- #1

- 19

- 0

## 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