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Chemistry AP

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It was determined that a 0.10 M solution of an acid was only 2.5% ionkzed. Find the Ka and pKa for the acid.

  • Chemistry AP - ,

    HA + H2O---> H3O+ + A-

    Ka=[H3O+][A-]/[HA]


    ..........HA I H3O+ I A-

    I.........I 0.1 M I 0 I 0 I
    C........I-2(0.1*0.025) I (0.1*0.025)I (0.1*0.025) I
    E.........I 0.1-2(0.1*0.025) I (0.1*0.025) I (0.1*0.025)


    Ka=[(0.1*0.025][(0.1*0.025]/[0.1-2(0.1*0.025]


    pka=-log(ka)

  • Chemistry AP - ,

    Opps,


    HA + H2O---> H3O+ + A-

    Ka=[H3O+][A-]/[HA]


    ..........HA I H3O+ I A-

    I.........I 0.1 M I 0 I 0 I
    C........I-2(0.1*0.025) I (0.1*0.025)I (0.1*0.025) I
    E.........I 0.1-(0.1*0.025) I (0.1*0.025) I (0.1*0.025)


    Ka=[(0.1*0.025][(0.1*0.025]/[0.1-(0.1*0.025]


    pka=-log(ka)

    I apologize about that one.

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