Chem

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in a 0.40 M solution of diprotic acid H2A(Ka=7.4x10^-5, Ka2=5.0x10^-10 at 25 celsius), what is the equilibrium concentration of A^2-? A)0.40m b)0.80M C)5.4x10^-3m d)1.4x10^-5M E)5.0x10^-10M

  • Chem -

    H2A+ H2O----H3O+ HA^-2

    HA^-2+ H2O-----> H3O + A-


    Let x=HA^-2
    Let y=A-

    So, for the following reaction

    H2C2O4 + H2O----H3O+ HC2O4

    H2C2O4...........H3O....HC2O4
    I0.40M..............0 ........ 0
    C-x.....................x............ x
    E0.40-x.............x.............x


    Solve for x,

    ka1=[x][x]/[0.40-x], which turns into,


    ka1=[x][x]/[0.40]

    sqrt*(ka1*0.40)=x=5.4 x 10^-3

    HC2O4 + H2O-----> H3O + C2O4


    HC2O4...........H3O....C2O4
    I.......Y..............0 ............ 0
    C....-z...............z.............z
    E....y-z..............z.............z



    But since it is a diprotic acid and x=H3O+ y= H3O+, the charts become


    H2C2O4 + H2O----H3O+ HC2O4

    H2C2O4...........H3O....HC2O4
    I0.4M..............0 ........ ..0
    C-x..............,,..x............ .x
    E0.4-x.........x+Y..........x-y


    HC2O4 + H2O-----> H3O + C2O4
    HC2O4..................H3O...............C2O4
    I...5.4 x 10^-3.........x.....................0
    C..-y......................x+y.....................y
    E...5.4 x 10^-3-y...x+y.....................y

    So, ka1=[5.4 x 10^-3+y][5.4 x 10^-3-y]/[0.370]

    and

    ka2=[5.4 x 10^-3+y][y]/[5.4 x 10^-3]


    We solved for x, so solve for y.


    I believe this is how you tackle this problem.

  • Chem -

    See the above.

  • Chem -

    Change H2C2O4 to H2A, HC2O4 to HA^2, and C2O4 to A-. Everything else is okay.

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