Unformatted text preview: The #1 Rule
The relative acidity of a pair of acids is dominated by the stability of the ionic species in the acid/conjugate base couples. This works when both ionic species have the same sign of charge, and when the proton is lost from an atom in the second row, For example: Circle the stronger acid of the following pair: CH4 NH 3 Solution using the #1 rule: a. Write the acid/conjugate base couples for the pair of acids. b. There are two ionic species; determine which is more stable using the basic concepts of Octets, Electronegativity and/or Coulomb’s law. c. These ions can be either acids or bases. In either case, the more stable ion is weaker, and the less stable ion is stronger. d. The stronger base is always the conjugate base of the weaker acid, and the weaker base is the conjugate base of the stronger acid. CH4 Pair of ac ids NH 3 CH3 + "H+" NH2 + "H+" Ionic s pecies in the acid/conjugate base couples
In this case the ions are the bases. NH2y is more stable than CH3y because nitrogen is more electronegative than carbon. Therefore, NH2y is a weaker base than CH3y, and NH3 is a stronger acid than CH4! So, the answer is: CH4 NH 3 There are several ways to ask this question. For example, for the following pair of compounds, circle the stronger base. CH3 NH2 Often, this type of question is asked in this way: For the following equilibrium, is Kequ greater than 1 or less than 1? C H3 + NH3 NH2 + C H4 One way of thinking about the answer to this question is to consider that a stronger acid and stronger base always react to favor a weaker acid and weaker base. Since NH3 is the stronger acid, the products (compounds on the right) in this equation must be favored (they’re more stable). Therefore, Kequ > 1 for this reaction. In fact, Kequ is much greater than 1 [10(60-36) = 1024]. The reaction of the methyl anion with ammonia is a great way to make the amide anion (NH2y). The First Exception to the # 1 Rule
The relative acidity of a pair of hydrogen halides is dominated by the stability of the neutral species in the acid/conjugate base couples. Specifically, the X-H bond strength is the dominant factor determining the relative acidity of a pair of hydrogen halides. Bond strength is always an important factor when comparing acids in the same column of the periodic table. So, circle the stronger acid:
H —F H —I Using the same approach, write the two half-reactions: H—F F + "H+" H—I I + "H+" This time, since we’re dealing with hydrogen halides (or more generally, atoms in a column of the periodic table), we determine the relative stability of the neutral species. HF is more stable than HI, since the bond is weak in HI. So, HI is less stable, which means it is stronger. Since the neutrals are the acids, HI is the stronger acid: H —F H —I Furthermore, the Kequ for the following reaction is greater than 1 (favors products), since the weaker (more stable) acid is favored in the equilibrium. F + H—I I + H—F One last point. Any acid stronger than H3O⊕ is a “strong acid.” The pKa of H3O⊕ is about –2. All of the hydrogen halides except HF are strong acids. This means that the following equilibrium favors products for all the hydrogen halides except HF. H —X + H2O X + H3O You can’t get this by application of the rules, since the ions don’t have the same sign of charge! ...
View Full Document