00:01
First we recognize that when we add the sodium hydroxide, the hydroxide is going to react with the acid in the solution, which is acetic acid.
00:10
I'll represent that as just ha, creating more acetate, a-, and water.
00:18
As long as we don't add more hydroxide than we have acetic acid, we'll still have a buffer solution, and ph can be calculated using the henderson -hasselbalch equation, where ph equals pka, 4 .757, plus the log of, a lot of people will do molarity of the base over molarity of the acid, but it's easier in most cases and equivalent to do a ratio of moles, moles acetate over moles acetic acid.
00:51
The moles of acetate will be the moles of acetate we start with.
00:56
We have 150 milliliters of the buffer, which is 0 .150 liters, at a concentration of 0 .350 moles per liter acetate in the form of sodium acetate...