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Homework Help: Need Help with Acid/Base Stoic problem!

  1. Nov 16, 2017 #1
    1. The problem statement, all variables and given/known data

    What is the final concentration of hydroxide (OH) ions when 2.00 L of 0.100 M Ba(OH)2 is combined with 1.00 L of 5.00*10^-3 M H3PO4 and allowed to react to completion?




    2. Relevant equations

    Chemical Equation that's balanced is:

    3Ba(OH)2 + 2H3PO4 --> Ba3(PO4)2 +6H20



    3. The attempt at a solution

    To start off, i found that H3PO4 is the limiting reactant so used the mols of that to calculate hydroxide ions.

    Therefore, 0.005 mols H3PO4 * (6 mols OH/2 mols H3PO4) = 0.15 mols OH
    so .15 / 3 liters = 0.05 M (but the answer is 0.128 M!)

    Can someone help me!
     
  2. jcsd
  3. Nov 16, 2017 #2

    Borek

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    Staff: Mentor

    Can you explain what have you calculated here?
     
  4. Nov 16, 2017 #3
    Now that I think about it. I calculated the mols of initial hydroxide .
     
  5. Nov 16, 2017 #4

    Borek

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    Staff: Mentor

    No. Initial hydroxide is 2 L of 0.1 M (times two).
     
  6. Nov 16, 2017 #5
    How can I calculate final concentration of hydroxide if I don't see hydroxide in the product side then?

    I'm not sure what OH to calculate since I can use the mol ratio and the mols of PO4 to calculate OH ions too.
     
  7. Nov 16, 2017 #6

    Borek

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    Staff: Mentor

    What can you tell about hydroxide if the acid was the limiting reagent?
     
  8. Nov 16, 2017 #7
    Hi Lori,

    Firstly, 0.005 times 3 equals 0.015.

    Secondly, think about Ba(OH)2. Does the remaining Ba(OH)2 play a role in (OH) final concentration?
     
  9. Nov 16, 2017 #8
    There would be 0.015 mols because 0.005 x (3).
     
  10. Nov 16, 2017 #9
    Yes it does. It would become water right
     
  11. Nov 16, 2017 #10
    The remaining Ba(OH)2 will dissociate as follows (in water) : Ba(OH)2 ## \to ## Ba2+ + 2 OH-.

    Hint: Find the remaining quantity of Ba(OH)2 and find the mols of OH from the reaction stoichiometry.
     
  12. Nov 16, 2017 #11
    I'm confused cause I thought that the chemical equation I gave is correct. Why is there suddenly dissociation for barium hydroxide??
     
  13. Nov 16, 2017 #12
    The chemical equation you gave is correct. But as you said, this equation does not give OH-.

    In addition, the acid is the limiting reagent (see post #6), and thus some quantity of Ba(OH)2 will remain in the final solution (this quantity of Ba(OH)2 did not react with H3PO4). This remaining quantity will dissociate in the solution. The dissociation reaction is what you need to calculate the final concentration of OH-.
     
  14. Nov 16, 2017 #13
    So the only OH is coming from the dissociation?
     
  15. Nov 16, 2017 #14
    Yes, but you have to calculate the remaining quantity of Ba(OH)2 that dissociates, so you have to take into consideration both chemical equations.
     
  16. Nov 16, 2017 #15
    Oh I seee. So we started off with 0.4 mols of OH but the acid limits the OH so there will be OH ions left . 0.4 - 0.015 = .128. Ions left that dissociate and so divide by 3 is .128 M
     
  17. Nov 16, 2017 #16
    To be more precise, 0.015 out of the initial 0.4 moles were consumed (in the reaction you gave initially), so the remaining OH- are 0.385 moles. If you divide them by the volume, the result gives you the final concentration, right.
     
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