Question

Concentrated HCL is 37.5% by weight and has a density of 1.19g/ml. Calculate the (a) molarity of the concentrated acid (b) mL of concentrated HCL required to make 500mL of 0.2 HCl (c) ml of concentrated HCL required to make 350 ml of 0.5 N HCL

Answer 1

(a)

Let us say that the total mass of HCl solution = 100 g

Thus, mass of HCl = 37.5 g

Molar mass of HCl = 36.5 g/mol

Hence, moles of HCl = mass of HCl/molar mass of HCl

                                  = 37.5 g/36.5 g/mol

                                  = 1.03 mol

Volume of conc. HCl solution = mass of conc. HCl/density of HCl

                                               = 100 g/1.19 g/mL

                                               = 84.0 mL

                                               = 0.084 L               ( 1 mL = 0.001 L)

Molarity = moles of solute/liter of solution

Hence, the molarity of HCl = 1.03 mol/0.084 L

                                           = 12.3 M

(b)

Initial volume = V₁

Initial concentration (C₁) = 12.3 M

Final volume (V₂) = 500 mL

Final concentration (C₂) = 0.2 M

Equating initial and final moles of HCl to get

V₁C₁ = V₂C₂

or, V₁ x 12.3 M = 500 mL x 0.2 M

or, V₁ = 8.1 mL

Hence, the volume of conc. HCl required = 8.1 mL

(c)

HCl is a monoprotic acid. Hence, 0.5 N = 0.5 M.

Initial volume = V₁

Initial concentration (C₁) = 12.3 M

Final volume (V₂) = 350 mL

Final concentration (C₂) = 0.5 M

Equating initial and final moles of HCl to get

V₁C₁ = V₂C₂

or, V₁ x 12.3 M = 350 mL x 0.5 M

or, V₁ = 14.2 mL

Hence, the volume of conc. HCl required = 14.2 mL