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The pH of a 0.02 M solution of an unknown weak base is 8.1. What is the pKb of the unknow solution
(I think it is referred to pKb=-logKb, but I am not sure how to solve it)

2006-07-18 18:58:49 · 6 answers · asked by Xiang W 1 in Science & Mathematics Chemistry

6 answers

First of all there is no such thing as pKb of a solution (only of a base).
Second it is impossible to solve if you don't know the valency of your base. Assuming it is monovalent:

________________MOH <=> M(+) + OH(-)
initial concentration__C
Dissociating________x
Producing_________________x______x
at Equilibrium ______C-x_____x______x

Kb=[M(+)]/[OH(-)]/[MOH]=x*x/(C-x)=x^2/(C-x)

since pH=8.1 pOH=14-8.1=5.9 and
[OH(-)]=10^(-pOH)=1.26*10^-6

But we said that [OH(-)]=x and know that C=0.02 M

Thus pKb=-logKb=-log(x^2/(C-x))=
=-log((1.26*10^-6)^2/(0.02-1.26*10^-6)=
=10.1

Stupid site doesn't show everything I am writing; just put the pointer over "..." to read the missing parts

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2016-12-10 09:58:01 · answer #2 · answered by ? 4 · 0 0

okay, I could have probably done this question a few months ago. But I've forgotten a lot of chem, so my advice would be to look up the Henderson-Hasselbalch equation and try using that

2006-07-18 19:06:23 · answer #3 · answered by Spreet 2 · 0 0

Yea my calculator also packed up.

2006-07-18 19:01:36 · answer #4 · answered by Anonymous · 0 0

and that is y i cheated in HS chem.

2006-07-18 19:01:25 · answer #5 · answered by me 2 · 0 0

i can

2006-07-18 19:01:33 · answer #6 · answered by Maulik 1 · 0 0

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