Question

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

Answer 1

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`.