# How do I make 100 ml of .3 M HCl from 1 M?

May 7, 2016

Here's how you could do that.

#### Explanation:

Your strategy here will be to use the molarity and volume of the target solution to determine how many moles of solute, which in your case is hydrochloric acid, $\text{HCl}$, it contains.

Once you know that, you can use the moalrity of the stock solution to determine the volume of this solution that will contain that many moles of hydrochloric acid.

So, molarity is defined as moles of solute per liter of solution. This means that you have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The target solution will contain

n_"HCl" = "0.3 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{\text{HCl" = "0.030 moles HCl}}$

Since this is exactly how many moles of solute you must take from the stock solution, you can say that

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies V_"solution" = n_"solute}} / c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

${V}_{\text{stock" = (0.030 color(red)(cancel(color(black)("moles"))))/(1 color(red)(cancel(color(black)("mol"))) "L"^(-1)) = "0.030 L}}$

Expressed in milliliters, this will be equivalent to

V_"stock" = color(green)(|bar(ul(color(white)(a/a)"30 mL"color(white)(a/a)|)))

So, to prepare your target solution, take $\text{30 mL}$ of a $\text{1-m}$ hydrochloric acid solution and add enough water to get the total volume of the solution to $\text{100 mL}$.