K instab

  • Thread starter XTEND
  • Start date
  • #1
19
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Homework Statement



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


Homework Equations



K instab

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
 

Answers and Replies

  • #2
90
0
Does 'K instab' simply mean 'Keq?
 
  • #3
90
0
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?
 
  • #4
19
0
I thought I set up the ICE method correctly?

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

Yes same as K instab
 
  • #5
90
0
Redo the ice table with Hg(CN)2 .
 
  • #6
Borek
Mentor
28,473
2,871
1042 is a range for stability constant for Hg(CN)4.

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