Many substances behave as weak bases in water. Such substances react with water, removing protons from H2O, thereby forming the conjugate acid of the base and OH– ions:
The most commonly encountered weak base is ammonia:
The equilibrium-constant expression for this reaction can be written as
Because the concentration of water is essentially constant, the [H2O] term is incorporated into the equilibrium constant, giving
The constant Kb is called the base-dissociation constant, by analogy with the acid-dissociation constant, Ka, for weak acids. The constant Kb always refers to the equilibrium in which a base reacts with H2O to form the conjugate acid and OH–. Table 16.4 lists the names, formulas, Lewis structures, equilibrium reactions, and values of Kb for several weak bases in water. Appendix D of the textbook includes a more extensive list. Notice that these bases contain one or more lone pairs of electrons. A lone pair is necessary to form the bond with H+. Notice also that in the neutral molecules the lone pairs are on nitrogen atoms and that the other bases are anions derived from weak acids.
Calculate the concentration of OH– in a 0.15 M solution of NH3.
SOLUTION We use essentially the same procedure here as used in solving problems involving the ionization of weak acids. The first step is to write the ionization reaction and the corresponding equilibrium-constant (Kb) expression:
We then tabulate the equilibrium concentrations involved in the equilibrium:
(Notice that we ignore the concentration of H2O because it is not involved in the equilibrium-constant expression.) Inserting these quantities into the equilibrium-constant expression gives the following:
Because Kb is small, we can neglect the small amount of NH3 that reacts with water, as compared to the total NH3 concentration; that is, we can neglect x in comparison to 0.15 M. Then we have
Notice that the value obtained for x is only about 1 percent of the NH3 concentration, 0.15 M. Therefore, our neglect of x in comparison with 0.15 is justified.
Which of the following compounds should produce the highest pH as a 0.05 M solution: pyridine, methylamine, or nitrous acid? Answer: methylamine
How can we recognize from a chemical formula whether a molecule or ion is able to behave as a weak base? Weak bases fall into two general categories. The first category contains neutral substances that have an atom with a nonbonding pair of electrons that can serve as a proton acceptor. Most of these bases, including all the uncharged bases listed in Table 16.4, contain a nitrogen atom. These substances include ammonia and a related class of compounds called amines. In organic amines, one or more of the N H bonds in NH3 is replaced with a bond between N and C. Thus, the replacement of one N H bond in NH3 with a N CH3 bond gives methylamine, NH2CH3 (usually written CH3NH2). Like NH3, amines can extract a proton from a water molecule by forming an additional N H bond, as shown here for methylamine:
The chemical formula for the conjugate acid of methylamine is usually written CH3NH3+.
The second general category of weak bases is composed of the anions of weak acids. Consider, for example, an aqueous solution of sodium hypochlorite, NaClO. This salt dissolves in water to give Na+ and ClO– ions. The Na+ ion is always a spectator ion in acid-base reactions. However, the ClO– ion is the conjugate base of a weak acid, hypochlorous acid. Consequently, the ClO– ion acts as a weak base in water:
A solution is made by adding solid sodium hypochlorite, NaClO, to enough water to make 2.00 L of solution. If the solution has a pH of 10.50, how many moles of NaClO were added to the water?
SOLUTION NaClO is an ionic compound consisting of Na+ and ClO– ions. As such, it is a strong electrolyte that completely dissociates in solution into Na+, which is a spectator ion, and ClO– ion, which is a weak base with Kb = 3.3 10–7 (Equation 16.35). We wish to determine the concentration of ClO– in solution that would generate enough OH– ion to raise the pH to 10.50.
We first calculate the concentration of OH–(aq) at equilibrium. We can calculate [OH–] by using either Equation 16.14 or Equation 16.17; we will use the latter method here:
This concentration is high enough that we can assume that Equation 16.35 is the only source of OH–; that is, we can neglect any OH– produced by the autoionization of H2O. We now assume a value of x for the initial concentration of ClO– and solve the equilibrium problem in the usual way:
We now use the expression for the base-dissociation constant to solve for x:
We say that the solution is 0.31 M in NaClO, even though some of the ClO– ions have reacted with water. Because the solution is 0.31 M in NaClO and the total volume of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was added to the water.
A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution? Answer: 0.0058 M