What is the initial concentration of KF salt if it has pH 8.35?

Question

So I had an exam today, and I was wondering if this is how you solve that question:

KF dissolves to K and F-.

Then F- undergoes hydrolysis to make HF and OH-.

Then you make a chart, but is the concentration of OH- at equilibrium measured from pH 8.35?

So the OH- concentration at eq. is 10^(-8.35)?

And then the initial concentration of F- would be x - (10^(-8.35))?

Then using Kb of HF, you would figure out x?

If x negligible? Is anything negligible?

Would the initial concentration turn out to be 0.70M?

Also, when they say "molten", is water involved? For example the question was "what formed at anode and cathode for electrolysis of molten KBr"?

And one last question, if there are 4 choices, P, N, Ba, and Zn, and the question goes, which oxide of the following elements always acts as a base? and I know that a nonmetal oxide acts as and acid, and metal oxide acts as a base right? But which one always acts and a base? Ba or Zn? I didn't know how to figure that out, or I never read anything about that.

Okay. that's a lot of questions, sorry, but I really want to know if I did them right! Thanks for helping!!!

Answer 1

KF -->  K+  +  F-

X______0____0_

0______X____X_

F- + H2O <=> HF + OH-

X_____/______0_____0_

X-Y___/______Y_____Y_

Y = 10^(14-8.35) = 10^-5.65 = 2.24*10^-6

Ki = 10^-14/Ka(HF) =

= Y^2 / (X-Y) = (2.24*10^-6)^2 / (X-2.24*10^-6)

so

X = ( (2.24*10^-6)^2 / Ki ) + 2.24*10^-6 = ...