Dissociation Constants for Acids Revision Notes for HSC SSCE Chemistry

Dissociation Constants for Acids

Introduction to acid dissociation constants

When weak acids dissolve in water, only a small proportion of the acid molecules ionise. This creates an equilibrium between the un-ionised acid molecules and the ions produced. The extent of this ionisation can be quantified using a special equilibrium constant called the acid dissociation constant, represented by the symbol $K_a$ .

The general equation for a weak acid ionising in water is:

$\text{HA(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{A}^-\text{(aq)}$

where HA represents any weak acid and A⁻ represents its conjugate base.

The equilibrium expression for weak acids

The acid dissociation constant expression follows the standard equilibrium format. For the general weak acid equation above, the equilibrium expression is:

$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$

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Water is not included in this expression because it is the solvent and its concentration remains essentially constant throughout the reaction.

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Worked Example: Hydrofluoric Acid Equilibrium

For the weak acid HF ionising in water:

$\text{HF(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{F}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

The equilibrium expression becomes:

$K_a = \frac{[\text{H}_3\text{O}^+][\text{F}^-]}{[\text{HF}]}$

For hydrofluoric acid, $K_a = 7.6 \times 10^{-4}$ . This small value indicates that only a small percentage of HF molecules ionise in water.

Understanding the magnitude of $K_a$

The size of the acid dissociation constant tells us about the strength of the acid:

Large $K_a$ values (close to 1 or greater): Strong acids that ionise almost completely in water. The numerator (products) is much larger than the denominator (reactants).
Small $K_a$ values (much less than 1): Weak acids that only ionise to a small extent. The denominator (reactants) is much larger than the numerator (products).

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Comparison with Strong Acids

When a strong acid like hydrochloric acid ionises in water, the reaction goes essentially to completion:

$\text{HCl(g)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_3\text{O}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$

The $K_a$ for strong acids is extremely large because virtually all the acid molecules ionise, making the concentration of un-ionised acid negligible. Notice we use a single arrow rather than equilibrium arrows for strong acids.

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Critical Distinction for pH Measurements

For strong acids, the pH gives a direct measurement of the original acid concentration because all molecules ionise. However, for weak acids, the pH only tells us about the hydronium ion concentration, not the total acid concentration. To find the acid concentration, we need to use the $K_a$ value in conjunction with the pH measurement.

$K_a$ values of common weak acids

Below is a table showing the acid dissociation constants for several common weak acids:

Acid name	Formula	$K_a$
Ammonium ion	NH₄⁺	$5.6 \times 10^{-10}$
Ethanoic	CH₃COOH	$1.7 \times 10^{-5}$
Hydrocyanic	HCN	$6.3 \times 10^{-10}$
Hydrofluoric	HF	$7.6 \times 10^{-4}$
Lactic	HC₃H₅O₃	$1.4 \times 10^{-4}$
Methanoic	HCOOH	$1.8 \times 10^{-4}$
Nitrous	HNO₂	$7.2 \times 10^{-4}$

Notice that hydrofluoric acid has the largest $K_a$ value ( $7.6 \times 10^{-4}$ ), making it the strongest weak acid in this table. Ammonium ion and hydrocyanic acid have the smallest values ( $\approx 10^{-10}$ ), making them the weakest acids listed.

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Temperature Dependence

Acid dissociation constants are temperature-dependent and are usually quoted at 25°C. They also only apply when the acid is dissolved in water.

Determining $K_a$ from experimental data

The acid dissociation constant of a weak acid can be determined experimentally by measuring the pH of a solution of known concentration. The pH allows us to calculate the hydronium ion concentration, which can then be used to find $K_a$ .

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Key Assumptions for Calculating $K_a$

When calculating $K_a$ for weak acids, we make two critical assumptions:

At equilibrium, [HA] is approximately equal to the initial concentration because only a small proportion of the weak acid ionises.
The [H₃O⁺] produced by the self-ionisation of water is negligible compared to that produced by the acid, so we can ignore it in our calculations.

