00:01
Let us find out answer to the question.
00:02
Monoprotic weak acid is dissolved to water to make a final concentration of 0 .0134 molar.
00:16
Ph of resulting solution is given 2 .51.
00:26
We have to find k.
00:29
Let us see the solution part.
00:32
So what is the weak monoprotic acid? weak acid will dissociate partially so let us say h a is weak acid and it will dissociate to form h positive plus a negative so the concentration of this is given as 0 .0134 initially these are 0.
00:53
So finally some amount of this h a will undergo dissociation and this let us say x is got dissociated this will become x and this will become x here h positive is equal to x which is equal to p which is equal to concentration of so p h is called minus log of h positive we can write so ph is 2 .51 minus log of h positive.
01:38
So if we solve this, h positive will become by taking anti -low 10 raise to the power minus 2 .51.
01:46
So this will become our concentration of h positive.
01:50
If we solve this, this will come out to be 0 .003.
01:59
So this is the concentration of h positive, which is equal to x...