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26 March, 21:13

If the ka of a monoprotic weak acid is 5.8 * 10-6, what is the ph of a 0.45 m solution of this acid?

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  1. 26 March, 23:11
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    Kₐ = 5.8*10⁻⁶

    C = 0.45 mol/L

    [H⁺] = √ (KₐC)

    pH = - lg[H⁺] = - lg{√ (KₐC) }

    pH = - lg{√ (5.8*10⁻⁶*0.45) } = 2.79
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