Question

What is the pH of a solution that is \(2.5 \times 10^{-9} \mathrm{M}\) in \(\mathrm{NaOH}\) ? Does your answer make sense? What assumption do we normally make that is not valid in this case?

Short answer

The pH of a solution with a given concentration of \(2.5 \times 10^{-9} \mathrm{M}\) in \(\mathrm{NaOH}\) is found to be 5.4, which appears to be slightly acidic, which doesn't make sense since we started with a base solution of NaOH. The reason for this discrepancy is our assumption that the concentration of \(\mathrm{OH^-}\) only comes from the dissociation of \(\mathrm{NaOH}\). In reality, the self-ionization of water should also be considered in such cases when the concentration of dissolved ions is relatively small and comparable to the self-ionization concentration.

Step-by-step solution

Determine the hydroxide ion concentration

Since \(\mathrm{NaOH}\) is a strong base, it dissociates completely into its ions in the solution: \[ \mathrm{NaOH} \rightarrow \mathrm{Na^+} + \mathrm{OH^-} \] The concentration of \(\mathrm{OH^-}\) ions in the solution will be equal to the concentration of \(\mathrm{NaOH}\). So, we have: \[ \mathrm{[OH^-]} = 2.5 \times 10^{-9} \,\mathrm{M} \]

Calculate pOH

Now, we will use the relation: \[ \mathrm{pOH} = -\log\mathrm{[OH^-]} \] to find pOH: \[ \mathrm{pOH} = -\log{(2.5 \times 10^{-9})} \] Using a calculator, we get: \[ \mathrm{pOH} \approx 8.6 \]

Calculate pH

Now, we will use the relationship between pH and pOH: \[ \mathrm{pH} + \mathrm{pOH} = 14 \] to find pH: \[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 8.6 \] We get: \[ \mathrm{pH} = 5.4 \]

Analyze the result

The pH value we obtained is 5.4, which is slightly acidic. However, this doesn't make sense since we started with a base solution of \(\mathrm{NaOH}\). The reason for this discrepancy is our assumption that the concentration of \(\mathrm{OH^-}\) only comes from the dissociation of \(\mathrm{NaOH}\). In practice, the water present in the solution self-ionizes, and this self-ionization must be considered when the concentration of dissolved ions is relatively small and comparable to the self-ionization concentration. Since \(\mathrm{[OH^-]}\) is similar to the self-ionization concentration of water (\(K_w/[H^+]\)), the assumption that the concentration of \(\mathrm{OH^-}\) only comes from the dissociation of \(\mathrm{NaOH}\) is not valid in this case. To accurately calculate the pH, we would need to take into account the self-ionization of water and solve for the equilibrium concentrations of \(\mathrm{OH^-}\) and \(\mathrm{H^+}\).

Key concepts

Hydroxide Ion Concentration

Understanding how the concentration of hydroxide ions (\(\mathrm{OH^-}\)) in a solution is calculated is crucial. When you have a strong base like sodium hydroxide (\(\mathrm{NaOH}\)), it dissociates entirely into its ions in water. This means that for every molecule of \(\mathrm{NaOH}\) dissolved, you get an equal number of \(\mathrm{OH^-}\) ions.

- For instance, if the \(\mathrm{NaOH}\) concentration is \(2.5 \times 10^{-9} \mathrm{M}\), then: \(\mathrm{[OH^-]} = 2.5 \times 10^{-9} \mathrm{M}\)

Complete dissociation is characteristic of strong bases, meaning all added base molecules split to form ions.

Self-Ionization of Water

Water is a unique molecule because it can ionize on its own. In pure water, a slight ionization occurs, forming equal concentrations of hydrogen ions (\(\mathrm{H^+}\)) and hydroxide ions (\(\mathrm{OH^-}\)). This is known as the self-ionization of water:

\[\mathrm{2H_2O \rightleftharpoons H_3O^+ + OH^-} \]

In neutral solutions at 25°C, the concentration of \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) are both \(1.0 \times 10^{-7} \mathrm{M}\). The ion product of water (\(K_w\)) is \(1.0 \times 10^{-14}\) at this temperature.

This self-ionization impacts pH calculations when the concentration of added ions, like \(\mathrm{OH^-}\) from \(\mathrm{NaOH}\), is low. In such cases, the ions from water's self-ionization become significant. Always consider this ionization when dissolved ion concentration is comparable to \(1.0 \times 10^{-7} \mathrm{M}\).

Strong Base Dissociation

Strong base dissociation is the process by which strong bases like \(\mathrm{NaOH}\) dissolve completely in water and break into their ions. This makes strong bases excellent sources of \(\mathrm{OH^-}\) ions.

- \(\mathrm{NaOH}\) dissociates completely into \(\mathrm{Na^+}\) and \(\mathrm{OH^-}\) ions: \[\mathrm{NaOH \rightarrow Na^+ + OH^-}\]

Understanding this complete dissociation helps us calculate the ion concentration exactly based on how much base is present. It's essential for predicting the basic nature of solutions. However, in highly dilute solutions, with concentrations comparable to the self-ionization of water, other factors like water's self-ionization need consideration.

Such complete dissolution ensures that pH levels can be accurately calculated for everyday concentrations, as long as they are not too low as to warrant consideration of the self-ionization effect. Consider the self-ionization impact whenever you work with very low concentration solutions.