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How could you prepare a buffer of pH=4.50 from a 1M of NaOH and a 1M...

How could you prepare a buffer of pH=4.50 from a 1M of NaOH and a 1M of Acetic Acid (pka=4.63) describe how and do the calculations.

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Answer #1

pka of Acetic Acid = 4.63

pH Of acidic buffer = pka + log(salt(or)base/acid)

let us assume we would like to prepare 1 L buffer

total no of mole of buffer = base + acid = 1*1 = 1 mole

no of mole of Acetic Acid = x mol

   no of mole of NaOH = 1-x mol

    4.5 = 4.63+log((1-x)/x)

    x = 0.574

no of mole of Acetic Acid = x = 0.574 mole

   no of mole of NaOH = 1-0.574 = 0.426 mole

volume of 1 M NaOH required = n/M = 0.574/1 = 0.574 L

volume of Acetic Acid must take = n/M =0.426/1 = 0.426 L

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