To `1.0L` solution containing `0.1mol` each of `NH_(3)` and `NH_(4)C1,0.05mol NaOH` is added. The change in `pH` will be `(pK_(a)` for `CH_(3)COOH = 4.74)`

Options

A. `0.30`
B. `-0.30`
C. `0.48`
D. `-0.48`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Initial Solution
We have a solution containing:
- 0.1 moles of NH₃ (weak base)
- 0.1 moles of NH₄Cl (conjugate acid)

### Step 2: Calculate Initial pH
Using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
Here, \(\text{pK}_a\) for acetic acid is given as 4.74. Since we have equal concentrations of the base and the acid:
\[ \text{pH} = 4.74 + \log\left(\frac{0.1}{0.1}\right) = 4.74 + \log(1) = 4.74 + 0 = 4.74 \]
Thus, the initial pH is 4.74.

### Step 3: Add NaOH
Now, we add 0.05 moles of NaOH (a strong base) to the solution. NaOH will react with NH₄⁺ ions from NH₄Cl:
\[ \text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O} \]

### Step 4: Calculate New Concentrations
- Initial concentration of NH₄⁺ = 0.1 moles
- After reaction with 0.05 moles of NaOH:
\[ [\text{NH}_4^+] = 0.1 - 0.05 = 0.05 \text{ moles} \]
- The concentration of NH₃ will increase by 0.05 moles:
\[ [\text{NH}_3] = 0.1 + 0.05 = 0.15 \text{ moles} \]

### Step 5: Calculate Final pH
Using the Henderson-Hasselbalch equation again with the new concentrations:
\[ \text{pH}_{\text{final}} = 4.74 + \log\left(\frac{0.15}{0.05}\right) \]
Calculating the ratio:
\[ \frac{0.15}{0.05} = 3 \]
Thus:
\[ \text{pH}_{\text{final}} = 4.74 + \log(3) \]
Using \(\log(3) \approx 0.477\):
\[ \text{pH}_{\text{final}} = 4.74 + 0.477 \approx 5.22 \]

### Step 6: Calculate Change in pH
Now, we find the change in pH:
\[ \Delta \text{pH} = \text{pH}_{\text{final}} - \text{pH}_{\text{initial}} = 5.22 - 4.74 = 0.48 \]

### Final Answer
The change in pH is **0.48**.

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