Question

The \(\mathrm{pH}\) of a \(1 \times 10^{8} \mathrm{M}\) aqueous solution of \(\mathrm{HCl}\) is slightly less than 7 because(1) the ionization of \(\mathrm{HCl}\) is incomplete (2) the ionization of water is negligible (3) the ionization of water at such a low concentration of \(\mathrm{HCl}\) is significant (4) the ionization of both \(\mathrm{HCl}\) and water are negligible

Short answer

Option (3)

Step-by-step solution

Understand the Concepts

To solve this problem, first understand the concept of pH. The pH of a solution is calculated using the formula: \( \text{pH} = -\log_{10} [\text{H}^+] \) where \( [\text{H}^+] \) is the concentration of hydrogen ions in the solution.

Ionization of \( \text{HCl} \)

Hydrochloric acid (HCl) is a strong acid, meaning it fully dissociates in water. Therefore, the concentration of \( [\text{H}^+] \) produced by HCl in the solution is equal to the initial concentration of HCl.

Calculate the pH from HCl Alone

Given the concentration of HCl is \( 1 \times 10^{-8} \) M, the initial estimate of pH would be: \[ \text{pH} = -\log_{10}(1 \times 10^{-8}) = 8 \] since HCl dissociates completely.

Consider Water Ionization

In extremely dilute solutions like this, the contribution of ionization from water cannot be ignored. The concentration of \( [\text{H}^+] \) from water itself, even in the absence of HCl, is \( 1 \times 10^{-7} \) M since: \[ 2\text{H}_2\text{O} \leftrightharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]

Determine Total Ion Concentration

Adding the contributions from both ionizations: \[ [\text{H}^+]_{\text{total}} = (1 \times 10^{-8}) + (1 \times 10^{-7})= 1.1 \times 10^{-7} \]

Calculate the pH

Finally, calculate the pH with this combined concentration of \( [\text{H}^+] \): \[ \text{pH} = -\log_{10}(1.1 \times 10^{-7}) \approx 6.96 \]

Choose the Correct Option

Since the pH is slightly less than 7 due to the significant contribution of water ionization, the correct answer is option (3).

Key concepts

pH calculation

Calculating the pH of a solution is essential in chemistry. The pH defines how acidic or basic a solution is. You determine pH using the formula: \[ \text{pH} = -\text{log}_{10} [\text{H}^+] \]
The \( [\text{H}^+] \) represents the concentration of hydrogen ions in the solution.
For strong acids like HCl, which dissociate completely in water, this calculation becomes straightforward. However, in very dilute solutions, additional factors, such as water ionization, must be considered.
Therefore, understanding how to use the logarithmic function to find the pH and knowing how to account for all sources of hydrogen ions, is crucial for accurate pH calculations.

strong acid dissociation

Strong acids, like Hydrochloric acid (HCl), fully dissociate in water.
This means each molecule of HCl splits into one hydrogen ion \( (\text{H}^+) \) and one chloride ion \( (\text{Cl}^-) \) in solution.
For example, if you have a solution with a concentration of \( 1 \times 10^{-8} \) M of HCl, it will produce \( 1 \times 10^{-8} \) M of \( \text{H}^+ \).
This complete dissociation is a key characteristic of strong acids and plays a significant role in pH calculation.
But remember, in very dilute solutions like this, the effect of water ionization can't be ignored, unlike in more concentrated solutions.

water ionization

Water naturally undergoes a slight ionization process, even in the absence of other substances.
Pure water has a \( [\text{H}^+] \) concentration of \( 1 \times 10^{-7} \) M because water molecules dissociate slightly into hydrogen ions \( (\text{H}_3\text{O}^+) \) and hydroxide ions \( (\text{OH}^-) \).
This process is represented by the equation: \[ 2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]
In very dilute solutions of strong acids, such as \( 1 \times 10^{-8} \) M HCl, the ionization of water has a significant impact.
The total hydrogen ion concentration needs to consider both contributions from HCl and water.

hydrogen ion concentration

The total hydrogen ion concentration \( [\text{H}^+] \) in a solution is key in determining the pH.
For dilute solutions, both the dissociation of the added acid (like HCl) and the natural ionization of water contribute to \( [\text{H}^+] \).
In our example, a \( 1 \times 10^{-8} \) M HCl solution contributes \( 1 \times 10^{-8} \) M hydrogen ions.
Water contributes \( 1 \times 10^{-7} \) M hydrogen ions.
Thus, the total \( [\text{H}^+] \) is: \[ (1 \times 10^{-8}) + (1 \times 10^{-7}) = 1.1 \times 10^{-7} \]
Using this combined \( [\text{H}^+] \) concentration, you can calculate the pH: \[ \text{pH} = -\text{log}_{10}(1.1 \times 10^{-7}) \]
This comprehensive understanding ensures accurate pH determination, especially in very dilute solutions.