Systematic Treatment of Molar Solubility

In summary: Can you please elucidate?The molar solubility is the concentration of the cation added to the solution. It is equal to the cation's mass balance, which would include all species containing the cation multiplied by the stoichiometric coefficient of the cation in those species, and this total sum of concentrations is the initial concentration of the cation.
  • #1
Big-Daddy
343
1
How do we systematically calculate the molar solubility of a substance with regards to its Ksp? In other words, where does the value of "molar solubility" fit into the equilibrium calculations?

This should build up so that I can understand how to systematically treat the common ion effect, among other equilibria, if necessary. For example, if I want to know what the molar solubility is of my salt when, if it dissolves, it will undergo hydrolysis of the anion, or complex formation, or both (so we would need to consider acid/base equilibria and Kf at the same time as Ksp), how do I set up the equilibrium expressions to solve to get my molar solubility? This would come from an understanding of how exactly to calculate "molar solubility" in relation to equilibria in general.

I suspect that a major step is to note that Ksp = the equilibrium concentrations of the two ions of the salt, to the power of their coefficients in the molecular formula respectively, multiplied together, ONLY when the molar solubility is actually reached - until then, their equilibrium concentrations will multiply to a lower value than the Ksp only. That still leaves me unclear as to exactly what "molar solubility" is and how we would go about calculating it in situations with many equilibria going around.
 
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  • #2
This is not different from the general equilibrium calculations that we discussed in the past. Ksp is just one of the equilibrium equations, that has to be used together wit all other equilibria, mass balances and charge balance. Molar solubility is something that you can easily calculate AFTER you know all equilibrium concentrations. If, for any reason, solution is not saturated then yes, Q < Ksp.
 
  • #3
Borek said:
This is not different from the general equilibrium calculations that we discussed in the past. Ksp is just one of the equilibrium equations, that has to be used together wit all other equilibria, mass balances and charge balance. Molar solubility is something that you can easily calculate AFTER you know all equilibrium concentrations. If, for any reason, solution is not saturated then yes, Q < Ksp.

1.) OK, so if the solution is not saturated then we cannot use the Ksp expression in the equilibrium calculations. However, this would not affect our attempt to find molar solubility anyway as the solution will be saturated and only then do we work out molar solubility (and then we can also use the Ksp expression in our equilibrium calculations).

2.) Will the molar solubility simply be the analytical concentration of the cation (in the same way as defined for the mass balance) divided by the coefficient of the cation in the molecular formula of the salt (and also the same thing, for the anion, both of these methods giving the same value)?

3.) Let's say I've added Ag2SO4 to solution and this has established an equilibrium not according to Ksp (the Ksp value is known but the solution is not saturated). Now I want to find out how much AgCl (with a known Ksp value) I can add, i.e. molar solubility of AgCl. Is there any way/how would we do this calculation?
 
  • #4
Big-Daddy said:
2.) Will the molar solubility simply be the analytical concentration of the cation (in the same way as defined for the mass balance) divided by the coefficient of the cation in the molecular formula of the salt (and also the same thing, for the anion, both of these methods giving the same value)?

Can be, but doesn't have to. Your cation can be complexed and then to calculate the molar solubility you will need to sum concentrations of all forms of the cation (followed by division by the stoichiometric coefficient).

3.) Let's say I've added Ag2SO4 to solution and this has established an equilibrium not according to Ksp (the Ksp value is known but the solution is not saturated). Now I want to find out how much AgCl (with a known Ksp value) I can add, i.e. molar solubility of AgCl. Is there any way/how would we do this calculation?

If the solution is not saturated, the only way to calculate concentration of Ag+ is from the mass balance - you should know how much Ag2SO4 were put into the mix. If you don't know it, you can't solve the problem
 
  • #5
Borek said:
Can be, but doesn't have to. Your cation can be complexed and then to calculate the molar solubility you will need to sum concentrations of all forms of the cation (followed by division by the stoichiometric coefficient).

OK so let's redefine as the molar solubility is the initial concentration of the cation added to the solution (which is equal to the cation's mass balance, which would include all species containing the cation multiplied by the stoichiometric coefficient of the cation in those species, and this total sum of concentrations is the initial concentration of the cation), which can be divided by the stoichiometric coefficient of the cation in the original salt's molecular formula to reach the molar solubility. Or the same thing for the anion.

Borek said:
If the solution is not saturated, the only way to calculate concentration of Ag+ is from the mass balance - you should know how much Ag2SO4 were put into the mix. If you don't know it, you can't solve the problem

You can have how many moles of Ag2SO4 were originally put in the solution, and the Ksp of Ag2SO4, but Ag2SO4 is not being added until saturation. Can you work it out then? Since Ksp ≠ [Ag+]2*[SO42-] unless the solution is saturated, which it here is not ...
 
  • #6
Help?
 
  • #7
If I am correctly understanding [Ag+] before saturation as part of your question :
Then can you get at this with electrode potentials ? Where E saturated =
[Ag+(aq)] ,E value,[?]/Ag electrode + .059 lg[Ag+aq]
Where [Ag(+)aq] on right side is silver concentration in saturated solution.
From the Nernst equation. E = E + (RT/zF) ln[Mz+]
E=E+(.059)/z) lg[Agz+]
E at electrode potential at 1mole/liter
Each 10x lowering of [Ag+] lowers potential by (.059/z) volts
 
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  • #8
Big-Daddy said:
You can have how many moles of Ag2SO4 were originally put in the solution, and the Ksp of Ag2SO4, but Ag2SO4 is not being added until saturation. Can you work it out then? Since Ksp ≠ [Ag+]2*[SO42-] unless the solution is saturated, which it here is not ...

If you know initial moles of Ag2SO4 you put them into the mass balance for silver and if the solution is not saturated, you simply drop Ksp from the set of equations.
 
  • #9
Borek said:
If you know initial moles of Ag2SO4 you put them into the mass balance for silver and if the solution is not saturated, you simply drop Ksp from the set of equations.

Thank you.
 

Related to Systematic Treatment of Molar Solubility

1. What is the concept of systematic treatment of molar solubility?

The systematic treatment of molar solubility is a method used to determine the solubility of a substance in a given solvent at different temperatures. It involves plotting the molar solubility of the substance against the temperature, and using various equations and calculations to determine the solubility at any given temperature.

2. Why is the systematic treatment of molar solubility important?

This method is important because it allows scientists to accurately determine the solubility of a substance at any temperature, rather than relying on experimental data at specific temperatures. It also helps to understand the factors that affect solubility, such as temperature and the nature of the solvent and solute.

3. What are the key equations used in the systematic treatment of molar solubility?

The main equations used in this method are the van't Hoff equation, which relates the change in solubility with temperature, and the Gibbs-Helmholtz equation, which relates the solubility at different temperatures.

4. How does the nature of the solvent and solute affect molar solubility?

The nature of the solvent and solute can greatly impact molar solubility. Substances with similar polarities are more likely to have higher solubility, while substances with different polarities may have lower solubility. The size and structure of the molecules can also affect solubility, as well as the presence of hydrogen bonding or other intermolecular forces.

5. How can the systematic treatment of molar solubility be used in practical applications?

This method can be used to predict the solubility of a substance in a given solvent at any temperature, which is important in industrial processes such as pharmaceuticals, cosmetics, and food production. It can also help in understanding the behavior of solutions and designing experiments in the laboratory.

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