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