K instab

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

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)

??? can't get any further, think this calculation may be off

Related Biology and Chemistry Homework Help News on Phys.org
Does 'K instab' simply mean 'Keq?

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?

I thought I set up the ICE method correctly?

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

Yes same as K instab

Redo the ice table with Hg(CN)2 .

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

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