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

Aug 15, 2017

You want us to make $1 \cdot m o l \cdot {L}^{-} 1$ hydrochloric acid from conc. $H C l$? There is a safety issue here.

#### Explanation:

This site gives me the following data with respect to conc. $H C l$: $\rho = 1.17 \cdot g \cdot m {L}^{-} 1$; "concentration"=36.5%(w/w).

We want $\text{Moles of HCl"/"Volume of solution (L)}$ to assess the molar concentration. And so.....we assume a $1 \cdot m L$ volume of stuff:

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

We want a $1 \cdot L$ volume of $1 \cdot m o l \cdot {L}^{-} 1$ $H C l$, i.e. we want a $1 \cdot m o l$ quantity of HCl.

And so we take a $85.5 \cdot m L$ volume of conc. $H C l$, and we add this to approx. $900 \cdot m L$ 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 $1 \cdot L$.

And so $\text{concentration}$ $=$ $\text{Moles of HCl"/"Volume of solution}$

$= \frac{85.5 \times {10}^{-} 3 \cdot L \times 11.7 \cdot m o l \cdot {L}^{-} 1}{1 \cdot L} = 1.00 \cdot m o l \cdot {L}^{-} 1$.