Question

The pH of an aqueous solution of 0.112 M potassium fluoride, KF (aq), is ___?

This solution is _________acidicbasicneutral (acidic, neutral, or basic)

Answer 1

Ka for HF = 3.5*10^-4


use:
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/3.5*10^-4
Kb = 2.857*10^-11
F- dissociates as

F-        + H2O   ----->     HF +   OH-
0.112                        0         0
0.112-x                      x         x


Kb = [HF][OH-]/[F-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((2.857*10^-11)*0.112) = 1.789*10^-6

since c is much greater than x, our assumption is correct
so, x = 1.789*10^-6 M



use:
pOH = -log [OH-]
= -log (1.789*10^-6)
= 5.75


use:
PH = 14 - pOH
= 14 - 5.75
= 8.25
Answer: 8.25

since, pH > 7
so, solution is basic

Answer : 8.25 , basic