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
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???)
3 answers
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.
And treat analine as the conjugate base after its protonated form was deprotanated.
I apologize, I have a typo.
kb=(OH concentration)^2/B
Solving for B,
B=(OH concentration)^2/kb
I apologize about that one.
kb=(OH concentration)^2/B
Solving for B,
B=(OH concentration)^2/kb
I apologize about that one.