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A 25.0mL sample of butanoic acis is titrated with 0.150M NaOH solution. What is the pH...

A 25.0mL sample of butanoic acis is titrated with 0.150M NaOH solution. What is the pH before any base is added? The Ka of butanoic is 1.5x10^-5.
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Answer #1

pka of butanoicacid = -logka

                   = -log(1.5*10^-5)

                   = 4.82

pH of butanoicacid =1/2(pka-logC)

      pka = 4.82

     C = concentration of butanoicacid = 0.15 M

pH = 1/2(4.82-log0.15)

     = 2.82

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