00:01
All right, so this question requires the use of an ice table for aniline, all right? so we have this ionization equation, which is 6.
00:16
This must be h5 and then h2.
00:30
To give c6 h5, we have an h3 ion, all right? this is an ion and hydroxyl ion.
00:43
So what happens here is that we are given.
00:47
The kb over 4 .3 times 10 to power negative 10 and the concentration is 0 .28 molar and so the first thing to do is perhaps figure out what these elements mean here right so given the ac e table for the concentration provided here we will therefore conclude that perhaps zero 2 .280 minus x.
01:26
All right, given these elements, because we assume that these two will be positive x, all right, as by the initial knowledge on that.
01:40
And then these two will be x and x, of course.
01:45
So if you have this formulae that links kb to their concentrations, c6h5, enh, marked that by the hydroxyl, divide that by the whole concentration.
02:17
You'll have some sort of 4 .3 10 % to negative 10 must be equal to x and x over 0 .28 minus x.
02:33
So this takes us back to math.
02:35
All right so uh this would be 4 .3 times 10 to negative 10 equaling x squared over 0 .28 okay we can assume that so it means that x squared equals to 4 .3 times 10 to 0 negative 10 multiply that by 0 .28 okay and we have x squared as equal to 1 .204 times 10 to power negative 10 which means x is a square root of that.
03:27
All right and this is around 1 .0 1 .10 times 2 .5...