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Equilibrium Reaction of Insoluble Water

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Lead(II) carbonate is very insoluble in water, meaning that the equilibrium constant for the dissociation reaction is much less than 1:

PbCO_3 (s) <-> Pb(2+) (aq) + CO_3(2-) (aq)

Write reactions for each of the following and explain how the equilibrium reaction for the dissociation of lead(II) carbonate is affected.

a. H+ from stomach acid reacts with the carbonate ion

b. Enzymes react with the lead ions to form Pb(enzyme) complexes

c. Lead poisoning can be treated with ethylenediamine (EDTA) which binds with Pb(2+) to form the complex ion [Pb(EDTA)](2=)

d.In order for EDTA administration to be effective, the size of K_eq must be very large for the binding reaction of Pb(2+) with EDTA. Explain why this is.

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This question deals with equilibrium. More specifically, this problem is asking about the topic of Le Chatelier's principle. Le Chatelier's principle states that when there is a "stress" or change made to a system at equilibrium, the system will shift to form more equilibrium. Addition or removal of a solid or pure liquid to a system should not have an affect on the equilibrium.

Your question asks for you to consider how addition of different things will affect the equilibrium.

The system that you have given at equilibrium is:

PbCO_3 (s) <-> Pb(2+) (aq) + CO_3(2-) (aq)

For part a:

The question informs you that H+ reacts with the carbonate ion. The carbonate ion is the CO_3(2-). The REACTION between the two would be:

2 H+(aq) + CO_3(2-) (aq) <-----> H2CO3 (aq)

Remember that because the hydrogen ion is ...

Solution Summary

The equilibrium reactions of lead(II) carbonate being insoluble in water is determined.