What is the pOH of a 0.01743 M solution of LiHS? The Ka of H2S is 8.9×10^–8.
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)/8.9*10^-8
Kb = 1.124*10^-7
HS- dissociates as
HS- + H2O -----> H2S + OH-
0.0174 0 0
0.0174-x x x
Kb = [H2S][OH-]/[HS-]
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 ((1.124*10^-7)*1.743*10^-2) = 4.425*10^-5
since c is much greater than x, our assumption is correct
so, x = 4.425*10^-5 M
use:
pOH = -log [OH-]
= -log (4.425*10^-5)
= 4.354
Answer: 4.35
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