00:01
Question number 13 is essentially the same as question number 11.
00:06
In question number 11, i already identified the conjugate acid -based pairs, but i will do it again here.
00:17
The chemical reaction that we are considering is h .s .o4 -1, reacting with c -2 -2 -1, going to s -o -4 -2 minus, and h -c -2 -04 -1.
00:34
So in the forward reaction, we see that h .s .o4 minus has a hydrogen ion it can donate, so it will serve as the acid, and c2, 042 minus, is accepting it, so it is behaving as the base.
00:49
But when h .s .o .4.
00:51
Minus donates its hydrogen ion, it creates s .o .42 minus.
00:56
So s .o .42 minus, which had a hydrogen ion as h .s .o .4 minus, this now can serve as a base...