To pH value of decinormal solution of `NH_(4)OH` which is 20% ionised is

Options

A. `13.30`
B. `14.70`
C. `12.30`
D. `12.95`

Answer

To find the pH value of a decinormal solution of ammonium hydroxide (NH₄OH) that is 20% ionized, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Concentration**:
- The solution is decinormal, which means the initial concentration (C) of NH₄OH is 0.1 N (or 0.1 M since NH₄OH is monovalent).

2. **Determine the Degree of Ionization**:
- Given that the solution is 20% ionized, we can express this as α = 20/100 = 0.20.

3. **Calculate the Concentration of OH⁻ Ions**:
- The concentration of OH⁻ ions at equilibrium can be calculated using the formula:
\[ [OH⁻] = C \times \alpha = 0.1 \times 0.20 = 0.02 \, \text{M} \]

4. **Calculate pOH**:
- The pOH can be calculated using the formula:
\[ pOH = -\log[OH⁻] = -\log(0.02) \]
- We can express 0.02 as \(2 \times 10^{-2}\):
\[ pOH = -\log(2 \times 10^{-2}) = -\log(2) - (-2) \]
- This simplifies to:
\[ pOH = 2 - \log(2) \]

5. **Find the Value of log(2)**:
- The value of log(2) is approximately 0.3010.
- Therefore:
\[ pOH = 2 - 0.3010 = 1.6990 \approx 1.7 \]

6. **Calculate pH**:
- The pH can be calculated using the relation:
\[ pH = 14 - pOH \]
- Substituting the value of pOH:
\[ pH = 14 - 1.7 = 12.3 \]

### Final Answer:
The pH value of the decinormal solution of NH₄OH which is 20% ionized is approximately **12.3**.