00:01
To calculate the ph of the buffer, we'll use the henderson -hasselbalch equation, where ph equals pka plus the log of...
00:08
Often they'll use molarity of the base over the molarity of the acid, but in most cases it's easier to use the moles of the base, which in this case will be the benzoate, divided by the moles of the acid, which will be the benzoic acid.
00:26
So ph will be equal to pka, which they've given to us at 4 .20, plus the log of the moles of the benzoate, is 40 millimoles, or 0 .040 moles, divided by the moles of benzoic acid, which is 60 millimoles, or 0 .060 moles, and we get a ph of 4 .02.
01:00
Then to calculate the volume of sodium hydroxide required to reach a ph of 4 .93, we need to recognize that when the sodium hydroxide is added, which i'll represent as just oh-, it's going to react with the benzoic acid, which i'll represent as ha, producing the conjugate base, benzoate, represented as a-, and water.
01:35
Every mole of the sodium hydroxide i add creates a mole of benzoate and consumes a mole of benzoic acid.
01:45
So if i want a ph of 4 .93, i'll set that equal to pka, 4 .20, plus the log of the moles of the benzoate i start with will be 0 .040, plus every mole of sodium hydroxide i add will make another mole of benzoate, so i'll just say plus x.
02:14
We'll then divide that by the 0 .060 minus x, because every mole of sodium hydroxide i add will consume a mole of the acid...