# How do we make a 1*L volume of a solution that is 0.5*mol*L^-1 with respect to HCl from a solution that is 10.6*mol*L^-1 with respect to HCl(aq)?

Jun 15, 2017

Given proper attire, with conc. acid THE PRIMARY PRACTICAL CONSIDERATION IS $\text{ADD YOUR ACID TO WATER}$ and NEVER $\text{VICE VERSA}$. We need approx. a $50 \cdot m L$ volume of conc. acid.

#### Explanation:

And why so?

$\text{Because if you spit in acid, it spits back!}$ I kid you not......

And other practical considerations. Wear a pair of SAFETY SPECTACLES to protect your mince pies; and wear a lab coat to protect your clothing......You should do this automatically in a lab (if you are a speccy, your prescription glasses are an adequate protection). Most chemists upon entering a lab will IMMEDIATELY put on a pair of safety spex from a tray, or put on the safety spec that they carry in their pocket. This is certainly a habit to develop.

And to make a $1 \cdot L$ volume of $0.5 \cdot m o l \cdot {L}^{-} 1$ $H C l$; we use the relationship.....

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, and we solve for ${V}_{1}$.....

${V}_{1} = \frac{{C}_{2} {V}_{2}}{C} _ 1 = \frac{0.5 \cdot m o l \cdot {L}^{-} 1 \times 1 \cdot L}{10.6 \cdot m o l \cdot {L}^{-} 1} \cong 50 \cdot m L$.

And remember to add the $\text{WATER to YOUR ACID...............}$