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Gen Chem Question from Kaplan

Discussion in 'DAT Discussions' started by esperi, Apr 17, 2007.

  1. esperi

    esperi New Member

    6
    0
    Feb 26, 2004
    Question from Kaplan Practice Test:

    How many moles of AgIO3 (Ksp=3.1x10^-8) will dissolve in 1 Liter of a 10^-5 M solution of NaIO3?

    a. sqrt(3.1x10^-8)
    b. [sqrt(3.1x10^-8)] [10^-5]
    c. [sqrt(3.1x10^-8)] / [10^-5]
    d. [10^-5] - [sqrt(3.1x10^-8)]
    e. [sqrt(3.1x10^-8)] - [10^-5]

    Kaplan says the answer is E, but I don't get why.

    I get how Ksp=[Ag+][IO3-]= X^2= 3.1x10^-8, thus X= sqrt(3.1x10^-8). But why subtract [10^-5] from it ??? Anyone care to explain? Would appreciate it very much!
     
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  3. jdmsamurai

    jdmsamurai 2+ Year Member

    237
    1
    Feb 11, 2007
    common ion effect. in the solution you already have 10^-5 Na and IO3, so when you are trying to dissolve AgIO3 there is already IO3 in it...roughly 10^-5 IO3 ions. so solving for X will give you the concentration of the ions in a pure aq. solution, but since you have IO3 in it you substract it from the sqrt(X). this gives you the concentration due to common ion effect. just imagine trying to dissolve AgIO3 but there is already IO3 in it...so then the reverse reactions would be Ag + IO3 --> AgIO3 again.

    Remember at equilibrium there is a forward reactions and backwards reactions going on...
     
  4. Greensalad

    Greensalad 10+ Year Member

    111
    0
    Apr 16, 2007
    So then how would you write the equation? AlIO3+NAIO3--> Na+ Al + 2IO3? Just realized, what are the oxidation numbers?
     
  5. jdmsamurai

    jdmsamurai 2+ Year Member

    237
    1
    Feb 11, 2007
    i would probably use the "ICE" method

    Initial, Change, Equilbrium...and since AgIO3 is probably a solid you dont need it in the final equilibrium equation

    so it would looke like this

    AgIO3 --> Ag + IO3
    1) Initial 0 10^-5
    2) X X X
    3) Initial-x X X + 10^-5

    so...
    ksp = [X][X + 10^-5] and i am guessing to make it easy they made the fact that Ag concentration would also be X + 10^-5 due to the fact of the common ion effect of IO3 in the water would reverse the reaction back into a precipitate therefore making the full potential of AgIO3 to dissolve into its ksp not true. so then your new equilibrium equation is

    ksp = [x + 10^-5]^2
    and you can solve it from there


    about your oxidation number...it doesnt play a crucial role in this one because it looks like its a neutral compound and your two major players are IO3 and Ag. and Ag has a positive charge while our compound IO3 has a negative charge. O usually has an oxidation number of -2 so then thats like -6, but I am guessing I has a positive oxidation number lets pretend and say +1 so then IO3 has a negative 5 oxidation and then Ag would have a +5
     

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