Ch 13A_ Normality by stariya

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									                                CHAPTER 13A
                         EQUIVALENTS AND NORMALITY

                                      INTRODUCTION

Chemists often find it convenient to use ―equivalents‖ instead of ―moles‖ to quantify the amount of a
substance. Equivalents are also used in biology, environmental science, pharmacy and other health
professions. Solution concentrations of equivalents are expressed in ―normality‖ instead of
―molarity‖. In this introduction to equivalents and normality, the discussion will be limited to
protonic (hydrogen ion) acids and hydroxide bases in acid-base reactions. Equivalents are also used
for amounts of oxidizing agents and reducing agents in redox reactions. An example of this use of
equivalents and normality will be demonstrated in a laboratory experiment involving the titration of
sodium oxalate with potassium permanganate.

The purpose of using equivalents instead of moles is that equivalents are defined, so that one
equivalent of any acid will react with one equivalent of any base. Recall that this is not true for
moles. One mole of any acid will not always react with one mole of any base. Consider the
following two chemical equations:

                       HCl + NaOH ——>             NaCl    + H2O

       In this first reaction one mole of HCl reacts with one mole of NaOH.

                       H2SO4 + 2 NaOH ——> Na2SO4 + 2 H2O

       In this second reaction one mole of H2SO4 reacts with two moles of NaOH.

                       EQUIVALENTS OF ACIDS AND BASES

In the context of acid-base reactions, an equivalent is the number of reactive hydrogen ions in one
mole of a protonic acid or the number of hydroxide ions in one mole of a hydroxide base.

The number of equivalents in one mole of a acid can be determined from its chemical formula.

               a. HCl has only one acidic hydrogen. Thus one mole of HCl is equal to one
               equivalent    (abbreviated ―eq‖), since one mole of HCl donates one mole of H+
               ions. 1 mol HCl = 1 eq HCl.

               b. H2SO4 has two acidic hydrogens. Thus one mole of H2SO4 donates two moles of
               H+ ions and contains two equivalents of H2SO4. 1 mol H2SO4 = 2 eq H2SO4.

               c. Consider acetic acid HC2H3O2. Chemical formulas for acids show the acidic
               hydrogens first. There is only one acidic hydrogen in acetic acid and three non-
               acidic hydrogens. 1 mol HC2H3O2 = 1 eq HC2H3O2.

A generalized equation is: 1 mol acid = n eq acid, where n is the number of acidic hydrogens in the
chemical formula for the acid.
The number of equivalents in one mole of base can also be determined from its chemical formula,
since one mole of hydroxide ions reacts with one mole of hydrogen ions, i.e., H+ + OH– —>
H2O. 1 mol base = n eq base, where n is the number of hydroxides in the chemical formula, e.g.

                               Base           Mol            Eq

                               NaOH           1 mol         1 eq
                               Ca(OH)2        1 mol         2 eq
                                                      Al(OH)3        1 mol          3 eq

Note that the # of equivalents of an acid or base is always  # moles of an acid or base.

In conclusion reconsider the balanced chemical equation for the reaction of H2SO4 and NaOH.

                                      H2SO4 + 2 NaOH ——> Na2SO4 + 2 H2O
                       1 mol       2 mol
                       2 eq        2 eq

We see that the # of equivalents of acid equals the # of equivalents of base.

Equivalent Weight

Molecular weight MW (molar mass) is defined as the weight (mass) in grams of one mole of a
substance. Similarly, ―equivalent weight‖ EW is defined as the weight in grams of one equivalent of
a substance, e.g.

               a. For HCl, 1 mol HCl = 1 eq HCl. MW of HCl = 36.46 g/mol.
               EW of HCl = 36.46 g/eq.

               b. For H2SO4, 1 mol H2SO4 = 2 eq H2SO4. MW of H2SO4 = 98.08 g/mol.
               EW of H2SO4 = 49.04 g/eq.

For acids, EW = MW / n, where n is the number of acidic hydrogens in the chemical formula.
For hydroxide bases, EW = MW / n, where n is the number of hydroxide ions in the chemical formula.
The EW of a substance is always less than or equal to the MW. EW  MW.

             Compound             MW                   EW

             HNO3                 63.01 g/mol          63.01 g/eq
             H3PO4                98.00 g/mol          32.67 g/eq
             NaOH                 40.00 g/mol          40.00 g/eq
             Ca(OH)2              74.09 g/mol          37.05 g/eq
             Al(OH)3              78.00 g/mol          26.00 g/eq
Calculating the Number of Equivalents

So far our discussion has been limited to how many theoretical equivalents there are in exactly one
mole of an acid or a base. Oftentimes, however, the quantity on hand will be less or more than one
mole of an acid or a base. How then do we calculate the exact number of equivalents of a given
quantity of an acid or a base?

