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2 June, 14:57

Find the pH of a 0.310 M aqueous benzoic acid solution. For benzoic acid, Ka=6.5⋅10-5.

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  1. 2 June, 14:58
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    C=0.310 mol/L

    Ka=6.5·10⁻⁵

    pH=-lg√ (C·Ka)

    pH=-lg√ (0.310·6.5·10⁻⁵) = 2.35
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