# If 16.0 mL of water are added to 31.5 mL M Ba(OH)_2(aq), what is the new solution molarity?

May 21, 2017

Here's what I got.

#### Explanation:

Since you don't know the initial molarity of the barium hydroxide solution, you can only determine the new molarity in relation to the initial value.

Now, when you dilute a solution, you are decreasing its concentration by increasing its volume while keeping the number of moles of solute constant.

In a dilution, the concentration of the solution will decrease and the volume of the solution will increase by the same factor called the dilution factor, $\text{DF}$.

$\text{DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted}$

In your case, the volume of the solution after the dilution will be equal to

${V}_{\text{diluted" = "16.0 mL" + "31.5 mL" = "47.5 mL}}$

This means that the volume increases by a factor of

"DF" = (47.5 color(red)(cancel(color(black)("mL"))))/(31.5color(red)(cancel(color(black)("mL")))) = color(blue)(1.508)

This implies that the concentration of the solution decreased by the same factor

${c}_{\text{diluted" = c_"concentrated}} / \textcolor{b l u e}{1.508}$

Let's say that the molarity of the $\text{31.5-mL}$ sample is $x$ $\text{M}$. The new molarity of the solution will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{molarity diluted" = (1/1.508 * x)color(white)(.)"M}}}}$