Problem 150 - Chemistry

Question

When the concentration of a strong acid is not substantially higher than \(1.0 \times 10^{-7} M\), the ionization of water must be taken into account in the calculation of the solution's \(\mathrm{pH}\). (a) Derive an expression for the \(\mathrm{pH}\) of a strong acid solution, including the contribution to \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) from \(\mathrm{H}_{2} \mathrm{O}\). (b) Calculate the pH of a \(1.0 \times 10^{-7} M \mathrm{HCl}\) solution.

Step-by-Step Solution

Answer

For a \(1.0 \times 10^{-7} \text{ M} \text{ HCl} \) solution, \(\text{pH} \approx 6.7\).

Step 1: Understanding the Ionization of Water

Water self-ionizes to produce hydronium and hydroxide ions. The ionization of water is represented by the equation: \[ \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^{+}\text{(aq)} + \text{OH}^{-}\text{(aq)} \] Water's ionic product, \( K_w \), at 25°C is equal to \( 1.0 \times 10^{-14} \text{ M}^2 \). This equilibrium must be considered alongside the contribution from the strong acid.

Step 2: Derive Total Hydronium Ion Concentration

The total concentration of \( \text{H}_3\text{O}^{+} \) ions in solution is the sum of contributions from the acid and from water. If \( C \) is the concentration of the strong acid, then: \[ [\text{H}_3\text{O}^{+}]_{\text{total}} = C + [\text{H}_3\text{O}^{+}]_{\text{water}} \] Since \( [\text{H}_3\text{O}^{+}] \approx [\text{OH}^{-}] \) for pure water:\[ [\text{H}_3\text{O}^{+}]_{\text{water}} = \sqrt{K_w} \]

Step 3: Expression Derivation for pH

To find the pH expression including water ionization, use:\[ [\text{H}_3\text{O}^{+}]_{\text{total}} = C + \sqrt{K_w} \]The \( \text{pH} \) is calculated as:\[ \text{pH} = -\log([\text{H}_3\text{O}^{+}]_{\text{total}}) \]

Step 4: Calculate the pH for Given Concentration

Given \( C = 1.0 \times 10^{-7} \text{ M} \), we calculate:\[ [\text{H}_3\text{O}^{+}]_{\text{water}} = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \text{ M} \]Thus, \[ [\text{H}_3\text{O}^{+}]_{\text{total}} = 1.0 \times 10^{-7} + 1.0 \times 10^{-7} = 2.0 \times 10^{-7} \text{ M} \]Now, find the pH:\[ \text{pH} = -\log(2.0 \times 10^{-7}) \approx 6.7 \]

Key Concepts

Ionization of Water

Water, even in its pure form, has a fascinating ability to ionize. This means that water molecules can split into hydronium ions (\(\text{H}_3\text{O}^+\)) and hydroxide ions (\(\text{OH}^-\)). This is known as the ionization of water. The reaction can be represented by the equation: \[ \text{H}_2\text{O} (l) \rightleftharpoons \text{H}_3\text{O}^+ (aq) + \text{OH}^- (aq) \] Even though this occurs to a very small extent, it is crucial to account for this process. Particularly when calculating pH values for acidic solutions at extremely low concentrations.
The degree to which water ionizes is indicated by its ion product constant, which is often referred to as \(K_w\). This value is constant at a specific temperature, for instance, at 25°C, \(K_w = 1.0 \times 10^{-14} \text{ M}^2\). Remembering this equilibrium helps us appreciate why there is always some level of \(\text{H}_3\text{O}^+\) present in water. This becomes particularly vital when we consider that in very dilute strong acid solutions, the ionization of water is not negligible.
To sum up:

• The ionization of water produces both \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\).
• The \(K_w\) value at 25°C is \(1.0 \times 10^{-14} \text{ M}^2\).
• This process affects pH calculations greatly, especially in very dilute conditions.

Hydronium Ion Concentration

Hydronium ion concentration in a solution is critical for understanding its acidity. In solutions diluted with a strong acid, the concentration of hydronium ions (\(\text{H}_3\text{O}^+\)) comes from both the acid itself and the water.
When a strong acid dissolves, it fully dissociates, meaning it adds a specific concentration of \(\text{H}_3\text{O}^+\) ions to the solution. However, with very dilute solutions, ionization of water must also be counted. Thus, the \(\text{H}_3\text{O}^+\) concentration has two components:

• Concentration from the strong acid, denoted as \(C\).
• Contribution from the water itself, due to its ionization
  (\([\text{H}_3\text{O}^+]_{\text{water}} = \sqrt{K_w}\)).

Combining these, the total hydronium ion concentration can be expressed as: \[ \text{[H}_3\text{O]}^+_{\text{total}} = C + \sqrt{K_w} \] Understanding these contributions ensures that the hydronium ion concentration accurately reflects all sources in a solution, which is crucial for truly understanding a liquid's acidity.

Ionic Product of Water

The ionic product of water, \(K_w\), is a measure of water's tendency to ionize and form ions. This small value, \(1.0 \times 10^{-14} \text{ M}^2\) at 25°C, is crucial for understanding the nature of water and its equilibrium balance. \(K_w\) comes from the concentrations of the \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\) ions in water.
In pure water, these concentrations are equal, typically representing \(\sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \text{ M}\) for each ion. The equation can be expressed as: \[ K_w = ([\text{H}_3\text{O}^+]) ([\text{OH}^-]) \] Knowing \(K_w\) allows chemists to predict the behavior of acids and bases in water. This is pivotal especially when acid and base concentrations are extremely low, and the effects of water's ionization become noticeable in calculating the pH of solutions.

pH Expression Derivation

Developing an expression for the pH of a solution containing a strong acid needs understanding of hydronium ions. When a strong acid is not substantially higher than the ionized form of water, we derive a pH expression that acknowledges all \(\text{H}_3\text{O}^+\) contributions.
The equation derived: \[ \text{pH} = -\log([\text{H}_3\text{O}^+]_{\text{total}}) \] captures both the strong acid's and water's contribution to \(\text{H}_3\text{O}^+\). In essence, this expression comprehensively encompasses:

• Contribution from the strong acid.
• Square root of the ionic product of water to account for ionization.

Such considerations are especially relevant when determining the true pH for very dilute solutions. By acknowledging contributions from both avenues, we get a complete view of the system's acidity, a consideration that becomes particular valuable in industrial and laboratory settings.