00:01
If we're preparing a buffer solution, we can carry out our calculation with the henderson -hasselbalch equation, where the ph of the buffer that we want, 5 .60, will be equal to pka, 4 .76, plus the log of the moles of the weak base, acetate, over the moles of the weak acid, acetic acid.
00:26
Sometimes they'll do a ratio of molarities, but in this case it's going to be easier to do a ratio of moles for this log value.
00:39
When we add the hydroxide, the hydroxide is going to react with the acetic acid and produce acetate and water.
00:51
So every mole of hydroxide we add produces a mole of the weak base, acetate.
00:56
We don't know how many moles that we need to add yet, so that's going to be our x value.
01:03
We'll then divide that by the moles of acetic acid still left in the solution.
01:09
We're starting with 580 ml, which is 0 .580 l, at a concentration of 0 .783 moles per liter.
01:21
That's how many moles we start with of acetic acid, but every mole of potassium hydroxide we add consumes a mole of acetic acid.
01:32
So we'll subtract off x the moles of potassium hydroxide that we're going to add...