00:01
Okay so what we have is 35 .6 ml of 0 .255 molar hydrogen fluoride solution and 0 .2 molar sodium hydroxide.
00:23
So the first question is about equivalence.
00:28
Equivalence is when we have similar number of moles.
00:32
So therefore we're looking at moles of hydrogen ions equalling to moles of hydroxyl ion.
00:43
Okay that means that molarity multiplied by the volume equals to molarity multiplied by the volume of the ions respectively.
00:52
Alright so in other words we're trying to say that 0 .255 multiplied by 35 .6 must give you 0 .2 multiplied by volume.
01:09
Okay you can choose to convert that into liters but let's leave it.
01:15
Alright so let's convert into liters here okay.
01:21
So therefore volume of hydroxyl ions in this case will be equal to 45 .39 ml.
01:36
Okay so that is the equivalence we're looking at.
01:42
Then the other bit is asking about the moles of the acid neutralized.
01:49
So therefore we know that the pka of hydrogen fluoride is normally 3 .14.
01:59
So therefore 8 .90 ml of the base added will give us, i mean will give us moles you know given the molarity and the volume.
02:10
So therefore we're looking at 0 .2 multiplied by 8 .9 and we'll have 1 .78 mmol.
02:22
Now initial moles of the acid normally was, so this is initial moles was to be 35 .6 multiplied by 0 .255 and this was to give you 9 .078 mmol.
02:45
Now moles of the acid neutralized will be equal to the moles of the acid added.
02:52
And then therefore the moles of the conjugate base will be equal to the mole of the acid neutralized which is basically 1 .78.
03:03
Okay so that's one way to look at it.
03:09
And so proceeding the remaining moles of the acid can be the difference.
03:22
So you can say remaining moles here will be 9 .078 minus 1 .78 and we're looking at 7 .298 mmol.
03:35
Now according to the henderson -hartzberg equation ph will be equal to pka plus log of, that will be the log of f minus divided by the log of hf that is hydrogen fluid.
03:57
So in other words we're looking at ph being equal to pka 3 .14 plus log of 1 .78 divided by 7 .298.
04:14
And this is 2 .53 as the ph.
04:19
Now moving on to the third bit at halfway through equivalence that means that the volume of the base will be, the volume of the base that is 45 .39 divided by 2.
04:45
So we're looking at 22 .695 ml.
04:51
So therefore moles of the conjugate base will be equal to 22 .695 minus, or let me say multiply because we're looking at the moles, multiplied by 0 .2 molar.
05:10
So this is 4 .539 mmol.
05:17
Now from here it's easier to proceed because the remaining moles of the acid therefore will be equal to the initial moles we calculated as 9 .078 minus 4 .539.
05:35
And this is going to be 4 .539 mmol...