The base dissociation constant of ethylamine (c2h5nh2) is 6.4×10^-4.?

The [H] in a 1.6×10^-2 M solution of ethylamine is ____?

The answer is supposed to be 3.5 x 10^-12, but somehow I keep getting 3.1×10^-12. So, if you can show your work that would be a huge help!

1 Answer

  • 1 month ago

    C2H5NH2 + H2O <=> C2H5NH3+ + OH-

    start

    1.6 x 10^-2

    change

    -x . .. . . . . . . . . . . . .. . .+x .. . . . .+x

    at equilibrium

    1.6 x 10^-2-x. . . . .. . . . . .x. . . .. . . x

    6.4 x 10^-4 = x^2 / 1.6 x 10^-2-x

    we must solve by quadratic formula because x is not negligible compared to 1.6 x 10^-2

    1.0 x 10^-5 – 6.4 x 10^-4 x = x^2

    x^2 + 6.4 x 10^-4 – 1.0 x 10^-5 =0

    x = – 6.4 x 10^-4 + sq.rt ( 4.1 x 10^-7 + 4.0 x 10^-5)/2 = 0.0028 M = [OH-]

    [H+]= Kw/[OH-]= 1.0 x 10^-14/ 0.0028 = 3.5 x 10^-12 M

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