00:01
To calculate the ph of the buffer solution, we'll use the henderson -hassel -baltz equation, where ph is going to be equal to p -k -a, that's provided at 3 .86, plus the log of, it could be the moles of base over moles acid, or concentration base over concentration acid.
00:24
It's often easier to just use moles, so that's what i will use.
00:29
The base in this case is the lactate, and we have 0 .050 moles of sodium lactate.
00:40
We divide that by the moles of the acid, 0 .010 moles of lactic acid, and we get a ph of 4 .59 or 4 .56.
01:01
Then if we're to add 5 milliliters of 0 .5 molar hcl to the buffer, a liter of the buffer, what's going to happen? well, the lactate, which i'll represent as a minus, is going to react with the hcl and make lactic acid, represented as h .a.
01:28
And chloride.
01:31
Every mole of hcl we add consumes a mole of lactate and makes a mole of lactic acid.
01:37
So if we have the full liter, then we have the full amounts of moles.
01:42
We go back to the henderson -hasselbalch equation, where ph will be equal to pca 3 .86 plus the log of the moles of lactate that will remain.
01:59
After the addition of 5 milliliters of 0 .5 molar hcl.
02:05
Well, we will start with 0 .050 moles.
02:09
We then need to subtract off the moles that are consumed.
02:13
Every mole of hcl we add consumes a mole of a minus.
02:18
So how many moles of hcl did we add? we added 5 milliliters, which is 0 .005 liters, at a concentration of 0 .5 moles.
02:29
Per liter.
02:32
We then divide that by the moles of lactic acid that will be present.
02:38
It'll be the moles that we started with plus the moles of hcl added.
02:45
Again, every mole of hcl that we add creates a mole, an additional mole of lactic acid...