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Chapter 17: Acid-Base Equilibria

July 9, 2021 by Admin

Chapter 17
Acid-Base Equilibria

Introduction

There are three theories of Acids and Bases, which are commonly overlooked by students. The most familiar theory to us would be the Arrhenius Theory. Arrhenius acids form hydrogen ions in aqueous solution and Arrhenius bases form hydroxide ions in aqueous solution. An acid such as HCl, which we learnt since secondary school fall under this theory. We know that as HCl dissolves in aqueous solution, it ionises to form hydrogen ions, H⁺.

1. HCl → H⁺ + Cl⁻

But more accurately, since the hydrogen ion has a large charge density, it tends to form a coordinate bond with a neighbouring water molecule instead of existing independently. A lone pair of electrons from a water molecule is donated to the empty orbital of the H⁺ ion.

Hence,

2. H₂O + H⁺ → H₃O⁺

Thus overall,

HCl + H₂O → H₃O⁺ + Cl⁻

Unfortunately, the Arrhenius Theory has its limitations. It cannot explain why certain compounds such as ammonia behave like bases as it does not contain hydroxide ions.

Hence the next model, the Brønsted-Lowry Theory comes in.
A brønsted acid is a proton donor, while a brønsted base is a proton acceptor. Hence ammonia would be considered a brønsted base, as it is able to accept a proton (H⁺).

Yet again, the Arrhenius Theory and Brønsted-Lowry Theory are still unable to explain the behaviour of substances such as BF₃ and AlCl₃, which have no hydrogen atom but behave as acids.

The Lewis Theory, the third theory, provides an explanation for these substances' behaviour.
According to the Lewis Theory, a Lewis acid is an electron-pair acceptor while a Lewis base is an electron-pair donor.
Hence in the case of BF₃, it accepts an electron pair from NH₃, hence it is a lewis acid and NH₃ is a Lewis base.

AD H2 Chem Articles - Acid-Base Equilibria (26)

Don’t worry, the Lewis theory does not contradict the Arrhenius Theory and Brønsted-Lowry.
For example,

Conjugate Acid-Base Pairs

Firstly, let’s refer back to the definitions of Brønsted-Lowry acid and base shown above. When a Brønsted-Lowry acid (HX) loses a proton(H⁺), it becomes the conjugate base of HX. Similarly, when a Brønsted-Lowry base (Y) accepts a proton(H⁺), it becomes the conjugate acid of Y.
Let’s look at a simple example below:

CH₃CH₂OH + HCl ⇌ CH₃CH₂OH₂⁺ + Cl⁻

The first step is to identify the conjugate pairs from the equation through the movement of H⁺.
We can clearly see that CH₃CH₂OH accepted a proton to form CH₃CH₂OH₂⁺, so CH₃CH₂OH₂⁺ is the conjugate acid of CH₃CH₂OH. As for HCl, it lost a proton to form Cl⁻, so Cl⁻ is the conjugate base of HCl.
This is how we easily identify conjugate pairs.

Test Yourself!

(i) Identify the two acids and the two bases present according to Brønsted-Lowry theory.

(a) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
(b) C₆H₅O⁻ + CH₃CO₂H ⇌ C₆H₅OH + CH₃CO₂⁻

(ii) Identify the acid and base present according to Lewis theory.

(a) BF₃ + F⁻ ⇌ BF₄⁻
(b) AlCl₃ + CH₃Cl ⇌ AlCl₄⁻ + CH₃⁺

Answer:
(i) (a) Acids: H₂O, NH₄⁺
Bases: NH₃, OH⁻
Since H₂O donates a proton to NH₃ it is a Brønsted acid, while NH₃ is a brønsted base. OH⁻ is the conjugate base of H₂O, and NH₄⁺ is the conjugate acid of NH₃.

(b) Acids: CH₃CO₂H, C₆H₅OH
Bases: C₆H₅O⁻, CH₃CO₂⁻
Since CH₃CO₂H donates a proton to C₆H₅O⁻, it is a Brønsted acid, while C₆H₅O⁻ is a brønsted base. CH₃CO₂⁻ is the conjugate base of CH₃CO₂H, and C₆H₅OH is the conjugate acid of C₆H₅O⁻.

