00:01
So we know that the sum of the two volumes needs to equal 60 ml.
00:06
So we'll say x plus y equals 60, where x we can say is the volume of the acid, and y will be equal to the volume of the base.
00:24
If you're assigned a ph of 9 .15, then you're going to want to use the ammonium chloride and ammonium hydroxide buffer solution, because it has a, it's actually pka, they say it's pkb, but they're wrong, because it has a pka closest to your ph.
00:46
Then we can use the henderson -hasselbalch equation as a second equation for these two unknowns.
00:56
The ph we want will be equal to pka plus the log of the moles of the base over the moles of the acid.
01:10
A lot of times people will do a ratio of molarity here, but it's equivalent and easier to do a ratio of moles.
01:19
So what is the moles? well, moles can always be calculated by taking volume multiplied by concentration.
01:31
So we've got volume of the base, which is y, multiplied by the concentration, 0 .10 molar, and then we've got the moles of the acid, which is the volume of the acid, multiplied by its concentration, and then these are going to cancel.
01:55
So this will just be y over x.
01:59
The ph we want to achieve is 9 .15...