# What is the molarity of a solution prepared by combining 125 mL of 0.251 molar HCl with 250 mL of pure water?

Nov 9, 2015

$\text{0.084 M}$

#### Explanation:

So, you know that you're mixing $\text{125 mL}$ of a $\text{0.251-M}$ hydrochloric acid solution with $\text{250 mL}$ of pure water.

In essence, you're diluting the initial solution by keeping the number of moles of solute, which in your case is hydrochloric acid, constant and increasing the amount of solvent.

This means that you can expect the molarity of the second solution to be smaller than that of the initial solution, which would be characteristic of a diluted solution.

Use the volume and the molarity of the initial solution to determine how many moles of hydrochloric acid you get in that volume

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{\text{HCl" = "0.251 M" * 125 * 10^(-3)"L" = "0.031375 moles HCl}}$

The total volume of the final solution will be equal to

${V}_{\text{total" = "125 mL" + "250 mL" = "375 mL}}$

The molarity of the second solution will thus be

c = "0.031375 moles"/(375 * 10^(-3)"L") = color(green)("0.0874 M")

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the target solution.