00:01
This question says a buffer contains 0 .1 molar acetic acid.
00:08
So we have 0 .10 molar acetic acid.
00:17
And 0 .13 molar sodium acetate 0 .13 molar.
00:24
I think i should just put it in like the formula, sodium acetate.
00:31
What is the ph of the buffer? so we know ph is p -k -h plus log of the sodium -hacetate concentration divided by the acetic acid concentration.
00:51
Okay? so what is the khe of acetic acid? so the acid dissociation constant for acetic acid is 1 .8.
01:14
8 times 10x to power minus 5.
01:18
Therefore, p .k.
01:19
He is a negative logarithm of the 1 .8 times 10 to power minus 5.
01:28
And therefore, pkk will be equals to 4 .74.
01:48
Therefore, ph will be equals to 4 .74 plus logarithm of 0 .13 divided by 0 .1.
02:01
So that's log .13 plus 4 .74.
02:17
So the ph will be equals to 3 .85.
02:30
Then the next question said, what is the ph of the buffer after the addition of 0 .02 mole of koh.
02:44
So after adding 0 .6.
02:51
02 mole of potassium hydroxide.
02:58
So how did potassium my drugs? that would lead to reaction with the acidic acids...