How 8.6 ml is calculated for preparing 0.1 n hcl in 1000ml?

Answer 1

Mol Wt of HCl = 36.461 G/Mole = 36.46 G HCl/Liter

0.2 X 36.461 = 7.2922 Grams HCl/Liter

Since HCl is fuming hydrochloric acid, this would be most difficult to weigh out, so doing it volumetrically is probably a more practical way of doing it. Typical Reagent Grade HCl, called fuming hydrochloric acid, is 36% HCl and has a density of 1.179 g/mL

So 7.2922 Grams HCl X 1.179 Grams/mL = 8.597 mL HCl

So the practical way to do this would be to add some water, perhaps 200 mL, to a 1 Liter Class "A" volumetric flask, then slowly add approximately 8.6 mL of 36% reagent grade HCl.

Swirl the flask and continue to add distilled water to the meniscus.

This is how you would make it up in a lab. Afterwards, you would do a titration against 0.1 N NaOH and do a calculation to determine if you need to adjust the HCl for slight errors. You would then use the corrected N value (normality or Molarity) in using the reagent.

Answer 2

To prepare 0.1N HCl in 1000 ml, you first calculate the normality by multiplying the molarity of HCl (0.1N) with its equivalent factor (1 equivalent for HCl). Next, you use the formula M1V1 = M2V2, where M1 = molarity of stock solution, V1 = volume of stock solution to be taken, M2 = final molarity, and V2 = final volume. Therefore, for 0.1N HCl, the volume required would be (0.1 * 8.6) / 0.1 = 8.6 ml.

Answer 3

http://wiki.answers.com/Q/Why_8.5_ml_requires_to_prepare_1_N_HCl"