pOH=pKb+log[[NH4OH][NH4+]]
pOH=−log(1.8×10−5)+log[0.250.4]
pOH=−log(1.8)+5+log(1.6)
pOH=5+log(1.81.6)
pOH=5+log(16)−log(18)
pOH=5+4log(2)−log(9×2)
pOH=5+4log(2)−log(2)−2log(3)
pOH=5+3×0.301−2×0.477
pOH=4.949
pH=9.051
(b)pOH=pKb+log[[NH4OH][NH4+]]
pOH=−log(1.8×10−5)+log[0.150.5]
pOH=−log(1.8×10−5)+log[0.150.5]
pOH=5−log(1.8)+log[310]
pOH=5−log(1.8)+log[10]−log(3)
pOH=5−log(18)+2−log(3)
pOH=5−log(2)+2−3log(3)
pOH=5.27
pH=8.73
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