00:01
Over here we want to determine the molar concentration of sodium fluoride that we're going to have given that a solution is going to have a ph equal to 8 .51 and the ka associated to hf is going to be equal to 6 .1 times 10 to the minus 4.
00:20
So over here what we need to consider is that f minus is going to have a reaction with water which is going to produce hf plus oh minus.
00:33
This reaction is associated to the value for kb and kb is equal to 10 to the minus 14 divided by ka by the definition of the kw constant.
00:44
So 10 to the minus 14 divided by this value is going to give us the result for kb which is 1 .64 6393 times 10 to the minus 11.
00:57
So this is going to be a very very small value.
00:59
So if we start with an initial concentration c0 of the fluorine ion, in this case sodium fluoride is going to dissociate completely, so we can assume that this concentration is going to be equal to the initial concentration that we need to solve for.
01:14
What we know is that the amount of change would be minus x plus x plus x, thus at equilibrium we would get c0 minus x, x and x.
01:25
Thus, our value for kb would be equal to the concentration of hf times the concentration of oh - at equilibrium, divided by the concentration of f-.
01:42
Replacing we get x squared divided by c0 -x, but the value for kb is going to be very small, compared to my initial concentration, thus we can expect that c0 is going to be much much larger than the value for x that we're going to find.
02:02
Thus, c0 minus x can be approximated as c0.
02:06
So over here, we're going to get x squared over c0.
02:11
Solving for c0, we're going to get c0 is going to be equal to x squared divided by kb...