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anonymous

  • 4 years ago

A solution is made by dissolving 5g of Ba(OH)2 to a total volume of 250ml. Calculate the molarity of hydroxide ion in this solution.

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  1. anonymous
    • 4 years ago
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    \[Ba(OH )_{2}\]

  2. anonymous
    • 4 years ago
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    is it asking to only find the molarity of the OH or the entire Ba(OH)2?

  3. anonymous
    • 4 years ago
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    I would do this in 2 steps first calculate how many moles are in 5 grams of baoh2 and then figure out how many moles it is per per ml then multiply that times 1000

  4. anonymous
    • 4 years ago
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    did u get the molarity of 0.1167?

  5. anonymous
    • 4 years ago
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    I didn't do it, whats the molar mass?

  6. anonymous
    • 4 years ago
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    171.33

  7. anonymous
    • 4 years ago
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    yea thats what i got also

  8. anonymous
    • 4 years ago
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    ok so isnt that the molarity of the entire Ba(OH)2? why is it saying to find the molarity of the hydroxide ion? are they the same question?

  9. anonymous
    • 4 years ago
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    ok take molarity of baoh2 and do a little stoichometry using the conversion factor 1 mole Ba(OH)2 = 2 mole OH Can you pull that one off?

  10. anonymous
    • 4 years ago
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    \[.1167 moles Ba(OH)_2 (\frac{2 mole OH}{1 mole Ba(OH)_2})=\]

  11. anonymous
    • 4 years ago
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    ok

  12. anonymous
    • 4 years ago
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    Sorry I missed that when i first told you how to solve it.

  13. anonymous
    • 4 years ago
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    ok cool, do you mind me giving you the next part of the question?

  14. anonymous
    • 4 years ago
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    whats the next part?

  15. anonymous
    • 4 years ago
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    a student took 20ml of the previous solution and added 50ml of water. Calculate the new hydroxide ion concentration

  16. anonymous
    • 4 years ago
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    now i know you use m1v1=m2v2

  17. anonymous
    • 4 years ago
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    right?

  18. anonymous
    • 4 years ago
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    Yea use that

  19. anonymous
    • 4 years ago
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    so (0.02L)(0.233)=(0.05L)(x) then solve for x?

  20. anonymous
    • 4 years ago
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    Yes that looks like it should work to me.

  21. anonymous
    • 4 years ago
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    ok last question

  22. anonymous
    • 4 years ago
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    i have two equations: pH=-log[H] and [H][OH]=1.00x10^-14. (The symbol "[ ]" indicates concentration/molarity). Using the previous the last molarity of OH (0.0932) to calculate the pH

  23. anonymous
    • 4 years ago
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    so [H]x0.0932=1.00x10^(-14)

  24. anonymous
    • 4 years ago
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    [H]=1.073x10^(-13) right?

  25. anonymous
    • 4 years ago
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    so then the pH is -log(1.073x10^(-13))

  26. anonymous
    • 4 years ago
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    12.97?

  27. anonymous
    • 4 years ago
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    you got me on this on, i'm stumped.

  28. anonymous
    • 4 years ago
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    ok thanks anyway?

  29. anonymous
    • 4 years ago
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    sorry i did not mean to put the question mark lol

  30. anonymous
    • 4 years ago
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    thank you

  31. anonymous
    • 4 years ago
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    No problem sorry i couldn't help with that last part.

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