`50.0` mL of `0.10` M ammonia solution is treated with `25.0` mL of `0.10M HCI`. If `K_(b)(NH_(3))=1.77xx10^(-5)`, the pH of the resulting solution will be

Options

A. `11.12`
B. `8.75`
C. `10.34`
D. `9.24`

Explanation

To solve the problem step by step, let's break down the process of calculating the pH of the resulting solution after the reaction between ammonia and hydrochloric acid.

### Step 1: Calculate the moles of ammonia and hydrochloric acid
First, we need to calculate the number of moles of ammonia (NH₃) and hydrochloric acid (HCl) present in the solution.

- For ammonia:
\[ \text{Moles of NH}_3 = \text{Volume (L)} \times \text{Molarity (M)} = 0.050 \, \text{L} \times 0.10 \, \text{M} = 0.005 \, \text{mol} \]

- For hydrochloric acid:
\[ \text{Moles of HCl} = \text{Volume (L)} \times \text{Molarity (M)} = 0.025 \, \text{L} \times 0.10 \, \text{M} = 0.0025 \, \text{mol} \]

### Step 2: Determine the limiting reactant and remaining moles
Next, we need to determine how much of each reactant remains after the reaction.

The reaction between ammonia and hydrochloric acid is:
\[ \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ + \text{Cl}^- \]

Since 1 mole of HCl reacts with 1 mole of NH₃, we can see that:
- 0.0025 mol of HCl will react with 0.0025 mol of NH₃.

Now, we calculate the remaining moles:
- Remaining moles of NH₃:
\[ 0.005 \, \text{mol} - 0.0025 \, \text{mol} = 0.0025 \, \text{mol} \]
- Remaining moles of HCl:
\[ 0.0025 \, \text{mol} - 0.0025 \, \text{mol} = 0 \, \text{mol} \]

### Step 3: Calculate the moles of ammonium chloride produced
Since the reaction produces ammonium chloride (NH₄Cl), the moles of NH₄Cl produced will be equal to the moles of HCl that reacted:
\[ \text{Moles of NH}_4^+ = 0.0025 \, \text{mol} \]

### Step 4: Calculate the total volume of the solution
The total volume of the solution after mixing is:
\[ 50.0 \, \text{mL} + 25.0 \, \text{mL} = 75.0 \, \text{mL} = 0.075 \, \text{L} \]

### Step 5: Calculate the concentrations of NH₃ and NH₄⁺
Now we can calculate the concentrations of NH₃ and NH₄⁺ in the solution:
- Concentration of NH₃:
\[ [\text{NH}_3] = \frac{0.0025 \, \text{mol}}{0.075 \, \text{L}} = 0.0333 \, \text{M} \]

- Concentration of NH₄⁺:
\[ [\text{NH}_4^+] = \frac{0.0025 \, \text{mol}}{0.075 \, \text{L}} = 0.0333 \, \text{M} \]

### Step 6: Calculate pKₐ from Kb
Given \( K_b \) for ammonia:
\[ K_b = 1.77 \times 10^{-5} \]
We can find \( K_a \) using the relation:
\[ K_w = K_a \times K_b \quad \text{(where \( K_w = 1.0 \times 10^{-14} \))} \]
Thus,
\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.77 \times 10^{-5}} \approx 5.65 \times 10^{-10} \]

### Step 7: Calculate pKₐ
Now, we calculate \( pK_a \):
\[ pK_a = -\log(K_a) \approx -\log(5.65 \times 10^{-10}) \approx 9.24 \]

### Step 8: Calculate pH using the Henderson-Hasselbalch equation
Since we have a buffer solution, we can use the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) \]
Here, the base is NH₃ and the acid is NH₄⁺:
\[ pH = 9.24 + \log\left(\frac{0.0333}{0.0333}\right) = 9.24 + \log(1) = 9.24 \]

### Final Answer
Thus, the pH of the resulting solution is approximately **9.24**.

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