PH calculations...?

Don't make this harder than it has to be. This is a solution of two weak acids, but one acid is a LOT weaker than the other. The Ka of HF is 7.2E-4 which corresponds to a pKa of 3.14. However, the pKa of NH4+ is 9.24. That means that the acidity of HF is 1.2 million times that of NH4+. Since the concentrations of the two are pretty close, we can ignore the contribution of NH4+ to the pH. Do the pH problem for HF as if there was nothing else there. When I did that, I got pH = 1.71.