00:01
In this question, it asks to first write the dissociation equation for ammonia at 0 .5 p .m.
00:05
Then we have to also write the dissociation equation for ammonia at 5 p .m.
00:09
Then we have to calculate ph.
00:10
So here the dissociation equation for ammonia is n .s3 aque plus h2 gives h2 n .4 plus oh man.
00:17
This is the dissociation equation.
00:19
Now to calculate psh, write ic table first.
00:22
So i see table.
00:23
Here initially, 0 .5m of nh3 is present and 0 of nh4 and 0 of h minus minus 6.
00:30
Is present.
00:31
Now, initially, then a change will be occurred.
00:34
Let minus x will be changed.
00:36
So it will be plus x plus x.
00:37
So at equilibrium, the concentration of nh3 will be 0 .5 minus x, and nh4 will be x, and it will be x.
00:44
So the kb formula for kv is nh4 plus into os minus divided by nh3 plus.
00:49
Now, given the value of kb is 1 .8 in 10 to the 5, and nh4 plus concentration is x and oh manage, concentration is x so x into x is x squared and n s 3 concentration is 0 .5 minus x so which is 0 .5 minus is but 0 .x is very very small than 0 .5 so we can write 0 .5 minus x are equals to 0 .5.
01:19
Okay now now let's see after solving this we will get the value of x is 3 .3 into 10 to the per minus 3 but you can see the value of x is nothing but this is the oh h minus concentration the x value is nothing about the concentration of oh h minus minus okay now since it is a concentration of oh h minus concentration so we can write uh x is equal to o h minus consensus equal to 3m so p o h is equal to minus log of o h minus so minus concentration so p oh will get 2 .52 now we know p h plus p .o .h is equal to 14.
01:58
So ph will be 14 minus p .o .h which is 14 minus 2 .5...