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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???)

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V

  1. 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

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    Devron

  2. 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.

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    Devron

  4. 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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    Devron

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