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    Stuck on this confusing buffer question! Q4eiii) from the f325(EEE) June 15 paper. Student adds 6.075g Mg to 1.00dm^3 of the buffer solution, calculate new Ph of buffer?. Previous info: Buffer solution contatins propanoic acid, C2H5COOH and propanoate ions. The concentrations are both 1.00 mol dm-3. Ka for C2H5COOH is 1.35*10^-3. ANY HELP WOULD BE APPRECIATED!
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    (Original post by Ss0)
    Stuck on this confusing buffer question! Q4eiii) from the f325(EEE) June 15 paper. Student adds 6.075g Mg to 1.00dm^3 of the buffer solution, calculate new Ph of buffer?. Previous info: Buffer solution contatins propanoic acid, C2H5COOH and propanoate ions. The concentrations are both 1.00 mol dm-3. Ka for C2H5COOH is 1.35*10^-3. ANY HELP WOULD BE APPRECIATED!
    pH = pKa + log([salt]/[acid])

    So, find the concs and get the pKa from Ka (using -log_10).
    The salt is Mg(C2H5O)2. Find its conc. Consider the reaction equation for ratios.

    Hopefully, that's a start.
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    This paper is still secure, i.e. it should not be available to the public.

    It really shouldn't be discussed.
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    Oh thankyou, i'll try this method.
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    (Original post by Ss0)
    Stuck on this confusing buffer question! Q4eiii) from the f325(EEE) June 15 paper. Student adds 6.075g Mg to 1.00dm^3 of the buffer solution, calculate new Ph of buffer?. Previous info: Buffer solution contatins propanoic acid, C2H5COOH and propanoate ions. The concentrations are both 1.00 mol dm-3. Ka for C2H5COOH is 1.35*10^-3. ANY HELP WOULD BE APPRECIATED!
    I was wondering since our upcoming chemistry exams are coming up if people wanted to be part of a chemistry whatsapp revision group. I guessing most of us use whatsapp so I guess it will be more suitable for everyone then. If you are interested either pm me or responded to this message











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