How Does Adding Water Affect the Equilibrium in the BiOCl Dissolution Reaction?

[Declan McKeown]

Hey everyone,

At the moment I am stuck with an example and was wondering if someone could please explain it to me:

You have the equation:

BiOCl(s) + 2H+(aq) -><- Bi3+(aq) +Cl-(aq) +H2O(l)

Why is it that when you add extra water to the system, there is no change to the equilibrium? I would have thought that the equilibrium would shift left and the reverse reaction would be favoured? (The answers from the book in which the question came does not give a reason why).

Thanks a tonne

 

[Ygggdrasil]

Since the reaction is happening in water, the concentration of water is already very high (~55M). Adding more water to the solution does not appreciably change the concentration of water in the reaction. Adding more water will, however, dilute the concentration of aqueous ions, so it's important to determine whether the dilution of H+, Bi3+ and Cl- will not affect the position of the equilibrium.

 

[Declan McKeown]

Since the reaction is happening in water, the concentration of water is already very high (~55M). Adding more water to the solution does not appreciably change the concentration of water in the reaction. Adding more water will, however, dilute the concentration of aqueous ions, so it's important to determine whether the dilution of H+, Bi3+ and Cl- will not affect the position of the equilibrium.

Aaah I see now. I did not think of that, thanks so much :D

 

[epenguin]

Well if you were confused there would be no shame in it. I was and am not sure I'm 100% unconfused.

Firstly I don't remember encountering this reaction but I looked it up and apparently it is a standard experiment for schools http://www.nuffieldfoundation.org/practical-chemistry/le-chatelier’s-principle-effect-concentration-equilibrium

But then I don't like the explanation given in this link - they are talking of BiCl₃ like a soluble covalent compound. I would expect Bi³⁺ as in your formulation. (Although if you allow both, in equilibrium, probably it doesn't make any difference to form of the resulting equations. But I would stick with your formulation for all your purposes).

Another possible source of confusion is you have Cl^- on the RHS lf your equilibrium so shouldn't adding the acid HCl drive the reaction leftwards?

No, because the BiOCl, and and hence any Cl^- coming from it corresponds to only about 4mM in the above quoted experiment. The Cl^- then approximately all comes from the added HCl. So we can say that to good enough approximation

[H⁺] = [Cl^-] = [HCl]. Where the [HCl] means molarity of total added HCl

So the equilibrium equation [BiOCl][H⁺]² = K [Bi³⁺][Cl^-] (the [H₂O] being left out of this equation because it is constant as already explained) becomes

[BiOCl][HCl]² = K [Bi³⁺][HCl]
i.e.
[BiOCl][HCl] = K [Bi³]

So molarity of dissolved Bi just depends straightforwardly on that of HCl. BiOCl is insoluble so solution [BiOCl] above can be assumed constant (small).

According to the above nitric or sulphuric acid should be more effective in dissolving because their concentrations would enter the equation to a higher power.

But then again to complicate it I read that Bi can form a variety of chloro complexes like BiCl₆^3- so who knows whole story? Probably not your teachers nor are you expected to - however if you get the chance to do the experiment try these other acids - at lower concentration than the [HCl] that dissolves the stuff. And if you can do that tell us.