Answer to Question #163598 in Chemistry for yara

Question

how to prepare 5 L of 0.5N hcl from 37% hydrochloric acid stock with density 1.19kg/L

Answer 1

According to the equation, acid normality equals acid molarity and can be calculated as:

1) w/v(HCl) = % w/w × density = 37% × 1.19kg/L =  0.44 kg/L HCl

2)  N(HCl) = M(HCl) = w/v(HCl) / Mr(HCl) =  0.44 kg/L / 36.5 g/mol = 12 mol/L

From here:

M₁V₁ = M₂V₂

(0.5 M)(5000 mL) = (12 M) (x L)

x = 0.5 M × 5000 mL / 12 M = 208.3 ml

To make 5 L of a 0.5 N solution, slowly add 208.3 mL of HCl stock solution to 1250 mL deionized water.

Adjust the final volume of solution to 5 L with deionized water.