrations and we know what the pga is.
11:33
So ph is equal to 1 .3 plus the log of 0 .5 over 0 .5 .5.
11:43
Well, 0 .50 .5 equals 1, log 1 is equal to 0.
11:49
So this whole component cancels out to 0.
11:54
And so what we have is ph is equal to pca1, so ph is equal to 1 .3.
12:03
And this is a really cool point.
12:06
This is one of my favorite points about titration chemistry, is if you have equal amounts of your acids, dense conjugate base your ph is always equal to your pka and so what we call this is the half equivalence point so this is the half equivalence point and so on graph that is right here so the half equivalence point is when you added half as much k -o -h as you do your acid or it also works in biversa if you're adding hcl to a base same thing happens but the half -equittance point is right in the middle of your buffer region and at that point ph is equal to one point, or ph is equal to your bka for that point.
13:00
And so this holds true for any half equivalence point and we'll see it again in two steps here.
13:09
So next is the volume at 50 millimeters.
13:12
So since this is a 1 .5 molar h3p -o4 and 1 .5 molar k -o -h, if we have 50 millilit of this, and we have 50 mill years of this, well, we've added the same amount of k -o -h as the moles of h3 -p -o -4 is equal to the moles of k -o -h we added, which means we've, based on this equation here, we've neutralized this completely, right? we have, this would be the same as putting one -to -one, and that becomes zero, and so we only have h -2 -p -o -4 left -in solution.
13:56
To do that mathematically, oh, i have run out of room.
14:03
To do that mathematically, we're going to, yeah, i'll carry it out.
14:11
H -3 -p -o -4 plus oh -minesis makes h -2p -o -4.
14:18
We've diluted each of these by half 1 .5, or initial volume divided by the total volume of 50 plus 50...