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CHAP. 17]                            ACID-BASE THEORY                                 249


                     The value of the equilibrium constant is given by
                                                                           −3
                                                                 −3
                                         [C 2 H 3 O 2 ][H 3 O ]  (1.64 × 10 )(1.64 × 10 )
                                                     +
                                               −
                                    K a =              =                      = 1.82 × 10 −5
                                            [HC 2 H 3 O 2 ]       0.148
               EXAMPLE 17.7. Calculate the hydronium ion concentration of a 0.250 M solution of acetic acid, using the equilibrium
               constant of Example 17.6.
                Ans.  In this example, we will use x for the unknown concentration of hydronium ions. We will solve in terms of x.
                                                      HC 2 H 3 O 2 + H 2 O ←− C 2 H 3 O 2 + H 3 O +
                                                                    −→
                                                                              −
                                        Initial       0.250             0        0
                                        Change       −x                 x        x
                                        Equilibrium   0.250 − x         x        x

                                               [C 2 H 3 O 2 ][H 3 O ]  (x)(x)    −5
                                                           +
                                                      −
                                           K a =              =         = 1.82 × 10
                                                  [HC 2 H 3 O 2 ]  0.250 − x
                     Since the acid is weak, we can solve this equation most easily by assuming that x is small enough to neglect when
                     added to or subtracted from a greater concentration. Therefore,
                                                         (x)(x)        −5
                                                    K a =     = 1.82 × 10
                                                         0.250
                                                     2
                                                    x = 4.55 × 10 −6
                                                     x = 2.13 × 10 −3  = [H 3 O ]
                                                                        +
                     In cases where x is too large to neglect from the concentration from which it is subtracted, a more exact method is
                     required. The quadratic formula is used (Appendix).
                                              (x)(x)
                                        K a =        = 1.82 × 10 −5
                                             0.250 − x
                                                    −5
                                                                                   −5
                                         2
                                        x = 1.82 × 10 (0.250 − x) = 4.55 × 10 −6  − 1.82 × 10 x
                                                    −5
                                         2
                                        x + 1.82 × 10 x − 4.55 × 10 −6  = 0
                                                                   −5 2
                                                                                  −6
                                             −1.82 × 10 −5  +    (1.82 × 10 ) − 4(−4.55 × 10 )
                                         x =
                                                                2
                                         x = 2.12 × 10 −3
                     In this case, the approximate solution gives almost the same answer as the exact solution. In general, you should
                     use the approximate method and check your answer to see that it is reasonable; only if it is not should you use the
                     quadratic equation.

               17.4. AUTOIONIZATION OF WATER
                   Since water is defined as both an acid and a base (Sec. 17.2), it is not surprising to find that water can react
               with itself, even though only to a very limited extent, in a reaction called autoionization:
                                                         −→
                                                                 +
                                              H 2 O + H 2 O  ←−  H 3 O + OH −
                                               acid  base     acid    base
               An equilibrium constant for this reaction, called K w , does not have terms for the concentration of water; otherwise
               it is like the other equilibrium constants considered so far.
                                                   K w = [H 3 O ][OH ]
                                                             +
                                                                  −
                   The value for this constant in dilute aqueous solution at 25 Cis1.0 × 10 −14 . Thus, water ionizes very little
                                                                 ◦
               when it is pure, and even less in acidic or basic solution. You must remember this value.
                   The equation for K w indicates that there is always some H 3 O and always some OH in any aqueous
                                                                                          −
                                                                      +
               solution. Their concentrations are inversely proportional. A solution is acidic if the H 3 O concentration exceeds
                                                                                     +
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