Buffer L

CH3CO2H: 0.109M & 20ml

NaOH : 0.109M& 10ml


Buffer Problem, when I solve it the ratio becomes the same and the pH is the same as the pka, is that right?

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  1. Yes, pH = pKa. WHY? because the log(base)/(acid) = 0
    pH = pKa + log(base)/(acid)
    pH = pKa + log (0.109/0.010)/(0.109/0.010)
    pH = pKa + log (1)
    pH = pKa + 0 = pKa.

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