Self-Ionisation of Water Revision Notes for HSC SSCE Chemistry

Self-Ionisation of Water

Water is not just a simple solvent - it has some fascinating chemical properties! One of the most important is its ability to react with itself through a process called self-ionisation (also known as autoionization). Understanding this concept is crucial for working with acids, bases, and pH calculations.

Water's amphiprotic nature

In earlier sections, you learned that water is amphiprotic, meaning it can act as both an acid and a base. This special property allows water molecules to react with each other in an equilibrium reaction that produces ions. Even in pure water, a small number of water molecules are constantly participating in this reaction:

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

In this reaction, one water molecule donates a proton (acting as an acid) while another water molecule accepts that proton (acting as a base). The result is the formation of a hydronium ion ( $\text{H}_3\text{O}^+$ ) and a hydroxide ion ( $\text{OH}^-$ ).

This equilibrium is present in all aqueous solutions, not just pure water. However, only a tiny fraction of water molecules actually undergo this ionisation - the equilibrium lies far to the left, favouring the undissociated water molecules.

infoNote

The equilibrium position lies so far to the left that only about 2 out of every billion water molecules are ionised at any given moment! This is why pure water is such a poor conductor of electricity despite containing ions.

The self-ionisation constant ( $K_w$ )

Since the self-ionisation of water is an equilibrium reaction, it has an equilibrium constant. However, this constant has a special name and symbol: the self-ionisation constant or ionic product constant for water, represented as $K_w$ .

The expression for $K_w$ is:

$K_w = [\text{OH}^-][\text{H}_3\text{O}^+]$

At $298\text{ K}$ (which is $25°\text{C}$ , standard room temperature):

$K_w = 1.0 \times 10^{-14} \text{ (mol L}^{-1})^2$

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Why isn't water included in the expression?

You might wonder why $[\text{H}_2\text{O}]$ doesn't appear in the $K_w$ expression. This is because the concentration of water is extremely large (approximately $55\text{ M}$ ) and remains essentially constant during the reaction. Since only a minuscule amount of water molecules actually ionise, the change in water concentration is negligible. For this reason, the water concentration is incorporated into the constant itself, giving us the simplified $K_w$ expression.

chatImportant

The value of $K_w = 1.0 \times 10^{-14}$ is only valid at $298\text{ K}$ ( $25°\text{C}$ ). Like all equilibrium constants, $K_w$ changes with temperature. Using this value at other temperatures will lead to incorrect results!

Ion concentrations in pure water

In pure water at $298\text{ K}$ , something special happens: the concentration of hydroxide ions exactly equals the concentration of hydronium ions. This makes sense when you look at the equation - for every $\text{H}_3\text{O}^+$ ion formed, exactly one $\text{OH}^-$ ion is also formed.

Therefore, in pure water:

$[\text{OH}^-] = [\text{H}_3\text{O}^+]$

We can use the $K_w$ expression to calculate these concentrations:

$K_w = [\text{OH}^-][\text{H}_3\text{O}^+] = 1.0 \times 10^{-14}$

Since the two concentrations are equal:

$[\text{H}_3\text{O}^+]^2 = 1.0 \times 10^{-14}$

Taking the square root:

$[\text{H}_3\text{O}^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}\text{ M}$

Therefore, at $298\text{ K}$ , pure water has:

$[\text{H}_3\text{O}^+] = 1.0 \times 10^{-7}\text{ mol L}^{-1}$
$[\text{OH}^-] = 1.0 \times 10^{-7}\text{ mol L}^{-1}$
$\text{pH} = 7$

This is why pH 7 is considered neutral - but remember, this is only true at $25°\text{C}$ !

Defining acidic, alkaline and neutral solutions

The relationship between hydronium and hydroxide ion concentrations allows us to classify any aqueous solution as acidic, alkaline, or neutral. The key is comparing the concentrations of these two ions:

Acidic solution:

$[\text{H}_3\text{O}^+] > [\text{OH}^-]$
At $298\text{ K}$ : $[\text{H}_3\text{O}^+] > 10^{-7}\text{ mol L}^{-1}$
$\text{pH} < 7$

Alkaline (basic) solution:

$[\text{H}_3\text{O}^+] < [\text{OH}^-]$
At $298\text{ K}$ : $[\text{H}_3\text{O}^+] < 10^{-7}\text{ mol L}^{-1}$
$\text{pH} > 7$

Neutral solution:

$[\text{H}_3\text{O}^+] = [\text{OH}^-]$
At $298\text{ K}$ : both concentrations $= 10^{-7}\text{ mol L}^{-1}$
$\text{pH} = 7$

The pH scale shown below illustrates these relationships visually:

infoNote

Notice how the exponents of the hydronium and hydroxide ion concentrations always add up to $-14$ . This is a direct consequence of $K_w = 1.0 \times 10^{-14}$ . For example, if $[\text{H}_3\text{O}^+] = 10^{-3}$ , then $[\text{OH}^-]$ must be $10^{-11}$ , and $-3 + (-11) = -14$ .

