1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solubility product

  1. May 13, 2006 #1

    egg

    User Avatar

    Hi need a bit of help with a question!!
    the solubility product of magnesium carbonate is 10^ -7.5. Calculate the concentration of Mg2+ and CO3 2- expected in a saturated solution assuming activity of both ions is 1.0
    cheers in advance!!
     
  2. jcsd
  3. May 13, 2006 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Can you start by writing an expression for the solubility product of [itex]MgCO_{3}[/itex]? I'll give you a hint as well;

    What does this tell you about the relative concentrations of the [itex]Mg^{2+}[/itex] and [itex]CO_{3}^{2-}[/itex]. It may also helpful if you where to write out an equation of the dissociation of the ions.

    ~H
     
  4. May 14, 2006 #3

    egg

    User Avatar

    cheers for the reply hootenanny,
    i wrote the initial question in a bit of a hurry yesterday and forgot to say i have NO idead what to do with the info!
    I know the relative concentrations of the ions will be equal and think that the dissociation is
    MgCO3 <=> Mg^2+ & CO3^2-
    Think the expression you mean is Ksp=(aMg^2+)(aCO3^2-)/aMgCO3 but not too sure where to go now, do i replace the ions and the Ksp with the values i know and rearrange to find activity of MgCO3? if i do this what does it tell me?
    This is a question in a past exam paper we've been given and I'm sure we weren't shown how to do it! My exams next week and this is the only question thats been causing us grief!
     
  5. May 14, 2006 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah, this is correct. You are also correct with your Ksp expression;

    [tex]K_{sp} = \frac{[Mg^{2+}][CO_{3}^{2-}]}{[MgCO_{3}]}[/tex]

    This is something of a trick question really. The square brackets mean concentration of whatever is inside the square brackets. Now, as you correctly said in your equation, once dissociated the concentration of [itex]Mg^{2+}[/itex] and [itex]CO_{3}^{2-}[/itex] will be equal, so we can re-write the above Ksp expression as;

    [tex]K_{sp} = \frac{[Mg^{2+}]^2}{[MgCO_{3}]}[/tex]

    Do you follow this?

    As [itex]MgCO_{3}[/itex] is only slightly soluble, once the above equation has reached equilibrium adding further solid will not cause any equilibrium shift as the solution is saturated. Therefore, the equation becomes;


    [tex]K_{sp} = [Mg^{2+}]^{2}[/tex]

    Do you follow?

    Now, consider how the concentration relates to the solubility product. What happens if the concentrations are greater, less than and equal to Ksp?


    ~H
     
    Last edited: May 14, 2006
  6. May 14, 2006 #5

    egg

    User Avatar

    think i'm starting to understand a bit better now!! can i now rearrange Ksp=[Mg^2+]^2 to
    SQRT Ksp=[Mg^2+] or SQRT10^-7.5 in which case the concentration of the ions will be 10^-3.75 ???????
     
  7. May 14, 2006 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your logic is correct, but you might want to check your arithmetic, I get it to be 1.78x10-4 mol.dm-3

    ~H
     
  8. May 14, 2006 #7

    egg

    User Avatar

    yeah thats it, 1.78x10^-4 is the same as 10^-3.75! Thank you loads for the help, just need to revise a ton of other stuff for this exam!!
    Thanks again!
     
  9. May 14, 2006 #8

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    O yeah, didn't spot that :rolleyes: No problem.

    ~H
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Solubility product
  1. Solubility product (Replies: 2)

  2. Solubility Product (Replies: 8)

Loading...