How do we prepare #1.00*mol*L^-1# #HCl(aq)# given #36.5%"(w/w)"# #HCl(aq)#?

1 Answer
Aug 15, 2017

Answer:

You want us to make #1*mol*L^-1# hydrochloric acid from conc. #HCl#? There is a safety issue here.

Explanation:

This site gives me the following data with respect to conc. #HCl#: #rho=1.17*g*mL^-1#; #"concentration"=36.5%(w/w)#.

We want #"Moles of HCl"/"Volume of solution (L)"# to assess the molar concentration. And so.....we assume a #1*mL# volume of stuff:

#((1.17*gxx36.5%)/(36.46*g*mol^-1))/(1xx10^-3*L)=11.7*mol*L^-1#

We want a #1*L# volume of #1*mol*L^-1# #HCl#, i.e. we want a #1*mol# quantity of HCl.

And so we take a #85.5*mL# volume of conc. #HCl#, and we add this to approx. #900*mL# 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 #1*L#.

And so #"concentration"# #=# #"Moles of HCl"/"Volume of solution"#

#=(85.5xx10^-3*Lxx11.7*mol*L^-1)/(1*L)=1.00*mol*L^-1#.