00:01
To calculate the molar mass of the base, we need to know the grams of the base and the moles of the base.
00:06
The grams is provided at 0 .239.
00:09
To calculate the moles of the base, we need to know the moles of hydrochloric acid required to reach the equivalence point.
00:15
Then, knowing that the stoichiometry is 1 to 1, 1 mole of hcl added will be equal to 1 mole of the organic base that was present.
00:27
So the moles of hcl added is the moles of the base that was present.
00:32
So we'll take the mass of the base divided by the moles of the base, equivalence point being 39 .24 milliliters or 0 .0394 liters times the molarity .135.
00:46
And we get 45 .12 grams per mole.
00:50
If we are pre -equivalents at 18 .35 mill liters, then we have a buffer, so we'll use the henderson -hasselbalch equation to ultimately solve for kb.
01:03
The henderson -hslebalch equation has ph, the ph that we've measured, 10 .73, will be equal to pca plus the log of the moles of base that are left.
01:14
Once we've added 18 .35 milliliters of the acid, that will be the grams of the base divided by its molar mass.
01:25
That's the moles of base we start with.
01:27
Then we need to subtract off the moles of base that reacted...