00:01
Polyprotic acids have multiple hydrogen atoms that they can lose, so they can act as buffers in multiple ph ranges.
00:10
Carbonic acid, or h2 -co -3, loses its first hydrogen to become h -c -o -3 -1, which is carbonic acid's conjugate base.
00:22
But since h -c -o -3 -1 still has a hydrogen to lose, it can act as an acid, where in the second reaction, it loses its last hydrogen to form co3 -2 minus h -c -o -3 minuses conjugate base.
00:42
Carbonic acid has two k -a values because it has two hydrogen atoms.
00:49
So k -a -1 is representing this first equation, the loss of the first hydrogen here, and k -a -2 is representing the loss of the second hydrogen.
01:03
Let's look at a buffer that contains kh -co3 and k -2 -co3.
01:14
Both of these will completely dissociate in water into potassium ions and their respective ionically bonded polyatomic ions.
01:34
H -c -o -3 -minus is in both the first and the second equation, but co3 -2 -minus is only in the second.
01:44
So we know that we're dealing with the second equation here, and therefore, k -a -2.
01:55
So now that we know which k -a value to use, we can solve for the ph of the buffer using the henderson -hasselbach equation...