20mL of `0.1M(H_(2)SO_(4))`solution is added to 30mL of `0.2M(NH_(4)OH)` soulution. The `._(p)H` of the resultant mixture is : `[._(p)k_(b) " of " NH_(4)OH=4.7]`.

Options

A. 9.4
B. 9
C. 5
D. 5.2

Explanation

To find the pH of the resultant mixture when 20 mL of 0.1 M H₂SO₄ is added to 30 mL of 0.2 M NH₄OH, we can follow these steps:

### Step 1: Calculate the number of millimoles of H₂SO₄ and NH₄OH

1. **For H₂SO₄:**
\[ \text{Millimoles of H₂SO₄} = \text{Volume (mL)} \times \text{Molarity (M)} = 20 \, \text{mL} \times 0.1 \, \text{M} = 2 \, \text{mmol} \]

2. **For NH₄OH:**
\[ \text{Millimoles of NH₄OH} = \text{Volume (mL)} \times \text{Molarity (M)} = 30 \, \text{mL} \times 0.2 \, \text{M} = 6 \, \text{mmol} \]

### Step 2: Determine the moles of H⁺ produced by H₂SO₄

Since H₂SO₄ is a strong acid, it dissociates completely:
\[ \text{1 mole of H₂SO₄} \rightarrow \text{2 moles of H⁺} \]
Thus, for 2 mmol of H₂SO₄:
\[ \text{Moles of H⁺} = 2 \, \text{mmol} \times 2 = 4 \, \text{mmol} \]

### Step 3: Reaction between H⁺ and NH₄OH

The reaction between H⁺ and NH₄OH can be represented as:
\[ \text{NH₄OH} + \text{H⁺} \rightarrow \text{NH₄⁺} + \text{H₂O} \]

Initially, we have:
- NH₄OH: 6 mmol
- H⁺: 4 mmol

After the reaction:
- Remaining NH₄OH = \(6 \, \text{mmol} - 4 \, \text{mmol} = 2 \, \text{mmol}\)
- Produced NH₄⁺ = 4 mmol

### Step 4: Calculate concentrations in the final solution

The total volume of the mixture is:
\[ \text{Total Volume} = 20 \, \text{mL} + 30 \, \text{mL} = 50 \, \text{mL} = 0.050 \, \text{L} \]

1. **Concentration of NH₄⁺:**
\[ [\text{NH₄⁺}] = \frac{4 \, \text{mmol}}{50 \, \text{mL}} = \frac{4 \times 10^{-3} \, \text{mol}}{0.050 \, \text{L}} = 0.08 \, \text{M} \]

2. **Concentration of NH₄OH:**
\[ [\text{NH₄OH}] = \frac{2 \, \text{mmol}}{50 \, \text{mL}} = \frac{2 \times 10^{-3} \, \text{mol}}{0.050 \, \text{L}} = 0.04 \, \text{M} \]

### Step 5: Calculate pOH using the Henderson-Hasselbalch equation

Using the formula for a basic buffer:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Where:
- \(\text{pK}_b = 4.7\)
- \([\text{Salt}] = [\text{NH₄⁺}] = 0.08 \, \text{M}\)
- \([\text{Base}] = [\text{NH₄OH}] = 0.04 \, \text{M}\)

Substituting the values:
\[ \text{pOH} = 4.7 + \log\left(\frac{0.08}{0.04}\right) = 4.7 + \log(2) \]
Using \(\log(2) \approx 0.3\):
\[ \text{pOH} = 4.7 + 0.3 = 5.0 \]

### Step 6: Calculate pH from pOH

Using the relation:
\[ \text{pH} + \text{pOH} = 14 \]
Thus:
\[ \text{pH} = 14 - 5.0 = 9.0 \]

### Final Answer:
The pH of the resultant mixture is approximately **9.0**.

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