Problem 108 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

The pH of the solution is calculated as \(pH = 14 - (-\log (2.5 \times 10^{-9}))\), which results in a value much higher than 7, indicating a basic solution. The answer makes sense, as it's a diluted solution of a strong base. However, the typical assumption that the \(\mathrm{OH^-}\) ion concentration comes only from the dissolution of the base may not be valid in this case due to the potential contribution from the autoionization of water in very dilute solutions.

Step 1: Calculate the concentration of \(\mathrm{OH^-}\) ions

Since \(\mathrm{NaOH}\) is a strong base, it completely dissociates in water. Therefore, the concentration of \(\mathrm{OH^-}\) ions is the same as the concentration of \(\mathrm{NaOH}\), which is given as \(2.5 \times 10^{-9}\,\mathrm{M}\).

Step 2: Calculate the \(\mathrm{pOH}\) of the solution

The \(\mathrm{pOH}\) of a solution is the negative logarithm of the \(\mathrm{OH^-}\) ion concentration in base 10. Hence, \[ \mathrm{pOH} = -\log (2.5 \times 10^{-9}) \]

Step 3: Calculate the \(\mathrm{pH}\) of the solution

Since the sum of \(\mathrm{pH}\) and \(\mathrm{pOH}\) is 14 at 25°C, we can find the \(\mathrm{pH}\) of the solution as follows: \[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - (-\log (2.5 \times 10^{-9})) \] Now, you can calculate the numerical value of \(\mathrm{pH}\).

Step 4: Does the answer make sense?

The \(\mathrm{pH}\) of the solution will likely be much higher than 7, indicating it's a basic solution. Also, since the \(\mathrm{pOH}\) value is significantly smaller compared to that of a neutral solution, you can assume the solution is very dilute.

Step 5: Discuss the assumption that may not be valid

The assumption we usually make is that the \(\mathrm{OH^-}\) ion concentration only comes from the dissolution of the base. However, in very dilute solutions, we cannot neglect the effect of the autoionization of water, which also contributes to the \(\mathrm{OH^-}\) ion concentration. In such cases, our normal assumption may not be valid, and our calculated \(\mathrm{pH}\) may not completely represent the acidity/basicity of the solution.

Key Concepts

Strong Base Dissociation

When a strong base like sodium hydroxide (NaOH) is dissolved in water, it completely dissociates into ions. This means that each molecule of NaOH separates into a sodium ion (\text{Na}^+) and a hydroxide ion (\text{OH}^-). Unlike weak bases that only partially dissociate, the complete dissociation of strong bases results in a solution where the concentration of hydroxide ions is equal to the initial concentration of the base.

For instance, if a solution contains a concentration of NaOH at \(2.5 \times 10^{-9}\)M, the concentration of hydroxide ions will be exactly the same, assuming no other sources of \text{OH}^{-}. This can be expressed as: \[ [\text{OH}^-] = [\text{NaOH}] \
\] The full dissociation is a key concept because it simplifies the process of calculating the pH of strong base solutions, meaning that we don't have to consider equilibrium constants or partial reactions as we would with weak bases.

POH Calculation

pOH is an important concept in chemistry that provides a measure of the alkalinity of a solution. It's the negative logarithm to the base 10 of the hydroxide ion concentration (\text{OH}^-). The formula to calculate pOH is given by: \[ \text{pOH} = -\log[\text{OH}^-] \
\] In the case of our example with a NaOH concentration of \(2.5 \times 10^{-9}\)M, the pOH calculation directly translates the hydroxide ion concentration into a scale that is easier to compare across different solutions. The pOH scale generally runs from 0 to 14, where lower values indicate higher alkalinity. Remember that pOH and pH are inversely related in a way that allows us to find the pH of the solution with the additional knowledge that at 25°C, the sum of the pH and pOH always equals 14.

Autoionization of Water

Water is a unique solvent that undergoes autoionization, a process where two water molecules react to form a hydronium ion (\text{H}_3\text{O}^+) and a hydroxide ion (\text{OH}^-). This reaction is described by the self-dissociation equation: \[ 2\text{H}_2\text{O} \leftrightarrow \text{H}_3\text{O}^+ + \text{OH}^- \
\] Under standard conditions, the concentrations of hydronium and hydroxide ions from water’s autoionization are very low (around \(10^{-7}\)M). However, in the context of very dilute solutions of strong bases, like the example with NaOH at a concentration of \(2.5 \times 10^{-9}\)M, the contribution of hydroxide ions from water's autoionization can no longer be ignored.

In these cases, simply adding the base's hydroxide ion concentration to that of autoionized water can offer a more accurate measure of the total hydroxide ion concentration. The presence of additional hydroxide ions can slightly alter the pH calculation, leading to a higher pH value than that calculated considering only the strong base's contribution. Thus, forgetting autoionization of water in dilute solutions can lead to incorrect pH estimations.