Asked by David
calculate the the pH of a buffer solution containing 0.246 M NH3 and 0.0954 M NH4Cl (for NH3 Kb = 1.74x 10-5)
Answers
Answered by
Whitney
Henderson-Hasselbalch equation:
pH=pKa+log((base)/(acid))
Kb*Ka=1.0*10^-14
1.74*10^-5*Ka=1.0*10^-14
Ka=1.0*10^-14/1.74*10^-5
Ka=5.75*10^-10
pKa=-log(5.75*10^-10)
pKa=9.24
pH=9.24+log((.246)/(.0954))
pH=9.65
pH=pKa+log((base)/(acid))
Kb*Ka=1.0*10^-14
1.74*10^-5*Ka=1.0*10^-14
Ka=1.0*10^-14/1.74*10^-5
Ka=5.75*10^-10
pKa=-log(5.75*10^-10)
pKa=9.24
pH=9.24+log((.246)/(.0954))
pH=9.65
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