Given that ( `ohm^(-1) cm^(2)eq^(-1)), T=298 K`
`{:(lambda_(E)^(oo) f o r Ba(OH)_(2)=228.8),(lambda_(E)^(oo) fo r BaCl_(2)=120.3),(lambda_(E)^(oo) for NH_(4)Cl=129.8):}|{:("specific conductance"),("for "0.2N NH_(4)OH" solution"),(is 4.766xx10^(-4)ohm^(-1)cm^(-1)):}`
then value of pH of the solution of `NH_(4)OH` will be nearly

Options

A. `9.2`
B. `11.3`
C. `12.1`
D. `7.9`

Explanation

To find the pH of a 0.2 N NH₄OH solution, we will follow these steps:

### Step 1: Calculate the Equivalent Conductivity (Λ) of the NH₄OH solution
The equivalent conductivity (Λ) can be calculated using the formula:

\[ \Lambda = \frac{k \times 1000}{N} \]

Where:
- \( k \) = specific conductance of the solution = \( 4.76 \times 10^{-4} \, \text{ohm}^{-1} \text{cm}^{-1} \)
- \( N \) = normality of the solution = 0.2 N

Substituting the values:

\[ \Lambda = \frac{4.76 \times 10^{-4} \times 1000}{0.2} = \frac{4.76 \times 10^{-1}}{0.2} = 2.38 \, \text{s cm}^2 \text{eq}^{-1} \]

### Step 2: Calculate the Equivalent Conductivity at Infinite Dilution (Λ°)
The equivalent conductivity at infinite dilution for NH₄OH can be calculated using the given data:

\[ \Lambda^\circ_{\text{NH}_4\text{OH}} = \frac{1}{2} \Lambda_{\text{Ba(OH)}_2} + \Lambda_{\text{NH}_4\text{Cl}} - \frac{1}{2} \Lambda_{\text{BaCl}_2} \]

Substituting the values:

- \( \Lambda_{\text{Ba(OH)}_2} = 228.8 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \)
- \( \Lambda_{\text{NH}_4\text{Cl}} = 129.8 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \)
- \( \Lambda_{\text{BaCl}_2} = 120.3 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \)

Calculating:

\[ \Lambda^\circ_{\text{NH}_4\text{OH}} = \frac{1}{2} \times 228.8 + 129.8 - \frac{1}{2} \times 120.3 \]
\[ = 114.4 + 129.8 - 60.15 = 184.05 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \]

### Step 3: Calculate the Degree of Dissociation (α)
The degree of dissociation (α) can be calculated using the formula:

\[ \alpha = \frac{\Lambda}{\Lambda^\circ} \]

Substituting the values:

\[ \alpha = \frac{2.38}{184.05} \approx 0.01293 \approx 0.013 \]

### Step 4: Calculate the Concentration of OH⁻ ions
From the dissociation of NH₄OH:

\[ \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

Let the initial concentration of NH₄OH be \( C = 0.2 \, \text{N} \). At equilibrium, the concentration of OH⁻ will be:

\[ [\text{OH}^-] = C \cdot \alpha = 0.2 \times 0.013 = 2.6 \times 10^{-3} \, \text{N} \]

### Step 5: Calculate pOH and then pH
Using the concentration of OH⁻ ions, we can find pOH:

\[ \text{pOH} = -\log([\text{OH}^-]) = -\log(2.6 \times 10^{-3}) \approx 2.58 \]

Now, using the relation \( \text{pH} + \text{pOH} = 14 \):

\[ \text{pH} = 14 - \text{pOH} = 14 - 2.58 = 11.42 \]

### Final Answer
The pH of the 0.2 N NH₄OH solution is approximately **11.42**.

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