In: Chemistry
What is the pH of a 0.186 M aqueous solution of sodium fluoride, NaF? (Ka for HF = 7.2×10-4)
NaF --------------> Na+ + F-
0.186M 0.186M
F- + H2O --------------> HF + OH-
I 0.186 0 0
C -x +x +x
E 0.186-x +x +x
Kb = [HF][OH-]/[F-]
Kb = KW/Ka
= 1*10^-14/7.2*10^-4 = 1.4*10^-11
Kb = [HF][OH-]/[F-]
1.4*10^-11 = x*x/0.186-x
1.4*10^-11*(0.186-x) = x^2
x = 1.6*10^-6
[OH-] = x = 1.6*10^-6M
POH = -log[OH-]
= -log1.6*10^-6
= 5.7958
PH = 14-POH
= 14-5.7958 = 8.2042>>>>>.answer