How do i solve this Chemistry equation?

[needanansnow]

The question is:

Municipal drinking water frequently has fluoride added to improve the dental health of the citizens. The "fluoride" is often present the range of 1.4ppm. But in reality, because the HF is a weak acid, this total "Fluoride" is a mixture of F- and HF, ir [HF] + [F-] = 1.4 ppm. Assuming the water is buffered sufficiently to assure that its pH is 4.4 and the concentration is 1.4 ppm "Fluoride", what percentage of the "fluoride" exists as fluoride ion, F-, and what percentage exists as HF molecules?

How do I solve this?

[Shan]

The dissociation of HF follows this equation:

HF <------> H+ + F- Ka = 7.2 * 10^(-4) = .00072

By the definition of Ka, this means that:
[H+]*[F-] / [HF] = .00072
where [x] means concentration of x.

You can use the pH, which is given to be 4.4, to figure out the concentration of H+. If the pH is 4.4, the concentration of H+ is always 10^(-4.4).

Plug that in to the first equation:
[10^-4.4]*[F-] / [HF] = .00072

And you get:
[F-] / [HF] = 18.1


Another way to do this is by using the Henderson-Hasselbach equation, which says that:
pH = pKa + log([A-]/[HA])
If you use the pKa of HF and set F- as A- and HF as HA, you'll get the same answer for [F-]/[HF].


Now that you know the ratio between the HF and F-, you can easily figure out the percentage:
18.1/19.1 = 94.7% F-
1/19.1 = 5.3% HF