Asked by V

A solution of aniline(C6H5NH2, Kb=4.2x10^-10) has a pH of 8.69 at 25 Celsius. What was the initial concentration of aniline?
I have so far don't know if it's right:
*. [H3O]=10^-8.69=2x10^-9
* 4.2x10^-10=(2x10^-9)^2 / x
X=1.2x10^-8 (correct???)
Answered by Devron
Aniline is a weak base, so I would tackle the problem this way.

14-pH=pOH

pOH=-log[OH]

Solving for OH concentration,

10^(-pOH)=OH concentration.


B + H2O----> BH + OH

Where B=aniline and BH= conjugate acid of aniline

So, Kb=[BH][OH]/B

The reaction shows that the concentration of OH= the concentration of BH, so the equation above becomes


kb=(OH concentration)^2/BH

Solving for BH,

BH=(OH concentration)^2/kb
Answered by Devron
You could have done it that way, but you would have to solve for ka=kw/kb.


And treat analine as the conjugate base after its protonated form was deprotanated.
Answered by Devron
I apologize, I have a typo.




kb=(OH concentration)^2/B

Solving for B,

B=(OH concentration)^2/kb

I apologize about that one.
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