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NH3 Kb=1.8x10^-5
CH3NH2 Kb=4.38x10^-4
C2H5NH2 Kb=5.6x10^-4
C6H5NH2 Kb=3.8x10^-10
C5H5N Kb=1.7x10^-9

2006-10-26 16:09:32 · 1 answers · asked by pixela007 2 in Science & Mathematics Chemistry

1 answers

Think in terms of the weak conjugate acid.
The best buffering effect is for pH=pKa +/- 1

Ka=Kw/Kb and pKa=-logKa

So

NH4+ pKa=9.26
CH3NH3+ pKa=10.64
C2H5NH3+ pKa=10.75
C6H5NH3+ pKa=4.58
C5H5NH+ pKa=5.23

The best seems to be C5H5N since the pKa is closest to 5.00 (C6H5NH2 looks also good). You don't define the concentration you want. Just dissolve the approriate amount in water (e.g. in 700-800 ml) in a 1L volumetric cylinder and adjust the pH to 5.00 with HCl, using a pH-meter. Then add water up to 1L.

2006-10-27 01:07:14 · answer #1 · answered by bellerophon 6 · 0 0

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