Inorganic Chemistry Help: Solubility Transformation and Equilibrium Constants

In summary: In equilibrium, there is no reaction taking place because the concentrations of reactants and products are equal.
  • #1
broegger
257
0
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:

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

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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  • #2
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.
 

1. What is the difference between organic and inorganic chemistry?

Organic chemistry is the study of carbon-containing compounds, while inorganic chemistry focuses on compounds that do not contain carbon. Inorganic chemistry also deals with the behavior and properties of non-living matter, such as metals, minerals, and gases.

2. What are some common applications of inorganic chemistry?

Inorganic chemistry has many practical applications, including the production of medicines, fertilizers, and plastics. It is also used in the development of new materials for electronics, energy storage, and environmental remediation.

3. How does inorganic chemistry relate to other branches of chemistry?

Inorganic chemistry is closely related to other branches of chemistry, such as analytical chemistry, physical chemistry, and bioinorganic chemistry. It also overlaps with fields like materials science, geochemistry, and environmental chemistry.

4. What techniques are commonly used in inorganic chemistry research?

Inorganic chemists use a variety of techniques to study and analyze compounds, including spectroscopy, X-ray crystallography, and chromatography. They also use computational methods to model and predict the properties of inorganic compounds.

5. How does inorganic chemistry contribute to our understanding of the natural world?

Inorganic chemistry is essential for understanding and explaining many natural phenomena, such as the formation of minerals, the composition of the Earth's atmosphere, and the behavior of elements in living organisms. It also plays a crucial role in the development of new technologies and materials that improve our quality of life.

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