Question

Calculate the pH (at SATP) of a solution of HF at which [HF] = [F-] given the following data:

Species G°f (kCal/mole)

HF -71.68
H+ 0
F- -67.34

Answer 1

If [HF] = [F-] is true

then

the equation

pH = pKa + log(F-/HF) is valid

since this is abuffer so

F-/HF --> F- = HF

then

log(1) = 0

pH = pKa

claculate pKa via

HF <-> H+ + F- Ka

we need Ka

so

dG = -RT*lnKa

dG = Gproducts - Greactants

dG = (-67.34) - (-71.68+0) = 4.34 kJ/mol = 4340 J/mol

dG = -RT*lnKa

Ka = exp(-dG/(RT))

Ka =exp(-4340/(8.314*298))

Ka = 0.17347

pKa = -log(Ka) = -log(0.17347) = 0.76195

pH = 0.76195

At standard conditions, i.e. concentrations are 1M each