00:01
A buffer is a solution formed from the combination of a weak acid or a weak base and its salt as one that contains both acid and base.
00:10
Then the buffer can resist change in its ph in addition with a small amount of acid or base.
00:18
So for the combination of a weak acid and its salt, the henderson -hasselback equation is used to solve for the ph.
00:25
This is of the form ph equals pca plus log of the most.
00:30
Of the salt over the moles of the acid so for this problem we're given with an acetic acid sodium acetate buffer system which concentrations of which has a volume of one liter and the concentrations are 1 .8 molar of acetic acid so h -o -ac and 1 .2 molar sodium acid acid so n -a -o -a -c now we want to find the most of n .a .o .h that must be added to this buffer system so that the ph equals 5 .22.
01:15
Assume that the volume does not change in addition of the most of the naoh that we will find.
01:24
So find most of naoh.
01:31
So let us note that when base is added to the buffer, it will react with the acid component of the buffer.
01:37
So that n -a -o -h will react with acetic acid to form sodium acetate and water.
01:49
So that as a result, the most of the acid will decrease by the most of the base that's added, whereas the most of the salt will increase by the same most of the base that's added.
02:02
So initially, the most of the acid and the salt are as follows.
02:09
So we're just going to multiply their concentration so for ethytic acid that's 1 .8 times the volume of the buffer solution which is 1 so that's 1 .8 for the for the salt you have 1 .2 times 1 that's 1 .2 so say we have x most of n aoha h added so the most of say again the most of the acid decreases by the most of the most of the base so we have 1 .8 minus x while the moles of the salt increases by the moles of the base added so that is 1 .2 plus x so we also take note that the acid dissociation constant k a for acetic acid at standard condition is 1 .8 times 10 to the negative 5 so we can solve for pk a as negative log of the k a and that gives as 4 .74.
03:16
So using the henderson heslbuck equation you have above, your ph is 5 .22 as desired...