 # Hydrolysis of Acids and Bases (Ka × Kb = Kw) 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)), Ka, and the hydrolysis (dissociation) of its conjugate base (A-(aq)), Kb, is given by the following expression:

Ka × Kb = Kw

• At 25°C, Kw = 10-14 therefore:

Ka × Kb = 10-14

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

 Kb = KwKa = 10-14Ka

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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:

acid + water oxidanium ion
(oxonium ion)
+ conjugate
base
HA(aq) + H2O(l) H3O+(aq) + A-(aq)

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:

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

Ka 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 Ka

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) ...

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## 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:

base + water hydroxide
ion
+ conjugate
acid
A-(aq) + H2O(l) OH-(aq) + HA(aq)

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:

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

Kb 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 Kb

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## Relationship Between Ka and Kb

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:

water oxidanium ions
(oxonium ions)
+ hydroxide ions
2H2O(l) H3O+(aq) + OH-(aq)

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:

Kw = [H3O+(aq)][OH-(aq)]

Kw is the equilibrium constant for the self-dissociation of water molecules (Kw = 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 Kw.

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 Ka to find [H3O+(aq)] as shown below:

 Ka = [H3O+(aq)][A-(aq)][HA(aq)] multiply both sides of the equation by [HA(aq)] [HA(aq)] × Ka = [HA(aq)] × [H3O+(aq)][A-(aq)][HA(aq)] [HA(aq)] × Ka = [H3O+(aq)][A-(aq)] divide both sides of the equation by [A-(aq)] [HA(aq)] × Ka [A-(aq)] = [H3O+(aq)][A-(aq)][A-(aq)] [HA(aq)] × Ka [A-(aq)] = [H3O+(aq)]

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 Kb as shown below:

 Kb = [OH-(aq)][HA(aq)][A-(aq)] Multiply both sides of the equation by [A-(aq)] [A-(aq)] × Kb = [A-(aq)] × [OH-(aq)][HA(aq)][A-(aq)] [A-(aq)] × Kb = [OH-(aq)][HA(aq)] Divide both sides of the equation by [HA(aq)] [A-(aq)] × Kb [HA(aq)] = [OH-(aq)][HA(aq)][HA(aq)] [A-(aq)] × Kb [HA(aq)] = [OH-(aq)]

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:

 Kw = [H3O+(aq)] × [OH-(aq)] = [HA(aq)] × Ka [A-(aq)] × [A-(aq)] × Kb [HA(aq)] Kw = Ka × Kb

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

Kw = Ka × Kb

Kw = equilibrium constant for the self-dissociation of water (at 25°C Kw = 10-14)
Ka = equilibrium constant for the dissociation of the acid HA(aq) (tabulated values usually at 25°C)
Kb = 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 Kw, Ka and Kb to calculate the value of Kb (the value that is not usually tabulated).

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

 Kw Ka = Ka × Kb Ka Kw Ka = Kb

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## Worked Example of Calculating Kb

Calculate the value of Kb for the acetate ion (ethanoate ion), CH3COO-(aq), at 25°C.
Ka 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:

 CH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq)

Step 2: Confirm that CH3COOH(aq) is the conjugate acid of the base CH3COO-(aq)

base + water conjugate acid + hydroxide ions
CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

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

Kw = 10-14 (at 25°C)

Ka = 1.8 × 10-5 (given in question)

Kb = ?

Step 4: Calculate Kb using Kw = Ka × Kb

 Kw = Ka × Kb Substitute the values into the expression: 10-14 = 1.8 × 10-5 × Kb Divide both sides of the expression by 1.8 × 10-5 10-14 1.8 × 10-5 = 1.8 × 10-5 × Kb1.8 × 10-5 Solve for Kb 5.6 × 10-10 = Kb

Step 5: Check that your answer makes sense, that is, Ka × Kb = Kw

Kw = (1.8 × 10-5) × (5.6 × 10-10) = 10-14 so value of Kb looks right

Kb = 5.6 × 10-10

## Problem Solving : Ka × Kb = Kw

The Problem: Chris the Chemist needs the value for the dissociation constant of CN-(aq) at 25°C to use in a calculation.

