if equal volume of O.1 M `NH_4OH` and 0.1 M HCl are mixed, then the pH of resulting mixture will be `[Given pK_b (NH_4 OH)= 4.75, log 5=0.7)]`

Options

A. 8.72
B. 6.28
C. 7.28
D. 5.28

Answer

To find the pH of the resulting mixture when equal volumes of 0.1 M NH₄OH (ammonium hydroxide) and 0.1 M HCl (hydrochloric acid) are mixed, we can follow these steps:

### Step 1: Identify the Reaction
When NH₄OH, a weak base, is mixed with HCl, a strong acid, they react to form NH₄Cl (ammonium chloride) and water:
\[ \text{NH}_4\text{OH} + \text{HCl} \rightarrow \text{NH}_4\text{Cl} + \text{H}_2\text{O} \]

### Step 2: Determine Moles of Reactants
Since both solutions are 0.1 M and equal volumes are mixed, we can assume:
- Volume of NH₄OH = 1 L
- Volume of HCl = 1 L

Thus, the moles of each reactant are:
- Moles of NH₄OH = 0.1 mol
- Moles of HCl = 0.1 mol

### Step 3: Determine Moles of Products
The reaction will consume all of the NH₄OH and HCl, producing:
- Moles of NH₄Cl = 0.1 mol

### Step 4: Calculate the Concentration of NH₄Cl
The total volume after mixing is:
\[ \text{Total Volume} = 1 \text{ L} + 1 \text{ L} = 2 \text{ L} \]

Thus, the concentration of NH₄Cl in the resulting solution is:
\[ C = \frac{\text{Moles of NH}_4\text{Cl}}{\text{Total Volume}} = \frac{0.1 \text{ mol}}{2 \text{ L}} = 0.05 \text{ M} \]

### Step 5: Use the pH Formula for Acidic Salts
For acidic salts like NH₄Cl, we can use the formula:
\[ \text{pH} = \frac{1}{2} \left( pK_w - pK_b - \log C \right) \]

Where:
- \( pK_w = 14 \)
- \( pK_b = 4.75 \)
- \( C = 0.05 \text{ M} \)

### Step 6: Calculate the pH
First, we need to calculate \( \log C \):
\[ \log C = \log(0.05) = \log(5 \times 10^{-2}) = \log 5 + \log(10^{-2}) = 0.7 - 2 = -1.3 \]

Now substituting into the pH formula:
\[ \text{pH} = \frac{1}{2} \left( 14 - 4.75 - (-1.3) \right) \]
\[ = \frac{1}{2} \left( 14 - 4.75 + 1.3 \right) \]
\[ = \frac{1}{2} \left( 10.55 \right) \]
\[ = 5.275 \]

### Step 7: Final pH Value
Rounding this value gives us:
\[ \text{pH} \approx 5.28 \]

### Conclusion
The pH of the resulting mixture is approximately **5.28**.

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