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What CH3COONa concentration is required to prepare a buffer solution with a pH of 4.80 if [CH3COOH] = 0.150 M? Ka of CH3COOH = 1.8 10-5.

  • chem -

    4.80 = pKa + log(x)/(0/150)
    Substitute pKa and solve for x.

  • chem -

    PH=-log(ka)+log(base/acid)
    4.8=-log(1.8*10^-8)+log(x/.15)
    4.8=7.7447+log(x/.15)
    -2.9447=log(x/.15)
    .001135=x/.15
    x=1.875*10^-4

  • chem -

    If I use your numbers I come out with 1.7E-4. The difference arises from 0.001135*0.15 = 1.7E-4 and I assume you just punched a wrong button; however, none of that counts because you didn't use the right Ka.
    Ka = 1.8E-5 so pKa = 4.74

  • chem -

    sorry read it wrong, doing this now in chem

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