00:01
First we recognize the stoichiometry will be one -to -one.
00:05
When we have a monoprotic acid reacting with sodium hydroxide, the hydroxide of sodium hydroxide will create the conjugate base of the acid and water.
00:21
So to determine the molar mass we need to know two things.
00:27
We need to know the mass of the acid and we need to know the moles of the acid.
00:33
The mass of the acid was provided.
00:35
We titrated 0 .1687 grams.
00:40
We then divide that by the moles of the acid which we can calculate from the titration information.
00:50
The acid required 15 .5 milliliters of the sodium hydroxide.
00:55
That is 0 .0155 liters.
01:01
We then multiply that by the concentration of sodium hydroxide 0 .1150 moles per liter.
01:10
And then from the stoichiometry we see that one mole sodium hydroxide reacts with one mole of the acid.
01:21
So now we have the grams and the moles of the acid and we get a molar mass of 94 .6 grams per mole.
01:36
Then for part b we know that 15 .5 milliliters was required to reach the equivalence point.
01:49
So if we add half of that 15 .5 divided by 2 that gives us 7 .75 milliliters...