# Question 0e130

Feb 11, 2015

You can't prepare $\text{10 mL}$ of a $\text{5 mM}$ solution from a stock solution of $\text{10 mL}$, $\text{10 mM}$, you'd have to mix a separate solution.

The idea behind using stock solutions to prepare solutions of various concentrations is that the number of moles of solute remains constant, but the volume of the solution changes.

This is how you dilute solutions - you keep the same amount of solute but change the volume of the solution.

For constant number of moles of solute

$\text{LESS volume = MORE concentrated solution}$

$\text{MORE volume = LESS concentrated solution}$

In your case, the volume is kept constant, which means that the number of moles of solute must change. For the stock solution, you have

C = n/V => n_("stock") = C * V = 10 * 10^(-3)"L" * 10 * 10^(-3)"M"
n_("stock") = 0.1 * 10^(-3)"moles" = "0.1 mmol"

For the target solution, you would need

n_("needed") = C * V = 10 * 10^(-3)"L" * 5 * 10^(-3)"M"
n_("needed") = 0.05 * 10^(-3)"moles" = "0.05 mmol"#

This means that you would have to remove $0.1 - 0.05 = 0.05$ moles of solute from the stock solution, which is more difficult to do than to simply make a new solution altogether.