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What is the pH of a solution made up of 50cm^3 of 1M Ethanoic Acid and 25cm^3 of 1M sodium ethanoate? Assume Ka for Ethanoic acid is 1.8 x 10^-5 mol dm^-3

I'm not quite sure how to do this. I was going to use the equation pH = pKa + log[A-]/[HA] but it doesn't seem to work

  • Chemistry -

    You're right. The Henderson-Hasselbalch equation is the one to use. If you want to post your work someone will take a look and try to find the error. You should be substituting 25/75 for for the base and 50/75 for the acid. The denominators cancel, of course, and you have just 25 for (A^-) and 50 for (HA)

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