If the Kb of a weak base is 2.8 × 10-6, what is the pH of a 0.30 M solution of this base?
let the weak base is BOH
BOH dissociates as:
BOH +H2O -----> B+ + OH-
0.3 0 0
0.3-x x x
Kb = [B+][OH-]/[BOH]
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 ((2.8*10^-6)*0.3) = 9.165*10^-4
since c is much greater than x, our assumption is correct
so, x = 9.165*10^-4 M
so.[OH-] = x = 9.165*10^-4 M
use:
pOH = -log [OH-]
= -log (9.165*10^-4)
= 3.04
use:
PH = 14 - pOH
= 14 - 3.04
= 10.96
Answer: 10.96
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