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Chemistry

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Aspirin is a weak acid with a Ka of 3.0x10^-5. Find the pH of a solution by dissolving .65g of aspirin in water and diluting it to 50mL. You can use RCOOH to represent aspirin. Molecular weight of aspirin is 180g/mol

  • Chemistry -

    moles aspirin = grams/molar mass = 0.65/180 = approximately 0.004 mole.
    .........RCOOH ==> H^+ + RCOO^-
    initial....0.004....0......0
    change.....-x......x........x
    equil...0.004-x.....x........x

    Ka = (H^+)(RCOO^-)/(RCOOH)
    Ka = (x)(x)/(0.004-x)
    What is x. Solve for x = (H^+), then convert to H.
    pH = -log (H^+).

  • Chemistry -

    The answer that was in my answer key was 2.83 but I got 3.48. Is there something wrong in the calculations?

  • Chemistry -

    2.83 is right (and I obtained 3.46 using 0.004, also). Apparently you took my "apparoximately 0.004) and assumed that was gospel. I had to stick something there and didn't want to calculate it so I guessed at it and wrote approximately. I assumed you would substitute the correct value although I didn't say you should confirm my approximate number (but you should have). The molarity of the aspirin is 0.0722. (NOTE: 0.65g x (1 mol/180g) x (1000 mL/50 mL)

  • Chemistry -

    For the ICE box, you are supposed to use the molarity of the aspirin. I did not realize that. Thank you so much for your help!

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