Kb for NH3 is 1.8x10^-5. What is the pH of a 0.35-M aqueous solution of NH4Cl at 25°C?
use:
Ka = Kw/Kb
Kw is dissociation constant of water whose value is 1.0*10^-14 at
25 oC
Ka = (1.0*10^-14)/Kb
Ka = (1.0*10^-14)/1.8*10^-5
Ka = 5.556*10^-10
NH4+ + H2O
-----> NH3 +
H+
0.35
0 0
0.35-x
x x
Ka = [H+][NH3]/[NH4+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.556*10^-10)*0.35) = 1.394*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.394*10^-5 M
So, [H+] = x = 1.394*10^-5 M
use:
pH = -log [H+]
= -log (1.394*10^-5)
= 4.8556
Answer: 4.86
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