00:01
In stoichiometry, we relate the quantities of the substances that are involved in a chemical reaction.
00:07
Here we have the neutralization reaction between acetic acid with sodium hydroxide to form sodium acetate and water according to this balance equation.
00:23
If we have 14 .7 ml of 0 .1 molar of the acetic acid reacting with 3 .40 ml of 0 .15 molar of the base, we wish to find the ph of the resulting solution.
00:42
So we should start by finding the millimoles of the acid and the base that are being reacted.
00:48
So to do that, we have to multiply the molarity by the volume in ml.
00:54
So for the acid we have 0 .1 times 14 .7 and this gives us 1 .47.
01:03
For the base, we have 0 .150 times 3 .40 and this gives us 0 .51.
01:17
So we see that the acid comes in excess and the base is completely consumed.
01:24
Now since for every one mole of the acid, there is one mole of the base that is that also reacts and one of the salt that forms, it follows that aside from the excess acid that is still present, we have its own salt that is formed and it's equal to the millimoles of the base that has been completely consumed.
01:45
So that's 0 .51...