Question

How to prepare 250 mL 2.0 N solution of HCl from its commercial reagent (purity of 38% w/w, with specific gravity = 1.19).

Answer 1

The equivalent factor of HCl is 1. Therefore, the value of the molarity of the solution needed equals the value of the normality:

$c =\frac{n}{V} = 2.0$ M.

As the volume of the solution needed is 250 mL, the number of the moles $n$ of HCl needed is (remember that molarity unit is mol/L and 1 mL is 1/1000 L):

$n = cV = 2\cdot250\cdot10^{-3} = 0.5$ mol.

This number of the moles must be multiplied by molar mass $M = 36.46$ g/mol in order to obtain the needed mass:

$m = nM = 0.5\cdot36.46 = 18.23$ g

The commercial reagent content of HCl is 38% w/w, so the relation between the mass of HCl and the mass of the commercial reagent is:

$m_r = \frac{m_{HCl}}{0.38} = \frac{18.23}{0.38} = 47.97$ g.

Finally, the volume of the commercial reagent is its mass divided by its specific gravity:

$V_r =\frac{m_r}{d} = \frac{47.97}{1.19} =40.31$ mL.

Answer: in order to prepare 250 mL 2.0 N solution of HCl from its commercial reagent (purity of 38% w/w, with specific gravity = 1.19), one needs to take 40.31 mL of the commercial reagent and add water (or the solvent) up to 250 mL.