Determine the pH of a 0.4 M NaF solution at 25°C. The Ka of HF is 3.5x10-5.
use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/3.5*10^-5
Kb = 2.857*10^-10
F- dissociates as
F- + H2O -----> HF + OH-
0.4 0 0
0.4-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^-10)*0.4) = 1.069*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.069*10^-5 M
use:
pOH = -log [OH-]
= -log (1.069*10^-5)
= 4.971
use:
PH = 14 - pOH
= 14 - 4.971
= 9.029
Answer: 9.03
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