Question

I'm drawing a blank as to where to start this question, I've done some working but I'm not getting plausible numbers.

Determine solutions. places) of the following solutions. A table of pKa values can be found here. If the 5% approximation is valid, use the assumption to compute pH following 3.25x10 M hydrofluoric acid Submit Answer Tries 0/3 2.63x10-1 M sodium hydrogen sulfate Submit Answer Tries 0/3 1.41x10-2 M sodium phosphate Submit Answer Tries 0/3 4.27x10-3 M sodium acetate Submit Answer Tries 0/3

Answer 1

For hydrofluoric acid

pKa = 3.17 = -log Ka

Ka= 6.76+10-4

then

Ka = [H+][F-]/[HF]

Here HF = H+ + F-

hence,[H+] = [F-]

then Ka = [H+]2 / [HF]

Now,

6.76*10-4 = [H+]²/3.25*10^-3

[H+] = 1.48*10^-3 M

pH= -log [H+]= -log1.48*10^-3 = 2.83 for HF