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Calculate [H+]?

  1. Nov 5, 2004 #1
    Calculate [H+]?????

    Calculate [ H + ] of a 0.220 M solution of Aniline C6H5NH2 Kb = 7.4e-10
  2. jcsd
  3. Nov 5, 2004 #2
    Consider that for a reaction
    we can use
    How can you find [itex]K_{a}[/itex] when given [itex]K_{b}[/itex]?
  4. Nov 5, 2004 #3


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    Sirus is right; an alternative point of view may be using dissociative approach. In this, you start with 0.220 M of aniline, but only [itex]\displaystyle x[/itex] of it is ionized to give some [itex]\displaystyle H^+[/itex]. We know the equilibrium constant of this reaction, i.e., [tex]\displaystyle \frac {10^{-14}}{K_b}[/tex].

    [tex]\displaystyle \underbrace{C_6H_5NH_2_{(aq)}}_{0.220-x} \leftrightharpoons \underbrace{C_6H_5NH^-_{(aq)}}_{x}+\underbrace{H^+_{(aq)}}_{x}[/tex]

    Now it is better for you to consider the magnitude of [tex]\displaystyle \frac {10^{-14}}{K_b}[/tex]; for the sake of simplification, you may omit the [itex]\displaystyle x[/itex] in [itex]\displaystyle 0.220-x[/itex], as we may omit values generally less than 5%. Of course, you may also want to solve the quadratic equation to find the exact answer, but believe me this is not very necessary.

    What you'll do next is to find the [itex]\displaystyle -\log x[/itex].
  5. Nov 6, 2004 #4


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    Aniline is a base, not an acid. Hydrogen will stick to the lone pair on the nitrogen so the reaction is like this:

    [tex]C_6H_5NH_2 + H_2O \rightarrow C_6H_5NH_3^+ + OH^-[/tex]
  6. Nov 6, 2004 #5
    In that case, for a reaction
    we can use
  7. Nov 6, 2004 #6


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    Dear ShawnD, you are right about aqueous reactions, but don't forget that there are very powerful bases for using in non-aqueous phases like sodium hydride, lithium diisopropylamide, etc, which can take a proton to give anilinide anion.

    In most cases, your reaction is sufficient, and same things may be said for Sirus' last post.
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