Problem 32 - Chemistry The Molecular Science

Question

The \(\mathrm{pH}\) of a \(\mathrm{Ba}(\mathrm{OH})_{2}\) solution is 10.66 at \(25^{\circ} \mathrm{C}\). Calculate the hydroxide ion concentration of this solution. If the solution volume is \(250 . \mathrm{mL},\) calculate the mass \((\mathrm{g})\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) that was used to make this solution.

Step-by-Step Solution

Answer

[OH⁻] = 4.57×10⁻⁴ M; Mass of Ba(OH)₂ = 9.77×10⁻³ g.

Step 1: Finding the pOH

Use the relationship between pH and pOH: \( \text{pH} + \text{pOH} = 14 \). Given that the pH is 10.66, calculate pOH as follows:\[ \text{pOH} = 14 - 10.66 = 3.34 \]

Step 2: Calculating Hydroxide Ion Concentration

The hydroxide ion concentration \([\text{OH}^-]\) is calculated from the pOH using the formula \([\text{OH}^-] = 10^{-\text{pOH}}\). Substitute the value of pOH calculated in Step 1:\[ [\text{OH}^-] = 10^{-3.34} \approx 4.57 \times 10^{-4} \, \text{M} \]

Step 3: Adjusting for Ba(OH)2 Ionization

Each \( \text{Ba(OH)}_2 \) dissociates into one \( \text{Ba}^{2+} \) and two \( \text{OH}^- \) ions in solution. So, the concentration of \( \text{Ba(OH)}_2 \) is:\[ [\text{Ba(OH)}_2] = \frac{1}{2} \times [\text{OH}^-] = \frac{1}{2} \times 4.57 \times 10^{-4} = 2.28 \times 10^{-4} \, \text{M} \]

Step 4: Finding Moles of Ba(OH)2

Use the molarity and volume of the solution to calculate moles. Moles \( n = [\text{Ba(OH)}_2] \times \text{Volume in liters} \). The solution volume is 250 mL, which is 0.250 L:\[ n = 2.28 \times 10^{-4} \, \text{M} \times 0.250 \, \text{L} = 5.70 \times 10^{-5} \, \text{mols} \]

Step 5: Calculating Mass of Ba(OH)2

Find the molar mass of \( \text{Ba(OH)}_2 \) by adding up the atomic masses: Ba (137.3 g/mol), O (2×16.00 g/mol), H (2×1.01 g/mol). Total molar mass = 137.3 + 32.00 + 2.02 = 171.32 g/mol.Calculate the mass:\[ \text{Mass} = n \times \text{Molar Mass} = 5.70 \times 10^{-5} \, \text{mols} \times 171.32 \, \text{g/mol} = 9.77 \times 10^{-3} \, \text{g} \]

Key Concepts

pH Calculations

The concept of pH is central to the study of acid-base equilibria, providing insight into how acidic or basic a solution is. pH stands for 'potential of Hydrogen' and is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It is calculated using the formula:

• \( \text{pH} = -\log[\text{H}^+] \)

where \([\text{H}^+]\) represents the concentration of hydrogen ions in the solution.
Lower pH values signify acidic solutions (high concentration of hydrogen ions), and higher pH values indicate basic solutions (low concentration of hydrogen ions). A neutral solution has a pH of 7, such as pure water at 25°C. The pH scale typically ranges from 0 to 14, although values beyond this range are possible under certain conditions.

In the given problem, the pH of the solution is 10.66 suggesting that it is basic. To further explore the concentration of ions in the solution, we need to calculate the pOH using the relationship:

• \( \text{pH} + \text{pOH} = 14 \)

This equation helps to determine the pOH from a known pH, guiding us to the concentration of hydroxide ions in the solution.

Hydroxide Ion Concentration

Determining the hydroxide ion concentration is a pivotal step in understanding the basic nature of a solution. In a basic solution, the concentration of OH⁻ ions is crucial as it dictates the solution's pH balance. To calculate the concentration of hydroxide ions, we first need to calculate pOH, which is obtained from the equation:

• \( \text{pOH} = 14 - \text{pH} \)

Once pOH is known, the hydroxide ion concentration \([\text{OH}^-]\) can be calculated using the formula:

• \([\text{OH}^-] = 10^{-\text{pOH}}\)

In this exercise, after finding that the pOH is 3.34, we substitute it into the formula to find \([\text{OH}^-] = 10^{-3.34}\), which equals approximately \(4.57 \times 10^{-4} M\).
This method leverages the logarithmic nature of the pH and pOH scales, which means that even small changes in ion concentration imply significant changes in pH and pOH numbers. These calculations are essential for predicting the behavior of chemical solutions.

Molarity

Molarity is a fundamental concept in chemistry used to express the concentration of a solution. It is defined as the number of moles of a solute per liter of solution, calculated with the formula:

• \( M = \frac{n}{V} \)

where \(n\) is the number of moles of the solute and \(V\) is the volume of the solution in liters.
In acid-base chemistry, molarity allows calculation of how concentrated a solution is. For instance, knowing the concentration of \(\text{Ba(OH)}_2\) can help determine the amount of \(\text{OH}^-\) ions it contributes upon dissociation. This information is crucial for understanding the solution's chemical properties.

In our example, with a 250 mL solution, we convert it into liters (0.250 L), and then use the molarity equation to find that the moles of \(\text{Ba(OH)}_2\) are \(2.28 \times 10^{-4} M\). The mass of \(\text{Ba(OH)}_2\) can then be calculated by using its molar mass (171.32 g/mol). Understanding molarity bridges the gap between theoretical calculations and practical applications, such as preparing solutions in laboratory settings.