Problem 14 - Chemistry and Chemical Reactivity

Question

The \(\mathrm{pH}\) of a solution of \(\mathrm{Ba}(\mathrm{OH})_{2}\) is 10.66 at \(25^{\circ} \mathrm{C}\) What is the hydroxide ion concentration in the solution? If the solution volume is \(125 \mathrm{mL},\) what mass of \(\mathrm{Ba}(\mathrm{OH})_{2}\) must have been dissolved?

Step-by-Step Solution

Answer

[OH^-] = 4.57 × 10^-4 M; dissolved mass = 4.89 mg.

Step 1: Understand the pH and pOH Relationship

The relation between pH and pOH in a solution is given by the equation: \( \text{pH} + \text{pOH} = 14 \) at \(25^{\circ} \text{C}\). Since the pH is 10.66, we can find the pOH by calculating \( 14 - 10.66 = 3.34 \).

Step 2: Calculate Hydroxide Ion Concentration

The hydroxide ion concentration \([\text{OH}^-]\) can be found using the formula \( \text{pOH} = -\log[\text{OH}^-] \). Thus, \([\text{OH}^-] = 10^{-\text{pOH}} = 10^{-3.34} \approx 4.57 \times 10^{-4} \text{ M}\).

Step 3: Relate to Barium Hydroxide Dissociation

Barium hydroxide \( \text{Ba(OH)}_2 \) dissociates as \( \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^- \). As a result, 1 mole of \( \text{Ba(OH)}_2 \) produces 2 moles of \( \text{OH}^- \). Thus, the concentration of \( \text{Ba(OH)}_2 \) is half the concentration of \( \text{OH}^-\): \( \text{[Ba(OH)}_2] = \frac{[\text{OH}^-]}{2} = \frac{4.57 \times 10^{-4}}{2} = 2.285 \times 10^{-4} \text{ M}\).

Step 4: Calculate Moles of Ba(OH)_2

Now, determine the moles of \( \text{Ba(OH)}_2 \) in 125 mL of solution: \( 0.125 \text{ L} \times 2.285 \times 10^{-4} \text{ M} = 2.85625 \times 10^{-5} \text{ moles} \).

Step 5: Determine Mass of Ba(OH)_2

Calculate the mass of \( \text{Ba(OH)}_2 \) using its molar mass \( (\text{137.33 (Ba)} + 2(16.00 \text{(O)}) + 2(1.01 \text{(H)}) = 171.35 \text{ g/mol}) \). Thus, mass \( = 2.85625 \times 10^{-5} \times 171.35 \approx 4.89 \times 10^{-3} \text{ g} \).

Key Concepts

pH calculation

The concept of pH is essential in understanding the acidity or basicity of a solution. The pH scale ranges from 0 to 14, where lower values indicate acidic solutions, and higher values suggest basic or alkaline solutions. The equation relating pH and pOH is given by:\[ \text{pH} + \text{pOH} = 14 \]This relationship holds true at 25°C, a common temperature for most laboratory conditions. Essentially, knowing the pH of a solution allows us to calculate the pOH. In the exercise, we find that a pH of 10.66 indicates a basic solution. By substituting the pH into the equation:\[ \text{pOH} = 14 - 10.66 = 3.34 \]This calculation is the first step in determining the concentration of hydroxide ions, which indicates how many hydroxide ions are present in the solution.

dissociation in water

Chemical compounds like barium hydroxide \((\text{Ba(OH)}_2)\) undergo a process called dissociation when dissolved in water. This is where these compounds break into their ionic components. For barium hydroxide, the dissociation can be expressed as:\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\, \text{OH}^- \]This equation shows that each mole of \(\text{Ba(OH)}_2\) produces one mole of \(\text{Ba}^{2+}\) ions and two moles of hydroxide ions \(\text{OH}^-\). Therefore, the concentration of hydroxide ions is twice that of the barium hydroxide concentration.Using the hydroxide ion concentration calculated from the pOH, we can infer the concentration of barium hydroxide:\[ \text{[Ba(OH)}_2] = \frac{[\text{OH}^-]}{2} \]It's crucial to understand this dissociation because it helps predict the behavior of the solution and calculate how much compound is necessary to achieve a desired concentration.

molar mass calculation

Molar mass is the weight of one mole of a given substance and is expressed in g/mol. To calculate the molar mass of barium hydroxide \(\text{Ba(OH)}_2\), we sum the atomic masses of its constituent atoms:- Barium (Ba): 137.33 g/mol- Oxygen (O): 16.00 g/mol (since there are two, we multiply by 2)- Hydrogen (H): 1.01 g/mol (similarly, multiply by 2)Thus, the molar mass calculation becomes:\[ 137.33 + (2 \times 16.00) + (2 \times 1.01) = 171.35 \text{ g/mol} \]This number is used to convert moles to grams in the final step of the exercise. Knowing the molar mass, we can easily calculate the mass of the compound in a specific volume of solution using this formula:\[ \text{mass} = \text{moles} \times \text{molar mass} \]In this way, students can translate between amounts of substance in moles and measurable quantities in the laboratory context.