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 gasous HCI.
Assume no change in volume. `K_(NH_(3))=1.8xx10^(-5)`

Answer

To solve the problem of calculating the change in pH of a buffer solution containing 0.1 mole each of NH₃ and NH₄Cl upon the addition of 0.02 mole of dissolved gaseous HCl, we will follow these steps:

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

The buffer solution consists of a weak base (NH₃) and its conjugate acid (NH₄Cl). 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 the pKₐ from Kb:

\[ K_b = 1.8 \times 10^{-5} \]

Using the relation \( K_w = K_a \times K_b \) (where \( K_w = 1.0 \times 10^{-14} \) at 25°C):

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

Now, calculate pKₐ:

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

Now we can find the initial pH:

\[ \text{pH}_{\text{initial}} = 9.255 + \log\left(\frac{0.1}{0.1}\right) = 9.255 + \log(1) = 9.255 \]

### Step 2: Calculate the change in concentrations after adding HCl.

When 0.02 moles of HCl are added to the buffer, it reacts with NH₃ to form NH₄Cl:

\[ \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl} \]

Before the reaction:
- [NH₃] = 0.1 moles
- [NH₄Cl] = 0.1 moles

After the reaction:
- [NH₃] = 0.1 - 0.02 = 0.08 moles
- [NH₄Cl] = 0.1 + 0.02 = 0.12 moles

### Step 3: Calculate the new pH after the addition of HCl.

Using the Henderson-Hasselbalch equation again:

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

Substituting the new concentrations:

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

Calculating the logarithm:

\[ \log\left(\frac{0.08}{0.12}\right) = \log(0.6667) \approx -0.1761 \]

Thus,

\[ \text{pH}_{\text{final}} = 9.255 - 0.1761 \approx 9.079 \]

### Step 4: Calculate the change in pH.

The change in pH is given by:

\[ \Delta \text{pH} = \text{pH}_{\text{initial}} - \text{pH}_{\text{final}} = 9.255 - 9.079 \approx 0.176 \]

### Final Answer:

The change in pH upon the addition of 0.02 moles of HCl is approximately **0.176**.

---