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K instab

  1. Aug 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the concentration of Hg+2 at equilibrium in a solution that is initially 0.10M in Hg+2 and 0.50M in CN-.

    K instab for [Hg(CN)4]2- = 3.0 x 10 -42


    2. Relevant equations

    K instab

    3. The attempt at a solution

    [Hg(CN)4]2- --> Hg2+ + 4 CN- Initial Conc (M) 0 0.10 0.50 Change x -x -4x Equilibrium x 0.10 - x 0.50 – 4x Kinstability = [Hg2+][CN-]4 [Hg(CN)4]2- 3.0 x 10-42 = (0.10 - x)(0.50 – 4x)4 x 3.0 x 10-42 x = (0.10 - x)(0.50 – 4x)4 3.0 x 10-42 x = (0.10 - x)[(0.50 – 4x)2]2 3.0 x 10-42 x = (0.10 - x)(0.25 – 8x – 16 x2)(0.25 – 8x – 16 x2) 3.0 x 10-42 x = (0.10 - x)(0.25 – 8x – 16 x2)(0.25 – 8x – 16 x2)

    answer so far: x= 0.029507

    ??? can't get any further, think this calculation may be off
     
  2. jcsd
  3. Aug 10, 2009 #2
    Does 'K instab' simply mean 'Keq?
     
  4. Aug 10, 2009 #3
    The empirical formula of the simplest combination of the Hg2+ ion and the CN- ion is Hg(CN)2. Are you sure you've got the right product in that equilibrium?
     
  5. Aug 10, 2009 #4
    I thought I set up the ICE method correctly?

    Is it not [Hg(CN)4]2???

    Yes same as K instab
     
  6. Aug 11, 2009 #5
    Redo the ice table with Hg(CN)2 .
     
  7. Aug 11, 2009 #6

    Borek

    User Avatar

    Staff: Mentor

    1042 is a range for stability constant for Hg(CN)4.

    --
     
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