A buffer solution contains `0.25M NH_(4)OH` and `0.3 NH_(4)C1`.
a. Calculate the `pH` of the solution.
`K_(b) =2xx10^(-5)`.

Explanation

To calculate the pH of a buffer solution containing 0.25 M NH₄OH and 0.3 M NH₄Cl, we will follow these steps:

### Step 1: Calculate pK_b
First, we need to calculate the pK_b from the given K_b value.

Given:
- K_b = 2 x 10^(-5)

Using the formula:
\[ \text{pK}_b = -\log(K_b) \]

Calculating pK_b:
\[ \text{pK}_b = -\log(2 \times 10^{-5}) \]
Using logarithmic properties:
\[ \text{pK}_b = -(\log(2) + \log(10^{-5})) \]
\[ \text{pK}_b = -(\log(2) - 5) \]
\[ \text{pK}_b = 5 - 0.301 \quad (\text{using } \log(2) \approx 0.301) \]
\[ \text{pK}_b \approx 4.699 \approx 4.7 \]

### Step 2: Calculate pOH using the Henderson-Hasselbalch equation
Now we will use the Henderson-Hasselbalch equation for bases:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]

In our case:
- Salt = NH₄Cl (0.3 M)
- Base = NH₄OH (0.25 M)

Substituting the values:
\[ \text{pOH} = 4.7 + \log\left(\frac{0.3}{0.25}\right) \]
Calculating the ratio:
\[ \frac{0.3}{0.25} = 1.2 \]
Now calculate the logarithm:
\[ \text{pOH} = 4.7 + \log(1.2) \]
Using the approximation \(\log(1.2) \approx 0.079\):
\[ \text{pOH} = 4.7 + 0.079 \approx 4.779 \]

### Step 3: Calculate pH
We know that:
\[ \text{pH} + \text{pOH} = 14 \]
Thus:
\[ \text{pH} = 14 - \text{pOH} \]
Substituting the pOH value:
\[ \text{pH} = 14 - 4.779 \approx 9.221 \]

### Final Answer
The pH of the buffer solution is approximately **9.22**.

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