How would you prepare 2.00 L of a .250 M NaOH solution from 1.00 M stock solution of NaOH?

Question

abhisar · Answer

We are given the molarity of a more concentrated solution (1.0 M stock solution) and the volume and molarity of a more dilute one containing the same solute (2 L of 0.250 M solution). We are asked to calculate the volume of the concentrated solution needed to prepare the dilute solution.

Our plan is to use Equation $ \tt M_1V_1=M_2V_2$ where $\tt M_1$ is the molarity of the more concentrated solution. Because $ \tt M_1V_1 = M_2V_2$, we can write

$\tt  \Large  V_1= \huge \frac{M_2\times V_2}{M_1} \ \Large  V_1= \huge \frac{2\times 0.250 }{1} = ?$

We see that if we start with $\tt V_1~mL$ of 1.0 M NaOH and dilute it to a total volume of 2 L, the desired 0.250 M solution will be obtained.

The calculated volume seems reasonable because a small volume of concentrated solution is used to prepare a large volume of dilute solution (that is, $\tt V_1 < V_2$ and $ \tt M_1 > M_2$.