AP®ChemistryCollege BoardStudy GuidesUnit 8: Acids & BasesAcids & BasesIonization of Weak Acids & Weak Bases

Ionization of Weak Acids & Weak Bases (College Board AP® Chemistry)

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Alexandra Brennan

Written by: Alexandra Brennan

Reviewed by: Stewart Hird

Ionization of a Weak Acid

  • Weak acids will only partially ionize in aqueous solutions 

    • Examples include ethanoic acid and carbonic acid 

  • An equilibrium is established between the un-ionized acid and its conjugate base (HA represents the acid molecule):

HA (aq) + H2O (l) ⇌  A- (aq) + H3O+ (aq)

OR

HA (aq) ⇌  A- (aq) + H+ (aq)

  • The position of the equilibrium lies to the left whereby most of the acid molecules remain unionized 

 Diagram to show the ionization of a weak acid

Diagram showing how weak acids partially dissociate

The diagram shows the partial dissociation of a weak acid in aqueous solution

  • The equations above can be used to write the following equilibrium expression:

Ka = [A-][H+][HA]

  • Ka is called the acid dissocation constant 

  • In the context of ethanoic acid, CH3COOH, the equation for the ionization of the acid would be:

CH3COOH (aq) + H2O (l) ⇌  CH3COO- (aq) + H3O+ (aq)

OR

CH3COOH (aq) ⇌  CH3COO- (aq) + H+ (aq)

  • These could be used to deduce an equilibrium expression of:

straight K subscript straight a space equals space fraction numerator left square bracket CH subscript 3 COO to the power of minus right square bracket left square bracket straight H to the power of plus right square bracket over denominator left square bracket CH subscript 3 COOH right square bracket end fraction

  • When writing the equilibrium expression for weak acids, we assume that the concentration of H+ (aq)  due to the ionisation of water is negligible

  • Ka values are very small so we often use pKa values when we are calculating pH values

  • To convert between Ka and pKa we do the following: 

pKa = –logKa   

Ka= 10–pKa

  • The larger the Ka value the stronger the acid

  • The larger the pKa value the weaker the acid

  • The pH of a weak acid solution can be determined from the initial acid concentration and the pKa

Worked Example

Phenol, a weak acid, partially ionizes in water according to the equation below.

C6H5OH (aq)  +   H2O (l)     ⇌     C6H5O– (aq)   +   H3O+ (aq)

What is the pH of a 0.85 M C6H5OH (aq) solution at 25 °C?

Answer: 

  • Start by drawing an ICE table and inputting the initial concentrations

 

C6H5OH 

+ 

H2O (l)

⇌

C6H5O- (aq)  

+   

H3O+

Initial Conc.

0.85

 

-

 

0.00

 

0.00

Change

 

 

-

 

 

 

 

Equilibrium 

 

 

-

 

 

 

 

  • In the change row we use x's to represent the relationship between the reactants and products

  • Any x's under reactants will be negative and under the products will be positive

 

C6H5OH 

+ 

H2O (l)

⇌

C6H5O- (aq)  

+   

H3O+

Initial Conc.

0.85

 

-

 

0.00

 

0.00

Change

- x

 

-

 

+ x

 

+ x

Equilibrium 

 

 

-

 

 

 

 

  • The sum of the initial concentration and change rows are used to fill in the equilibrium row 

 

C6H5OH 

+ 

H2O (l)

⇌

C6H5O– (aq)  

+   

H3O+

Initial Conc.

