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Predicting a precipitate

  1. Jan 22, 2009 #1
    If 1.5g of [tex]Al_{2}(SO_{4})_{3)_{(s)}[/tex] is added to a beaker containing 1125 mL of 0.015M NaOH solution will a precipitate occur ? Ksp of [tex]Al(OH)_{3}[/tex] = [tex]2x10^{-8}[/tex]
     
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  3. Jan 22, 2009 #2

    Borek

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    You have to try by yourself first.
     
  4. Jan 22, 2009 #3
    15g/342g/mol = 0.00438 mol/1.125 L = 0.00389 M

    [tex]Al_{2}(SO_{4})_{3)_{(s)} --> 2Al^{3+} + 3SO_{4}^{-2}[/tex]
    [tex][Al^{3+}]=0.00389 mol/l divided by 2 moles of 2Al^{3+} produced=0.001945M[/tex]
    [tex]NaOH_{(aq)} --> Na^{+}_{(aq)} + OH^{-}_{(aq)}[/tex]
    [tex][OH] = 0.015 M 1:1 mole ratio[/tex]

    Q=(0.001945)(0.015)^3
    Q=6.56x10^-9
    Q<Ksp
    therefore precipitate forms.
     
  5. Jan 22, 2009 #4

    Borek

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    1.5 or 15?

    Why do you divide by two?
     
  6. Jan 22, 2009 #5
    Because 2 moles of Al3+ are produced for every Al2(SO4)3 that dissolved
     
  7. Jan 22, 2009 #6
    and Al(OH)3 produces 1 mole of Al3+ ions
     
  8. Jan 22, 2009 #7
    This problem is a little different instead of having 2 solutions being mixed together and findting the concentration of the diluted participants its rather putting a mass of something in this case Al2(SO4)3(s) into a beaker full of 1125ml = 1.125 L
     
  9. Jan 22, 2009 #8

    Borek

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    Once again - if two moles of Al3+ are produced per each mole of salt dissolved, why do you divide by two?

    Once again: was it 1.5 g, or 15 g, as you started with one number, but you used th eother in your calculations.
     
  10. Jan 22, 2009 #9
  11. Jan 22, 2009 #10
    i just did my exam and surprisingly this question was on it. Let me set up the problem properly
     
  12. Feb 21, 2009 #11
    so is anyone going to help me out on this or not ?
     
  13. Feb 21, 2009 #12
    Ive trie and tried and i tried what the ****
     
  14. Feb 22, 2009 #13
    If 1.5 grams of Al2(SO4)3(s) is added to a beaker containing 1125 mL of 0.015M NaOH solution will a precipitate occur? Ksp of Al(OH)3(s)=2.0x10^-8
     
  15. Feb 22, 2009 #14

    Borek

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    Show your calculations. You have already tried once, but you were wrong - and I have pointed to two problems in your calculations. You have ignored my post so far.
     
  16. Feb 22, 2009 #15
    Thats the problem i dont know how to do the calculation!! When i say i dont know how i reaaaaaally dont know how to do the problem !!!!!!!!!!
     
  17. Feb 22, 2009 #16
    What dont u understand!?!?!?
     
  18. Feb 22, 2009 #17
    it looks like u ignored my my POST so far when u asked me if it was 15 or 1.5 grams I ALREADY TOLD YOU its 1.5 grams man ****
     
  19. Feb 22, 2009 #18

    Borek

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    You have correctly started solving the question by calculating concentration of Al3+. However, you calculated it wrong.
     
  20. Feb 22, 2009 #19
    Al2(SO4)3(s) [tex]\Longleftrightarrow[/tex] 2Al3+(aq) + 3SO42-(aq)

    The question says i have 1.5 grams so im gonna take 1.5g/342.14g/mol molar mass = 0.00438 mol

    Then since the Al2(SO4)3(s) is being submerged into the NaOH solution that has volume 1125 mL (1.125 L) im assuming it goes in the beaker so it's concentration becomes:

    0.00438 mol/1.125 L = 0.00389 mol/L
     
  21. Feb 22, 2009 #20

    Borek

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    This is concentration of aluminum sulfate, but not of Al3+. How many moles of Al3+ are introduced into the solution per each mole of dissolved sulfate?
     
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