Chemical Forums

Please login or register.

Login with username, password and session length

Sponsored links

Pages: [1]   Go Down

Author Topic: Blank Titration *delete me*!  (Read 4179 times)

0 Members and 1 Guest are viewing this topic.


  • Guest
Blank Titration *delete me*!
« on: June 01, 2005, 05:19:00 PM »

Hi everyone, I am a Nursing student taking my 3rd quarter of General Chemistry and I am having quite some trouble figuring out how to calculate the moles of iodine absorbed by this solution:

10 mL of dichloromethane, 20.0 mL of Hanus reagent, 10 mL of 15% KI solution and 20 mL of distilled water;  Mixed and titrated with ~0.2 M sodium thiosulfate solution from a 50 mL buret.
Once titrated I added 2mL of 1% starch solution and titrated until the blue color was gone.

The way my instructions have me calculating this is:

   moles of I atoms absorbed = (B - S) x M

   where:  B = liters of Na2S2O3 for blank titration

   S = liters of Na2S2O3 needed to titrate halogen unabsorbed by the sample

   (B - S) = liters of Na2S2O3 equivalent to halogen absorbed by the sample

   M = molarity of Na2S2O3, moles/liter

The mass of iodine absorbed by the sample is:

   grams of iodine adsorbed = moles of iodine atoms absorbed  
   x  molar weight of iodine atom

(The molar weight of iodine atom is 126.9 grams/mole)

The Iodine Number is defined as the number of grams of iodine absorbed by 100 grams of fat:

   Iodine Number = grams of iodine absorbed by sample x 100
   sample weight of fat in grams

I started out with 50mL of of titrate in my buret, and once the titration was complete I had 46.36mL left, which means 3.64mL is the volume of Na2S2O3 used to titrate blank, I'm guessing.

I know this may be clear as daylight for a few of you, but its exam week and I am TOTALLY burnt out over this. Someone help me please. If you can, I would be forever grateful.

Thank You

« Last Edit: June 01, 2005, 05:54:43 PM by chardonay86 »


  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Mole Snacks: +1594/-393
  • Offline Offline
  • Gender: Male
  • Posts: 24262
  • I am known to be occasionally wrong.
    • Chembuddy
Re:Blank Titration *delete me*!
« Reply #1 on: June 02, 2005, 12:31:27 AM »

And what is your question?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation,,, PZWT_s1


  • Chemist
  • Full Member
  • *
  • Mole Snacks: +77/-37
  • Offline Offline
  • Gender: Male
  • Posts: 786
Re:Blank Titration *delete me*!
« Reply #2 on: June 02, 2005, 01:43:13 AM »

the question was in the first paragraph

reffering to the ionic equation of the reduction of iodine,

2S2O32- + I2  --> 2I- + S4O62-

thus, 2 mol of Na2S2O3 will reduce a mol of iodine.

from your titration results, u have the amount of Na2S2O3, so u can find the mol of it.

the Mr is quite easy to find, just look up the individual Mr of each element at  that, or just refer to the periodic table for their mass no. (Na=23, S=32, O=16) in the exam.

so after finding the moles of Na2S2O3, divide it by 2 to get the mol of I2 and u're done! :)
one learns best by teaching


  • Guest
Re:Blank Titration *delete me*!
« Reply #3 on: June 02, 2005, 03:18:30 AM »

Wow, thank you so much!! I never knew it could be so easy. Have a lovely day :)

Charde'(Who now understands how to calculate her titration)
Pages: [1]   Go Up

Mitch Andre Garcia's Chemical Forums 2003-Present.

Page created in 0.081 seconds with 23 queries.