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March 28, 2017

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HF + NaOH =NaF + H2O
What will be the pH of a solution produced when 40 mL of 0.25 M HF is titrated with 80 mL of 0.125M NaOH? Ka for HF is 3.5*10^-4

  • Chemistry - ,

    40 mL x 0.25M = 10.0 millimols HF.
    80 mL x 0.125M = 10.0 mmols NaOH
    So you are at the equivalence point and the pH will be determined by the hydrolysis of the NaF.
    (NaF) = 10 mmols/120 mL - 0.0833M

    ..........F^- + HOH ==> HF + OH^-
    init..0.0833............0.....0
    change...-x.............x.....x
    equil.0.0833-x...........x....x

    Kb for F^- = (Kw/Ka for HF) = (HF)(OH^-)/(F^-)
    Substitute the equil line from the ICE chart into the K expression and solve for x = (OH^-) then convert to pH.

  • oops--Chemistry - ,

    See line 5.
    (NaF) = 10 mmols/120 mL - 0.0833M
    Replace the - with =

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