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Systematic Treatment of Molar Solubility

  1. Mar 17, 2013 #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 in to 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.
    Last edited: Mar 17, 2013
  2. jcsd
  3. Mar 17, 2013 #2


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    Staff: Mentor

    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.
  4. Mar 17, 2013 #3
    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?
  5. Mar 17, 2013 #4


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    Staff: Mentor

    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).

    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
  6. Mar 17, 2013 #5
    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.

    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 ...
  7. Mar 20, 2013 #6
  8. Mar 21, 2013 #7


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    If im 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
    Last edited: Mar 22, 2013
  9. Mar 22, 2013 #8


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    Staff: Mentor

    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.
  10. Mar 24, 2013 #9
    Thank you.
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