Sample Problem #1 and Solution:

       How many equivalents are there in 6.32 g of Ca(OH)2?

       Solution: Knowing there are 2 equivalents in every 1 mole of Ca(OH)2, provides us with
the conversion factor we need to solve the problem by dimensional analysis:

               1 mole Ca(OH)2 = 2 equivalents Ca(OH)2

       The problem solving sequence is: # grams  # moles  # equivalents.

              6.32 g Ca(OH)2 = [6.32 g Ca(OH)2] [1 mol Ca(OH)2] [2 eq Ca(OH)2]
                                                 [74.09 g Ca(OH)2] [1 mol Ca(OH)2]

                                 = 0.171 eq Ca(OH)2

                                      NORMALITY

The ―normality‖ N of a substance in solution is equal to the number of equivalents of that
substance in one liter of solution. Thus in general for any substance A, its normality

                              NA = # eq A
                                    LSOLN

Sample Problem #2 and Solution:

       Calculate the normality of a solution containing 6.32 g Ca(OH)2 in 585 mL of solution.

       Solution: First write the equation needed to solve the problem.

       The equation is NCa(OH)2 = # eq Ca(OH)2
                                     LSOLN

In sample problem #1, we calculated that 6.32 g Ca(OH)2 = 0.171 eq Ca(OH)2

       Thus,          NCa(OH)2 = 0.171 eq Ca(OH)2 = 0.292 eq Ca(OH)2
0.585 L   L
The solution 0.292 N is said to be a 0.292 ―normal‖ solution of Ca(OH)2. N is the symbol which
can stand for the noun ―normality‖ or the adjective ―normal‖, just as M stands for molarity and
molar. The units for N are eq per L, just as the units for the symbol M are mols per L.

Sample Problem #3 and Solution:

       How many equivalents are there in 585 mL of a 0.292 N Ca(OH)2 solution?

       Solution: This is also a normality problem. Start by writing the equation you need to use:

              NCa(OH)2 = # eq Ca(OH)2
                           LSOLN

       Rearrange the equation and solve for # eq Ca(OH)2:

                  # eq Ca(OH)2 = NCa(OH)2  LSOLN

                                 = 0.292 eq Ca(OH)2  0.585 L
                                           L

                                 = 0.171 eq Ca(OH)2

       Note that the symbol N was replaced with its units eq /L in the calculation.

Conversion of Normality to Molarity and Molarity to Normality

A problem that comes up frequently is to calculate the normality of a molar solution and vice
versa. To keep from getting confused, remember that # eq  # mol. Therefore, N  M.

For acids, N = nM where n is the number of acidic hydrogens in the chemical formula of the acid.
For hydroxide bases, N = nM where n is the number of hydroxide ions in the chemical formula of
the base.
                                    Examples

                                6.0 N HCl = 6.0 M HCl
                             6.0 N H2SO4 = 3.0 M H2SO4
                         0.075 N Al(OH)3 = 0.025 M Al(OH)3

Using dimensional analysis: 0.075 N Al(OH)3 = [0.075 eq Al(OH)3] [1 mol Al(OH)3]
                                                      L         [3 eq Al(OH)3]

                                               = 0.025 mols Al(OH)3 = 0.025 M Al(OH)3
                                                        L
Note again the key to these conversions is knowing how many theoretical equivalents there are
in one mole of a substance and using this as your conversion factor.

Note also the symbol N was converted to its units of eq / L to make the calculation by dimensional
analysis.

Titrations

Titrations are reactions in which we measure the volume of a standardized solution (e.g. NaOH)
that reacts stoichiometrically with an acid. Using the concept of equivalents in an acid-base
reaction, we know that at the end of a titration:

                            # eq base = # eq acid

As we have seen in sample problem #3 we can calculate the exact number of equivalents of any
quantity of an acidic or basic substance in solution, if both the volume of that solution and its
normality are known. In calculations involving titrations, the general equation

       NA = # eq A is used in rearranged form:        # eq A = NA  LSOLN.
            LSOLN


Sample Problem #4 and Solution:

     A 0.3725 g sample of a sulfuric acid solution is titrated with 28.63 mL of a 0.1095 N
NaOH solution. Calculate the % weight (% mass) of sulfuric acid in this aqueous solution.

        Solution: Since this is a normality problem, we will start by using the normality equation
but in rearranged form: NNaOH  LSOLN = # eq NaOH. The problem solving sequence is to
calculate in this order:
        # eq NaOH  # eq H2SO4  # mols H2SO4  # grams H2SO4  % weight

      # eq NaOH = 0.1095 eq NaOH  0.02863 L = 0.003135 eq NaOH
                          L

0.003135 eq NaOH = [0.003135 eq NaOH] [1 eq H2SO4] [1 mol H2SO4] [98.08 g H2SO4]
                                      [1 eq NaOH] [2 eq H2SO4] [1 mol H2SO4]

                    = 0.1537 g H2SO4

       % weight = 0.1537 g H2SO4  100 = 0.1537 g H2SO4  100 = 41.27 %
                  total grams soln       0.3725 g soln
Sample Problem #5 and Solution:

     It took 33.6 mL of 0.225 N NaOH to titrate a 0.958 g solution containing acetic acid
HC2H3O2. Calculate the % weight (% mass) of acetic acid in the solution.

