What is the pH of the solution containing 0.20 M NH3 and 0.15 M NH4Cl?
Kb (NH3) = 1.8•10-5
pKb= 4.74
pOH = pKb - log (NH4Cl/NH3)
pOH = 4.74 - log(0.2/0.15)= 4.62
pH = 14 - pOH = 14 - 4.62 = 9.38
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