00:01
Hello students, consider the reaction nh3 plus h2 is nh4 plus o h minus.
00:07
Concentration of nh3 is given as 0 .015 molar, concentration of nh4 plus is 0 .037 molar and kb is given as 1 .8 into 10 x to minus 5.
00:20
We have to find ph of the buffer solution.
00:24
First of all, let's write down the expression for kb.
00:29
Kb is equal to of nh 4 plus multiplied by concentration of o h minus divided by concentration of n h 3 now we can rearrange the equation as concentration of o h minus is equal to k b multiplied by concentration of n h 3 divided by concentration of nh 4 plus.
01:21
Now let's substitute the values.
01:24
This is equal to 1 .8 into 10 raised minus 5 multiplied by 0 .015 whole divided by 0 .037.
01:43
This is equal to 2 .48 into 10 .8 into 10 .8 into 10.
01:50
10 raised to minus 5.
01:53
Now let's find out poh.
01:58
Poh is equal to negative logg concentration of oh -h minus...