Weak Acids and Bases Revision Notes for Leaving Cert Chemistry

Weak Acids and Bases

What are Weak Acids and Bases?

A weak acid or weak base only partially dissociates into ions when dissolved in water.

This means that, unlike strong acids and bases (which fully dissociate), weak acids and bases establish an equilibrium between the dissociated ions and the undissociated molecules.

For weak acids, the equilibrium lies more towards the undissociated acid, meaning fewer hydrogen ions $H^+$ are produced in the solution.
For weak bases, fewer hydroxide ions $OH^-$ are produced compared to strong bases. This partial dissociation is what gives weak acids and bases their higher pH (compared to strong acids and bases at the same concentration).

Examples of Weak Acids and Bases

Weak Acids

Ethanoic acid (acetic acid) $CH_3COOH$
Carbonic acid $H_2CO_3$
Citric acid (found in citrus fruits)

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Example: The dissociation of ethanoic acid

CH_3COOH \rightleftharpoons H^+ + CH_3COO^-

In this case, only a small fraction of $CH_3COOH$ molecules release $H^+$ ions, resulting in a weakly acidic solution.

Weak Bases

Ammonia $NH_3$
Methylamine $CH_3NH_2$

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Example: The dissociation of ammonia in water

NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-

Again, only a small amount of ammonia forms hydroxide ions OH^-, resulting in a weakly basic solution.

pH of Weak Acids and Bases

The pH of weak acids and bases depends on both their concentration and their dissociation constant ( $K_a$ for acids and $K_b$ for bases).

Since weak acids and bases only partially dissociate, their pH values will be higher for acids and lower for bases compared to strong counterparts of the same concentration.

Formula for Weak Acid Calculations

For weak acids, we use an approximation to calculate the concentration of $H^+$ ions.

The formula is derived from the dissociation constant $K_a$ :

[H^+] = \sqrt{K_a \times M_a}

Where:

$K_a$ is the acid dissociation constant (a measure of how much the acid dissociates in water).
$M_a$ is the concentration (molarity) of the acid in moles per litre (mol/L).

This formula can be applied to many weak acids, but for specific exam questions, you would replace the $K_a$ and $M_a$ values according to the acid being used.

Example Calculations

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Example: Calculate the pH of a 0.01 M Solution of Ethanoic Acid Given:

Concentration of ethanoic acid $M_a = 0.01 \, \text{M}$
Dissociation constant $K_a = 1.8 \times 10^{-5}$

Step 1: Write the dissociation equation

Ethanoic acid $CH_3COOH$ is a weak acid, so it dissociates partially:

CH_3COOH \rightleftharpoons H^+ + CH_3COO^-

Step 2: Use the formula to find the concentration of $H^+$ ions

Substitute the given values into the formula:

[H^+] = \sqrt{(1.8 \times 10^{-5}) \times (0.01)}

= \sqrt{1.8 \times 10^{-7}}

\approx 1.34 \times 10^{-4} \, \text{M}

Step 3: Calculate the pH

Now, use the $[H^+]$ to calculate the pH:

pH = -\log [H^+]

= -\log (1.34 \times 10^{-4}) \approx 3.87

Conclusion:

The pH of a 0.01 M solution of ethanoic acid is approximately 3.87, which indicates that the solution is weakly acidic.

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Example: Find the pH of Vinegar Labelled as 6% (w/v) Ethanoic Acid Given:

The concentration of vinegar is 6% w/v, meaning 6 grammes of ethanoic acid per 100 cm³.
The dissociation constant $K_a = 1.8 \times 10^{-5}$

Step 1: Convert concentration to molarity

First, convert 6% w/v to molarity (moles per litre):

6 grams of ethanoic acid in 100 cm³ is equivalent to 60 grams in 1 litre (1000 cm³).
Molar mass of ethanoic acid ( CH_3COOH ) is 60 g/mol, so:

\text{Molarity} = \frac{60 \, \text{g}}{60 \, \text{g/mol}} = 1 \, \text{M}

Step 2: Use the formula to find the concentration of $H^+$ ions

Substitute the given values into the formula:

[H^+] = \sqrt{(1.8 \times 10^{-5}) \times (1.0)}

= \sqrt{1.8 \times 10^{-5}} \approx 4.24 \times 10^{-3} \, \text{M}

Step 3: Calculate the pH

Use the $[H^+]$ to find the pH:

pH = -\log [H^+]

= -\log (4.24 \times 10^{-3}) \approx 2.37

Conclusion:

The pH of the vinegar solution is approximately 2.37, which is quite acidic.

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Example: Find the pH of a Solution Containing 1.7 Grammes of Ammonia Dissolved in 500 cm³ of Water Given:

The amount of ammonia $NH_3$ is 1.7 grammes.
The volume of water is 500 cm³.
Base dissociation constant $K_b = 1.7 \times 10^{-5}$

Step 1: Convert mass to moles

The molar mass of ammonia $NH_3$ is 17 g/mol.

Convert grammes to moles:

\text{Moles of } NH_3 = \frac{1.7 \, \text{g}}{17 \, \text{g/mol}} = 0.1 \, \text{mol}

Step 2: Calculate molarity

Now, convert the volume of water to litres (500 cm³ = 0.5 L), and use it to calculate molarity:

\text{Molarity} = \frac{0.1 \, \text{mol}}{0.5 \, \text{L}} = 0.2 \, \text{M}

Step 3: Set up the formula for weak bases

For weak bases, we use a similar formula to find the concentration of $OH^-$ ions:

[OH^-] = \sqrt{K_b \times M_b}

Substitute the given values:

[OH^-] = \sqrt{(1.7 \times 10^{-5}) \times (0.2)}

= \sqrt{3.4 \times 10^{-6}} \approx 1.84 \times 10^{-3} \, \text{M}

Step 4: Calculate pOH and then pH

Now, use $[OH^-]$ to calculate the pOH:

pOH = -\log [OH^-]

= -\log (1.84 \times 10^{-3}) \approx 2.73

Finally, use the relationship pH + pOH = 14 to find the pH:

pH = 14 - 2.73 = 11.27

Conclusion:

The pH of the ammonia solution is approximately 11.27, indicating a basic solution.

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Key Points to Remember:

For weak acids, use $[H^+] = \sqrt{K_a \times M_a}$
For weak bases, use $[OH^-] = \sqrt{K_b \times M_b}$
pH is calculated using $pH = -\log [H^+]$ and for bases
First calculate $pOH$ and use pH + pOH = 14

Weak Acids and Bases (Leaving Cert Chemistry): Revision Notes