Let's figure out the pH of a very dilute hydrochloric acid (HCl) solution. The concentration of HCl is given as \(1 \times 10^{-8}\) M.

Normally, for a strong acid like HCl, we might think the hydrogen ion concentration (\([\text{H}^+]\)) is simply equal to the acid concentration because it dissociates completely: \(\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-\). If we just used the concentration of HCl, \([\text{H}^+] = 1 \times 10^{-8}\) M. Then the pH would be calculated as:

\(\text{pH} = -\log[\text{H}^+]\)

\(\text{pH} = -\log(1 \times 10^{-8})\)

\(\text{pH} = -(-8)\)

\(\text{pH} = 8.00\)

However, a pH of 8 is basic. An acidic solution, even a very dilute one, cannot be basic. This tells us that when the concentration of the strong acid is very low, close to or less than the concentration of \(\text{H}^+\) ions in pure water (\(1 \times 10^{-7}\) M at 25°C), we must consider the contribution of \(\text{H}^+\) ions from the autoionization of water itself.

Understanding Water Autoionization and pH Calculation

Water undergoes autoionization according to the equilibrium:

\(\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-\)

The ion product of water, \(K_w\), relates the concentrations of \(\text{H}^+\) and \(\text{OH}^-\) ions in any aqueous solution:

\(K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}\) at 25°C.

In our dilute HCl solution, there are two sources of \(\text{H}^+\) ions:

• From the complete dissociation of HCl: \([\text{H}^+]_{\text{from HCl}} = 1 \times 10^{-8}\) M
• From the autoionization of water: \([\text{H}^+]_{\text{from water}}\)

There is one source of \(\text{OH}^-\) ions:

• From the autoionization of water: \([\text{OH}^-]_{\text{from water}}\)

The total concentration of \(\text{H}^+\) ions, \([\text{H}^+]_{\text{total}}\), is the sum of contributions from HCl and water. Similarly, the total \([\text{OH}^-]\) concentration is from water autoionization.

We can use the principle of electroneutrality. In any solution, the total concentration of positive charges must equal the total concentration of negative charges. In this case, the positive ions are \(\text{H}^+\) and the negative ions are \(\text{Cl}^-\) (from HCl) and \(\text{OH}^-\) (from water):

\([\text{H}^+]_{\text{total}} = [\text{Cl}^-]_{\text{total}} + [\text{OH}^-]_{\text{total}}\)

Since HCl is a strong electrolyte, \([\text{Cl}^-]_{\text{total}} = [\text{HCl}]_{\text{initial}} = 1 \times 10^{-8}\) M.

From the water equilibrium, we know \([\text{OH}^-]_{\text{total}} = [\text{OH}^-]_{\text{from water}}\). Also, \(K_w = [\text{H}^+]_{\text{total}}[\text{OH}^-]_{\text{total}}\). So, \([\text{OH}^-]_{\text{total}} = \frac{K_w}{[\text{H}^+]_{\text{total}}}\).

Let \(x = [\text{H}^+]_{\text{total}}\). Substituting the values into the electroneutrality equation:

\(x = 1 \times 10^{-8} + \frac{1 \times 10^{-14}}{x}\)

To solve for \(x\), we can multiply the entire equation by \(x\) (assuming \(x \neq 0\), which it must be):

\(x^2 = (1 \times 10^{-8})x + 1 \times 10^{-14}\)

Rearrange this into a quadratic equation:

\(x^2 - (1 \times 10^{-8})x - (1 \times 10^{-14}) = 0\)

We can solve this quadratic equation using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-1 \times 10^{-8}\), and \(c=-1 \times 10^{-14}\).

