Problem 92 - Chemistry An Atoms First Approach

Question

What are the major species present in the following mixtures of bases? a. \(0.050 M \mathrm{NaOH}\) and \(0.050 \mathrm{M} \mathrm{LiOH}\) b. \(0.0010 M \mathrm{Ca}(\mathrm{OH})_{2}\) and \(0.020 \mathrm{M} \mathrm{RbOH}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

Step-by-Step Solution

Answer

The major species present in the mixtures are: a. \(\mathrm{Na}^+\), \(\mathrm{Li}^+\), \(\mathrm{OH}^-\) b. \(\mathrm{Ca}^{2+}\), \(\mathrm{Rb}^+\), \(\mathrm{OH}^-\) The hydroxide ion concentrations are: a. \(\left[\mathrm{OH}^-\right] = 0.100\ \mathrm{M}\) b. \(\left[\mathrm{OH}^-\right] = 0.022\ \mathrm{M}\) The pH values of the mixtures are: a. pH ≈ 13 b. pH ≈ 12.34

Step 1: Find the major species in each mixture

In each mixture, we have to consider the dissociation of each base in water. The base dissociation reaction for a general base is given by: \[ B + H_2O \rightarrow BH^+ + OH^- \] We can apply this reaction for both the mixtures separately to find the major species present. a. Mixture with 0.050 M NaOH and 0.050 M LiOH \[ \mathrm{NaOH} \rightarrow \mathrm{Na}^+ + \mathrm{OH}^- \] \[ \mathrm{LiOH} \rightarrow \mathrm{Li}^+ + \mathrm{OH}^- \] Major species present: \(\mathrm{Na}^+\), \(\mathrm{Li}^+\), \(\mathrm{OH}^-\) b. Mixture with 0.0010 M Ca(OH)2 and 0.020 M RbOH \[ \mathrm{Ca(OH)_2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^- \] \[ \mathrm{RbOH} \rightarrow \mathrm{Rb}^+ + \mathrm{OH}^- \] Major species present: \(\mathrm{Ca}^{2+}\), \(\mathrm{Rb}^+\), \(\mathrm{OH}^-\)

Step 2: Find the hydroxide ion concentration \(\left[ OH^- \right]\) for each mixture

Now, we need to find the total concentration of hydroxide ions for each mixture by summing up the contributions from each base. a. For the first mixture, Total \(\left[\mathrm{OH}^-\right] = [\mathrm{OH}^-]_{\mathrm{NaOH}} + [\mathrm{OH}^-]_{\mathrm{LiOH}} = 0.050 + 0.050 = 0.100\ \mathrm{M}\) b. For the second mixture, Total \(\left[\mathrm{OH}^-\right] = [\mathrm{OH}^-]_{\mathrm{Ca(OH)_2}} + [\mathrm{OH}^-]_{\mathrm{RbOH}} = 2 (0.0010) + 0.020 = 0.022\ \mathrm{M}\)

Step 3: Calculate the pOH and pH values for each mixture

Now, we can use the relationship between pOH and \(\left[ \mathrm{OH}^- \right]\) as well as the relationship between pH and pOH to find the pH of each mixture. The formula for pOH is: \[ \mathrm{pOH} = -\log_{10}[\mathrm{OH}^-] \] The relationship between pH and pOH is: \[ \mathrm{pH} + \mathrm{pOH} = 14 \] a. For the first mixture, pOH = \(-\log_{10}(0.100) = 1\) pH = \(14 - 1 = 13\) b. For the second mixture, pOH = \(-\log_{10}(0.022) \approx 1.66\) pH = \(14 - 1.66 \approx 12.34\) So, the pH values for the mixtures are approximately 13 and 12.34, respectively.

Key Concepts

Hydroxide Ion Concentration

Hydroxide ion concentration, denoted as \([\text{OH}^-]\), is a crucial aspect in understanding the strength and behavior of basic (alkaline) solutions. When bases dissolve in water, they dissociate to release hydroxide ions. This concentration indicates how many hydroxide ions are present in a solution. To calculate it, you must consider each base's contribution in a mixture.
In the given exercise, for instance, a mixture of \(0.050 \text{M NaOH}\) and \(0.050 \text{M LiOH}\) means that both compounds fully dissociate in water to release \([\text{OH}^-]\) ions. Therefore, to find the total \([\text{OH}^-]\), you simply add up the concentrations coming from each base:

• From NaOH: \(0.050\text{ M [OH]^-}\)
• From LiOH: \(0.050\text{ M [OH]^-}\)

Adding these provides a total hydroxide ion concentration: \(0.100\text{ M [OH]^-}\).
Similarly, for the mixture of \(0.0010 \text{M Ca(OH)}_2\) and \(0.020 \text{M RbOH}\), you account for the fact that one calcium hydroxide molecule provides two hydroxide ions. Hence, \(\text{Ca(OH)}_2\) contributes \(2 \times 0.0010 = 0.0020\text{ M [OH]^-}\), while RbOH contributes \(0.020\text{ M [OH]^-}\). The total \([\text{OH}^-]\) becomes \(0.022\text{ M}\).
Understanding hydroxide ion concentration is fundamental for predicting pH and understanding the solution's basic nature.

Base Dissociation

Base dissociation refers to the process where a base dissolves in water and separates into its constituent ions. This results in the release of hydroxide ions, which characterizes the solution as basic. For instance, sodium hydroxide (NaOH) and lithium hydroxide (LiOH) are strong bases, meaning they dissociate completely in aqueous solutions.
In detail, when NaOH dissolves in water, it breaks apart into \(\text{Na}^+\) and \(\text{OH}^-\). Similarly, LiOH dissociates into \(\text{Li}^+\) and \(\text{OH}^-\). These dissociations contribute directly to the basic nature of the solution because of the hydroxide ions produced:

• Strong bases dissociate entirely, offering full ionization and making them straightforward to handle in calculations for concentration.

Calcium hydroxide, \(\text{Ca(OH)}_2\), behaves slightly differently. Though it is not as soluble as NaOH, any part of it that does dissolve releases twice the amount of \(\text{OH}^-\) ions per molecule due to its \(\text{OH}\) groups. This needs to be accounted for when calculating total hydroxide ion concentration in solutions.
Knowing the dissociation properties of bases helps predict the ion presence in solutions and their relative strengths, which are essential for chemical reactions involving acids and bases.

pH Calculation

The pH of a solution is a measure of its acidity or basicity. It is inversely related to the pOH, which is calculated directly from the hydroxide ion concentration. To find the pH of a basic solution, you first determine the pOH using the formula: \[ \text{pOH} = -\log_{10}[\text{OH}^-] \]
Once pOH is known, pH is calculated using: \[ \text{pH} + \text{pOH} = 14 \]
For example, in a solution containing \(0.100 \text{ M [OH]^-}\), the pOH is calculated as:

• \(\text{pOH} = -\log_{10}(0.100) = 1\)

Then, using the relationship between pH and pOH:

• \(\text{pH} = 14 - 1 = 13\)

Therefore, this solution is highly basic.
In the case where \( [\text{OH}^-] = 0.022\text{ M} \), the calculation would be a bit different:

• \(\text{pOH} = -\log_{10}(0.022) \approx 1.66\)
• \(\text{pH} = 14 - 1.66 \approx 12.34\)

This pH indicates a less basic solution compared to the first.
The calculation steps demonstrate how changes in \([\text{OH}^-]\) impact pH, making this knowledge critical for applications ranging from laboratory experiments to industrial processes.