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How many mL of .200M acetic acid are mixed with 13.2mL of.200M sodium acetqte to give a buffer with pH=4.2

  • Sara -

    Use the Henderson-Hasselbalch equation.

  • Sara -

    I'm making a mistake in using that

    n (CH3COO-)= .0132*.2 = .oo264

    ..does that mean that CH3COOH also has the same n?

    ph= pKa +log(Conjugate Base/ Weak Acid)
    4.2= pKa + log (.00264/ .00264)

    since i have n the same it doesn't work. How do you calculate n?

  • Sara -

    pKa = 4.76 are thereabout. Look it up if you know it.
    4.2 = 4.76 + log (base/acid)
    -0.56 = log (B/A)
    -0.56 = log (0.00264/A)
    0.275 = 0.00264/A
    So the acid (acetic acid) must be
    0.00264/0.275 = 0.00960 moles (technically molar but the volume always cancels and you can ingore that for the moment.)
    M x L = moles to find the L necessary.
    Check my arithmetic. My calculator rounded here and there so you need to go through it again to make sure.

  • Sara -

    how did you calculate pKa because this is a practise question for my test and we won't be given the value. so how would you calculate it?

  • Sara -

    Don't you have Ka for acetic acid = 1.75 x 10^-5 or something like that.
    Then pKa = -log Ka = -log 1.75 x 10^-5 = -(-4.75696) = 4.75696 which I rounded to 4.76. And I get something like 48 mL of the acetic acid required but check my math.

  • Sara -

    Neither Ka nor Pka was given

  • Sara -

    On tests I have seen, they usually give Ka or pKa in the problem OR there will be an appended set of tables that will contain constants necessary to take the exam. I don't know of any test in which you are expected to know those constants.

  • Sara -

    okie thank you =)

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