Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 14.10–14.12 Kw, pKa and pKw definitions-Study Notes - New Syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.10–14.12 Kw, pKa and pKw definitions- Study Notes- New syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 14.10–14.12 Kw, pKa and pKw definitions- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.
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Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics
14.10 Ionic Product of Water (\( \mathrm{K_w} \))
Water undergoes slight self-ionisation, producing hydrogen ions and hydroxide ions. The equilibrium constant for this process is known as the ionic product of water, \( \mathrm{K_w} \).
Self-Ionisation of Water
\( \mathrm{2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)} \)
(Often simplified as:)
\( \mathrm{H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)} \)
Definition
The ionic product of water, \( \mathrm{K_w} \), is the product of the equilibrium concentrations of hydrogen ions and hydroxide ions in water.
\( \mathrm{K_w = [H^+][OH^-]} \)
Value at 298 K
\( \mathrm{K_w = 1.0 \times 10^{-14}\ mol^2\ dm^{-6}} \)
- Applies to pure water at \( \mathrm{298\ K} \).
- Also applies to any aqueous solution at this temperature.
Important Notes
- \( \mathrm{K_w} \) is temperature-dependent.
- In pure water:
\( \mathrm{[H^+] = [OH^-] = 1.0 \times 10^{-7}\ mol\ dm^{-3}} \)
- Therefore, pure water is neutral at pH 7 (at 298 K).
Key Features
- \( \mathrm{K_w = [H^+][OH^-]} \).
- Constant for a given temperature.
- Determines relationship between acidity and alkalinity.
- Basis for pH and pOH calculations.
Example 1:
Define the ionic product of water and give its value at \( \mathrm{298\ K} \).
▶️ Answer/Explanation
\( \mathrm{K_w} \) is the product of the concentrations of \( \mathrm{H^+} \) and \( \mathrm{OH^-} \) ions in water.
\( \mathrm{K_w = [H^+][OH^-]} \)
At \( \mathrm{298\ K} \), \( \mathrm{K_w = 1.0 \times 10^{-14}\ mol^2\ dm^{-6}} \).
Example 2:
Calculate \( \mathrm{[OH^-]} \) in pure water at \( \mathrm{298\ K} \).
▶️ Answer/Explanation
In pure water:
\( \mathrm{[H^+] = [OH^-]} \)
\( \mathrm{K_w = [H^+]^2 = 1.0 \times 10^{-14}} \)
\( \mathrm{[H^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}} \)
Therefore:
\( \mathrm{[OH^-] = 1.0 \times 10^{-7}\ mol\ dm^{-3}} \)
14.11 Calculating the pH of a Strong Base (using \( \mathrm{K_w} \) or \( \mathrm{pK_w} \))
Strong bases completely dissociate in aqueous solution, so the hydroxide ion concentration (\( \mathrm{[OH^-]} \)) can be determined directly from the base concentration. The pH is then calculated using the relationship between \( \mathrm{[H^+]} \) and \( \mathrm{[OH^-]} \).
Key Relationships
\( \mathrm{K_w = [H^+][OH^-]} \)
\( \mathrm{pK_w = pH + pOH} \)
At \( \mathrm{298\ K} \):
\( \mathrm{K_w = 1.0 \times 10^{-14}} \), \( \mathrm{pK_w = 14} \)
Step-by-Step Method
- Calculate \( \mathrm{[OH^-]} \) from base concentration.
- Calculate \( \mathrm{pOH} \):
\( \mathrm{pOH = -\log_{10}[OH^-]} \)
- Calculate pH:
\( \mathrm{pH = 14 – pOH} \)
Alternative Method (Using \( \mathrm{K_w} \))
- Calculate \( \mathrm{[H^+]} \) using:
\( \mathrm{[H^+] = \frac{K_w}{[OH^-]}} \)
- Then calculate pH using:
\( \mathrm{pH = -\log_{10}[H^+]} \)
Important: Stoichiometry of Base
- Some bases release more than one \( \mathrm{OH^-} \):
- \( \mathrm{NaOH} \) → 1 \( \mathrm{OH^-} \)
- \( \mathrm{Ca(OH)_2} \) → 2 \( \mathrm{OH^-} \)
So:
\( \mathrm{[OH^-] = \text{concentration} \times \text{number of } OH^-} \)
Key Features
- Strong bases fully dissociate.
- Use \( \mathrm{pOH} \) as an intermediate step.
- Relationship: \( \mathrm{pH + pOH = 14} \).
- Account for number of \( \mathrm{OH^-} \) ions released.
Example 1:
Calculate the pH of \( \mathrm{0.020\ mol\ dm^{-3}} \) \( \mathrm{NaOH} \).
▶️ Answer/Explanation
\( \mathrm{[OH^-] = 0.020} \)
\( \mathrm{pOH = -\log_{10}(0.020)} = 1.70 \)
\( \mathrm{pH = 14 – 1.70 = 12.30} \)
Example 2:
Calculate the pH of \( \mathrm{0.015\ mol\ dm^{-3}} \) \( \mathrm{Ca(OH)_2} \).
▶️ Answer/Explanation
Each molecule gives 2 \( \mathrm{OH^-} \):
\( \mathrm{[OH^-] = 2 \times 0.015 = 0.030} \)
\( \mathrm{pOH = -\log_{10}(0.030)} = 1.52 \)
\( \mathrm{pH = 14 – 1.52 = 12.48} \)
14.12 Definitions of \( \mathrm{pK_a} \) and \( \mathrm{pK_w} \)
The terms \( \mathrm{pK_a} \) and \( \mathrm{pK_w} \) are logarithmic forms of equilibrium constants. They are used to simplify calculations and make comparisons of acid strength and water ionisation more convenient.
(i) Definition of \( \mathrm{pK_a} \)
\( \mathrm{pK_a = -\log_{10}K_a} \)
- \( \mathrm{K_a} \) is the acid dissociation constant.
- \( \mathrm{pK_a} \) is a measure of acid strength.
Key Interpretation
- Small \( \mathrm{pK_a} \) → large \( \mathrm{K_a} \) → stronger acid.
- Large \( \mathrm{pK_a} \) → small \( \mathrm{K_a} \) → weaker acid.
(ii) Definition of \( \mathrm{pK_w} \)
\( \mathrm{pK_w = -\log_{10}K_w} \)
- \( \mathrm{K_w} \) is the ionic product of water.
- \( \mathrm{pK_w} \) relates \( \mathrm{pH} \) and \( \mathrm{pOH} \).
At 298 K
\( \mathrm{K_w = 1.0 \times 10^{-14}} \)
\( \mathrm{pK_w = 14} \)
Relationship
\( \mathrm{pH + pOH = pK_w} \)
Key Features
- Both are negative logarithms of equilibrium constants.
- \( \mathrm{pK_a} \) indicates acid strength.
- \( \mathrm{pK_w} \) links \( \mathrm{pH} \) and \( \mathrm{pOH} \).
- Both are temperature-dependent.
Example 1:
Explain what is meant by a weak acid having a high \( \mathrm{pK_a} \) value.
▶️ Answer/Explanation
A high \( \mathrm{pK_a} \) means \( \mathrm{K_a} \) is small.
This indicates the acid dissociates only slightly.
Therefore, the acid is weak.
Example 2:
Define \( \mathrm{pK_w} \) and state its value at \( \mathrm{298\ K} \).
▶️ Answer/Explanation
\( \mathrm{pK_w} \) is the negative logarithm of \( \mathrm{K_w} \).
\( \mathrm{pK_w = -\log_{10}K_w} \)
At \( \mathrm{298\ K} \), \( \mathrm{pK_w = 14} \).