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Concentration ion problem

  1. Sep 4, 2007 #1
    1. The problem statement, all variables and given/known data

    A solution is found to be 0.0330% by weight AlBr3. What is the concentration of the Br1- ion expressed in ppm?

    2. Relevant equations

    3. The attempt at a solution

    Assume .0330 % => .0330 g AlBr3

    .0330 g AlBr3*(1 mol AlBr3/266.697 g AlBr3)*(3 mol Br^-1 /1 mol AlBr3) =.0003712 mol Br^-1

    10^6 g sol'n *(1 mL/1.0 g sol'n)*(1 L/1000^3) = 1000 L

    (.0003712 mol Br^-1) /(1000 L) = 3.712*10^-6
  2. jcsd
  3. Sep 4, 2007 #2


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    Start over.

    The percent of AlBr3 is grams per 100 grams of solution. You need to make an assumption that this low concentration (0.0330%) will yield a solution with a density of approximately 1.00 g/mL. (so 0.00330 g AlBr3/100g solution = 0.0330 g AlBr3/100 mL)

    Since the concentration is given with a specific number of significant digits, use the same number of significant digits in your answer.
  4. Sep 4, 2007 #3
    don't you mean 0.00330 g AlBr3/100g solution = 0.00330 g/100 mL
    AlBr3 not 0.00330 g AlBr3/100g solution = 0.0330/100 mL g AlBr3

    Anywh, would you now multily (.00330 g AlBr3/100 g AlBr3)*(1 mol AlBr3 /266.68 g AlBr3)*(3 mol Br1-/1 mol AlBr3) to get the number of moles of Br1- ions? If I followed the correct procedure, would divide my number of bromine moles by 10^6 g to calculate the concentration?
  5. Sep 4, 2007 #4


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    No. I mean 0.0330 g aluminum bromide per one hundred milliters of solution. In shorthand it looks like this: 0.0330 g AlBr3/100 mL solution.

    Nope. If there are 0.0330 g AlBr3 per 100 mL of solution, how many grams of AlBr3 will there be in a liter? You need to express the number of moles of Br- in terms of moles per liter and then multiply by 10^6. You need to do a little work to convert percent to moles per liter... but you must do it. Try multiplying the percentage by ten and see where that leads. (since 10 X 100 mL is 1000 mL or one liter)
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