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Solubility product

  1. Jun 24, 2009 #1
    1. The problem statement, all variables and given/known data

    The ksp values of AgCl and Ag2CrO4 are 2*10-10 mol2dm-6 and 3*10-12mol3dm-9 respectively.100cm3 of an aqueous solution contains NaCl and K2CrO4 only,whose cocentrations are 0.1M and 0.1M respectively.To this solution,is added AgNO3 dropwise while mixing.

    (1) Calculate and show whether AgCl or Ag2CrO4 will be precipitated first?

    (2) When the second precipitate begins to appear,what is the concentration of the anion of the Ag salt that precipitated first ,which still remains in the solution?

    (3) Using the above answer,explain briefly the accuracy of K2CrO4 as an indicator when used in the above titration.

    2. Relevant equations



    3. The attempt at a solution

    I managed to do part (1) of the problem but m having trouble with the rest

    For(1),
    AgNO3[tex]\rightarrow[/tex] Ag++NO3-

    AgCl [tex]\rightarrow[/tex] Ag++ Cl-

    Ksp = [ Ag+] [Cl-]
    2*10-10/0.1 = [ Ag+]
    [ Ag+]=2*10-9M

    Ag2CrO4[tex]\rightarrow[/tex]2Ag++ CrO42-
    Ksp= [Ag+]2[CrO42-]
    3*10-12/0.1 = [Ag+]2
    [Ag+] = 5.477*10-6M

    since the minimum concentration of Ag+ needed to precipitate AgCl is less than that needed to precipitate Ag chromate,AgCl would precipitate 1st.

    (2) I'm not sure I understand the question,
    which means the [Ag+] present in the solution is now, 5.477*10-6M,right? but what do I do now?Do I just substitute this value to the ksp equation of AgCl and find the[chloride] in the precipitate and then substract this value from 0.1M?

    THANK YOU.
     
  2. jcsd
  3. Jun 24, 2009 #2

    Borek

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    Staff: Mentor

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    precipitation titration end point detection

    After all Cl- was precipitated in form of AgCl you still add more solution of AgNO3 till the concentration of Ag+ is high enough for the Ag2CrO4 to precipitate.

    At this moment you are done :smile:
     
  4. Jun 25, 2009 #3
    Thanx for the very helpful link.
    So the [chloride] of AgCl would be=2*10-10/5.477*10-6

    =0.365*10-4M

    Thanks again.
     
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