1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Another successive dilution problem
Loading...