I am trying to make 5N $\ce{HCl}$ solution. So far I understood that you need to get the molecular weight (36.5) and divide it by the number of $\ce{H+}$ that react in the acid-base reactions. The problem is that my $\ce{HCl}$ stock is 37 %. How should I take this into account?


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Hydrogen chloride in its standard state is a gas. We usually do not make solutions of $\ce{HCl}$ by the method of dissolving a known mass of the the solute in enough solvent to make the appropriate volume of the specific concentration since gaseous substances are difficult to weigh (and weight is one of the most direct ways we have to determine mass in the lab).

Hydrochloric acid (aqueous solutions of $\ce{HCl}$) is commercially available in several concentrations, usually $1\ \text{N}$, $0.1\ \text{N}$. and "concentrated".

"Concentrated" simply means that you have in that bottle the highest concentration of $\ce{HCl}$ water that is stable. For $\ce{HCl}$, a concentrated solution is 36-38% by mass. If you know the density of the solution (should be on the bottle), you can convert % by mass into moles of $\ce{HCl}$ per liter of solution (you also need the molar mass of $\ce{HCl}$).

For example, if you had a solution that was 20% $\ce{HCl}$ by mass with a density of $1.098 \ \text{g/mL}$, you can determine molarity by converting mass of $\ce{HCl}$ into moles and mass of solution into liters:

$$20\% \ \ce{HCl} \ \text{by mass}=\frac{20 \text{ g }\ce{HCl}}{100\ \text{g solution}}$$ $$20 \text{ g }\ce{HCl} \times \frac{1 \text{ mol }\ce{HCl}}{36.46 \ \text{g }\ce{HCl}}=0.5485\ \text{mol }\ce{HCl}$$ $$100\ \text{g solution}\times \frac{1\ \text{mL solution}}{1.098\ \text{g solution}}\times \frac{1\text{ L}}{1000\ \text{mL}}=0.09107\ \text{L solution}$$ $$\frac{0.5485\ \text{mol }\ce{HCl}}{0.09107\ \text{L solution}}=6.023 \ \text{M}$$ However, other folks have the done the math for you. The Wikipedia article of hydrochloric acid lists the concentrations (and densities) for several $\ce{HCl}$ solutions as a function of % by mass. This listing of concentrated reagents lists $37.2\% \ \ce{HCl}$.

With a concentration for $37.2\% \ \ce{HCl}$, you can proceed to work through the dilution process using $c_1 v_1 = c_2 v_2$.


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