00:01
Hello students, in this question the concentration of the weak monoprotic acid is given.
00:05
Let us consider the acid as ha and its ph value is given.
00:09
We have to find out the ka that is the dissociation constant for this acid.
00:15
Let us consider that acid dissociated as follow ha will be dissociated into h plus and a minus ion.
00:23
The initial concentration of the acid is 0 .06 m and the change in concentration will be minus x.
00:32
Here there will be no h plus ions and there will be no a minus ions and the change in concentration will be plus x plus x and the final concentration will be 0 .06 minus x x and x.
00:48
This will be approximately equal to 0 .06 because x is negligible as it is a infinitely small value.
00:58
Now we are going to calculate the concentration of the h plus ion that is x value from the ph.
01:04
We know that ph is equal to minus log of concentration of the h plus ion.
01:12
Rearranging this equation in order to get the concentration of the h plus ion which is equal to 10 to the power minus ph.
01:20
Substituting the value 10 to the power minus 3 .44 this will be equal to 3 .63 into 10 to the power minus 4 m...