\(x = \frac{-(-1 \times 10^{-8}) \pm \sqrt{(-1 \times 10^{-8})^2 - 4(1)(-1 \times 10^{-14})}}{2(1)}\)

\(x = \frac{1 \times 10^{-8} \pm \sqrt{(1 \times 10^{-16}) + (4 \times 10^{-14})}}{2}\)

\(x = \frac{1 \times 10^{-8} \pm \sqrt{(1 \times 10^{-16}) + (400 \times 10^{-16})}}{2}\)

\(x = \frac{1 \times 10^{-8} \pm \sqrt{401 \times 10^{-16}}}{2}\)

\(x = \frac{1 \times 10^{-8} \pm \sqrt{401} \times \sqrt{10^{-16}}}{2}\)

\(x = \frac{1 \times 10^{-8} \pm \sqrt{401} \times 10^{-8}}{2}\)

The value of \(\sqrt{401}\) is approximately 20.025. Since concentration \(x\) must be positive, we take the positive root:

\(x = \frac{1 \times 10^{-8} + 20.025 \times 10^{-8}}{2}\)

\(x = \frac{(1 + 20.025) \times 10^{-8}}{2}\)

\(x = \frac{21.025 \times 10^{-8}}{2}\)

\(x \approx 10.5125 \times 10^{-8}\)

\(x \approx 1.05125 \times 10^{-7}\) M

So, the total hydrogen ion concentration is approximately \(1.05125 \times 10^{-7}\) M.

Now, calculate the pH using the definition \(\text{pH} = -\log[\text{H}^+]_{\text{total}}\):

\(\text{pH} = -\log(1.05125 \times 10^{-7})\)

\(\text{pH} \approx -(\log(1.05125) + \log(10^{-7}))\)

\(\text{pH} \approx -(\log(1.05125) - 7)\)

\(\text{pH} \approx 7 - \log(1.05125)\)

Using a calculator, \(\log(1.05125) \approx 0.0217\).

\(\text{pH} \approx 7 - 0.0217\)

\(\text{pH} \approx 6.9783\)

Rounding to two decimal places, the pH is approximately 6.98.

This value makes sense because the solution is slightly acidic (pH less than 7), as expected from dissolving an acid in water, but very close to neutral water's pH (7) due to the extremely low concentration of HCl.

Summary of the pH Calculation Process

When dealing with very dilute strong acids or bases (concentrations close to or below \(10^{-7}\) M), remember to consider the contribution of \(\text{H}^+\) or \(\text{OH}^-\) ions from the autoionization of water.

Steps followed:

1. Identify sources of \(\text{H}^+\) and \(\text{OH}^-\) ions (acid/base and water).
2. Apply the principle of electroneutrality.
3. Use the \(K_w\) relationship (\([\text{H}^+][\text{OH}^-] = K_w\)).
4. Solve the resulting equation (often quadratic) for the total hydrogen ion concentration, \([\text{H}^+]_{\text{total}}\).
5. Calculate pH using \(\text{pH} = -\log[\text{H}^+]_{\text{total}}\).

Let's compare our calculated value to the options provided:

• 8.00 (Incorrect - does not account for water autoionization)
• 6.98 (Matches our calculated value)
• 7.01 (Incorrect - slightly above 7)
• 6.00 (Incorrect - would correspond to a higher acid concentration, like \(10^{-6}\) M)

Revision Table: Key Concepts for Dilute Acid pH

Additional Information: pH of Very Dilute Solutions

Calculating the pH of very dilute solutions of strong acids or bases highlights the importance of the water autoionization equilibrium. When the concentration of the added acid or base is around \(10^{-7}\) M or lower, the contribution of \(\text{H}^+\) and \(\text{OH}^-\) from water becomes comparable to or even greater than that from the solute. Simply using the solute concentration will give an incorrect pH value (e.g., pH > 7 for an acid or pH < 7 for a base).

The correct approach involves considering both sources of ions and using the electroneutrality principle along with the \(K_w\) expression. This leads to a more complex equation, often a quadratic one for strong acids/bases, which yields the accurate total \([\text{H}^+]\) or \([\text{OH}^-]\) concentration and consequently the correct pH.

For strong acids, the pH of a very dilute solution will approach 7 from below (e.g., 6.98, 6.99). For strong bases, the pH will approach 7 from above (e.g., 7.01, 7.02). The pH will never cross 7 for a pure acid or base solution, no matter how dilute, unless other substances are present.