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

Question

Naomi Aronson · Accepted Answer

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"#. #"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 #"31.5-mL"# sample is #x# #"M"#. The new molarity of the solution will be #color(darkgreen)(ul(color(black)("molarity diluted" = (1/1.508 * x)color(white)(.)"M")))#

Avery Adams · Answer

undefined Calculate the new molarity using the formula M1V1 = M2V2.