Example 1. Given that 0.4 mole sodium acetate (NaAc) plus 0.1 mole acetic acid (HAc) yields a pH of 5.36 when mixed, what is the pK_a of acetic acid?
   HAc º Ac^- + H⁺     pH = pK_a + log[Ac^-]/ [HAc]

                          5.36 = pK_a + log 0.4/0.1
                          5.36 = pK_a + log 4
                          pK_a = 5.36 - 0.602 = 4.76

Example 2. Mixing 0.4 moles of CH₃CH₂NH₃⁺ and 0.1 moles of CH₃CH₂NH₂ gives a pH of 8.3. What is the pK_a of ethylamine?

                       8.3 = pK_a + log 0.1/0.4 base/acid
                       8.3 = pK_a -0.602
                     pK_a = 8.9

If 0.2 mole of OH^- is added, what is the new pH?

                  CH₃CH₂NH₃⁺ º CH₃CH₂NH₂ + H⁺

       CH₃CH₂NH₃⁺ + OH^- º CH₃CH₂NH₂ + H₂O
 
           0.4 mole                        0.1 mole
                    to                                         to
          0.2 mole                         0.3 mole

                pH = pK_a + log base/acid
                pH = 8.9 + log 0.3/0.2
                pH = 8.9 +0.18 or pH = 9.08