How do we prepare 1.00molL1 HCl(aq) given 36.5%(w/w) HCl(aq)?

1 Answer
Aug 15, 2017

You want us to make 1molL1 hydrochloric acid from conc. HCl? There is a safety issue here.

Explanation:

This site gives me the following data with respect to conc. HCl: ρ=1.17gmL1; concentration=36.5%(ww).

We want Moles of HClVolume of solution (L) to assess the molar concentration. And so.....we assume a 1mL volume of stuff:

1.17g×36.5%36.46gmol11×103L=11.7molL1

We want a 1L volume of 1molL1 HCl, i.e. we want a 1mol quantity of HCl.

And so we take a 85.5mL volume of conc. HCl, and we add this to approx. 900mL of water in a volumetric flask. The order of addition is ALWAYS acid to water. Why? Because if you spit in conc. acid it spits back, I kid you not! After the initial dilution we can make this up to 1L.

And so concentration = Moles of HClVolume of solution

=85.5×103L×11.7molL11L=1.00molL1.