# What volume of 12 M HCl is needed to make 500 mL of .10 M HCl?

Nov 13, 2015

$\text{4.2 mL}$

#### Explanation:

To make the problem more interesting, let's assume that you don't know the formula for dilution calculations.

The idea with diluting a solution is that the number of moles of solute will remain constant after the initial solution is diluted. The only thing that changes in such cases is the volume of the solution. This means that if you know how many moles of solute you have in the target solution, you also know how many moles of solute were present in the stock solution sample.

Use the molarity and volume of the target solution to determine how many moles of hydrochloric acid, $\text{HCl}$, you need in that solution

$c = \frac{n}{V} \implies n = c \cdot V$

${n}_{\text{HCl" = "0.10 M" * 500 * 10^(-3)"L" = "0.050 moles HCl}}$

Now the question is - what volume of stock solution would contain this many moles of hydrochloric acid?

$c = \frac{n}{V} \implies V = \frac{n}{c}$

${V}_{\text{stock" = (0.050color(red)(cancel(color(black)("moles"))))/(12color(red)(cancel(color(black)("moles")))/"L") = "0.0041667 L}}$

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the volume of the target solution

V_"stock" = color(green)("4.2 mL")