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25 February, 07:42

Calculate the ph of a 0.20 m solution of kcn at 25.0 ∘

c.

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  1. 25 February, 09:32
    0
    CN^-1 is the conjugate base of the weak acid HCN (Ka 3 X 10^-9) so CN^-1 is a strong base and will hydrolyze, by the reaction:

    CN^-1 + H2O - - > HCN + OH^-1

    The amount is based on the hydrolysis constant which is

    Kw / KaHCN = 1 X 10^-14 / 3 X 10^-9

    = 3.333 X 10 ^-6

    0.2 M CN^-1 will ionize to a small extent (A) to produce A amounts of HCN and A amounts of OH^-1

    soo Kh = 3.333 X 10^-6 = [A][A] / 0.2 - A

    but A <<< 0.2 (much lesser than) so it will be omitted in the denominator

    A^2 = 0.2 x 3.333 X 10^-6

    A^2 = 6.67 X 10^-7

    A = 8.17 X 10^-4

    pOH = - log (8.17 X 10^-4) = 3.09

    then ph is

    pH = 14 - pOH

    so

    pH = 14 - 3.09

    pH = 10.91
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