Question

I would like to make 3.8 M of Perchloric acid and, i have no idea how to do it since it is not my field of study.

But have to make 3.8 M of Perchloric acid because we need to make sampling fluid with it.

Sampling fluid contains 0.11~0.13 grams of 2,4-DNPH+1L of Acetonitrile+0.15 mL of 3.8 M Perchloric acid.

The Perchloric acid that i have right now is 70%, 99.999% trace metals basis, Cas:7601-90-3 HCIO4, MW:100.46 g/mol, in a bottle of 50 mL.

My colleague told me that it contains 2009.2 M of HCIO4 in 50 mL from the calculation; 100.46/0.05 L = 2009.2 M HCIO4

If this it right, how do i make 2009.2 M of HCIO4 to 3.8 M HCIO4 ???

please help.

Answer 1

Your calculation is wrong. please follow the following procedure.

Your HCLO4 is of 70 % purity.

Also the density of this product is 1.664 g/ml

You have 50 ml of bottle so mass of the HCLO4 in the bottle = 50 * 1.664

Mass of 50 ml of solution = 83.2 g

Now purity of the solution is 70 % that means that 100 gm of this solution contain 70 gm HCLO4 .

So 83.2 gm of HCLO4 contain 0.7*83.2 gm of HCLO4 = 58.24 gm

Given Molecular weight of HCLO4 = 100.46 g/mol

So Moles of HCLO4 present in the bottle = 58.24/100.46

=0.5797 moles

So concentration of HCLO4 in the bottle = 0.5797/50 (mole/ml)

=0.0114 mole/ml = 11.4 mol/L or 11.4 M

So the molarity of your solution is 11.4 M.

Now you can use

V1= Volume of the 11.4 M solution to be used to make 3.8 M solution.

M1 = Molarity of given solution =11.4 M

M2 = Molarity of the solution to be prepared = 3.8 M

V2 = Volume of the solution to be prepared. ( let we want to prepare 100 ml of 3.8 M solution)

V2= 100 ml =0.1 L

So from the equation we get V1= 33.33 ml

So to prepare 100 ml solution of 3.8 M first you can take 33.33 ml the 11.4 M solution in a volumetric flask and than add water upto the level of 100 ml (So that the total volume of the 3.8 M solution to be 100 ml)