Simplified formula:

Since each acid molecule that ionises produces one H₃O⁺ ion and one A⁻ ion, we know that [A⁻] = [H₃O⁺]. This allows us to simplify the $K_a$ expression:

$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} = \frac{[\text{H}_3\text{O}^+]^2}{[\text{HA}]}$

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Worked Example: Determining $K_a$ from pH

Problem: A 0.10 M solution of hypobromous acid (HOBr) has a pH of 4.80. Determine $K_a$ for this acid.

Solution:

Step 1: Write the ionisation equation for the acid

$\text{HOBr(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{OBr}^-\text{(aq)}$

Step 2: Calculate the hydronium ion concentration from the pH

$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-4.80} = 1.58 \times 10^{-5} \text{ M}$

Step 3: Apply the $K_a$ expression

Since [OBr⁻] = [H₃O⁺], we can use:

$K_a = \frac{[\text{H}_3\text{O}^+]^2}{[\text{HOBr}]}$

Assuming [HOBr] at equilibrium equals the initial concentration:

$K_a = \frac{(1.58 \times 10^{-5})^2}{0.10} = 2.5 \times 10^{-9}$

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Worked Example: Calculating pH from $K_a$

Problem: Determine the pH of a 0.01 M solution of HF given that $K_a = 6.8 \times 10^{-4}$ .

Solution:

Step 1: Write the equilibrium expression

$K_a = \frac{[\text{H}_3\text{O}^+][\text{F}^-]}{[\text{HF}]} = 6.8 \times 10^{-4}$

Step 2: Simplify using [F⁻] = [H₃O⁺]

$K_a = \frac{[\text{H}_3\text{O}^+]^2}{[\text{HF}]}$

$[\text{H}_3\text{O}^+]^2 = K_a \times [\text{HF}] = 6.8 \times 10^{-4} \times 0.01$

$[\text{H}_3\text{O}^+] = \sqrt{6.8 \times 10^{-4} \times 0.01} = 2.6 \times 10^{-3} \text{ M}$

Step 3: Calculate pH

$\text{pH} = -\log[\text{H}_3\text{O}^+] = -\log(2.6 \times 10^{-3}) = 2.58$

Exam tip: Always check that your final pH value makes sense. For a 0.01 M weak acid solution, a pH around 2-3 is reasonable. A pH of 7 would indicate no ionisation, while a pH of 1 would suggest a strong acid.

Investigation: determining $K_a$ for acetic acid

This practical investigation allows you to experimentally determine the acid dissociation constant for acetic acid (ethanoic acid) by measuring pH values for different concentrations.

Aim: To determine $K_a$ for acetic acid

Materials needed:

100 mL of 0.100 mol L⁻¹ acetic acid (CH₃COOH)
Distilled water
Volumetric flasks (100 mL)
Beakers (150 mL)
Bulb pipettes (10 mL and 25 mL)
pH meter
Safety glasses

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Risk Assessment

What are the risks?	How can you stay safe?
Chemicals may splash onto skin or into eyes	Wear safety glasses and wash hands at the end of the experiment
Glassware may break	Keep glassware away from the edge of the bench. Place pipettes in the fold of a book to prevent rolling

Procedure:

You will prepare five solutions of different concentrations by serial dilution:

Solution	Volume/chemical
A	100 mL of 0.100 mol L⁻¹ acetic acid
B	25 mL solution A + 75 mL distilled water
C	10 mL solution A + 90 mL distilled water
D	25 mL solution C + 75 mL distilled water
E	10 mL solution C + 90 mL distilled water

Label five beakers A–E
Pour the stock acetic acid solution into beaker A
Measure and record the pH of solution A
Use appropriate pipettes to prepare solutions B–E according to the table
Measure and record the pH of each solution
Remember to rinse pipettes and the pH meter with distilled water between measurements

Analysis:

Calculate the initial concentration of CH₃COOH for each solution
Plot a graph of pH versus [CH₃COOH]
Use spreadsheet software to add a line of best fit and obtain its equation
Select a concentration value within your experimental range and use the line of best fit to determine its pH
Convert this pH to [H₃O⁺] using: $[\text{H}_3\text{O}^+] = 10^{-\text{pH}}$
Calculate $K_a$ using: $K_a = \frac{[\text{H}_3\text{O}^+]^2}{[\text{CH}_3\text{COOH}]}$

