You will learn how to write the net reaction, calculate the equilibrium constant, and determine the pH of reactions between strong acids and strong bases, strong acids and weak bases, weak acids and strong bases, and weak acids and weak bases.
First we remember that we can predict the direction of an acid-base reaction by using a table such as Table 17.3 and that reactions proceed in the direction of the weaker acid-base pair. After writing the net reaction for the reaction between the acids and bases, we can then determine the equilibrium constant. The net reaction is obtained by separately writing the acid-water reaction, the base-water reaction, and the reaction between the hydronium and hydroxide ions. These three reactions are written with their equilibrium constants and the three reactions are then summed to give the net reaction and the net equilibrium constant.
A. Strong acid-strong base
This reaction you may remember from high school for it is the most common reaction studied. It is the easiest to discuss because both the acid and base are totally ionized in solution and we don't have an equilibrium reaction. The net reaction between equal molar quantities of any strong acid and any strong base is always
H3O+ + OH- 2 H2O and the equilibrium constant is just 1/Kw = 1 x 1014
Because this solution is neutral, the reaction is called a neutralization reaction.
B. Strong acid-weak base
The strong acid can be assumed to be 100% ionized in water so the only equilibrium reaction will be that for the weak base. The only equilibrium reactions will involve the base-water reaction and the hydronium ion-hydroxide reaction, so only those two will be coupled. The acid reaction will give the hydronium ion and the anion from the acid, such as HCl:
HCl + H2O --> H3O+ + Cl-
Using ammonia as an example of a weak base, the equilibrium reactions will be
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Kb = 1.8 x 10-5
H3O+(aq) + OH-(aq) 2 H2O(l) K = 1/Kw = 1 x 1014
Summing these two reactions gives
H3O+(aq) + NH3(aq) NH4+(aq) + H2O(l) Knet = Kb/Kw = 1 x 109
This is a large equilibrium constant so the reaction proceeds nearly completely toward the products. If equal molar amounts of each acid and base are mixed, the solution will contain the salt NH4Cl, which in water solution give the ions NH4+ and Cl-. The Cl- ion does not contribute to the pH of the solution because it is the conjugate base of the strong acid HCl. However, NH4+ is the conjugate acid of the weak base ammonia so it will contribute to the pH of the solution. The solution will then have a pH less than 7 (acidic). Knowing the water reaction of NH4+ and its equilibrium constant we could calculate the pH of the solution if we knew the concentrations of the acid and base initially.
NH4+(aq) + H2O(l) --> H3O + NH3 Ka = 5.6 x 10-10
We can set up a table as we did in Chapter 17 to calculate the concentration of hydronium ion and then the pH if we know the initial concentrations of acid and base. Suppose that we used 0.100 L of a 0.100-M solution of HCl and 0.050 L of a 0.200-M solution of ammonia, what is the resultant pH of the solution? The acid and base amounts are both 0.010 mol, so 0.010 mol of ammonium chloride will be produced in the reaction. The product ions will then each have the concentration of [Cl-] = [NH4+] = 0.010 mol/0.150L = 0.067 M. We can then set up a table to calculate the concentration of hydronium ions in the resultant solution from the equation
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq).
|Before ammonia hydrolysis||0.067||0||0|
|At equilibrium||0.067 - x||x||x|
C. Weak acid-strong base
The base will now be assumed to be 100% ionized so we only have to be concerned with the equilibrium reaction of the weak acid. Taking the weak acid to be acetic acid we have
CH3COOH(aq) + H2O(l) H3O+(aq) + CH2COO-(aq) Ka = 1.8 x 10-5
H3O+(aq) + OH-(aq) 2 H2O(l) K = 1/Kw = 1 x 1014
Summing the above gives:
CH3COOH(aq) + OH-(aq) H2O(°) + CH3COO-(aq) K = Ka/Kw = 1.8 x 109
This equilibrium constant is also very large so the reaction proceeds mostly to products. If we start with equal molar concentrations of acid and base, the product will be sodium acetate, which is totally dissociated in water solution to the acetate ion and the sodium ion. The cation is the cation of a strong base and so yields a neutral solution. The pH of the solution is then determined by the concentration of acetate ion and the equilibrium:
H2O(l) + CH3COO-(aq) --> CH3COOH(aq) + OH-(aq) Kb = 5.6 x 10-10 (Again note that 1/Kb = K)
We can now set up a table as before and calculate the concentration of hydroxide ion at equilibrium and hence the pH of the solution. Here you would find that the solution is basic (pH > 7).
D. Weak acid-weak base
For an equal molar reaction between a strong acid and a strong base, the solution will be neutral. For an equal molar reaction between a strong acid and a weak base, the solution will be acidic. For an equal molar reaction between a weak acid and a strong base, the solution will be basic. But now neither reactant is strong so the situation is a little more complicated and we cannot just a priori predict whether the solution will be acidic or basic.
The equilibrium constant will be K = KaKb/Kw as both the weak acid and the weak base equilibrium must now be considered. We can then determine for particular weak acids and weak bases if the resultant equilibrium constant is large enough so that the reaction proceeds mostly to the right. For example, for acetic acid and ammonia, the resultant equilibrium constant is 3.2 x 104, which is still rather large so the reaction proceeds mostly to the products. The product would then be mainly ammonium acetate that dissociates into the acetate and ammonium ions. The pH of the resultant solution will depend upon the relative values of the equilibrium constants of the conjugate acid and the conjugate base. For this example the K values of the acetate ion and the ammonium ion are equal so the solution should be neutral if equal molar values of each were involved.
Web Author: Dr. Leon L. Combs
Copyright ©2000 by Dr. Leon L. Combs - ALL RIGHTS RESERVED