00:01
All right, it says to calculate the ph of a dihydrogen phosphate buffer prepared with each of the following concentrations.
00:08
What do you conclude about the ph of a buffer when equal concentrations of the acid and the conjugate base are used to prepare it? here's the ph of a buffer using anderson is the pka of the acid plus log the concentration of the base divided by the concentration of the acid.
00:29
Now in each case, right, so here's it.
00:33
If you have equal, before we answer the question, let me answer the last question.
00:37
If the concentration of the base and the acid are the same, of course it means that this equation will become ph equals the pka plus the log of one over one, right? and this is, that's one, right? basically.
01:01
And so this is like the ph equals the pka because this is zero.
01:07
So if you use equal number, equal concentration of the acid, the weak acid and the conjugate base to prepare a buffer, the ph of the solution will be the same as the pka of the acid.
01:18
Basically that's going to be the ph.
01:21
So if we look at the questions that we have, right, in each case, we have nah2po4, right? which if you ionize that's na plus h2po4 minus and then you have na2hpo4 and that's 2na plus for example and hpo4 2 minus.
01:58
Now these are the particular substances present.
02:03
So you have h2po4 minus and hpo4 2 minus...