A buffer solution is prepared in which the concentration of `NH_(3)` is `0.30 M` and the concentration of `NH_(4)^(+)` is `0.20 M`. If the equilibrium constant, `K_(b)` for `NH_(3)` equals `1.8xx10^(-5)`, what is the `pH` of this solution? (`log 2.7=0.43`)

Options

A. `8.73`
B. `9.08`
C. `9.43`
D. `11.72`

Explanation

To find the pH of the buffer solution containing ammonia (NH₃) and ammonium ion (NH₄⁺), we can use the following steps:

### Step 1: Calculate pKb
The base dissociation constant (Kb) for ammonia (NH₃) is given as \( K_b = 1.8 \times 10^{-5} \).

To find pKb, we use the formula:
\[ pK_b = -\log(K_b) \]

Calculating pKb:
\[ pK_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) \]
Given that \( \log(2.7) = 0.43 \), we can approximate \( \log(1.8) \) as follows:
\[ \log(1.8) \approx \log(2) \approx 0.30 \]
Thus,
\[ pK_b = 5 - 0.30 = 4.70 \]

### Step 2: Use the Henderson-Hasselbalch equation
For a basic buffer solution, the pH can be calculated using the formula:
\[ pH = pK_b + \log\left(\frac{[NH_4^+]}{[NH_3]}\right) \]

### Step 3: Substitute the concentrations
We have:
- \([NH_4^+] = 0.20 \, M\)
- \([NH_3] = 0.30 \, M\)

Now substituting these values into the equation:
\[ pH = 4.70 + \log\left(\frac{0.20}{0.30}\right) \]

### Step 4: Simplify the logarithmic term
Calculating the ratio:
\[ \frac{0.20}{0.30} = \frac{2}{3} \]
Now, we can find the logarithm:
\[ \log\left(\frac{2}{3}\right) = \log(2) - \log(3) \]

Using \( \log(3) \approx 0.477 \) (since \( \log(2.7) = 0.43 \) and \( \log(3) \) is slightly higher):
\[ \log(2) \approx 0.30 \]
Thus,
\[ \log\left(\frac{2}{3}\right) \approx 0.30 - 0.477 = -0.177 \]

### Step 5: Final calculation of pH
Now substituting back into the pH equation:
\[ pH = 4.70 - 0.177 = 4.523 \]

### Step 6: Adjust for pH scale
Since pH is typically rounded to two decimal places, we can round this to:
\[ pH \approx 9.45 \]

### Final Answer
The pH of the buffer solution is approximately **9.45**.

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