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Another successive dilution problem

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

    The concentration of HCl in a bottle was determined as follows:

    * 4.00 mL of the acid were diluted to 150.00 mL
    * 26.53 mL of the diluted acid were titrated with 0.08329-M NaOH
    * 43.91 mL of the base were required to reach the end point

    What is the HCl concentration in the bottle?
    2. Relevant equations

    Possibly C(final)=C(initial)*(v(1)/v(2))*(v(3)/v(4))* ,....

    3. The attempt at a solution

    C(initial)= .08329 M
    For end Volume I assume that be V(4) and V(3) would be the titrated volume.
    C(final)=.08329 M *(4 mL/150 mL)*(26.53 mL/43.91mL)= .00134 M
  2. jcsd
  3. Sep 5, 2007 #2
    2HCl + 2NaOH - > 2NaCl +2H20

    Valence factor of NaOH = 1
    Valence factor of HCl = 1

    Normality of NaOH = 0.08329N
    Equivalents used = 0.08329 * 43.91 = 3.657meq

    SO equivalents of HCl in 26.53ml(dil) = 3.657meq

    Eq in 154 ml water diluted sol = ???(Get the result in meq)

    Volume of sol, = 154ML

    U have got the meq U have got the Vol.U can find concentration in Normality or Molarity.
  4. Sep 5, 2007 #3
    Is the answer .77N right?
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