Page 268 - Theory and Problems of BEGINNING CHEMISTRY
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CHAP. 17]                            ACID-BASE THEORY                                 257


               17.24. Calculate the hydronium ion concentration of a 0.250 M acetic acid solution also containing 0.190 M
                                                 −5
                     sodium acetate.  K a = 1.81 × 10 .
                     Ans.                                HC 2 H 3 O 2 + H 2 O ←− H 3 O + C 2 H 3 O −
                                                                     −→
                                                                             +
                                                                                     2
                                           Initial       0.250           0      0.190
                                           Change      −x                x      x
                                           Equilibrium   0.250 − x       x      0.190 + x
                                                 [C 2 H 3 O 2 ][H 3 O ]  (x)(0.190 + x)  −5
                                                             +
                                                       −
                                            K a =              =             = 1.81 × 10
                                                   [HC 2 H 3 O 2 ]  0.250 − x
                                                 (x)(0.190)
                                              ∼          ∼        −5
                                            K a =        = 1.81 × 10
                                                  0.250
                                                               +
                                             x = 2.38 × 10 −5  = [H 3 O ]
               17.25. A solution contains 0.20 mol of HA (a weak acid) and 0.15 mol of NaA (a salt of the acid). Another
                     solution contains 0.20 mol of HA to which 0.15 mol of NaA is added. The final volume of each solution
                     is 1.0 L. What difference, if any, is there between the two solutions?

                     Ans.  There is no difference between the two solutions. The wording of the problems is different, but the final
                           result is the same.

               17.26. What are the concentrations in 1.00 L of solution after 0.100 mol of NaOH is added to 0.200 mol of
                     HC 2 H 3 O 2 but before any equilibrium reaction is considered? Of what use would these be to calculate the
                     pH of this solution?
                     Ans.  The balanced equation is
                                                  NaOH + HC 2 H 3 O 2 −→ NaC 2 H 3 O 2 + H 2 O
                           The limiting quantity is NaOH, and so 0.100 mol NaOH reacts with 0.100 mol HC 2 H 3 O 2 to produce
                           0.100 mol NaC 2 H 3 O 2 + 0.100 mol H 2 O. There is 0.100 mol excess HC 2 H 3 O 2 left in the solution. So far,
                           this problem is exactly the same as Problem 10.70. Now we can start the equilibrium calculations with these
                           initial quantities. We have a buffer solution problem.

                                                                                                −5
               17.27. Calculate the pH of a solution containing 0.250 M NH 3 plus 0.200 M NH 4 Cl. K b = 1.8 × 10 .
                     Ans.                              NH 3    +  H 2 O ←− NH 4  +  +  OH −
                                                                      −→
                                         Initial       0.250              0.200      0
                                         Change       −x                  x          x
                                         Equilibrium   0.250 − x          0.200 + x  x

                                                      [NH 4 ][OH ]  (x)(0.200)
                                                          +
                                                               −
                                                 K b =           =          = 1.8 × 10 −5
                                                         [NH 3 ]     0.250
                                                                    −
                                                  x = 2.25 × 10 −5  = [OH ]
                                                       K w    1.0 × 10 −14
                                                 +                             −10
                                              [H 3 O ] =    =          = 4.4 × 10
                                                      [OH ]   2.25 × 10 −5
                                                         −
                                                 pH = 9.35
                                            Supplementary Problems

               17.28. Identify each of the following terms: (a) hydronium ion, (b)Brønsted-Lowry theory, (c) proton (Brønsted sense),
                     (d) acid (Brønsted sense), (e) base (Brønsted sense), (f ) conjugate, (g) strong acid or base, (h) acid dissociation
                     constant, (i) base dissociation constant, (j) autoionization, (k) pH, and (l) K w .
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