Calculate the change in pH of 1 litre buffer solution containing `0.1` mole each of `NH_(3)` and `NH_(4)CI` upon addition of:
(i) `0.02` mole of dissolved NaOH.
Assume no change in volume. `K_(NH_(3))=1.8xx10^(-5)`

Answer

To solve the problem step by step, we will calculate the initial pH of the buffer solution, then determine the changes in the concentrations of the components after the addition of NaOH, and finally calculate the new pH and the change in pH.

### Step 1: Calculate the initial pH of the buffer solution

The buffer solution consists of 0.1 moles of NH₃ (base) and 0.1 moles of NH₄Cl (salt). We can use the Henderson-Hasselbalch equation to find the initial pH:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]

First, we need to calculate pKₐ from Kb:

Given:
- \( K_b (\text{NH}_3) = 1.8 \times 10^{-5} \)

Using the relation \( K_w = K_a \cdot K_b \) where \( K_w = 1.0 \times 10^{-14} \):

\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]

Now, calculate pKₐ:

\[ \text{pK}_a = -\log(5.56 \times 10^{-10}) \approx 9.255 \]

Now we can calculate the initial pH:

\[ \text{pH} = 9.255 + \log \left( \frac{0.1}{0.1} \right) = 9.255 + 0 = 9.255 \]

### Step 2: Determine the effect of adding NaOH

When we add 0.02 moles of NaOH to the buffer solution, it reacts with NH₄Cl to form NH₃ and water:

\[ \text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O} \]

#### Initial moles before reaction:
- Moles of NH₃ = 0.1
- Moles of NH₄Cl = 0.1
- Moles of NaOH = 0.02

#### Moles after reaction:
- Moles of NH₃ after reaction = \( 0.1 + 0.02 = 0.12 \)
- Moles of NH₄Cl after reaction = \( 0.1 - 0.02 = 0.08 \)

### Step 3: Calculate the new pH after addition of NaOH

Using the Henderson-Hasselbalch equation again:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]

Now, substituting the new concentrations:

\[ \text{pH} = 9.255 + \log \left( \frac{0.12}{0.08} \right) \]

Calculating the log term:

\[ \log \left( \frac{0.12}{0.08} \right) = \log(1.5) \approx 0.1761 \]

Now substituting back into the pH equation:

\[ \text{pH} = 9.255 + 0.1761 \approx 9.4311 \]

### Step 4: Calculate the change in pH

The change in pH is given by:

\[ \Delta \text{pH} = \text{pH}_{\text{new}} - \text{pH}_{\text{initial}} = 9.4311 - 9.255 \approx 0.1761 \]

### Final Answer

The change in pH upon the addition of 0.02 moles of NaOH is approximately **0.1761**.

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