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

##### 1 Answer

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**,

#"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_"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_"diluted" = c_"concentrated"/color(blue)(1.508)#

Let's say that the molarity of the

#color(darkgreen)(ul(color(black)("molarity diluted" = (1/1.508 * x)color(white)(.)"M")))#