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10 April, 17:55

What is the ph of a solution of 0.450 m kh2po4, potassium dihydrogen phosphate?

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  1. 10 April, 18:21
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    PKa = - log (Ka) = log [HPO4 (2-) ] - log[H+]^2 = - log (4.2*10^-13)

    pH = - log [H+]

    - log [H+]^2 = - 2 log [H+]

    2pH = - log (4.2*10^-13) - log [HPO4 (2-) ]

    2pH = - log (4.2*10^-13) - log (0.550)

    pH = 6.32
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