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Posted by on Wednesday, November 23, 2011 at 12:50am.

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54e-5), with 0.1000 M NaOH solution after the following additions of titrant: 0 mL, 14.00 mL, 19.95 mL, 20.00 mL, 20.05 mL, 24.00 mL.

  • Chemistry - , Wednesday, November 23, 2011 at 12:02pm

    The secret to doing this is to recognize what you have at each titration point. If we let HB stand for butanoic acid, then
    the equation is HB + NaOH ==> NaB + H2O

    a. at the beginning you have just HB. Set up an ICE chart
    ............HB ==> H^+ + B^-
    initial....0.1.......0....0
    change.....-x.........x....x
    equil.....0.1-x.......x.....x

    Substitute into the Ka expression, solve for H^+, and convert to pH.

    Next determine the equivalence point from mL HB x MHB = mLNaOH x MNaOH
    The pH at the equivalence point is determined by the concn of the salt produced. It is hydrolyzed. Set up an ICE chart for that.
    .........B^- + HOH ==> HB + OH^-
    initial..C.............0.....0
    change...-x............x......x
    equil....C-x............x.....x

    Kb = (Kw/Ka) = (HB)(OH^-)/(C)
    Kw you know. Ka you know.
    C is determined from the initial equation at the top of the page and is moles/L = M. Solve for HB = x= OH^-, convert to pOH, then to pH.

    All points leading up to the equivalence point are buffered solutions, partly HB an partly NaB and can be solved with the Henderson-Hasselbalch equation.

    All points after the equivalence point are due to the excess HCl. Calculate excess HCl (don't forget this is diluted by the volumes) and convert to pH.
    Post your work if you get stuck.

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