(ii) (a) Acid: BF₃

Base: F⁻
An electron pair from F⁻ is donated to form a dative bond with the B atom of BF₃

(b) Acid: AlCl₃
Base: CH₃Cl
An electron pair from Cl is donated to form a dative bond with the Al atom of AlCl₃.

Buffer

One of the fundamental concepts of acid-base is the buffer reactions.
A buffer is a solution which pH remains almost unchanged when a small amount of H⁺ or OH⁻ is added to it. This definition is extremely important and should be memorised. There are 2 types of buffers. The table below will provide more explanation of each type of buffer.

Acidic buffer	Basic buffer
A solution of a weak acid and its conjugate base Example: CH₃COOH (weak acid) and CH₃COONa (contains the conjugate base CH₃COO⁻) (I) On adding a small amount of H⁺, H⁺ + CH₃COO⁻ → CH₃COOH The small amount of added H⁺ is removed by the large amount of CH₃COO⁻ in the buffer. Therefore pH remains almost unchanged as [H⁺] does not change significantly (II) On adding a small amount of OH⁻, OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O The small amount of added OH⁻ is removed by the large amount of CH₃COOH in the buffer. Therefore pH remains almost unchanged as [OH⁻] does not change significantly.	A solution of a weak base and its conjugate acid Example: NH₃ (weak base) and NH₄Cl (contains the conjugate acid NH₄⁺) (I) On adding a small amount of OH⁻ OH⁻ + NH₄⁺ → NH₃ + H₂O The small amount of added OH⁻ is removed by the large amount of NH₄⁺ in the buffer. Therefore pH remains almost unchanged as [OH⁻] does not change significantly (II) On adding a small amount of H⁺ H⁺ + NH₃ → NH₄⁺ The small amount of added H⁺ is removed by the large amount of NH₃ in the buffer. Therefore pH remains almost unchanged as [H⁺] does not change significantly.
In order to calculate the pH of the buffer, we should utilise the Henderson-Hasselbalch Equation (HHE for short).

Acidic buffer

Basic buffer

A solution of a weak acid and its conjugate base

Example: CH₃COOH (weak acid) and CH₃COONa (contains the conjugate base CH₃COO⁻)

(I) On adding a small amount of H⁺,
H⁺ + CH₃COO⁻ → CH₃COOH The small amount of added H⁺ is removed by the large amount of CH₃COO⁻ in the buffer.
Therefore pH remains almost unchanged as [H⁺] does not change significantly

(II) On adding a small amount of OH⁻,
OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O The small amount of added OH⁻ is removed by the large amount of CH₃COOH in the buffer. Therefore pH remains almost unchanged as [OH⁻] does not change significantly.

A solution of a weak base and its conjugate acid

Example: NH₃ (weak base) and NH₄Cl (contains the conjugate acid NH₄⁺)

(I) On adding a small amount of OH⁻
OH⁻ + NH₄⁺ → NH₃ + H₂O
The small amount of added OH⁻ is removed by the large amount of NH₄⁺ in the buffer.
Therefore pH remains almost unchanged as [OH⁻] does not change significantly

(II) On adding a small amount of H⁺
H⁺ + NH₃ → NH₄⁺ The small amount of added H⁺ is removed by the large amount of NH₃ in the buffer.
Therefore pH remains almost unchanged as [H⁺] does not change significantly.

In order to calculate the pH of the buffer, we should utilise the Henderson-Hasselbalch Equation (HHE for short).

Acidic buffer
A solution of a weak acid and its conjugate base Example: CH₃COOH (weak acid) and CH₃COONa (contains the conjugate base CH₃COO⁻) (I) On adding a small amount of H⁺, H⁺ + CH₃COO⁻ → CH₃COOH The small amount of added H⁺ is removed by the large amount of CH₃COO⁻ in the buffer. Therefore pH remains almost unchanged as [H⁺] does not change significantly (II) On adding a small amount of OH⁻, OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O The small amount of added OH⁻ is removed by the large amount of CH₃COOH in the buffer. Therefore pH remains almost unchanged as [OH⁻] does not change significantly.
Basic buffer
A solution of a weak base and its conjugate acid Example: NH₃ (weak base) and NH₄Cl (contains the conjugate acid NH₄⁺) (I) On adding a small amount of OH⁻ OH⁻ + NH₄⁺ → NH₃ + H₂O The small amount of added OH⁻ is removed by the large amount of NH₄⁺ in the buffer. Therefore pH remains almost unchanged as [OH⁻] does not change significantly (II) On adding a small amount of H⁺ H⁺ + NH₃ → NH₄⁺ The small amount of added H⁺ is removed by the large amount of NH₃ in the buffer. Therefore pH remains almost unchanged as [H⁺] does not change significantly.
In order to calculate the pH of the buffer, we should utilise the Henderson-Hasselbalch Equation (HHE for short).

