00:01
So we're talking about h2s .o4 here.
00:05
If we simply look at the first ionization that makes h plus plus h .o .4 and minus.
00:15
We'll assume that we're going to ionize that completely, and i'll end up with 0 .15 molar h plus.
00:25
So if we take minus the log of that, the ph of our acid, our diprotic acid here, is 0 .82.
00:34
39 if we ignore the second ionization.
00:39
Well, if we do include the second ionization now, we have hso4 minus, breaking into h plus and sulfate.
00:52
We'll make an ice box to keep track of what's happening.
00:56
So i know i start with 0 .15 molar of that, but i also start with 0 .15 molar of this, and no sulfate yet.
01:06
Yet.
01:07
So minus x plus x plus x, 0 .15 minus x, 0 .15 plus x, and x.
01:18
So i know our k, which was 0 .01, is equal in the products over the reactants.
01:28
So 0 .15 plus x times x...