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15 April, 17:06

What is the ph of a 0.023 m solution of hcn (ka = 4.9 x 10-10) ?

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  1. 15 April, 18:13
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    The ionic equation is simply:

    HCN - - > H + + CN-

    Therefore the formula for Ka is:

    Ka = [H+] [CN-] / [HCN]

    [H+] = [CN-] = x

    4.9 x 10^-10 = x^2 / 0.023

    x^2 = 1.127 x 10^-11

    x = 3.357 x 10^-6 = [H+] = [CN-]

    pH = - log [H+]

    pH = - log 3.357 x 10^-6

    pH = 5.47
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