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Equivalence Point- Concentration

  1. Sep 22, 2008 #1


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    1. The problem statement, all variables and given/known data

    A 20.00 ml stock sample of HBr is diluted to 50 mL. If 18.76 ml of 0.1345 M KOH was required to reach the equivalence point, what is the concentration of the stock HBr solution?

    2. Relevant equations

    KOH + HBr --> KBr + H20

    M = conc. = mol./L

    n = V * conc.

    3. The attempt at a solution

    n-KOH= V-KOH * conc. KOH = (18.76*10^-3 L) * (0.1345 mol./L) = (2.52322*10^-3 mol.)

    => n-HBr= (2.52322*10^-3 mol.)

    V-HBr= 5.0*10^-4 L

    conc. HBr = (2.52322*10^-3 mol.)/ (5.0*10^-4 L) = 5.04644 M

    I dont think this the right answer but I seem to endup with the right units, can someone please tell me what I am doing wrong and how to go about this question the right way?
  2. jcsd
  3. Sep 22, 2008 #2


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    Staff: Mentor

    Generally you are on the right track, however:

    that's not true. 5*10-4 L = 0.5 mL.

    Also think whether you should use 20 mL or 50 mL. Question asks for the concentration of the stock solution. Note, that if you will not dilute your sample to 50, you will use exactly the same amount of KOH, so you can omit dilution calculations.
  4. Sep 24, 2008 #3


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    I did it over agian and used 20 ml and the answer i got was 0.1262 M

    can you please verify?

    btw, is there a thank you button on this fourm?
  5. Sep 24, 2008 #4


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    Staff: Mentor


    I think there was once, but it was getting too hot of continuous pressing.
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