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Chemistry

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Given that the Ka of benzoic acid is 6.50 x 10-5, how would one prepare 0.500 L of a benzoic acid/sodium benzoate buffer of a desired pH of 4.00? The starting concentration of the benzoic acid is 3.00 M and the molecular weight of sodium benzoate is 144.10 g/mol

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    I think the easiest way to do this is in one step after the preliminary information is set. For simplicity, let's call benzoic acid HB (instead of C6H5COOH) and the salt will be NaB.
    HB + NaOH ==> NaB + H2O
    Now we have 3.00M HB and we want 0.5L of the buffer so we use all of the HB (which is 3.0M x 0.5L = 1.5 moles HB.
    ..............HB + NaOH ==> NaB + H2O
    initial......1.5...0.........0.....0
    add................x................
    change.....1.5-x..-x.........+x.....+x
    equil......1.5-x...0..........x......x

    4.00 = pKa + log (B)/(HB)
    4.00 = pKa + log (x)/(1.50-x)
    Solve for x which equals moles NaOH (of course that is moles NaB but that's the same as moles NaOH. Then moles NaOH*molar mass NaOH = grams NaOH. If I didn't goof that should be close to 24 grams but you need to check that out closely. That's not an exact number.
    I STRONGLY suggest you do one more thing, as I always do, and that is to check the final pH. Start with the 24 g NaOH, or whatever it turns out to be, convert to moles, plug into the ICE chart and plug into the HH equation and see if the pH turns out to be 4.00. If so you're in good shape, if not, there is an error somewhere.

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