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March 26, 2017

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For a solution that is 0.280M HC3H5O2 (propionic acid, Ka= 1.3*10^-5) and 0.0894M HI , calculate the following:

[H3O+]
[OH-]
[C3H5O2-]
[I-]

  • chemistry - ,

    You should write out the equations to understand what is going on. Let's call propionic acid HP.
    .............HP ==> H+ + P-
    initial.....0.280M...0....0
    change......-x.......x.....x
    equil.....0.280-x....x......x

    ...........HI ==> H+ + I-
    initial..0.0894....0....0
    equil.......0..0.0894..0.0894

    Ka = (H+)(P-)/(HP)
    Substitute from the ICE charts and solve for (P-). (H+) = 0.0894+x and (HP) = 0.280-x

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