If 16.0 mL of water are added to 31.5 mL M #Ba(OH)_2(aq)#, what is the new solution molarity?
Here's what I got.
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")))#