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Discussion Points

Why is using the line of best fit better than using individual data points? (It averages out random errors and gives a more reliable result)
The accepted value for $K_a$ of acetic acid at 25°C is $1.8 \times 10^{-5}$ . How does your experimental value compare?
What sources of error might affect the accuracy of your result? (Temperature variations, pH meter calibration, measurement errors)

Percentage ionisation

Percentage ionisation tells us what proportion of the original acid molecules have ionised in water. This is an important concept because it helps us understand how valid our assumptions are when calculating $K_a$ .

For the general weak acid reaction:

$\text{HA(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{A}^-\text{(aq)}$

The percentage ionisation is calculated using:

$\text{Percentage ionisation} = \frac{[\text{A}^-]}{[\text{HA}]_{\text{initial}}} \times 100\%$

Since [A⁻] equals [H₃O⁺] for a monoprotic acid:

$\text{Percentage ionisation} = \frac{[\text{H}_3\text{O}^+]}{[\text{HA}]_{\text{initial}}} \times 100\%$

where [HA]initial is the original concentration of the acid before any ionisation occurred.

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Once you know the percentage ionisation, you can determine a more accurate value for [HA] at equilibrium by calculating how much acid has ionised. This is particularly important when the percentage ionisation is significant (more than a few percent), as our assumption that [HA]initial ≈ [HA]equilibrium becomes less valid.

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Worked Example: Percentage Ionisation

Problem: Hydrofluoric acid (HF) reacts with water according to:

$\text{HF(g)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{F}^-\text{(aq)}$

Given $K_a = 6.8 \times 10^{-4}$ . Determine the percentage ionisation of HF in a 0.1 M HF solution that has a pH of 2.1.

Solution:

Step 1: Calculate [H₃O⁺] from pH

$[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-2.1} = 0.008 \text{ mol L}^{-1}$

Step 2: Determine [F⁻]

From the equation, we can see that [F⁻] = [H₃O⁺]:

$[\text{F}^-] = [\text{H}_3\text{O}^+] = 0.008 \text{ mol L}^{-1}$

Step 3: Calculate percentage ionisation

$\text{Percentage ionisation} = \frac{[\text{F}^-]}{[\text{HF}]_{\text{initial}}} \times 100$

$= \frac{0.008}{0.1} \times 100 = 8\%$

This means that 8% of the original HF molecules have ionised in this solution. This is a relatively high percentage for a weak acid, which explains why HF is considered one of the stronger weak acids.

Polyprotic acids

Some acids can donate more than one proton. These are called polyprotic acids. Each proton donation step has its own equilibrium and therefore its own $K_a$ value. These are labelled $K_{a1}$ , $K_{a2}$ , $K_{a3}$ , etc.

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Key Principle for Polyprotic Acids

The first proton is always easier to remove than subsequent protons. This means that $K_{a1}$ is always larger than $K_{a2}$ , which is larger than $K_{a3}$ , and so on.

Strong polyprotic acids

Consider sulfuric acid (H₂SO₄), a strong diprotic acid that can donate two protons:

Step 1: (Strong acid behaviour)

$\text{H}_2\text{SO}_4\text{(l)} + \text{H}_2\text{O(l)} \rightarrow \text{HSO}_4^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

$K_{a1} = 10^9$

This first step goes essentially to completion (notice the single arrow). The very large $K_{a1}$ value indicates that virtually all H₂SO₄ molecules ionise.

Step 2: (Weak acid behaviour)

$\text{HSO}_4^-\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{SO}_4^{2-}\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

$K_{a2} = 10^{-2}$

The second proton is much harder to remove. The HSO₄⁻ ion acts as a weak acid (notice the equilibrium arrows). The $K_{a2}$ value is much smaller than $K_{a1}$ .

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What's Present at Equilibrium?

