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Homework Help: Ingredient in aspirin is acetylsalicylic acid

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data
    The active ingredient in aspirin is acetylsalicylic acid. A 2.51 g sample of acetysalicylic acid required 27.36 mL of a 0.5106 M NaOH for complete reaction. Addition of 13.68 mL of 0.5106 M HCl to the flask containing the aspirin and sodium hydroxide produced a mixture with pH=3.48. Find the molar mass of acetylasalicylic acid and Ka value.

    2. Relevant equations

    3. The attempt at a solution
    I'm not really sure what to do but first i found the number of moles of acetysalicylic acid and NaOH (which is OH-) which were both 0.014 moles and they both have an equivalence point of 0 since it goes to complete reaction...then i get kinda lost..i did a neutralization with C9H8O4 which is acetysalicylic acid with OH- and found the Molar mass of C9H8O4- to be 0.014/v M because we don't know the total volume...then i get lost...
  2. jcsd
  3. May 9, 2009 #2


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    Science Advisor
    Homework Helper
    Gold Member

    Re: Titration

    At the point where you have 'complete reaction' of the NaOH with ASA, the number of moles of sodium hydroxide equals the number of moles of ASA. So you know moles and the mass (2.51g). Calculate the FW from that. Volume for the base is 27.36 mL. That is the only important volume in your problem.

    Remember that pH = -log[H+] and that Ka = [ASA-][H+]/[ASA]

    Any help?
  4. May 9, 2009 #3
    Re: Titration

    well what i did is:
    C9H8O4 + OH- --> C9H8O4- +H20..from that i found that the moles is 0.015 and since they're equal, both reactants cancel and all that is left is 0.015 moles of C9H8O4- (which i then divided by 27.36 mL converted into liters).

    then what? do i do a buffer with C9H8O4- or do i use the HCl onto that? Thanks.
  5. May 10, 2009 #4


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    Staff: Mentor

    Re: Titration

    I can't see 0.015 moles here.

    Once you will know correct number of moles, combine it with the sample mass to calculate molar mass.

    Note, that HCl added has identical concentration with NaOH, and that volume used is exactly half that of NaOH. Do you know phH of acid neutralized only in half?

    Finally, when you will solve the question, you should forget it as fast as possible. ASA doesn't react with NaOH in the way whoever wrote the question imagines. It hydrolizes fast enough to make this titration difficult if not impossible. The most reliable method calls for back titration - addition of excess NaOH, hydrolyzis (they don't react 1:1 then) and titration of the excess HCl.
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