Chemical Equilibrium: Co(H2O)62+ + 4 Cl- ----> CoCl42- & 6 H2O

[corinnab]

I am totally confused with this question and have trying to figure it out for hours!

Co(H2O)62+ (aq) + 4 Cl- (aq) ----> CoCl42- (aq) + 6 H2O (l)

The equilibrium constant K = 4.82x 10-4
We begin by putting enough [Co(H2O)6]Cl2 (assume the salt dissociate completely into
ions) into 100.0 mL of water to make the formal concentration 0.182 mol/L

(a) Demonstrate that this system is not at equilibrium, initially.
(b) When the system comes to equilibrium, what are the final concentrations of each
of the species?

 

[Kushal]

(a) you should show that the concentration constant initially is not equal to the equilibrium constant.

(b) find the concentrations of each species in terms of x and then solve the equation to find it.
e.g. let final concentration of CoCl42- be x moldm-3. you may ignore the concentration of water since it is in excess.

 

[engware]

Hi there:

You could try to use some of the powerful software packages to help you do the calculations that eventually need to be done by hand -- software help is suggested to more efficiently get to the first numerical values and ease the process of getting meaningful numerical values quickly...

Thanks,

Gordan

 

[corinnab]

I am still not sure how to show the conc. of Cl- in terms of x. Can you show the conc. of Co(h20)62+ as 0.182-x?
And I know that intially Q doesn't equal K but how can I show this?

 

[epenguin]

Well write out what equations you are able. State what numbers from the problem you know for any terms or combination of terms. When you do that you may find the answer comes, or someone may be enable to make suggestions.

Check back in your book what it says about a convention regarding the "concentration" of water in the definition of equilibrium constants of this type, something essential to realize.

 

[Kushal]

you calculate Q(the initial value for the constant, when the reaction is not yet in equilibrium).

let the conc. of CoCl4 2- at equilibrium be x moldm-3
the conc of Co(H2O)6 2+ at equilibrium is (its initial concentration) - x
apply the same for Cl-

use K to solve for x

 

[Barfolumu]

Are you familiar with the http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/ICEchart.htm"

This often helps me in conceptually setting up the algebraic expression necessary to solve equilibrium problems.

EDIT: I also noticed that you did not mention http://www.chemguide.co.uk/physical/equilibria/lechatelier.html" , which also helps me in using ICE charts.

 

[corinnab]

I came up with x=9.625x10-8 hope this is right

 

[corinnab]

for the equilibrium part is it ok to say that Q doesn't equal K initially therefore reaction isn't at equilibrium. q is less than K therefore reaction moves towards products

 

[johnpotty]

did you manage to figure out how to get the initial concentrations for each of those ions corinnab? coz I am stuck on the exact same problem!

 

[Kushal]

(a) you calculate Q, show that that it is not equal to K. because if the reaction was at equilibrium, Q should have been equal to K.

(b) water is largely in excess(aqueous medium). the change in concentration is very very
small. therefore it can be ignored.

K = [CoCl42-]/[Co(H2O)62+][Cl-]^4

initial concentration of Co(H2O)6^2+ = 0.0182 mol/dm3
initial concentration of Cl- = 2*0.0182 = 0.0364 mol/dm3

let final concentration of Cl- be x mol/dm3
final concentration of Co(H2O)6^2+ = (x/2) mol/dm3

final concentration of CoCl42- = 0.0182-(x/2) mol/dm3

K = 0.0182-(x/2)/(x/2)x^4

i ended up with K*x^5 + 8x = 0.2912

i got x = 2.06 moldm-3
this cannot be the answer... i probably left an error somewhere

 

[Kushal]

if you do it the other way round, putting the final concentration of CoCl42- to be x mol/dm3 it's even more complicated to solve.

 

[Kushal]

THE ABOVE IS WRONG FOR PART (b)!

K = [CoCl42-]/[Co(H2O)62+][Cl-]^4

initial concentration of Co(H2O)6^2+ = 0.0182 mol/dm3
initial concentration of Cl- = 2*0.0182 = 0.0364 mol/dm3

let final concentration of CoCl42- be x mol/dm3

final concentration of Co(H2O)6^2+ = (0.0182-x) mol/dm3
final concentration of Cl- = (0.0364-4x) mol/dm3

K = x/(0.0182-x)(0.0364-4x)^4

This one should be ok