pH of a solution...?

Question

Calculate the pHof a solution prepared by mixing 20.5 mL of a .5M sodium fluoride solution with 32.5 mL of a .75 M hydrofluoric acid solution.

No Ka is given. Would I work this with an SRF table or what?
Using an SRF table, I got pH=3.34.

Answer 1

They should have given you the Ka (Otherwise it's impossible to solve)

HF has pKa=3.15

You can calculate the initial concentrations just after mixing (the volume changes, thus the concentrations also change) and set up a table for the reaction
HF <=> H+ + F- having initial concentrations for both HF and F- (coming from NaF). This way you would solve the quadratic and get an exact solution.Let's do that
The new concentrations just after mixing are
HF = M1V1/(V1+V2) = 0.75*32.5/(32.5+20.5) = 0.460
F- = M2V2/(V1+V2) = 0.5*20.5/(32.5+20.5) = 0.193

.. .. .. .. .. .. .. HF <=> H+ + F
Intial .. .. .. 0.460 .. .. .. .. . 0.193
Dissoc .. .. .. x
Produce .. .. .. .. .. .. .. x .. ..x
At Equi .. 0.460-x .. .. x .. 0.193+x

Ka=x(0.193+x)/(0.460-x) =10^-3.15 =7.08*10^-4 =>
0.193x+x^2 = -7.08*10^-4x+3.26*10^-4 =>

x^2 +0.193708x -3.26*10^-4 =0
x1= -0.19538<0 rejected
x2= 0.001668 =0.00167

pH= -logx =-log(0.00167) = 2.78

However this is a buffer system, so you could use the Henderson-Hasselbalch equation and approximations

pH= pKa+log[conj.base]/[acid] = pKa+log[F-]/[HF]

The approximation is that the changes of the intial amounts of HF and F- are neglegible so that the equilibrium concentrations are practically equal to the initial concentrations

Thus

pH= pKa+log[NaF]/[HF]
but the ratio of concentrations is equal to the ratio of mmole since volume and the conversion factor from mole to mmole are simplified by the ratio, thus

pH= pKa+log mmole NaF/mmole HF =
= 3.15+log (0.5*20.5/(0.75*32.5)) = 2.77 You see that the deviation from the exact solution is very small