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21 November, 12:34

What is the ph of a 0.35 m solution of anilinium nitrate (c6h5nh3no3) ? kb for aniline is 4.2 * 10-10. your answer must be within ± 0.4%?

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  1. 21 November, 13:05
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    When the balanced reaction equation is:

    and by using ICE table:

    C6H5NH3 + (aq) + H2O (l) ↔ C6H5NH2 (aq) + H3O + (aq)

    initial 0.35 0 0

    change - X + X + X

    Equ (0.35-X) X X

    so, the Ka expression = [C6H5NH2][H3O]/[C6H5NH3+]

    when Ka = Kw / Kb

    and we know that Kw = 1 x 10 ^-14 & Kb is given = 4.2 x 10^-10

    ∴ Ka = (1 x 10^-14) / (4.2 x 10^-10)

    = 2.4 x 10^-5

    by substitution on Ka expression:

    2.4 x 10^-5 = X*X / (0.35-X) by solving for X

    ∴ X = 0.00289 M

    ∴[H3O+] = X = 0.00289

    when PH = - ㏒[H3O+]

    = - ㏒ 0.00289

    ∴PH = 2.54
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