# 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 ($H_3O^+$ or $\mathrm{H_3O^+}$), and hydroxide ions ($OH^-$ or $\mathrm{OH^-}$). $$ \mathrm{H_2O} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{OH^-} $$ ### The Ionic Product Constant of Water ($K_w$) The equilibrium constant ($K$) for the auto-ionization of water is: $$ K = \frac{[\mathrm{H_3O^+}][\mathrm{OH^-}]}{[\mathrm{H_2O}]^{2}} $$ Since water is the solvent and its concentration is very large and essentially constant, we can combine $K$ and $[\mathrm{H_2O}]^{2}$ into a new constant, $K_w$, the **ionic product constant of water**. $$ K_w = K[\mathrm{H_2O}]^{2} = [\mathrm{H_3O^+}][\mathrm{OH^-}] $$ Since $[\mathrm{H_3O^+}]$ is often simplified to $[\mathrm{H^+}]$, the expression becomes: $$ K_w = [\mathrm{H^+}][\mathrm{OH^-}] $$ - At **25°C**, the value of $K_w$ is experimentally determined to be $1.0 \times 10^{-14}$. $$ K_w = [\mathrm{H^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14} $$ - **Temperature Dependence:** $K_w$ is an equilibrium constant and is temperature-dependent. For example, at **40°C**, $K_w = 3.8 \times 10^{-14}$. #### Acidity, Basicity, and Neutrality The relationship between $[\mathrm{H^+}]$ and $[\mathrm{OH^-}]$ determines the nature of an aqueous solution at 25°C. | Solution Type | Ion Concentration | | :------------ | :-------------------------------- | | **Neutral** | $[\mathrm{H^+}] = [\mathrm{OH^-}] = 1.0 \times 10^{-7} \, \mathrm{M}$ | | **Acidic** | $[\mathrm{H^+}] > [\mathrm{OH^-}]$ or $[\mathrm{H^+}] > 1.0 \times 10^{-7} \, \mathrm{M}$ | | **Basic** | $[\mathrm{OH^-}] > [\mathrm{H^+}]$ or $[\mathrm{H^+}] < 1.0 \times 10^{-7} \, \mathrm{M}$ | ### 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. $$ \mathrm{pOH} = -\log[\mathrm{OH^-}] $$ #### 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: $$ [\mathrm{H^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14} $$ 2. Take the negative logarithm of both sides: $$ -\log([\mathrm{H^+}][\mathrm{OH^-}]) = -\log(1.0 \times 10^{-14}) $$ 3. Apply logarithm rules: $$ (-\log[\mathrm{H^+}]) + (-\log[\mathrm{OH^-}]) = -(-14) $$ 4. Substitute the definitions of pH and pOH: $$ \mathrm{pH} + \mathrm{pOH} = 14 $$ #### pH Values for Different Solutions (at 25°C) | Solution Type | $[\mathrm{H^+}]$ Concentration | pH Value | | :------------ | :---------------------- | :------- | | **Acidic** | $> 1.0 \times 10^{-7} \, \mathrm{M}$ | pH < 7.00 | | **Neutral** | $= 1.0 \times 10^{-7} \, \mathrm{M}$ | pH = 7.00 | | **Basic** | $< 1.0 \times 10^{-7} \, \mathrm{M}$ | pH > 7.00 | ## Worked Examples ### Example 9.2 The concentration of $[\mathrm{OH^-}]$ ion in a household ammonia solution is 0.005 M. Calculate the concentration of $[\mathrm{H^+}]$ in it. **Solution:** 1. **Given Values:** - $[\mathrm{OH^-}] = 0.005 \, \mathrm{M}$ - $K_w = 1.0 \times 10^{-14}$ (at 25°C) 2. **Apply the Formula:** $$ K_w = [\mathrm{H^+}][\mathrm{OH^-}] $$ 3. **Calculation:** $$ 1.0 \times 10^{-14} = [\mathrm{H^+}] \times 0.005 $$ $$ [\mathrm{H^+}] = \frac{1.0 \times 10^{-14}}{0.005} $$ $$ [\mathrm{H^+}] = 2.0 \times 10^{-12} \, \mathrm{M} $$ ### 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. $$ \mathrm{HCl} + \mathrm{H_2O} \rightarrow \mathrm{H_3O^+} + \mathrm{Cl^-} $$ The concentration of $\mathrm{H_3O^+}$ (or $\mathrm{H^+}$) will be the same as 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:** $$ \mathrm{pH} = -\log(0.001) = -\log(10^{-3}) $$ $$ \mathrm{pH} = -(-3) = 3.00 $$ ### 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. $$ \mathrm{NaOH}_{(aq)} \rightarrow \mathrm{Na^+}_{(aq)} + \mathrm{OH^-}_{(aq)} $$ The concentration of $\mathrm{OH^-}$ will be the same as the initial concentration of NaOH. - $[\mathrm{OH^-}] = 0.062 \, \mathrm{M}$ 2. **Calculate pOH:** First, find the pOH using the $[\mathrm{OH^-}]$ concentration. $$ \mathrm{pOH} = -\log[\mathrm{OH^-}] $$ $$ \mathrm{pOH} = -\log(0.062) \approx 1.21 $$ 3. **Calculate pH:** Use the relationship between pH and pOH. $$ \mathrm{pH} + \mathrm{pOH} = 14 $$ $$ \mathrm{pH} = 14 - \mathrm{pOH} $$ $$ \mathrm{pH} = 14 - 1.21 = 12.79 $$ ## Possible Questions/Answers - **Q:** What is the pH of a solution containing 1.95 g pure $\mathrm{H_2SO_4}$ per $\mathrm{dm}^3$ of solution? - **A:** 1. **Molar Mass of $\mathrm{H_2SO_4}$**: $2(1.01) + 32.07 + 4(16.00) = 98.09 \, \text{g/mol}$. 2. **Molarity of $\mathrm{H_2SO_4}$**: $$ M = \frac{\text{moles}}{\text{volume (L)}} = \frac{1.95 \, \text{g} / 98.09 \, \text{g/mol}}{1 \, \text{dm}^3} \approx 0.0199 \, \mathrm{M} $$ 3. **Hydrogen Ion Concentration**: $\mathrm{H_2SO_4}$ is a strong acid that donates two protons. $$ \mathrm{H_2SO_4} \rightarrow 2\mathrm{H^+} + \mathrm{SO_4^{2-}} $$ $$ [\mathrm{H^+}] = 2 \times [\mathrm{H_2SO_4}] = 2 \times 0.0199 \, \mathrm{M} = 0.0398 \, \mathrm{M} $$ 4. **Calculate pH**: $$ \mathrm{pH} = -\log[\mathrm{H^+}] = -\log(0.0398) \approx 1.40 $$