Steps in tackling calculation questions regarding buffer solutions

❏ Identify the type of buffer (whether acidic or basic buffer)

❏ Determine [acid] or [base]

❏ Determine [salt] in the buffer

❏ Calculate pH using the appropriate Henderson-Hasselbalch equation

Let's take a look at worked example 1 to illustrate pH calculations in a buffer solution:

Worked Example 1

Calculate the pH of a 1 dm³ solution prepared by mixing 0.04 mol of NaOH and 0.10 mol of CH₃COOH. The pKₐ of CH₃COOH is 4.74.

Solution:
Upon mixing, an acid base reaction occurs following the equation
OH⁻ (aq) + CH₃COOH (aq) → CH₃COO⁻ (aq) + H₂O (l)
Since initial n(CH₃COOH) > n(OH⁻) added, some CH₃COOH remains unreacted and the resultant mixture contains both CH₃COOH and CH₃COO⁻ , hence an acidic buffer is formed.

After mixing,
n(CH₃COOH) left = initial n(CH₃COOH) - n(CH₃COOH) reacted

= 0.10 - 0.04
= 0.06 mol

n(CH₃COO⁻) = n(OH⁻) added = 0.04 mol
Hence [CH₃COOH] = 0.06 mol dm⁻³ and [CH₃COO⁻] = 0.04 mol dm⁻³

For this acidic buffer,
pH = pKₐ + lg([CH₃COO⁻]/[CH₃COOH])

= 4.74 + lg(0.04/0.06)
= 4.56

Acid-Base Titration Curves

The acid-base titration curves help us to explain and understand the changes in pH during
acid-base titrations better. Students are required to know how to draw the shapes of the titration
graphs based on the given scenarios in the question and/or practical paper 4.

Do note that only 4 possible types of titration curves can be derived from any acid -base titration experiment:

1. Strong acid titrated against strong base
2. Weak acid titrated against strong base
3. Strong acid titrated against weak base
4. Weak acid titrated against weak base

Checklist to ensure a perfect titration curve drawing:

❏ Labelling of axes (pH and Volume of titrant added/cm³)

❏ Indicate the equivalence point(s), and label the pH and volume of titrant added at that point

❏ Indicate the maximum buffer capacity and label volume of titrant added and pH of solution at that point

❏ Label the initial pH

❏ Label the pH of solution when the excess volume of acid or base is added (depends if the question provided the information)

The following is a summary of the Acid-Base Titration graphs:

Strong Acid-Strong Base	Strong base titrated against strong acid	Strong acid titrated against strong base
Weak Acid-Strong Base	Weak base titrated against strong acid	Strong base titrated against weak acid
Strong Acid-Weak Base	Strong base titrated against weak acid	Weak acid titrated against strong base

Strong Acid-Strong Base
Strong base titrated against strong acid
Strong acid titrated against strong base

Weak Acid-Strong Base
Weak base titrated against strong acid
Strong acid titrated against weak base

Strong Acid-Weak Base
Strong base titrated against weak acid
Weak acid titrated against strong base

We will explain to you in-depth using a sample worked example below to understand it better through application:

Worked Example 2
[2019 RVHS JC2 H2 Chemistry Paper 3 Qn2(b)(i)(ii)(iii)]

Phthalic anhydride is the acid anhydride of phthalic acid. It is an important industrial chemical for the synthesis of phthalic esters, which are used as plasticisers to soften plastics.

Hydrolysis of phthalic anhydride produces phthalic acid, which is an aromatic dicarboxylic acid that can ionise in stages. Some information on phthalic acid and another aromatic acid are given below.