Chris has found the following tabulated values for the dissociation constants of some acids and bases at 25°C

Equilibrium Constants at 25°C
weak acids weak bases
name Ka name Kb
hydrocyanic acid
(HCN)
4.0 × 10-10 phosphine
(PH3)
1 × 10-14
hypochlorous acid
(HOCl)
3.2 × 10-8 hydroxylamine
(NH2OH)
9.1 × 10-9
acetic acid
(CH3COOH)
1.8 × 10-5 ammonia
(NH3)
1.8 × 10-5
nitrous acid
(HNO2)
5.0 × 10-4 methanamine
(CH3NH2)
4.4 × 10-4

Use the information given to determine the value of the equilibrium constant for the dissociation of CN-(aq) in water at 25°C.

The Solution to the Problem

(using the StoPGoPS approach to problem solving)

STOP STOP! State the Question.
What is the question asking you to do?

Calculate the value of the equilibrium constant
K = ?

PAUSE PAUSE to Prepare a Game Plan
(1) What information (data) have you been given in the question?

Molecular formula of the species: CN-

Table of Ka and Kb values (a value for CN- is NOT included, but the value of Ka for HCN is included!)

For HCN, Ka = 4.0 × 10-10

(2) What is the relationship between what you know and what you need to find out?

(i) Write the chemical equation for the dissociation of CN-(aq)

(ii) Use the relationship Ka × Kb = Kw to calculate the value of the dissociation constant for CN-(aq)

At 25°C, Kw = 10-14, so Ka × Kb = 10-14

GO GO with the Game Plan
(i) Write the chemical equation for the dissociation of CN-(aq)

CN- + H2O ⇋ OH- + HCN       K = Kb = [OH-][HCN]/[CN-]

(ii) Kw = Ka × Kb = 10-14 (at 25°C)

Kb = equilibrium constant for the dissociation of the base, CN- = ?

Ka = equilibrium constant for the dissociation of the acid HCN (which is the conjugate acid of the base) = 4.0 × 10-10 (from the table)

 Ka × Kb = 10-14 (4.0 × 10-10) × Kb = 10-14 (4.0 × 10-10) × Kb (4.0 × 10-10) = 10-14     4.0 × 10-10 Kb = 2.5 × 10-5

PAUSE PAUSE to Ponder Plausibility

Yes, we have calculated the value of Kb

For HCN, Ka = 4 × 10-10, that is, Ka is very small and HCN is a very weak acid so the equilibrium position for its dissociation lies well to the left:

 HCN(aq) ↼ ⇁ H+(aq) + CN-(aq)

Therefore there is little CN- in solution. The ability of CN- to attract a proton to itself (whether from H+ or from H2O) is greater than the ability of the HCN to dissociate.
So the value for Kb (CN- attracting a proton) must be greater than the value for Ka (HCN donating a proton).

Since our calculated value for Kb (2.5 × 10-5) is greater than the value for Ka (4.0 × 10-10) we are confident our answer is plausible.

STOP STOP! State the Solution

Kb = 2.5 × 10-5

Footnotes

(1) It doesn't really matter if you write this dissociation of acid as HA(aq) ⇋ H+(aq) + A-(aq) with Ka = [H+(aq)][A-(aq)]/[HA(aq)] because H+(aq) is understood to mean H3O+

(2) In this case we must include the water molecule in the equation!

(3) H3O+ has the systematic IUPAC name of oxidanium, and the acceptable non-sytematic IUPAC name of oxonium.
In older textbooks you may also see this referred to as the hydronium ion, but this is no longer an acceptable IUPAC name.
Refer to the "Nomenclature of Inorganic Chemistry: IUPAC Recommendations 2005" (Red Book) for information about naming inorganic compounds.

(4) Strong bases are generally hydroxides of Group 1 or 2 elements, that is, the dissociation of the strong base produces hydroxide ions, OH-, and the conjugate acid of this base is water, H2O, so the hydrolysis of a strong base increases the concentration of OH-(aq) and decreases the concentration of H3O+ according the relationship Kw = [OH-][H3O+]

(5) An alternative derivation of this relationship starts with Ka for the hydrolysis (dissociation) of acid HA(aq):

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

and multplies the numerator and denominator by [OH-]:
 Ka = [OH-(aq)][H+(aq)][A-(aq)][HA(aq)][OH-(aq)]

Substitute Kw for [OH-(aq)][H+(aq)]:
 Ka = Kw[A-(aq)][HA(aq)][OH-(aq)]

Rearrange:
 Ka × [HA(aq)][OH-(aq)][A-(aq)] = Kw

Substitute Kb for [HA(aq)][OH-(aq)]/[A-(aq)] (that is the equilibrium constant expression for the hydrolysis of base A-: A- + H2O ⇋ HA(aq) + OH-(aq))
Ka × Kb = Kw