Question

Despite \(\mathrm{HCl}\) 's being a strong acid, the \(\mathrm{pH}\) of \(1.00 \times 10^{-7} \mathrm{M} \mathrm{HCl}\) is not exactly 7.00 . Can you suggest a reason why?

Short answer

The pH of the \(1.00 \times 10^{-7} \mathrm{M}\) HCl solution is not exactly 7.00 because the ionization of water contributes a significant amount of H+ ions in such a dilute solution. The total concentration of H+ ions is the sum of H+ ions from HCl and the ionization of water, which gives a pH of approximately 6.70 instead of 7.00. Therefore, we cannot ignore the contribution of H+ ions from the ionization of water while calculating the pH of this very dilute HCl solution.

Step-by-step solution

Write the chemical equation for the ionization of HCl and water

In this step, we'll write the chemical equation for the ionization of HCl in water and the ionization of water itself: \( \mathrm{HCl} \rightarrow \mathrm{H^+} + \mathrm{Cl^-} \) \( \mathrm{H_2O} \rightleftharpoons \mathrm{H^+} + \mathrm{OH^-} \)

Calculate the concentration of H+ ions

Using the given concentration of HCl, 1.00 x 10^-7 M, determine the concentration of H+ ions in the solution. Since HCl is a strong acid, it will dissociate completely. Thus, the concentration of H+ ions from HCl is also 1.00 x 10^-7 M. Also, consider the ionization of water: \( K_w = [\mathrm{H^+}] [\mathrm{OH^-}] = 1.0 \times 10^{-14} \) Since the solution is very dilute, the concentration of H+ ions coming from the ionization of water cannot be ignored. At 25°C, \(K_w = 1.0 \times 10^{-14}\) and pure water has equal concentrations of H+ and OH- ions, which are 1.00 x 10^-7 M each.

Calculate the total concentration of H+ ions

Add the H+ ion concentration from HCl and the ionization of water: \( [\mathrm{H^+}]_\text{total} = [\mathrm{H^+}]_\text{HCl} + [\mathrm{H^+}]_\text{H2O} \) \( [\mathrm{H^+}]_\text{total} = 1.00 \times 10^{-7} \mathrm{M} + 1.00 \times 10^{-7} \mathrm{M} \) \( [\mathrm{H^+}]_\text{total} = 2.00 \times 10^{-7} \mathrm{M} \)

Calculate the pH of the solution

Use the total concentration of H+ ions to calculate the pH: \( \mathrm{pH} = -\log_{10} [\mathrm{H^+}]_\text{total} \) \( \mathrm{pH} = -\log_{10} (2.00 \times 10^{-7}) \) \( \mathrm{pH} \approx 6.70\)

Explain the reason for the deviation from pH 7.00

From the above calculation, we see that the pH is approximately 6.70 instead of 7.00. The reason for this difference is that the ionization of water contributes a significant amount of H+ ions in such a dilute HCl solution. Therefore, we cannot ignore the contribution of H+ ions from the ionization of water while calculating the pH of this very dilute HCl solution.

Key concepts

ionization of water

When we talk about the ionization of water, we refer to water's ability to dissociate into ions. This process involves water molecules (\(\mathrm{H_2O}\)) splitting into hydrogen ions (protons, \(\mathrm{H^+}\)) and hydroxide ions (\(\mathrm{OH^-}\)). The equilibrium constant for this ionization is denoted as \(K_w\), which has a value of \(1.0 \times 10^{-14}\) at 25°C. This value is a result of multiplying the concentrations of \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) ions in pure water:

• At 25°C, in pure water, the concentrations of these ions are both \(1.0 \times 10^{-7}\) M.

Although water is a neutral solvent, these ion balances play a crucial role in determining the pH of solutions, especially in very dilute conditions. Understanding that water itself can contribute to the \(\mathrm{H^+}\) concentration is key in accurately predicting the pH of solutions.

strong acids

Strong acids, like hydrochloric acid (\(\mathrm{HCl}\)), are known for their ability to completely dissociate in water. This means when dissolved, they liberate their hydrogen ions fully into the solution, resulting in an equal molar concentration of \(\mathrm{H^+}\) ions as the original acid concentration:

• For example, a \(1.0 \times 10^{-7} \mathrm{M} \mathrm{HCl}\) solution results directly in a \(1.0 \times 10^{-7} \mathrm{M}\) concentration of \(\mathrm{H^+}\) ions from the acid.

However, in very dilute solutions, the \(\mathrm{H^+}\) ions from the water’s ionization cannot be neglected, as they have a similar concentration scale.
This interaction demonstrates why strong acids alone don't always determine a solution's pH. Instead, the overall ion balance, including contributions from water, must be considered to understand the full picture.

dilute solutions

Dilute solutions are those in which a small amount of solute is dissolved. Let's explore why very dilute solutions are unique, especially in the context of pH calculations:

• The \(\mathrm{H^+}\) concentration in dilute solutions of acids, like \(\mathrm{HCl}\), may closely match that of water’s self-ionization.
• Because of the similar concentration levels, both sources of \(\mathrm{H^+}\) ions — from the acid and from water ionization — must be added together to calculate the true \(\mathrm{H^+}\) concentration.

This scenario is highlighted when calculating the pH of very dilute \(\mathrm{HCl}\) solutions, where water’s ionization contributes measurably to the \(\mathrm{H^+}\) total.
Thus, it is essential to include all potential contributions to the \(\mathrm{H^+}\) concentration to determine an accurate pH. Ignoring the ionization of water in such cases could lead to incorrect calculations, as seen when the pH is significantly below 7, even for very dilute solutions of strong acids.