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Sollubility of BaSO4 in 4M HCL help

  1. Nov 14, 2012 #1
    Hi can anyone help me out with this problem?


    This is how far I came

    BaSO4 +H(+) <---> SO4+H2O +Ba
    s 4-s s s

    ksp = [SO4][Ba]/[H+] = s^2/[4-s]

    So I saturated the solution in s, so the H+ will be 4M-s at equilibrium,

    But it's not correct :(
    What am I doing wrong?
     
  2. jcsd
  3. Nov 14, 2012 #2

    Borek

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    Staff: Mentor

    Your reaction equation is nonsensical.

    Start with a correct Ksp formula. Then remember that SO42- gets protonated to HSO4- - so you have to take the second dissociation constant Ka2 into account.
     
  4. Nov 14, 2012 #3
    Oops forgot something

    BaSO4 +H(+) <---> HSO4+H2O +Ba(+2)

    This must be the right formula.
    So if we saturate BaSO4 with s, and we start with 4M HCl (= 4M H+) then we must get s HSO4 at equilibrium and s Ba. Or is my logic off?


    Ksp=[reactants ]/[products ]
     
  5. Nov 14, 2012 #4
    I'm good at sulubility products in water, but when it get's to acids, i lack a bit understanding, that's why i i want gain the right logic.
     
  6. Nov 14, 2012 #5

    Borek

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    Staff: Mentor

    It is still nonsensical and can't be balanced.

    That's not the way we usually write Ksp.
     
  7. Nov 14, 2012 #6
    So if i take a easy example

    Mg(OH)2 ---- Mg2+ +2OH
    s s 2s

    Ksp =[Mg][OH]^2 = (s)(2s)^2

    Is this not correct?

    Can you help me with the reaction? I'm not quite sure how to write it :(
     
  8. Nov 14, 2012 #7

    Borek

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    OK, now write Ksp for barium sulfate.
     
  9. Nov 14, 2012 #8
    Ksp = [Ba][SO4]
     
  10. Nov 14, 2012 #9
    And you said that "SO42- gets protonated to HSO4-"

    So perhaps the second Ksp= [HSO4]/[H+][SO4]?

    I've never done a problem like this, so It's hard for me to imagine what the method is (combining two ksp) But do you equate the two solubility product expressions?
     
  11. Nov 14, 2012 #10
    Woa, just saw your website! Perhaps I should begin reading some of the pH calculations you write now and then.
     
  12. Nov 14, 2012 #11

    Borek

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    There is no second Ksp, there is one Ksp that you already wrote.

    Your problem now is how to calculate what fraction of sulfuric acid is in the form of SO42- that is present in Ksp formula. This is where Ka2 comes into play.
     
  13. Nov 15, 2012 #12
    SO4 + H3O(+) <-----> HSO4
    ------- 4M-x -----------x

    Ka2 for this reaction is 10^-(1,99) (because pKa=1,99)

    So Ka2 =[HSO4]/[H3O+] = [x]/[4-x] Korrekt?
     
  14. Nov 15, 2012 #13

    Borek

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    Close, but no.

    Ka describes the dissociation, not protonation. Can you write reaction equation for the HSO4- dissociation, and then write correct reaction quotient? One that doesn't miss any important reactant nor product?
     
  15. Nov 15, 2012 #14
    Ahh yes that makes sense, Ka is the acid constant, so we need to have the acid as reactant. It's the reverse i assume?

    HSO4(-) + H2O <-----> SO4(2-) + H3O(+)
    ---x-------------------------------4M-x--

    So I know for sure that we start with 4M [H+], so we must end with 4M-x, which is the amount added to SO4 to form HSO4, right?

    Ka = [SO4][H+]/[HSO4]
     
    Last edited: Nov 15, 2012
  16. Nov 15, 2012 #15

    Borek

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    I am almost 100% sure you can safely assume 4M and ignore x. Even if not, you can easily check if the assumption was correct after finding the solution.
     
  17. Nov 15, 2012 #16
    Okay, so we have two reactions and two balance equations.
    BaSO4 <----> Ba(+2) + SO4(2-)
    ---0--------------s---------s+ (at equilibrium)



    HSO4(-) + H2O <-----> SO4(2-) + H3O(+)
    --x-------------------------------4M

    So how do we approach this HSO4 dissociation? Do we say, We start with with 4M of H+ from HCl dissociation. And we start with x moles of SO4 (from the BaSO4), and at equilibrium we get x moles of HSO4 (Or do we start with x moles of HSO4? (this is where is get confused)

    We have the following expressions from the equations.

    Ksp = [Ba][SO4]=s^2

    Ka = [SO4][H+]/[HSO4] =
     
  18. Nov 15, 2012 #17

    Borek

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    We start with 4M H+ and x moles of SO42-.

    You are missing mass balance - how does the amount of Ba2+ depend on the amount of SO42- and HSO4-?
     
  19. Nov 15, 2012 #18
    Thank you for taking this in steps, I can see how we need to approach this problem in parts now.
    If i look at the two reactions, this would make sense, no?

    [Ba] =[SO4] - [HSO4]

    So when you say we start with x moles of SO4 I assume "we" assume that we start with a solution of BaSO4 which dissolves into the x SO4, that we add to our HSO4 dissociation equation?
     
  20. Nov 15, 2012 #19

    Borek

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    Something like that. SO42- is consumed by protonation, which shifts the solubility equilibrium to the right.

    At this stage it is just a matter of solving system of three equations.
     
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