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48                                                   Chapter 5
       where  Kb  =  ([CH3NH3 +][OH-])/[CH3NH2] is  the equilibrium con-
       stant for base dissociation.
         Table  5.1  summarises  some common  strong and  weak  acids and
       bases, which need to be identified in problems (R  = an alkyl group
       e.g. CH3, CH3CH2 etc.)


       Table 5.1  Some examples of strong and weak acids and bases
       Strong acids:  1. HCl   2. HN03     3. HC104    4. H2S04
        Weak acids:   1. HN02   2. HC102   3. Carboxylic acids: RC02H
       Strong bases:  1. NaOH   2. KOH
        Weak bases:   1. NH3   2. Amines: 1" (RNH2), 2" (RzNH), 3" (R3N)


                          COMMON ION EFFECT
       A common ion is an ion (charged species) common to two substances
       in the same mixture, e.g. in a solution of ethanoic acid, CH3C02H and
       sodium  ethanoate,  CH3C02Na, the  common  ion  is  the  ethanoate
       anion, CH3CO:.  The common ion effect occurs when the presence of
       extra (common) ions in the solution represses or restrains the dissocia-
       tion of a species. To explain this, consider the following example:

         Example: A  solution  is prepared  by  adding  0.6 moles  of  sodium
         ethanoate, CH3C02Na, and 0.8 moles of ethanoic acid, CH3C02H,
         to water, to make  up  1 dm3. Determine  the concentration  of  all
         solute species, given that Ka(CH3C02H)  = 1.8 x   at 25 "C.


        Solution: To solve this problem, the working method of Chapter 4 is
        applied.
          1. Read the question Carefully-K,   problem!
         2.  Species present  in  solution:  CH3C02Na(,,),  CH3CO,H(,,)  and
            the corresponding ions (step 3 below).
         3. (a) CH3C02Na(,,)  is a salt and undergoes 100% dissociation into
            its anion and cation, CH3C02Na(,,)  + Na+(,,)  + CH,CO,(aq),
            i.e. the equilibrium lies completely to the ionic products!
            (b) CH3C02H(,,)  is a weak acid. Weak acids do not  dissociate
            completely  into  anions  and  cations.  Hence,  the  reaction  is  at
            equilibrium, i.e. CH3C02H + H20 + CH3C0;  + H30+.

              In  this  solution,  both  reactions  (a)  and  (b)  occur  simulta-
            neously,  and  so  both  must  be considered when  calculating the
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