9.4 Ionisation Constant of Water and Calculation of pH

This section explores the self-ionization of water, the concept of the ionic product constant ($K_w$), and how these relate to the pH scale for measuring acidity and basicity.

Auto-ionization of Water

Water is an amphoteric substance, meaning it can act as both an acid and a base. In a sample of pure water, a small fraction of water molecules can react with each other in a process called auto-ionization or self-ionization.

• One water molecule donates a proton (acts as a Brønsted-Lowry acid).
• Another water molecule accepts the proton (acts as a Brønsted-Lowry base).

This establishes an equilibrium between water molecules, hydronium ions (${\mathrm{H}}_3{\mathrm{O}}^{+}$), and hydroxide ions ($\mathrm{O}{\mathrm{H}}^{-}$).

${\mathrm{H}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+\mathrm{O}{\mathrm{H}}^{-}$

The Ionic Product Constant of Water ($K_w$)

The equilibrium constant ($K$) for the auto-ionization of water is:

$K=\frac{[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]}{[{\mathrm{H}}_2\mathrm{O}{]}^2}$

Since water is the solvent and its concentration is very large and essentially constant, we can combine $K$ and $[{\mathrm{H}}_2\mathrm{O}{]}^2$ into a new constant, $K_w$, the ionic product constant of water.

$K_w=K[{\mathrm{H}}_2\mathrm{O}{]}^2=[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]$

Since $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ is often simplified to $[{\mathrm{H}}^{+}]$, the expression becomes:

$K_w=[{\mathrm{H}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]$

• At 25°C, the value of $K_w$ is experimentally determined to be $1.0\times 10^{-14}$.

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

• Temperature Dependence: $K_w$ is an equilibrium constant and is temperature-dependent. For example, at 40°C, $K_w=3.8\times 10^{-14}$. Since auto-ionization is endothermic, increasing temperature increases $K_w$.

Acidity, Basicity, and Neutrality

The relationship between $[{\mathrm{H}}^{+}]$ and $[\mathrm{O}{\mathrm{H}}^{-}]$ determines the nature of an aqueous solution at 25°C.

The pH Scale

Working with very small concentrations like $1.0\times 10^{-7}\mathrm{M}$ can be inconvenient. The pH scale provides a more practical, logarithmic measure of acidity.

• Definition: pH is the negative base-10 logarithm of the hydrogen ion concentration.

$\mathrm{pH}=-\log [{\mathrm{H}}^{+}]$

• pOH: Similarly, pOH is defined for the hydroxide ion concentration.

$\mathrm{pOH}=-\log [\mathrm{O}{\mathrm{H}}^{-}]$

Relationship between pH and pOH

We can derive a simple relationship between pH, pOH, and $K_w$:

1. Start with the $K_w$ expression at 25°C: $[{\mathrm{H}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]=1.0\times 10^{-14}$
2. Take the negative logarithm of both sides: $-\log ([{\mathrm{H}}^{+}][\mathrm{O}{\mathrm{H}}^{-}])=-\log (1.0\times 10^{-14})$
3. Apply logarithm rules: $(-\log [{\mathrm{H}}^{+}])+(-\log [\mathrm{O}{\mathrm{H}}^{-}])=14$
4. Substitute the definitions of pH and pOH: $\mathrm{pH}+\mathrm{pOH}=14$

pH Values for Different Solutions (at 25°C)

Worked Examples

Example 9.2

The concentration of $[\mathrm{O}{\mathrm{H}}^{-}]$ ion in a household ammonia solution is 0.005 M. Calculate the concentration of $[{\mathrm{H}}^{+}]$ in it.

Solution:

1. Given Values:

   • $[\mathrm{O}{\mathrm{H}}^{-}]=0.005\mathrm{M}$
   • $K_w=1.0\times 10^{-14}$ (at 25°C)

2. Apply the Formula: $K_w=[{\mathrm{H}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]$

3. Calculation: $1.0\times 10^{-14}=[{\mathrm{H}}^{+}]\times 0.005$ $[{\mathrm{H}}^{+}]=\frac{1.0\times 10^{-14}}{0.005}=2.0\times 10^{-12}\mathrm{M}$

Example 9.3

Calculate the pH of 0.001 M aqueous hydrochloric acid solution.

Solution:

1. Ionization: Hydrochloric acid (HCl) is a strong acid and ionizes completely in water. $\mathrm{HCl}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_3{\mathrm{O}}^{+}+\mathrm{C}{\mathrm{l}}^{-}$ The concentration of ${\mathrm{H}}^{+}$ equals the initial concentration of HCl.

   • $[{\mathrm{H}}^{+}]=0.001\mathrm{M}=10^{-3}\mathrm{M}$

2. Apply the Formula: $\mathrm{pH}=-\log [{\mathrm{H}}^{+}]$

3. Calculation: $\mathrm{pH}=-\log (10^{-3})=3.00$

Example 9.4

Calculate the pH of 0.062 M NaOH solution.

Solution:

1. Dissociation: Sodium hydroxide (NaOH) is a strong base and dissociates completely. ${\mathrm{NaOH}}_{(aq)}\to {\mathrm{N}{\mathrm{a}}^{+}}_{(aq)}+{\mathrm{O}{\mathrm{H}}^{-}}_{(aq)}$ The concentration of $\mathrm{O}{\mathrm{H}}^{-}$ equals the initial concentration of NaOH.

   • $[\mathrm{O}{\mathrm{H}}^{-}]=0.062\mathrm{M}$

2. Calculate pOH: $\mathrm{pOH}=-\log [\mathrm{O}{\mathrm{H}}^{-}]=-\log (0.062)$ $\mathrm{pOH}=-\log (6.2\times 10^{-2})=-(\log 6.2+\log 10^{-2})$ $\mathrm{pOH}=-(0.792-2)=1.208$

3. Calculate pH: $\mathrm{pH}=14-\mathrm{pOH}=14-1.208=12.79$