Describe the preparation of 350 mL of 1.20 M from the commercial reagent that is 86% (w/w) and has a specific gravity of 1.71

Supposition:

Let's use HNO₃ as the reagent, because the question does not inform which one is. The solution will differ only on the molar mass of the reagent, but the step by step solution is the same.

Answer:

To prepare this solution, first, a little of water will be added to a volumetric flask of 350 mL, then, 18 mL of the acid will be measured by a pipet, and added to the water (the acid must be added to water to avoid explosion), then more water is added until the line of the flask and the mixture is agitated.

Explanation:

First, let's determine the molar concentration (M) of the commercial reagent. The unit w/w indicates the mass of the reagent by mass of the solution (so in 100 g of solution, 86 g are of the reagent), and the specific gravity is in a unit of g/mL. The molar mass of HNO₃ is 63.01 g/mol.

So, in 1 mL (0.003 L) of the acid, the concentration is:

M = (1.71 g/mL * 1 mL * 0.86) / (63.01 g/mol * 0.001 L)

M = 23.34 mol/L

The number of moles in the dilution remains constant, so the multiplication of the concentration by the volume is constant. If 1 is the commercial reagent, and 2 the solution prepared:

M1*V1 = M2*V2

23.34*V1 = 1.20*350

V1 = 18.00 mL

So, to prepare this solution, first, a little of water will be added to a volumetric flask of 350 mL, then, 18 mL of the acid will be measured by a pipet, and added to the water (the acid must be added to water to avoid explosion), then more water is added until the line of the flask and the mixture is agitated.