00:01
Hello, let's have a look on the question ch3cooh -x -plus -koh -x -plus gives out ch3cook -plus -h2o.
00:28
At t is equal to 0, concentration of ch3cooh is 0 .88 multiplied by 608.
00:39
Concentration of koh is 2 .28.
00:49
Concentration of ch3cook is 0.
00:53
H2o is also 0.
00:55
At t is equal to t equilibrium, concentration is equal to 598 .4 minus 2 .28 milligram.
01:15
Concentration of koh is 0.
01:19
Concentration of ch3cook is 2 .28 pm.
01:24
Concentration of h2o is 2 .28 pm.
01:35
It is cast of acetic buffer solution.
01:42
We know that ph is equal to pka plus log of concentration of salt divided by concentration of acid.
01:58
So, concentration of salt is equal to 2 .28 divided by 680 plus v.
02:25
This is mole per liter.
02:35
Concentration of acid that is equal to 598 .4 minus 2 .28 v divided by 680 plus v mole per liter.
02:57
Pka is equal to 4 .76.
03:07
So, ph is equal to 4 .76 plus log of 2 .28 v divided by 680 plus v divided by 598 .4 minus 2 .284 divided by 680 plus v.
03:42
This will get cancelled.
03:43
Here mole per liter...