Calculate the pH of a 0.25M solution of F− at 25∘C. The Ka of HF is 6.8×10−4.

Question

Calculate the pH of a 0.25M solution of F− at 25∘C. The Ka of HF is 6.8×10−4.

Answer 1

F^- acts as a base when in H2O

F^- + H2O ==> HF + OH^-

Kb for F^- is 10^-14/6.8x10^-4

So, solve for Kb. Then use Kb = [HF][OH-]/[F-] = x^2/0.25

Now solve for x which will be = [OH-]

From this calculate pOH (-log OH-)

Now find pH from 14 - pOH

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