0.85

 

-

 

0.00

 

0.00

Change

- x

 

-

 

+ x

 

+ x

Equilibrium 

0.85 - x

 

-

 

+ x

 

+ x

  • The equilibrium expression for the equation is:

straight K subscript straight a space equals fraction numerator space open square brackets straight C subscript 6 straight H subscript 5 straight O to the power of negative space end exponent close square brackets left square bracket straight H subscript 3 straight O to the power of plus right square bracket over denominator left square bracket straight C subscript 6 straight H subscript 5 OH right square bracket end fraction

  • The terms from the equilibrium row can be inputted into the equilibrium expression with the value for Ka you were given in the question: 

  • 1.12 space cross times space 10 to the power of negative 10 end exponent equals space fraction numerator straight x squared over denominator 0.85 space minus space straight x space end fraction 

  • straight x squared space equals space 9.52 space cross times 10 to the power of negative 11 end exponent

  • straight x space equals space square root of 9.52 space cross times 10 to the power of negative 11 end exponent end root

  • x = 9.56 x 10-6 M

  • The pH can be calculated using -log[H+] 

    • –log[9.56 x 10-6 ] = 5.01

Ionization of a Weak Base

  • Weak bases will also partially ionize in aqueous solutions 

    • Examples include ammonia and methylamine 

  • There will be a lower concentration of OH– ions so the pH of a weak base is lower than that of a strong base 

  • Again, an equilibrium is established between the un-ionized base and its conjugate acid with its position lying to the left so that a large proportion of base molecules remain un-ionizied 

 Diagram to show the ionization of a weak base

Equilibria Dissociation of a Weak Base, downloadable AS & A Level Chemistry revision notes

The diagram shows the partial dissociation of a weak base in aqueous solution

  • The equilibrium established is:

B (aq) + H2O (l) ⇌  BH+ (aq) + OH- (aq)

  • The base dissociation constant, Kb is deduced from this as: 

Error converting from MathML to accessible text.

  • In the context of 1-phenylmethanamine, C6H5CH2NH2 (aq) this dissociates as follows: 

C6H5CH2NH2 (aq) + H2O (l) ⇌ C6H5CH2NH3+ (aq) + OH– (aq)

  • This leads to the Kb  expression: 

Error converting from MathML to accessible text.

  • Kb values are very small so we often use pKb values when we are calculating pH values

  • To convert between Kb and pKb we do the following: 

pKb = –logKb   

Kb= 10–pKb

  • The pH of a weak base solution can be determined from the initial base concentration and the pKb

Examiner Tips and Tricks

To calculate the pH for a weak base, ICE tables can be used as they were to calculate the pH for a weak acid.

Percentage Ionization Calculations

  • The percentage ionization of a weak acid / base represents the extent to which the acid / base dissociates into ions in a solution 

  • To calculate the percentage ionization of a weak acid use:

percent sign space ionization space equals space fraction numerator stretchy left square bracket straight H subscript 3 straight O to the power of plus stretchy right square bracket over denominator HA end fraction space straight x space 100

  • To calculate the percentage ionization of a weak base (A-) use:

percent sign space ionization space equals space fraction numerator stretchy left square bracket OH to the power of minus stretchy right square bracket over denominator straight A to the power of minus end fraction space straight x space 100

Worked Example

Calculate the percentage ionization for a 0.1 M solution of acetic acid with a pH of 2.5. 

Answer:

Convert pH to [H3O+]

  • 2.5 = -log[H3O+]

  • [H+] = 10−pH

  • [H+] = 10-2.5

  • [H+] = 0.00316 M

Calculate the percentage ionization 

  • percent sign space ionization space equals space fraction numerator stretchy left square bracket straight H subscript 3 straight O to the power of plus stretchy right square bracket over denominator HA end fraction space straight x space 100

  • percent sign space ionization space equals space fraction numerator 0.00316 over denominator 0.10 end fraction space straight x space 100

  • % ionization = 3.16 %

This indicates that 3.16% of acetic acid molecules ionize into H3O+ ions in the solution

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    Alexandra Brennan

    Author: Alexandra Brennan

    Expertise: Chemistry

    Alex studied Biochemistry at Newcastle University before embarking upon a career in teaching. With nearly 10 years of teaching experience, Alex has had several roles including Chemistry/Science Teacher, Head of Science and Examiner for AQA and Edexcel. Alex’s passion for creating engaging content that enables students to succeed in exams drove her to pursue a career outside of the classroom at SME.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.