00:01
In the first part of this question, we're going to be looking at a reaction in which we've got the ch3 -c -o -o -0 -2 -o -produce ch3 -c -o -h and the oh -h negative ion.
00:18
Now looking at the initial concentration, we've got 0 .1, and before the ionization process takes place, we won't be having any of these in solution.
00:27
Looking at the change in the concentrations due to the ionization process if this decreases by x this will increase by x and this will also increase by x since they are in the ratios of 1 is to 1 so at equilibrium we've got 0 .1 minus x and we've got x here and x right there now if we are to use the kb expression recall that the kb it is going to be the concentration of the ph 3 c o o h multiplied by the o h divided by the concentration of ch3co negative.
01:03
And these concentrations are evaluated at equilibrium, and we've just determined those concentrations in the above step.
01:10
So we are saying kb is equal to x multiplied by x divided by 0 .1 minus x.
01:17
And this kb, it is equal to kw divided by ka, which is 10 to the power negative 17 divided by 1 .8, by 10 to the power negative 5 for the ascended i.
01:32
Therefore, if we isolate this and we solve for x, we've got our x being equal to 2 .37 by 10 to the power negative 6.
01:42
And this is equal to the concentration of the o h negative i.
01:46
So solving for the p .o .h, this is going to be negative log, 2 .37 by 10 to the power negative 6...