        Solution: This is a normality problem involving a titration. Thus we must first calculate
the # eq NaOH used in the titration. The problem solving sequence is to calculate in this order:
        # eq NaOH  # eq HC2H3O2  # mols HC2H3O2  # grams HC2H3O2  % weight

       # eq NaOH = NNaOH  LSOLN = 0.225 eq NaOH  0.0336 L
                                        L

                    = 0.00756 eq NaOH

       0.00756 eq NaOH = [0.00756 eq NaOH] [1 eq HC2H3O2] [1 mol HC2H3O2]
                                            [1 eq NaOH]    [1 eq HC2H3O2]

                           = 0.00756 mol HC2H3O2

       0.0756 mol HC2H3O2 = [0.00756 mol HC2H3O2] [60.05 g HC2H3O2]
                                                   [1 mol HC2H3O2]

                            = 0.454 g HC2H3O2

         % weight = 0.454 g HC2H3O2  100 = 47.4 %
                    0.958 g soln

Standardization of an Acid Solution by Titration with a Standardized Base

In titrations the term to standardize generally means finding the solution concentration of the acid
using an already known or standardized concentration of base. For example, HCl solutions are
usually standardized by titration with a standardized or a known concentration of NaOH. At the
endpoint, # eq HCl = # eq NaOH. Therefore, NHClVHCl = NNaOHVNaOH . Rearranging this
equation, the normality of the HCl solution can be calculated as follows:

                            NHCl = NNaOH  VNaOH
                                          VHCl

Sample Problem # 6 and Solution:

       It takes 32.75 mL of 0.1835 N NaOH to titrate 23.42 mL of an HCl solution. Calculate the
normality of the HCl solution.
         NHCl = NNaOH  VNaOH      = 0.1835 eq NaOH  32.75 mL = 0.2566 N NaOH
                       VHCl                 L        23.42 mL
                                      PROBLEMS

1.       Determine the equivalent weights of the following acids and bases:
         a. Ca(OH)2 ________ b. H2C2O4 ________ c. KOH ________ d. H2CO3 __________

2.       How many equivalents are there in 15.0 g of
         a. H2SO4 ____________ b. Ca(OH)2 ______________ c. Al(OH)3 ______________ ?

3.       Calculate the number of equivalents and the number of mols of solute in the following:
         a. 0.642 L of a 0.450 N Ba(OH)2 solution: ______________ eq; _____________ mols
         b. 150 mL of a 0.550 N KOH solution: ________________ eq; _____________ mols
         c. 34.0 mL of a 2.55 N H3PO4 solution: ________________ eq; _____________ mols

4.       What is the molarity of:
         a. 0.12 N H2SO4 _______ b. 0.55 N NaOH _______ c. 0.020 N Ca(OH)2 _______ ?

5.       What is the normality of
         a. 0.45 M Al(OH)3 _______ b. 0.122 M H2SO4 _______ c. 0.675 M H3PO4 ________ ?

6.       A 15.0 mL sample of an acid requires 37.3 mL of 0.303 N NaOH for neutralization.
         Calculate the normality of the acid.

7.       a. How many equivalents of NaOH are present in 305 mL of 0.150 N NaOH?
         b. How many equivalents of H2SO4 are present in 35.5 mL of 1.35 N H2SO4?

8.       A 0.243 g sample containing Ca(OH)2 was titrated with 22.7 mL of a 0.109 N HCl
                                                                         solution. Calculate the
                                                                         weight % (mass %) of
                                                                         Ca(OH)2 in the sample.


9.       A 0.336 g sample containing oxalic acid H2C2O4 was titrated with 17.7 mL of a 0.0996 N
         NaOH solution. Calculate the weight % (mass %) of oxalic acid in the sample.




     Answers:

     1a. 37.05 g, 1b. 45.02 g, 1c. 56.11 g, 1d. 31.01g; 2a. 0.306 eq, 2b. 0.405 eq,
     2c. 0.577 eq; 3a. 0.289 eq, 0.144 mols, 3b. 0.0825 eq, 0.0825 mols, 3c. 0.0867 eq, 0.0289
     mols, 4a. 0.060 M, 4b. 0.55 M, 4c. 0.010 M; 5a. 1.35 N, 5b. 0.244 N,
            5c. 2.03 N; 6. 0.753 N; 7a. 0.0458 eq, 7b. 0.0479 eq; 8. 37.7 %; 9. 23.6 %

								
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