Inorganic Chemistry Help: Solubility Transformation and Equilibrium Constants

AI Thread Summary
The discussion focuses on the transformation of sulfate ions to carbonate ions in a solution containing Ba^2+, Sr^2+, and Ca^2+ ions. The proposed reaction equation for the transformation is BaSO4(s) + CO3^2-(aq) --> BaCO3(s) + SO4^2-(aq). The solution is determined to be basic due to the presence of CO3^2-, which is a weaker base than SO4^2-. There is confusion regarding the calculation of equilibrium constants, as the user notes that the constants should differ for each cation but initially uses a single equation. The discussion concludes that BaSO4 can react to form BaCO3 despite its lower solubility because the system will reach equilibrium, allowing for the transformation to occur.
broegger
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Hi,

I'm doing this chemistry report, and I have absolutely no experience in chemistry.

We have this solution containg (among some other stuff) Ba^2+, Sr^2+ and Ca^2+ ions along with SO4^2-ions in excess. Now we add some Na2CO3 to transform the sulphates to carbonates. They ask me these questions:

1) Write the reaction equations for this transformation. My guess is:

BaSO4(s) + CO3^2-(aq) --> BaCO3(s) + SO4^2-(aq)​

and so on.

2) Is the solution acid or basic in this reaction? Since CO3^2- is a weaker base than SO4^2-, I would say that the answer is basic.

3) Calculate the equilibrium constants by using the solubility products. Ok, here I start by writing the equilibrium equation:

K=\frac{[{\text{SO}_4}^{2-}]_{eq}}{[{\text{CO}_3}^{2-}]_{eq}}​

This should be valid for both the Ba-, Sr- and Ca-case. But this doesn't make sense, since the constants should be different. What am I missing here?

4) Explain why BaSO4 reacts to form BaCO3 even though BaSO4 is sparingly soluble (I hope this is the phrase) to a higher degree than BaCO3. I'm lost at sea here.
 
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broegger said:
1) Write the reaction equations for this transformation. My guess is:

BaSO4(s) + CO3^2-(aq) --> BaCO3(s) + SO4^2-(aq)​
and so on.

2) Is the solution acid or basic in this reaction? Since CO3^2- is a weaker base than SO4^2-, I would say that the answer is basic.
Good so far.
broegger said:
3) Calculate the equilibrium constants by using the solubility products. Ok, here I start by writing the equilibrium equation:

K=[SO42−]eq[CO32−]eqK=[SO42−]eq[CO32−]eq
K=\frac{[{\text{SO}_4}^{2-}]_{eq}}{[{\text{CO}_3}^{2-}]_{eq}}​
This should be valid for both the Ba-, Sr- and Ca-case. But this doesn't make sense, since the constants should be different. What am I missing here?
You are missing/skipping/forgetting the cation concentrations.
broegger said:
4) Explain why BaSO4 reacts to form BaCO3 even though BaSO4 is sparingly soluble (I hope this is the phrase) to a higher degree than BaCO3. I'm lost at sea here.
The system(s) will eventually reach equilibrium.
 
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