Calculating pH using $K_w$

One of the most powerful applications of $K_w$ is that it allows you to calculate the pH of any solution, even alkaline solutions where you only know the hydroxide ion concentration.

The calculation pathway:

When you're given information about either acids or bases, you can use $K_w$ to find the missing ion concentration, and then calculate pH. Here's a helpful diagram showing the relationships:

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Key steps for pH calculations:

If you know $[\text{OH}^-]$ , use $K_w$ to find $[\text{H}_3\text{O}^+]$ : $[\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{[\text{OH}^-]}$
If you know $[\text{H}_3\text{O}^+]$ , calculate pH: $\text{pH} = -\log_{10}[\text{H}_3\text{O}^+]$
For strong acids and bases, remember they dissociate completely, so:
- For a strong acid: $[\text{H}_3\text{O}^+] = [\text{acid}]$
- For a strong base with one $\text{OH}^-$ : $[\text{OH}^-] = [\text{base}]$

chatImportant

Exam tip: Always check whether you're given the concentration of an acid or a base, and whether you need to find $[\text{H}_3\text{O}^+]$ , $[\text{OH}^-]$ , or pH. Draw a simple pathway diagram if needed!

Worked examples

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Worked Example 1: Finding pH of a strong base

Problem: Find the pH of a $0.02\text{ mol L}^{-1}$ solution of sodium hydroxide.

Step-by-step solution:

Step 1: Identify the species

Sodium hydroxide ( $\text{NaOH}$ ) is a strong base that dissociates completely in water:

$\text{NaOH}(\text{aq}) \rightarrow \text{Na}^+(\text{aq}) + \text{OH}^-(\text{aq})$

Step 2: Determine $[\text{OH}^-]$

Since $\text{NaOH}$ dissociates completely:

$[\text{OH}^-] = 0.02\text{ mol L}^{-1}$

Step 3: Use $K_w$ to find $[\text{H}^+]$

$K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}$

$[\text{H}^+] = \frac{1 \times 10^{-14}}{0.02} = 5 \times 10^{-13}\text{ mol L}^{-1}$

Step 4: Calculate pH

$\text{pH} = -\log(5 \times 10^{-13}) = 12.3$

Answer: The pH is 12.3, which confirms this is an alkaline solution.

lightbulbExample

Worked Example 2: Finding concentration from hydroxide ion concentration

Problem: If $[\text{OH}^-] = 3.98 \times 10^{-11}\text{ mol L}^{-1}$ in a solution of $\text{HNO}_3$ , calculate the concentration of the solution.

Step-by-step solution:

Step 1: Use $K_w$ to calculate $[\text{H}_3\text{O}^+]$

$K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1 \times 10^{-14}$

$[\text{H}_3\text{O}^+] = \frac{1 \times 10^{-14}}{3.98 \times 10^{-11}} = 2.51 \times 10^{-4}\text{ mol L}^{-1}$

Step 2: Identify the species

Nitric acid ( $\text{HNO}_3$ ) is a strong acid that dissociates completely:

$\text{HNO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_3\text{O}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$

Step 3: Determine stoichiometric relationship

From the equation, one mole of $\text{HNO}_3$ produces one mole of $\text{H}_3\text{O}^+$ :

$[\text{H}_3\text{O}^+] = [\text{HNO}_3]$

Step 4: Final answer

$[\text{HNO}_3] = 2.51 \times 10^{-4}\text{ mol L}^{-1}$

Answer: The concentration of the nitric acid solution is 2.51 × 10⁻⁴ mol L⁻¹.

chatImportant

Exam tip: When dealing with strong acids and bases, always remember they dissociate 100% in solution. This means the concentration of ions equals the original concentration of the acid or base. This is NOT true for weak acids and bases!

bookmarkSummary

Key Points to Remember:

Water undergoes self-ionisation to produce equal concentrations of $\text{H}_3\text{O}^+$ and $\text{OH}^-$ ions in pure water
The self-ionisation constant $K_w = [\text{OH}^-][\text{H}_3\text{O}^+] = 1.0 \times 10^{-14}$ at $298\text{ K}$ ( $25°\text{C}$ )
In pure water at $298\text{ K}$ : $[\text{H}_3\text{O}^+] = [\text{OH}^-] = 1.0 \times 10^{-7}\text{ mol L}^{-1}$ and $\text{pH} = 7$
A solution is acidic when $[\text{H}_3\text{O}^+] > [\text{OH}^-]$ , alkaline when $[\text{H}_3\text{O}^+] < [\text{OH}^-]$ , and neutral when $[\text{H}_3\text{O}^+] = [\text{OH}^-]$
You can use $K_w$ to convert between hydroxide and hydronium ion concentrations, enabling pH calculations for both acidic and alkaline solutions
Always check the temperature - $K_w$ values are temperature-dependent!

Self-Ionisation of Water (HSC SSCE Chemistry): Revision Notes