Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

The \(\mathrm{pH}\) of solution made by mixing \(50 \mathrm{~mL}\) of \(0.01 \mathrm{M}\) barium hydroxide solution with \(50 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) is (a) \(3.0\) (b) \(6.0\) (c) \(12.0\) (d) \(15.0\)

Step-by-Step Solution

Answer

(c) 12.0

Step 1: Identify the Initial Concentration

We start with \(0.01\, \text{M}\) barium hydroxide \((\text{Ba(OH)}_2)\). Since barium hydroxide dissociates completely in water, it provides 2 hydroxide ions \((\text{OH}^-)\) per molecule. Thus, the concentration of \(\text{OH}^-\) ions is \(0.02\, \text{M}\) in the original solution.

Step 2: Adjust for Volume Change

When mixed with 50 mL of water, the total volume is \(100 \text{ mL} = 0.1 \text{ L}\). Using dilution formula \(C_1V_1 = C_2V_2\), where \(C_1 = 0.02\, \text{M}\), \(V_1 = 0.05\, \text{L}\), \(V_2 = 0.1\, \text{L}\), find \(C_2\): \[ 0.02 \times 0.05 = C_2 \times 0.1 \] \[ C_2 = \frac{0.02 \times 0.05}{0.1} = 0.01 \, \text{M} \] The concentration of \(\text{OH}^-\) ions is now \(0.01 \text{ M}\).

Step 3: Calculate pOH

pOH is calculated using the formula \( \text{pOH} = -\log[\text{OH}^-] \). With \([\text{OH}^-] = 0.01 \text{ M}\), \[ \text{pOH} = -\log(0.01) = 2 \]

Step 4: Calculate pH

The relationship between \(\text{pH}\) and \(\text{pOH}\) is given by: \[ \text{pH} + \text{pOH} = 14 \] Substituting the value of \(\text{pOH}\), we get: \[ \text{pH} = 14 - 2 = 12 \] Therefore, \(\text{pH} = 12\).

Step 5: Select Correct Answer

The calculated \(\text{pH}\) is 12, which corresponds to option (c).

Key Concepts

Barium Hydroxide Dissociation

Understanding how barium hydroxide dissociates in water is the first step in calculating the pH of a solution. Barium hydroxide, represented as \( ext{Ba(OH)}_2\), is a strong base, which means that it dissociates completely when added to water. Each molecule of barium hydroxide releases two hydroxide ions (\( ext{OH}^-\)).
For instance, when \( 0.01 \, ext{M} \) barium hydroxide is dissolved, it results in a \( 0.02 \, ext{M} \) concentration of \( ext{OH}^-\) ions. This is because each formula unit of \( ext{Ba(OH)}_2\) provides two \( ext{OH}^-\) ions:

• Barium hydroxide contributes barium ions \((\text{Ba}^{2+})\) and hydroxide ions \((\text{OH}^-)\) to the solution.
• The complete dissociation into ions signifies that there are twice as many hydroxide ions available per mole of barium hydroxide.

Through understanding this dissociation, we can now proceed to calculate the concentration of hydroxide ions in any given situation involving barium hydroxide solutions.

Concentration Calculation

Finding the concentration after a volume change is a crucial skill in chemistry. When barium hydroxide is mixed with water, the solution's total volume changes, affecting the concentration of ions. Initially, we have \(50 \, ext{mL} \) of a \( 0.01 \, ext{M} \) barium hydroxide solution mixed with an equal volume of water, bringing the total volume to \(100 \, ext{mL}\).
To determine the new concentration of \( ext{OH}^-\), we use the dilution formula:
\[ C_1V_1 = C_2V_2 \]

• \(C_1\) is the initial concentration of \( ext{OH}^-\), which is \(0.02 \, ext{M}\).
• \(V_1\) is the initial volume, \(0.05 \, ext{L}\) (or \(50 \, ext{mL}\)).
• \(V_2\) is the final volume, \(0.1 \, ext{L}\) (or \(100 \, ext{mL}\)).

Performing the calculation gives us \(C_2\), the new concentration of \( ext{OH}^-\), as \(0.01 \, ext{M}\). This calculation ensures that the concentration is accurately adjusted for the increased volume.

pOH and pH Relationship

The relationship between \( ext{pOH}\) and \( ext{pH}\) is fundamental in chemistry, especially when dealing with acidic and basic solutions. Knowing the concentration of \( ext{OH}^-\) ions allows us to calculate \( ext{pOH}\) using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]

With the concentration of \( ext{OH}^-\) being \(0.01 \, ext{M}\), the \( ext{pOH}\) is calculated as 2. This leads to finding the \( ext{pH}\) through the relationship:
\[ \text{pH} + \text{pOH} = 14 \]

• Solving for \(\text{pH}\) gives us \(\text{pH} = 14 - 2 = 12\).
• The solution's \(\text{pH}\) of 12 indicates a basic solution, aligning well with the complete dissociation of barium hydroxide.

This calculation shows the interplay between \( ext{pH}\) and \( ext{pOH}\), illustrating how basic solutions have high \( ext{OH}^-\) concentrations reflected in lower \( ext{pOH}\) values, consequently raising the \( ext{pH}\).