A $\pu{1.1g}$ sample of an oil was treated with Wij’s solution and excess potassium iodide solution. The liberated iodine reacted with $\pu{7.3 cm3}$ of $\pu{0.1 mol/dm3}$ sodium thiosulfate solution. The blank titration required $\pu{43.5cm3}$ of sodium thiosulfate. What is the iodine value?

I've been asked this question but there are no examples of calculating iodine value in my textbook. I'm aware that the iodine value is the mass of iodine to react with $\pu{100g}$ fat/oil and the question gives a $\pu{1.1g}$ sample of an oil. Does this mean $1.1$ should be divided by $\pu{100g}$? I'm just a bit confused how to even start this question.

  • $\begingroup$ I assume the normality of sodium thiosulphate is 0.1 N since no explicit mass was given. $\endgroup$ – xavier_fakerat May 16 '17 at 17:16

Assuming you are given normality of sodium thiosulphate as 0.1 N (or sometimes called standard), you could use this formula to find the iodine value:

iodine number=$$\frac{(mL 0.1N Na_2S_2O_3 blank – mL 0.1 N Na_2S_2O_3 test) (12.7)(100)}{(1000) (1.1)}$$

  • Where “mL 0.1 N Na2S2O3 Blank” = the volume of sodium thiosulphate required to titrate the blank solution,

  • Where “mL 0.1 N Na2S2O3 Test” = the volume of sodium thiosulphate required to titrate the test solution that contained either solid fat or oil as the sample,

  • Where “12.7” is the amount of grams of iodine contained in one litre of 0.1 N iodine,

  • Where “1.1” is the amount of fat taken in grams and,

  • Where “100” converts the sample to per 100 grams of fat.

| improve this answer | |
  • $\begingroup$ is 12.7 just a standard to multiply by? Thanks for your help! $\endgroup$ – Amy McCulloch May 16 '17 at 18:18
  • $\begingroup$ @Amy Its a gram gram equivalent of weight iodine (iodine has molar mass 127), so its already accounted for in this formula. (in 0.1 litre there is 12.7g of iodine which gives 0.1N iodine) so if "standard" solutions are used then this value becomes a constant $\endgroup$ – xavier_fakerat May 16 '17 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.