Hydrolysis of Acids and Bases (K_a × K_b = K_w) Chemistry Tutorial

Key Concepts

• Hydrolysis is a reaction with water.
• Hydrolysis of an acid (HA_(aq)) produces the conjugate base (A^-_(aq)) of this acid.⁽¹⁾

  HA_(aq) + H₂O_(l) ⇌ H₃O⁺_(aq) + A^-_(aq)

  Ka =	[H3O+(aq)][A-(aq)] [HA(aq)]

• Hydrolysis of this conjugate base (A^-_(aq)) produces the conjugate acid (HA_(aq)) of this base.⁽²⁾

  A^-_(aq) + H₂O_(l) ⇌ OH^-_(aq) + HA_(aq)

  Kb =	[OH-(aq)][HA(aq)] [A-(aq)]

• The relationship between the equilibrium constants for the hydrolysis (dissociation) of the acid (HA_(aq)), K_a, and the hydrolysis (dissociation) of its conjugate base (A^-_(aq)), K_b, is given by the following expression:

  K_a × K_b = K_w

• At 25°C, K_w = 10^-14 therefore:

  K_a × K_b = 10^-14

• We can calcuate the value of K_b for aqueous solutions at 25°C given the value of K_a for the base's conjugate acid:

  Kb =

  Kw Ka

  =

  10-14 Ka

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Hydrolysis of Acids

Hydrolysis is a reaction with water.

When an acid undergoes hydrolysis it donates a proton to the water molecule to produce the conjugate base of the acid and the oxidanium ion (or oxonium ion)⁽³⁾ as shown by the general chemical equation below:

If the acid is a strong acid, then the reaction above goes to completion, but if the acid is a weak acid then molecules of undissociated acid are present in solution as well as oxidanium ions and the conjugate base anions.
In a closed system, equilibrium is established in which the rate at which the acid molecules dissociate is the same as the rate at which the oxidanium and conjugate base ions come together to form molecules of acid and water, that is, the concentrations of acid molecules, anions and oxidanium ions does not change with time.
Therefore we can write an expression to describe this equilibrium condition for the dissociation of a weak acid:

K_a is the acid dissociation constant for this acid (tables of values are available, usually at 25°C)
[H₃O⁺_(aq)] = equilibrium concentration of H₃O⁺_(aq) in solution
[A^-_(aq)] = equilibrium concentration of A^-_(aq) (the conjugate base of the acid HA_(aq)) in solution
[HA] = equilibrium concentration of undissociated HA_(aq) molecules in solution
Note that the equilibrium concentration of water, which is a liquid, is essentially constant and therefore said to be incorporated into the value for K_a

Why doesn't the conjugate base, A^-_(aq) undergo hydrolysis?
IF A^-_(aq) is the conjugate base of a strong acid then A^-_(aq) is an extremely weak base which will not react with water to any appreciable degree, so A^-_(aq) will not undergo hydrolysis.
IF A^-_(aq) is the conjugate base of a weak acid then it will react with water!
So, let's take a look at the hydrolysis of the base A^-_(aq) ...

Hydrolysis of Bases

Hydrolysis is a reaction with water.

When a base undergoes hydrolysis it accepts a proton from a water molecule to produce the conjugate acid of the base and a hydroxide ion as shown by the general chemical equation below:

If the base is a strong base, then the reaction above goes to completion⁽⁴⁾, but if the base is a weak base then the mixture will contain some base, some hydroxide ions and some conjugate acid.
In a closed system, equilibrium is established in which the rate at which the base hydrolyses is the same as the rate at which the hydroxide ions and conjugate acid molecules come together to form the base and water, so the concentrations of base, conjugate acid and hydroxide ions does not change over time.
Therefore we can write an expression to describe this equilibrium condition for the hydrolysis, or dissociation, of a weak base:

K_b is the base dissociation constant for this base (tables of values are not usually available for conjugate bases of weak acids)
[A^-_(aq)] = equilibrium concentration of the base A^-_(aq) in solution
[OH^-_(aq)] = equilibrium concentration of OH^-_(aq) in solution
[HA] = equilibrium concentration of HA_(aq) molecules (the conjugate acid of the base A^-_(aq)) in solution
Note that the equilibrium concentration of water, which is a liquid, is essentially constant and therefore said to be incorporated into the value for K_b

Relationship Between K_a and K_b

Hydrolysis of an acid or of a base is a reaction with water.

Water undergoes auto-dissociation, or self-dissociation, producing oxidanium ions (oxonium ions) and hydroxide ions according to the chemical equation below:

At equilibrium, the rate at which water molecules dissociate to produce ions is the same as the rate at which the ions react to form water molecules, so the concentration of hydroxide ions and oxidanium ions does not change over time.
Therefore we can write an expression for the equilibrium constant as shown below:

K_w = [H₃O⁺_(aq)][OH^-_(aq)]

K_w is the equilibrium constant for the self-dissociation of water molecules (K_w = 10^-14 at 25°C)
[H₃O⁺_(aq)] = equilibrium concentration of H₃O⁺ ions in solution
[OH^-_(aq)] = equilibrium concentration of OH^- ions in solution
Note that since water, H₂O_(l), is a liquid, its concentration is effectively constant and is therefore incorporated into the value for K_w.

So what is the concentration of oxidanium ions, H₃O⁺_(aq), in a solution of acid HA_(aq) at equilibrium?

We can rearrange the expression for K_a to find [H₃O⁺_(aq)] as shown below:

Similarly, we can find the concentration of hydroxide ions, OH^-_(aq), in a solution of the base A^-_(aq) (the conjugate base of the acid HA_(aq)) by rearranging the expression for K_b as shown below:

Then we can substitute these expressions for [OH^-_(aq)] and [H₃O⁺_(aq)] into the expression for the equilibrium constant for the self-dissociation of water as shown below:

So, for the hydrolysis of acid HA_(aq) which produces the conjugate base A^-_(aq):

K_w = K_a × K_b

K_w = equilibrium constant for the self-dissociation of water (at 25°C K_w = 10^-14)
K_a = equilibrium constant for the dissociation of the acid HA_(aq) (tabulated values usually at 25°C)
K_b = equilibrium constant for the dissociation of A^-_(aq), the conjugate base of the acid HA_(aq) (usually NOT tabulated)

We can use this relationship⁽⁵⁾ between K_w, K_a and K_b to calculate the value of K_b (the value that is not usually tabulated).

Given tabulated values for K_w and K_a at the same specified temperature (usually 25°C), we can calculate the value for K_b as shown below:

Worked Example of Calculating K_b

Calculate the value of K_b for the acetate ion (ethanoate ion), CH₃COO^-_(aq), at 25°C.
K_a for acetic acid (ethanoic acid), CH₃COOH_(aq), is 1.8 × 10^-5

Step 1: Write the balanced chemical equation for the hydrolysis of the base:

Step 2: Confirm that CH₃COOH_(aq) is the conjugate acid of the base CH₃COO^-_(aq)

Step 3: Extract the data (information from the question)

K_w = 10^-14 (at 25°C)

K_a = 1.8 × 10^-5 (given in question)

K_b = ?

Step 4: Calculate K_b using K_w = K_a × K_b

Step 5: Check that your answer makes sense, that is, K_a × K_b = K_w

K_w = (1.8 × 10^-5) × (5.6 × 10^-10) = 10^-14 so value of K_b looks right

Step 6: State the answer

K_b = 5.6 × 10^-10

Problem Solving : K_a × K_b = K_w

The Solution to the Problem

(using the StoPGoPS approach to problem solving)