In a sulfuric acid solution, you would find mainly:

Water (H₂O)
Hydronium ions (H₃O⁺)
Hydrogen sulfate ions (HSO₄⁻)

You would find very little:

Sulfuric acid (H₂SO₄) - nearly all ionised in step 1
Sulfate ions (SO₄²⁻) - only small amount formed in step 2

The HSO₄⁻ ion is amphiprotic, meaning it can both accept and donate protons. It appears as a product in step 1 and as a reactant in step 2.

Weak polyprotic acids

Consider sulfurous acid (H₂SO₃), a weak diprotic acid:

Step 1:

$\text{H}_2\text{SO}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{HSO}_3^-\text{(aq)}$

$K_{a1} = 1.7 \times 10^{-2}$

Step 2:

$\text{HSO}_3^-\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{SO}_3^{2-}\text{(aq)}$

$K_{a2} = 6.4 \times 10^{-8}$

Even though $K_{a1}$ is larger than $K_{a2}$ , both represent weak acid behaviour (both much less than 1). This means that at equilibrium, you would expect to find significant amounts of all three species:

Sulfurous acid (H₂SO₃)
Hydrogen sulfite ions (HSO₃⁻)
Sulfite ions (SO₃²⁻)

Understanding p $K_a$

Just as pH provides a convenient way to express hydrogen ion concentration, p $K_a$ provides a convenient way to express acid strength. The 'p' notation means taking the negative logarithm (base 10):

$\text{p}K_a = -\log_{10} K_a$

Key relationships:

The larger the $K_a$ , the stronger the acid
The smaller the p $K_a$ , the stronger the acid (inverse relationship)

This inverse relationship exists because of the negative sign in the p $K_a$ definition. Think of it like pH: a lower pH means a more acidic solution, and a lower p $K_a$ means a stronger acid.

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Example Comparison

Acid name	Formula	$K_a$	p $K_a$
Hydrofluoric	HF	$7.6 \times 10^{-4}$	3.12
Lactic	HC₃H₅O₃	$1.4 \times 10^{-4}$	3.85
Ethanoic	CH₃COOH	$1.7 \times 10^{-5}$	4.77
Hydrocyanic	HCN	$6.3 \times 10^{-10}$	9.20
Ammonium ion	NH₄⁺	$5.6 \times 10^{-10}$	9.25

Notice that:

Hydrofluoric acid has the largest $K_a$ (7.6 × 10⁻⁴) and the smallest p $K_a$ (3.12) - it's the strongest acid
Ammonium ion has the smallest $K_a$ (5.6 × 10⁻¹⁰) and the largest p $K_a$ (9.25) - it's the weakest acid

Connection to buffers:

The p $K_a$ value is particularly important when studying buffer solutions. There's a relationship between pH and p $K_a$ given by:

$\text{pH} = \text{p}K_a + \log_{10}\frac{[\text{A}^-]}{[\text{HA}]}$

This equation, called the Henderson-Hasselbalch equation, is essential for buffer calculations and will be explored further in the next chapter.

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Key Points to Remember:

The acid dissociation constant ( $K_a$ ) measures the strength of an acid in water. Strong acids have very large $K_a$ values (approaching or exceeding 1), while weak acids have small $K_a$ values (much less than 1).
The general equilibrium expression for a weak acid is: $K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$ . Water is not included because it's the solvent and its concentration doesn't change significantly.
Percentage ionisation tells us what proportion of acid molecules have ionised: $\text{Percentage ionisation} = \frac{[\text{A}^-]}{[\text{HA}]_{\text{initial}}} \times 100\%$ . For most weak acids, this is less than 5%.
Polyprotic acids donate protons in steps, with each step having its own $K_a$ value. The first proton is always easiest to remove, so $K_{a1} > K_{a2} > K_{a3}$ .
The p $K_a$ scale ( $\text{p}K_a = -\log_{10} K_a$ ) provides a convenient way to compare acid strengths. Remember: smaller p $K_a$ values indicate stronger acids.

Dissociation Constants for Acids (HSC SSCE Chemistry): Revision Notes