00:01
All right, so assume that we have a weak acid, which will represent as h .a.
00:05
And if we want to write the dissociation equation, that's going to be splitting up into ions.
00:10
It's an equilibrium because it's a weak acid, and it'll form h plus and the anion a minus.
00:16
And then we can write the equilibrium constant expression.
00:18
Ka is equal to concentration of h plus multiplied by that of a minus, concentrations with products on top, over concentration of ha, the reactant on the bottom like so.
00:33
Okay, and then we want the percent dissociation.
00:41
So that's going to equal the amount, the ratio basically, of h -plus over h -a.
00:55
So if we draw basically a table of values, assume that we start with x number of moles of h -a, and zero and zero.
01:18
And then we could say minus some change, y, and plus the unequal amount of h plus an a minus because they form in a one -to -one ratio.
01:28
So you have x minus y, some change in the initial concentration, is equal to, and then we have y and y.
01:35
So if we plug into the k -a expression, we have y and a y, so y squared, over x minus y.
01:45
But the problem says if the dissociation is negligible compared to the initial concentration, we can use the approximation that x minus y is the same thing as just x, that it hasn't really changed.
01:58
So it's y squared over x, which we can rewrite as the concentration of, i'll leave it as x.
02:13
Okay.
02:14
Now we can rewrite this...