chemistry 30 unit 2 solubility ch 16 in text n.
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Chemistry 30 – Unit 2 – Solubility – Ch. 16 in Text. 03 - Concentration - DILUTIONS. Stock Solutions.

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stock solutions
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
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
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
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.

Final moles of solute

Initial moles of solute

…is equal to

dilution equation
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
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
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
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
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
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 example1
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 example2
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