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Posted by on Saturday, June 30, 2012 at 10:02pm.

the ph of a 1.00*10^-2 M solution of cyanic acid (HOCN) is 2.77 at 25 C. Calculate Ka and pKa for HOCN from this result.

  • chemistry - , Saturday, June 30, 2012 at 11:18pm

    pH = 2.77 = -log(H^+)
    (H^+) = 1.70E-3

    .......HOCN ==> H^+ + OCN^-
    I....0.0100.....0.......0
    C......-x.......x.......x
    E.....0.0100-x..x........x

    Ka = (H^+)(OCN^-)/(HOCN)
    Substitute into the Ka expression and solve for Ka. You know (H^+), (OCN^-) and (HOCN)
    Then pKa = -log Ka.

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