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03 - CONCENTRATION -
DILUTIONSCHEMISTRY 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
Final moles of solute…is equal
to
DILUTION EQUATION:• The fact that the solute amount stays constant allows us to develop
calculation techniques.
• First, we write:
moles before dilution = moles after dilution
• From the definition of molarity, we know that the moles of solute equals the molarity times the volume.
• So we can substitute MV (molarity times volume) into the above equation, like this:
M1V1 = M2V2
• M1V1 means the molarity and the volume of the original solution
• M2V2 means the molarity and the volume of the second solution.
CAUTION!
• You may also see this equation written as
C1V1 = C2V2.
• “C” stands for concentration.
• We use the formula M1V1 = M2V2 when our
concentration is expressed in molarity.
• We would use C1V1 = C2V2 if our
concentration was expressed in something else, like g/L, or ppm, etc. ***More on those later**.
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:
(1.50 mol/L) (0.0534 L) = (0.800 mol/L) (x)
• Solving the equation gives an answer of 0.100 L.
• Notice that the volumes need not be converted to liters. 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:
(2.500 mol/L) (0.1000 L) = (0.5500 mol/L) (x)
• x = 0.4545 L
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. mL of .01 M HCl from 10. M HCl?
• Calculate the missing value using the dilution equation
• M1v1 = M2v2 10. ( vi ) = 0.200
(.01)
• v1 = 0.00020 L or 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.
• For example use 10. ml into 100. ml
• M1v1 = M2v2
• 10M ( 10mL ) = ( M2 ) 100mL M2 = 1.0 M
• Step 2: Now you have a 1.0 molar solution so redo the math
• M1v1 = M2v2
• (1.0)v1 = 200 ( .01 ) v1 = 2 mL
• 2 mL is a much more reasonable amount to measure than 0.2 mL
SERIAL DILUTIONS - EXAMPLE
• If 2 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 (.01) v1= 20 mL
• 20 mL is even easier than 2 mL to measure accurately