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Solubility problems

Discussion in 'DAT Discussions' started by Harley comic, Aug 9, 2006.

  1. Harley comic

    Harley comic Junior Member
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    I am having troubles with the solubility problems. I tried to study for it, but it didnt seem to promissing. I am taking the test very soon, do you guys think I should just forget about it and study something else instead? If you guys have a better way of doing it, please help!!!
    EX:
    How much (CrO4)2- must be addd to a liter of a saturated solution of AgCl in order to precipitate Ag2CrO4? (Ksp of AgCl=2.8 x 10^-10, Ksp of Ag2CrO4=1.4x10^-22)
    A. 2.5x10^-12
    B. 2.5x10^-13
    C.5.0x10^-13
    D.5.0x10^-12
    E.1.0x10^-13
    Kaplan practice test.
    Thanks a bunch
     
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  3. KiTmAn

    KiTmAn Kit
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    okay..first thinking logically first we need to set up teh equation using the second Ksp..
    equation 1 ..Ksp for Ag2CrO4 = [Ag2+]^2 [CrO4 2-]
    so now we need to know the concentration of Ag2+ which can be calculated by using the solubility for AgCl -->
    equation 2...2.8 x 10^-10 = x^2
    no need to calculate this..reason for this will become clear in a minute...
    if u think about it x value will be [Ag2+]..after taking the square root which u will have to square it later..when plugged into equation 1...so u can use the same value so in equation 1 ..
    1.4x10^-22 = (2.8 x 10^-10) (x)
    x = 1.4x10^-22 / 2.8 x 10^-10
    x = 0.5 x 10 ^-12
    x = 5 x 10^ -13
    now i m not sure if answer is C or A b/c adding amt of C will create a saturated solution ...so need to add more so answer can be A..
    let me know what the answer is ..
     
  4. Harley comic

    Harley comic Junior Member
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    the answer is C
    but you think we have enough time in the real test? This problem take me at least 5 minutes, do you think it is worth trying?
     
  5. KiTmAn

    KiTmAn Kit
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    now wouldnt C jus create a saturated solution Har..?
     
  6. KiTmAn

    KiTmAn Kit
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    i guess its C...i doubt we get qts like these on dat...it took me atleast 3 mins....
     
  7. keibee82

    keibee82 Blue_tooth...
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    in the second equation forming Ag2CrO4, there is two moles of Ag+ per
    one mole of CrO4. Therefore its 2x so you have to calculate the x value
    by square root from the first equation you had and then multiply by 2
    and then square. That gives you the value for the Ag+ concentration
    and divide Ksp for Ag2CrO4 by this concentration will give you concentration
    of CrO4.
     
  8. kkumalap

    kkumalap Member
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    that's wut I thought too. But Kaplan said otherwise...it's so weird. Can someone please clear this up.
     
  9. slayerdeus

    slayerdeus Member
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    If you are trying to find the concentration of Ag+ then you need to multiply it, but since you already know the concentration from the AgCl solution you don't need to multiply it by 2.
     
  10. KiTmAn

    KiTmAn Kit
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    yea...i knew i had to do that...jus dint kno the reason behind my using it...
     
  11. JohnDoeDDS

    JohnDoeDDS Senior Member
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    quick question. I understand this problem but...

    In this problem you got KSP= X*X

    But howcome sometimes you would say KSP = (2X)^2 *X
    Because you have 2 Ag+ so why doesnt it equal (2x)^2 but just X?
     
  12. fancymylotus

    fancymylotus A Whole New World
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    cause for AgCl it goes Ksp= [Ag][Cl]...no 2.
     
  13. DonExodus

    DonExodus Dentist in Virgin Islands
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    2.8e-10 = [Ag][Cl]. [Ag] = [Cl], so 2.8e-10= [Ag]² = 2.8e-10
    Therefore, [Ag] = square root of 2.8e-10

    For the second part, AgCl + CrO4 --> Ag2CrO4.
    The rate expression is
    1.4e-22 = [Ag]²[CrO4]
    [Ag]² = 2.8e-10
    Plug in, and you get
    1.4e-22 = (2.8e-10)[CrO4]

    Solving for [CrO4] yields .5e-12, or 5x10^-13
     
  14. KiTmAn

    KiTmAn Kit
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    k..i still have one last qt on this stupid problem...wouldnt 5 e -13 of CrO4 jus create a saturated solution...so wouldnt u need to add more.. e -12 or something? lemme know pleaseeeeee......
     

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