Posted by Henry on Friday, November 9, 2012 at 5:58pm.
.......NH4^+ + HOH ==> NH3 + H3O^+
I......0.1..............0.....0
C.......-x..............x.....x
E.....0.1-x.............x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(NH4^+)
Substitute and solve for x = (H3O^+) and convert to pH.
note: If pKb = 4.74 then Kb = 1.8E-5
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