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Quick Easy Question, takes 5 seconds

  1. Feb 12, 2007 #1
    How does the solubility of BaSO4 in a solution of Na2SO4 compare to its solubility in pure water?

    I said that it is greater in the water. I don't know if I'm right, or how to explain it however. I see that if those two compounds reacted, then the products would be the exact same as the reactants. Therefore there would be no dominant charges in gettting the reaction to move forward. Is this part of it, or all of it, help me out please.
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  3. Feb 13, 2007 #2
    with solubility, remember that like dissolves like (in terms of intermolecular forces)

    BaSO4 is quite similar to Na2SO4 but quite different from H2O (water has hydrogen bonding) so BaSO4 will be less soluble in water


    to understand why like dissolves like, consider a flask with two kinds of molecules, A and B. On the top layer is B molecules, on the bottom layer is A molecules. will they mix up, or stay apart (are they miscible, or immiscible, respectively)?

    well let's consider what happens when an A molecule moves into the region of B molecules... if the attraction between A molecules is much stronger than the attraction between an A and a B molecule, then the A molecule will get sucked back into the A region because it is attracted to all the other A molecules, and it is immiscible.

    however, if the A molecule is attracted to the B molecules at about the same strength as it is attracted to its own kind, nothing will stop them from mixing


    so, whenever faced with this kind of question, consider which molecules have which attractive forces (and how much of each force) and compare

    the three attractive forces you will want to consider, for each molecule, are: hydrogen bonding, london forces, and dipole dipole forces
  4. Feb 14, 2007 #3


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    for: BaSO4 <---> Ba+2 + SO4-2

    Keq = [Ba+2]*[SO4-2]/ [BaSO4]

    You see that the equilibrium constant is related to the concentration of both Barium and sulfate in solution divided by the concentration of Barium Sulfate in solid state (a constant value). Adding sulfate from sodium sulfate increases the concentration of [SO4] on the right. To reestablish equilibrium, The reaction proceeds from right to left so that the product of barium and sulfate remains constant. This is a description of LeChatlier's Principle.

    Another form for this solubility equilibrium is the Solubility Product Constant (Ksp).

    Ksp = [Ba+2] * [SO4-2]

    To remain in equilibrium, the product of these two species must equal to the Ksp. Increase one and the other is decreased. Lower one and the other is increased.

    The value for Barium Sulfate is 1.08 X 10-10, a very small number (that is nine zeroes in front of the 1).

    Adding Na2SO4 to a solution of BaSO4 at equilibrium adds [SO4-2] to the above expression and thus the concentration of [Ba+2] must change to remain a equilibrium.
  5. Feb 14, 2007 #4


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    qwedsa: This is incorrect - for one thing "similar" and "different" are ill-defined, but that's not all.

    This not how you want to answer this question, and furthermore, it is incorrect as it completely ignores any entropic contribution to the free energy.

    That is definitely not the right way to approach the kind of question raised by jerry. The right way is through the solubility product or equilibrium constant (as explained above by c-tree).
    Last edited: Feb 14, 2007
  6. Feb 14, 2007 #5


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    Google common ion effect (after reading chemisttree answer :wink: ).
  7. Feb 14, 2007 #6
    i defined them in terms of the strength of intermolecular forces. not ill-defined at all

    yes, i'm sorry, i forgot that we are dealing with ions, not molecules

    well it was nearly word for word from the solubility section of my chem textbook, it's not incorrect

    again, i forgot about the ion issue (dissociation). but for any question dealing with solubility of molecules (as opposed to ions), my method IS how you would address the problem
  8. Feb 15, 2007 #7
    OK so as i understand, it is more soluble in water? I see that with water it's ksp is just [Ba][SO4], but with the equation: BaSO4 <---> Ba + SO4; when you add Na2SO4, the SO4 conc. would increase, and the equilibrium would shift left. Now, this means that the Ksp would decrease, therefore its less soluble in water....right?
  9. Feb 15, 2007 #8


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    Ksp won't change (it is a constant), the concentration of barium in solution will change.

    When barium sulfate dissolves, it produces barium and sulfate ions. The process, at equilibrium, is denoted as the reaction:

    BaSO4(s) <----> Ba+2 + SO4-2

    This expression evokes the concept that barium and sulfate ions are continually being dissolved from the solid barium sulfate and barium sulfate is continually being precipitated from a solution containing barium and sulfate ions. At equilibrium these two processes (dissolution and precipitation) occur at exactly the same rate and to the observer it appears as though nothing further is happening... until you add something like sodium sulfate and disturb the equilibrium. After a time, equilibrium is reestablished such that the expression for Ksp is valid.

    You can think of the solubility of BaSO4 as the concentration of barium in solution. The more barium in solution, the more soluble BaSO4 must be. If something decreases the concentration of barium in solution, the less soluble BaSO4 must be.
  10. Feb 15, 2007 #9
    I meant trial Ksp, not Ksp.
  11. Feb 16, 2007 #10


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    Qwedsa's comments are valid for a so-called inert salt. If a salt that contained no common ions were added to water, the solubility of BaSO4 could be increased. The phenomenon is referred to as "Salting In".
    If the salt contains a common ion the Ksp argument is applied.
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