4. consider a 33.0 ml solution containing 0.0758 m naf and 0.0955 m hf. what is the total number of moles of hf after the addition of 8.00x105 mols hcl? the ka of hf is 6.8x104

Buffer solution resist the change in pH upon addition of small amount of strong acid or strong base.
Buffer consists of weak acid as HF / and its conjugate base NaF
When strong acid as HCl is added to buffer, it respond with its conjugate base to convert the strong acid to weak acid like this:
HCl (S.A) + NaF → NaCl + HF (W.A)
moles of HF we already have = M * V(in liters)
                                                = 0.0955 M * 0.033 L = 3.15 x 10⁻³ mole
moles of HCl added = 8.00 x 10⁻⁵ mole
one mole HCl reacts with 1 mole NaF to give 1 mole HF
so the amount added to HF = 8.00 x 10⁻⁵
Total moles of HF present = (3.15 x 10⁻³) + (8.00 x 10⁻⁵) = 3.23 x 10⁻³ mole

Tolerance to high levels of salinity