00:01
All right, so we have the strong acid, strong base titration here.
00:05
We're to consider the ph initially.
00:10
The initial ph would be a minus log of the concentration of the acid.
00:17
And this gives us 0 .74.
00:20
The volume, the b part says we should, the volume of the base required of koh would be 37 .0 times 0 .18 divided by 0 .205.
00:33
That's give you 32 .5 mules.
00:36
This is the volume required to get to equivalence point.
00:40
Now the next, when you've added some base, the number of moles of the base is 0 .0116, which is the volume added times 0 .025.
00:49
And this gives you 0 .00226.
00:53
Or the number of moles of the acid given that 0 .037 multiplied by 0 .0, 0 .180, the concentration, and this is 0 .00666, right? since the acid, the number of moles of the acid is more, it's in excess, right? so the concentration of the excess hydrogen ions, right? the excess would be 0 .00666 minus 0 .00226 divided by the total volume of the solution, which is 48 .6, right? and this would give you 0 .0905 molar...