Question

What is the pH of a 0.128 M aqueous solution of potassium fluoride, KF?

pH =

This solution is _________acidicbasicneutral.

Answer 1

Ka of HF = 6.6*10^-4

we have below equation to be used to find Kb for F-

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/6.6*10^-4

Kb = 1.515*10^-11

F- dissociates as

F- + H2O -----> HF + OH-

0.128 0 0

0.128-x x x

Kb = [HF][OH-]/[F-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.515*10^-11)*0.128) = 1.393*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.393*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.393*10^-6)

= 5.8562

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.8562

= 8.14

Answer: 8.14

since pH is greater than 7, the solution is basic