Ka, Kw, Kb - The Acid, Water, and Base Dissociation Constant for AP Chemistry (page 2)
Practice problems for these concepts can be found at:
- Equilibrium Multiple Choice Review Questions for AP Chemistry
- Equilibrium Free-Response Questions for AP Chemistry
Ka—The Acid Dissociation Constant
Strong acids completely dissociate (ionize) in water. Weak acids partially dissociate and establish an equilibrium system. But as shown in Figure 15.1 there is a large range of weak acids based upon their ability to donate protons. Consider the general weak acid, HA, and its reaction when placed in water:
An equilibrium constant expression can be written for this system:
The [H2O] is assumed to be a constant and is incorporated into the Ka value. It is not shown in the equilibrium constant expression.
Since this is the equilibrium constant associated with a weak acid dissociation, this particular Kc is most commonly called the acid dissociation constant, Ka. The Ka expression is then:
Many times the weak acid dissociation reaction will be shown in a shortened notation, omitting the water:
The greater the amount of dissociation is, the larger the value of Ka. Table 15.1 shows the Ka values of some common weak acids.
Here are a couple of tips: For every H+ formed, an A– is formed, so the numerator of the Ka expression can be expressed as [H+]2 (or [A–]2, although it is rarely done this way). Also, the [HA] is the equilibrium molar concentration of the undissociated weak acid, not its initial concentration. The exact expression would then be [HA] = Minitial – [H+], where Minitial is the initial concentration of the weak acid. This is true because for every H+ that is formed, an HA must have dissociated. However, many times if Ka is small, you can approximate the equilibrium concentration of the weak acid by its initial concentration, [HA] = Minitial.
If the initial molarity and Ka of the weak acid are known, the [H+] (or [A–]) can be calculated easily. And if the initial molarity and [H+] are known, Ka can be calculated.
For example, calculate the [H+] of a 0.300 M acetic acid solution.
For polyprotic acids, acids that can donate more than one proton, the Ka for the first dissociation is much larger than the Ka for the second dissociation. If there is a third Ka, it is much smaller still. For most practical purposes you can simply use the first Ka.
Kw—The Water Dissociation Constant
Before examining the equilibrium behavior of aqueous solutions of weak bases, let's look at the behavior of water itself. In the initial discussion of acid–base equilibrium above, we showed water acting both as an acid (proton donor when put with a base) and a base (proton acceptor when put with an acid). Water is amphoteric, it will act as either an acid or a base, depending on whether the other species is a base or acid. But in pure water the same amphoteric nature is noted. In pure water a very small amount of proton transfer is taking place:
This is commonly written as:
There is an equilibrium constant, called the water dissociation constant, Kw, which has the form:
Again, the concentration of water is a constant and is incorporated into Kw.
The numerical value of Kw of 1.0 × 10–14 is true for the product of the [H+] and [OH–] in pure water and for aqueous solutions of acids and bases.
In the discussion of weak acids, we indicated that the [H+] = [A–]. However, there are two sources of H+ in the system: the weak acid and water. The amount of H+ that is due to the water dissociation is very small and can be easily ignored.
Because the concentration of the hydronium ion, H3O+, can vary tremendously in solutions of acids and bases, a scale to easily represent the acidity of a solution was developed. It is called the pH scale and is related to the [H3O+]:
- pH = –log [H3O+] or –log [H+] using the shorthand notation
Remember that in pure water Kw = [H3O+][OH–] = 1.0 × 10–14. Since both the hydronium ion and hydroxide ions are formed in equal amounts, the Kw expression can be expressed as:
Solving for [H3O+] gives us [H3O+] = 1.0 × 10–7. If you then calculate the pH of pure water:
The pH of pure water is 7.00. On the pH scale this is called neutral. A solution whose [H3O+] is greater than in pure water will have a pH less than 7.00 and is called acidic. A solution whose [H3O+] is less than in pure water will have a pH greater than 7.00 and is called basic. Figure 15.2 shows the pH scale and the pH values of some common substances.
The pOH of a solution can also be calculated. It is defined as pOH = –log[OH–]. The pH and the pOH are related:
- pH+pOH= pKw =14.00 at 25°C
In any of the problems above in which [H+] or [OH–] was calculated, you can now calculate the pH or pOH of the solution.
You can estimate the pH of a solution by looking at its [H+]. For example, if a solution has an [H+] = 1 × 10–5, its pH would be 5. This value was determined from the value of the exponent in the [H+].
Kb—The Base Dissociation Constant
Weak bases (B), when placed into water, also establish an equilibrium system much like weak acids:
The equilibrium constant expression is called the weak base dissociation constant, Kb, and has the form:
The same reasoning that was used in dealing with weak acids is also true here: [HB+] = [OH–]; [HB] ≈ Minitially; the numerator can be represented as [OH–]2; and knowing the initial molarity and Kb of the weak base, the [OH–] can easily be calculated. And if the initial molarity and [OH–] are known, Kb can be calculated.
For example, a 0.500 M solution of ammonia has a pH of 11.48. What is the Kb of ammonia?
The Ka and Kb of conjugate acid–base pairs are related through the Kw expression:
- Ka × Kb = Kw
This equation shows an inverse relationship between Ka and Kb for any conjugate acid–base pair.
This relationship may be used in problems such as: Determine the pH of a solution made by adding 0.400 mol of strontium acetate to sufficient water to produce 2.000 L of solution.
The initial molarity is 0.400 mol/2.000 L = 0.200 M.
When a salt is added to water dissolution will occur:
The resultant solution, since strontium acetate is soluble, has 0.200 M Sr2+ and 0.400 M C2H3O2–
Ions such as SrSr2+, which come from strong acids or strong bases, may be ignored in this type of problem. Ions such as C2H3O2–, from weak acids or bases, will undergo hydrolysis. The acetate ion is the conjugate BASE of acetic acid (Ka = 1.8 × 10–5). Since acetate is not a strong base this will be a Kb problem, and OH– will be produced. The equilibrium is:
Determining Kb from Ka (using Kw = KaKb = 1.0 × 10–14 gives:
with x = 1.5 × 10–5 = [OH–], and pH = 9.180
Practice problems for these concepts can be found at: