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					Normality
Dr. Steve Badger
 Concentrations of Solutions

A brief review of molarity & molality

                 Number of moles of solute
    Molarity =
                     Liter of solution


                 Number of moles of solute
    Molality =        Kg of solvent
Preparing a Solution of Known Molarity
 Would you use a volumetric flask to
prepare a solution of known molality?




       How would you prepare a
      solution of known molality?
   Another way to express
    solution concentration
• Now that you understand molarity…

                Number of moles of solute
  Molarity =
                    Liter of solution

  …let’s consider normality!

                Number of equivalents of solute
  Normality =
                     Liter of solution
         What’s an equivalent ?

• An equivalent of a substance is the
  mass (grams) of that substance that
  will combine with one mole of another
  reactant.
• In an acid-base reaction, an equivalent
  is that amount of a substance that
  reacts with or liberates 1.0 mole of H+.
       What’s an equivalent ?

• An equivalent of a substance is the
  mass (grams) of that substance that
  will combine with one mole of another
  reactant.
• In a redox reaction, an equivalent is
  that amount of a substance that gains
  or loses 1.0 mole of e–s.
     Making a 1.00N Solution
• Calculate the mass of one equivalent
  of the substance, then measure that
  number of grams of the substance.
• Put that substance in…. ??
• Add how much solvent?
Preparing a Solution of Known Normality
          AbNormality?

• Why don’t many modern chemistry
  textbooks cover normality?
• If we take a dimensional analysis
  approach to problems solving,
  normality is a superfluous concept.

• Consider the following problem:
  What volume of a 0.500 M KOH
  solution is required to titrate 10.0
  mL of a 0.20 M H2SO4 solution?


   H2SO4 + KOH          H2O + K2SO4

Could we just use this equation?

              MaVa = MbVb
   Here’s what we’d get if we used

              MaVa = MbVb
0.20 M H2SO4 X 10.0 mL = 0.500 M KOH X ? mL


 Solving this, we get 4.0 mL of 0.500 M KOH.
 But is this correct? No, it’s wrong! Why?

 So let’s see how we solve this correctly.
              What volume of a 0.500 M KOH
              solution is required to titrate 10.0
              mL of a 0.20 M H2SO4 solution?


                BALANCE THE CHEMICAL EQUATION!
                H2SO4 + KOH                     H2O + K2SO4


               M                   rx                   M
volume acid          moles acid           moles base          volume base
              acid                coef.                base
       What volume of a 0.500 M KOH
       solution is required to titrate 10.0
       mL of a 0.20 M H2SO4 solution?

      H2SO4 + 2KOH                        2H2O + K2SO4

                 M                   rx                   M
volume acid            moles acid           moles base          volume base
                acid                coef.                base


            ___ mol H2SO4       __ mol KOH          ____ ml soln
10.0 mL x                   x                   x                  = ___ mL
            ____ mL soln        __ mol H2SO4        ____ mol KOH
       What volume of a 0.500 M KOH
       solution is required to titrate 10.00
       mL of a 0.20 M H2SO4 solution?

       H2SO4 + 2KOH                       2H2O + K2SO4


                 M                   rx                    M
volume acid            moles acid           moles base           volume base
                acid                coef.                 base


            0.20 mol H2SO4       2 mol KOH           1000 ml soln
10.0 mL x                    x                  x                   = 8.0 mL
             1000 mL soln        1 mol H2SO4        0.500 mol KOH
                             Notice this!



                 M                   rx                    M
volume acid            moles acid           moles base           volume base
                acid                coef.                 base


            0.20 mol H2SO4       2 mol KOH           1000 ml soln
10.0 mL x                    x                  x                   = 8.0 mL
             1000 mL soln        1 mol H2SO4        0.500 mol KOH




                             Notice this!
  This is what makes normality superfluous!
 Is the following statement
       true or false?
• Any volume of a base will completely
  react with (neutralize) that same
  volume of acid if the two solutions
  have the same normality (vice versa too).
• In other words, is this true:

            NaVa = NbVb
 Is the following statement
       true or false?
• Any volume of a reducing agent will
  completely react with that same
  volume of an oxidizing agent if the
  two solutions have the same normality.
• In other words, is this true:

        NoxVox = NredVred
    Let’s solve a few problems
          using normality


• Look at the handout that has sample
  problems and work the first one.
  And another thing…

• We also use equivalents and
  milliequivalents as an amount
  of a substance
• The same way that we use
  moles (mol) and millimoles
  (mmol) as an amount of a
  substance
    If you need more help:
Come by my office and I can give you
a few pages of worked examples from
a Schaum’s Solved Problems Series.
You can download this PowerPoint file
at my EU web site:
www.evangel.edu/Personal/badgers/Web/
        The End


Now wasn’t that fascinating?

				
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