# To what volume must a 250*mL volume of 10*mol*L^-1 HCl be diluted to give a 0.050*mol*L^-1 concentration?

Sep 11, 2017

Before we start remember the CARDINAL RULE:
$\text{Acid to water not water to acid.....}$ I get a final volume of $50 \cdot L$ of dilute acid.

#### Explanation:

$\text{May her rest be long and placid, she added water to the acid.}$

$\text{The other girl did as she oughter, she added acid to the water.}$

A small drop of water placed in a strong acid might spit as it heats up upon hydration (another old saw: $\text{if you spit in acid it spits back at you}$).

For your problem, we use the relationship.....

$\text{Concentration"="Moles of solute (moles)"/"Volume of solution (L)}$

We have a starting volume of $250 \cdot m L$ of $10.0 \cdot m o l \cdot {L}^{-} 1$ $H C l$ (the stuff you get out of the bottle is $10.6 \cdot m o l \cdot {L}^{-} 1$).

$\text{Moles of HCl} = 250 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 10 \cdot m o l \cdot {L}^{-} 1$

$= 2.50 \cdot m o l$ with respect to $H C l$.

Now this molar quantity is diluted up to give a $0.050 \cdot m o l \cdot {L}^{-} 1$ (and remember acid to water!).

We solve for $\text{volume}$ in $\text{concentration"="moles of HCl"/"volume of solution}$

i.e. $V = \frac{2.50 \cdot m o l}{0.050 \cdot m o l \cdot {L}^{-} 1} = 50 \cdot L$, why? Because $\frac{1}{L} ^ - 1 = \frac{1}{\frac{1}{L}} = L$ as required.

AND WHEN YOU WORK WITH CONC. ACIDS YOU MUST WEAR (i) SAFETY SPECTACLES to protect your mince pies; and (ii) a LAB-COAT to protect your clothing.