00:01
When we add sodium hydroxide to a solution of benzoic acid, the benzoic acid represented as ha, reacts one -to -one with sodium hydroxide represented as hydroxide, producing its conjugate base in water.
00:17
Because we've added 30 milliliters of 0 .150 molar sodium hydroxide to 25 milliliters of 0 .150 molar benz pero benzoic acid, and the stoichiometry is one to one, we have added more moles of sodium hydroxide.
00:36
Thus, we have gone past the point of neutralizing all of the benzodiaic acid.
00:42
Therefore, sodium hydroxide is in excess.
00:47
To calculate the concentration of hydroxide in excess, we need to take the moles of sodium hydroxide that we added, that is 30 milliliters or 0 .030 liters at a concentration of 0 .1 .50 moles per liter, and we subtract off the moles that reacted.
01:11
These will be the moles of benzoic acid that were present in the 25 milliliters, or 0 .0250 liters, at a concentration of 0 .1 .50 moles per liter.
01:23
We then divide that by the new volume...
