03 - CONCENTRATION DILUTIONS
CHEMISTRY 30 – UNIT 2 – SOLUBILITY – CH. 16 IN TEXT
STOCK SOLUTIONS
• A common situation in a chemistry laboratory occurs when a solution of a
particular concentration is desired, but only a solution of greater
concentration is available: this is called a stock solution.
WHAT IS A DILUTION?
• To dilute a solution means to add more solvent without the addition of
more solute.
• You may have heard it referred to as “watering down”
• This statement is somewhat accurate, as long as your solvent actually is
water
• Of course, the resulting solution is thoroughly mixed so as to ensure that
all parts of the solution are homogenous.
WE CAN DILUTE A SOLUTION BY
ADDING WATER TO IT:
BEFORE DILUTION:
M1 = INITIAL MOLES OF
SOLUTE
AFTER DILUTION:
M2 = FINAL MOLES OF
SOLUTE
FROM PREVIOUS SLIDE…
• NOTICE THAT WE DID NOT ADD ANY MORE SOLUTE AT ALL
• THE NUMBER OF MOLES WILL NOT CHANGE.
• THE VOLUME CHANGES, AND ALSO THE CONCENTRATION.
Initial moles of
solute
…is equal to
Final moles of
solute
DILUTION EQUATION:
• The fact that the solute amount stays constant allows us to develop
calculation techniques. First, we write:
𝑚𝑜𝑙𝑏𝑒𝑓𝑜𝑟𝑒 𝑑𝑖𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑜𝑙𝑎𝑓𝑡𝑒𝑟𝑑𝑖𝑙𝑢𝑡𝑖𝑜𝑛
• From the definition of molarity, we know that the moles of solute equals
the molarity times the volume.
𝑚𝑜𝑙 = 𝑀 ∗ 𝑉
• So we can substitute equation, like this:
𝑀1 𝑉1 = 𝑀2 𝑉2
• M1V1 means the molarity and the volume of the original solution
• M2V2 means the molarity and the volume of the second solution.
EXAMPLE #1
• 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some
0.800 M solution. How many mL of 0.800 M can you make?
• Using the dilution equation, we write:
𝑀1 𝑉1 = 𝑀2 𝑉2
1.50𝑀 00534𝐿 = 0.800𝑀 𝑉2
𝑉2
1.50𝑀 00534𝐿
=
0.800𝑀
𝑉2 = 0.100L=100.mL=1.00∗102 𝑚𝐿
• Any old volume measurement is fine, just so long as the same one is used
on each side.
EXAMPLE #2
• 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What
is the final volume of solution which results?
• Placing the proper values into the dilution equation gives:
𝑀1 𝑉1 = 𝑀2 𝑉2
(2.500 mol/L) (0.1000 L) = 0.5500 mol/L 𝑉2
𝑉2
(2.500 M) (0.1000 L)
=
0.5500 M
𝑉2 = 0.4545L=454.5mL
SERIAL DILUTIONS
• The dilution equation is fairly easy to understand – we
shouldn’t have any trouble calculating new or desired
concentrations or volumes
• Sometimes however, we will calculate out a value that is too
small or unrealistic – this makes it difficult to measure in a
practical lab-like setting
SERIAL DILUTIONS - EXAMPLE
• How could we prepare 200.0 mL of .010 M HCl from 10. M HCl?
• Calculate the missing value using the dilution equation
• M1v1 = M2v2
10. ( vi ) = 200.0 (.010)
• v1 = 0.20 mL
• It is unrealistic for us to measure .20 mL so we do a series of dilutions so
we can maintain our accuracy
SERIAL DILUTIONS - EXAMPLE
• How do we solve this?
• Step 1: First pick a practical dilution into an appropriately small volume.
This is usually determined by the tools that you have in the lab.
• For example use 10. ml into 200.0 ml
• M1V1 = M2V2
• 10.M ( 10.mL ) = ( M2 ) 100.mL
M2 = 1.0 M
• Step 2: Now you have a 1.0 molar solution so redo the math to find how
much of our NEW solution we need to make the concentration we want.
• M1V1 = M2V2
• (1.0)v1 = 200.0mL ( .010 )
v1 = 2.0 mL
• 2.0 mL is a much more reasonable amount to measure than 0.20 mL
SERIAL DILUTIONS - EXAMPLE
• If 2.0 mL is still too small and you don't have the equipment to do it, continue
to dilute:
• Step 1: Do the 10. ml into 100. ml again; this time your initial concentration will
be 1.0 M
• M1v1 = M2v2
•
(1.0) 10. = (M2) 100.
M2 = .10 M
• Step 2: Now you have a .10 molar solution so redo the math
• M1v1 = M2v2
• (.10)v1 = 200.0 (.010)
v1= 20. mL
• 20. mL is even easier than 2.0 mL to measure accurately
SERIAL DILUTION VISUAL
Stock Solution Volumetric Pipette
M1 = 10.M
V1 = 10.0mL
X
1st Dilution
M2 = ?M
Volumetric Flask
V2 = 100.0mL
M1V1=M2V2
Gives us 1.0M
1st Dilution
M1 = 1.0M
Graduated Pipette
V1=?mL
M1V1=M2V2
Gives us 2.0mL
X
2nd Dilution
M2 = 0.01M
Volumetric Flask
V2 = 200.0mL