What are the equilibrium concentrations of NH3, NH4+, and OH- in a 0.95 M solution of ammonia? Kb = 1.8x10^-5
What is the pH of the solution? pH =
NH3 dissociates as:
NH3 +H2O -----> NH4+ + OH-
0.95 0 0
0.95-x x x
Kb = [NH4+][OH-]/[NH3]
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 ((1.8*10^-5)*0.95) = 4.135*10^-3
since c is much greater than x, our assumption is correct
so, x = 4.135*10^-3 M
So, [OH-] = x = 4.135*10^-3 M
[NH4+] = x = 4.135*10^-3 M
[NH3] = 0.95 - x
= 0.95 - 4.135*10^-3 M
= 0.946 M
use:
pOH = -log [OH-]
= -log (4.135*10^-3)
= 2.3835
use:
PH = 14 - pOH
= 14 - 2.3835
= 11.6165
Answer:
[NH3] = 0.95 M
[NH4+] = 4.1*10^-3 M
[OH-] = 4.1*10^-3 M
PH = 11.62
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