Table 2.1

	pK₁	pK₂
Phthalic acid	2.89	5.51
Benzoic acid	4.20	-

(i) Suggest a reason why the pK₂ of phthalic acid is higher than its pK₁.

(ii) Calculate the pH of a solution that is obtained after 50 cm³ of 0.1 mol dm⁻³ of NaOH is added to 10 cm³ of 0.2 mol dm⁻³ phthalic acid.

(iii) Sketch the pH-volume added curve you would expect to obtain when 50 cm³ of 0.1 mol dm⁻³ of NaOH is added to 10 cm³ of 0.2 mol dm⁻³ phthalic acid. Include relevant details from (ii) on your graph.

Solution:
(i) For pK₁, the H⁺ is removed from a neutral molecule (referring to phthalic acid). For pK₂, the removal of a H⁺ from the anion that already carries a negative charge is electrostatically unfavourable.
OR
Favourable intramolecular hydrogen bonding in the anion will be disrupted when the second H⁺ dissociates in pK₂. (Refer to the diagram below for a better understanding of the second alternative answer.)

(ii) Phthalic acid is the limiting reagent / NaOH is in excess

Amount of excess NaOH =

₅₀

x 0.1 - (

₁₀

x 0.2 x 2) = 1.0 x 10⁻³ mol

pOH = lg(1.667 × 10⁻² ) = 1.78
pH = 14 - 1.78 = 12.2

(iii) Step-By-Step Guide:
Firstly, you need to recall what pKₐ is. pKₐ is a way to express the extent of dissociation. The lower the pKₐ, the stronger the acid.

Phthalic acid has 2 pKₐ values, meaning that this is a polyprotic acid-base titration and phthalic acid underwent two dissociations with 2 acidic protons.
In this case, the value of pK₁ of phthalic acid is smaller than the value of pK₂ of phthalic acid, so we can infer that the solution is getting more basic as we proceed on with the dissociations, and the titration curve will move upwards (increasing pH).

Next, the question mentioned “50 cm³ of 0.1 mol dm⁻³ of NaOH is added to 10 cm³ of 0.2 mol dm⁻³ phthalic acid”, so it is a weak acid-strong base titration. Recall the titration curve shape.

Do remember to label the pH values that are known to you already.
Initial pH = pH of phthalic acid = unknown
1ˢᵗ buffer pH = pK₁ = 2.89 (given)
2ⁿᵈ buffer pH = pK₂ = 5.51 (given)
pH of solution when 50 cm³ of NaOH is added: 12.2 [calculated in (ii)]

Maximum Buffer Capacity (half-equivalence point):
The horizontal parts of the curve are where Maximum Buffer Capacity (MBC) occurs, which is when [weak acid] = [conjugate base]. In this case, the buffers formed are acidic buffers, thus pH= pKₐ. You need to know that this point occurs at half the volume of base required to neutralise the acid completely.

At every equivalence point, all the weak acid in the conical flask would be completely neutralised. However, strong base NaOH will still be added, so what’s left in the flask is the salt formed and excess strong NaOH base.

Therefore, the titration curve should be as such:

Lastly, here is a summary table of the commonly used indicators in acid-base. This table would also be useful in tackling the practical portion of H2 Chemistry.

Indicator	Colour		pH range for colour change/working range
Indicator	Acid	Base	pH range for colour change/working range
Methyl orange	Red	Yellow	3.2 - 4.4
Screened methyl orange	Violet	Green	3.2 - 4.4
Phenolphthalein	Colourless	Pink	8.2 - 10.0
Thymolphthalein	Colourless	Blue	9.3 - 10.5

Indicator		Methyl orange
Colour	Acid	Red
Colour	Base	Yellow
pH range for colour change/working range		3.2 - 4.4

Indicator		Screened methyl orange
Colour	Acid	Violet
Colour	Base	Green
pH range for colour change/working range		3.2 - 4.4

Indicator		Phenolphthalein
Colour	Acid	Colourless
Colour	Base	Pink
pH range for colour change/working range		8.2 - 10.0

Indicator		Thymolphthalein
Colour	Acid	Colourless
Colour	Base	Blue
pH range for colour change/working range		9.3 - 10.5

Chapter 17: Acid-Base Equilibria

Chapter 17 Acid-Base Equilibria

Chapter 17
Acid-Base Equilibria