A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 200.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.


To determine the pH of the solution after the addition of 200.0 mL of \(0.10 \, \text{M}\) \( \text{HNO}_3 \) to a \(100.0 \, \text{mL}\) sample of \(0.10 \, \text{M}\) \( \text{NH}_3 \), let's follow these steps.

### Step 1: Initial Moles of \( \text{NH}_3 \) and \( \text{HNO}_3 \)
1. Calculate the initial moles of \( \text{NH}_3 \) in the \(100.0 \, \text{mL}\) solution.
   \[
   \text{moles of } \text{NH}_3 = 0.10 \, \text{M} \times 0.100 \, \text{L} = 0.010 \, \text{mol}
   \]

2. Calculate the moles of \( \text{HNO}_3 \) added.
   \[
   \text{moles of } \text{HNO}_3 = 0.10 \, \text{M} \times 0.200 \, \text{L} = 0.020 \, \text{mol}
   \]

### Step 2: Reaction Between \( \text{NH}_3 \) and \( \text{HNO}_3 \)
When \( \text{NH}_3 \) reacts with \( \text{HNO}_3 \), it forms \( \text{NH}_4^+ \) ions:
\[
\text{NH}_3 + \text{HNO}_3 \rightarrow \text{NH}_4^+
\]

Since there are more moles of \( \text{HNO}_3 \) than \( \text{NH}_3 \), all the \( \text{NH}_3 \) will be neutralized, leaving an excess of \( \text{HNO}_3 \) in the solution.

1. After the reaction, calculate the remaining moles of \( \text{HNO}_3 \):
   \[
   \text{Remaining moles of } \text{HNO}_3 = 0.020 - 0.010 = 0.010 \, \text{mol}
   \]

2. The moles of \( \text{NH}_4^+ \) formed will equal the moles of \( \text{NH}_3 \) that reacted:
   \[
   \text{Moles of } \text{NH}_4^+ = 0.010 \, \text{mol}
   \]

### Step 3: Total Volume of the Solution
The total volume of the solution after the addition of \( \text{HNO}_3 \) is:
\[
\text{Total volume} = 100.0 \, \text{mL} + 200.0 \, \text{mL} = 300.0 \, \text{mL} = 0.300 \, \text{L}
\]

### Step 4: Calculate the Concentration of \( \text{H}^+ \) and \( \text{NH}_4^+ \)
1. Concentration of excess \( \text{HNO}_3 \), which provides \( \text{H}^+ \):
   \[
   [\text{H}^+] = \frac{0.010 \, \text{mol}}{0.300 \, \text{L}} = 0.0333 \, \text{M}
   \]

2. Concentration of \( \text{NH}_4^+ \):
   \[
   [\text{NH}_4^+] = \frac{0.010 \, \text{mol}}{0.300 \, \text{L}} = 0.0333 \, \text{M}
   \]

### Step 5: Determine the pH
Since we have excess \( \text{H}^+ \) in solution, the pH is determined directly by the concentration of \( \text{H}^+ \):
\[
\text{pH} = -\log [\text{H}^+]
\]
\[
\text{pH} = -\log (0.0333) \approx 1.48
\]

### Answer
The pH of the solution after the addition of \(200.0 \, \text{mL}\) of \( \text{HNO}_3 \) is approximately **1.48**.

Would you like more details on any part of this calculation?

---

1. How would the pH change if less \( \text{HNO}_3 \) were added?
2. What role does \( K_b \) play if the titration hadn’t reached excess \( \text{HNO}_3 \)?
3. How would we approach the pH calculation if \( \text{NH}_4^+ \) and \( \text{NH}_3 \) were in equilibrium?
4. How does the initial volume of \( \text{NH}_3 \) solution affect the final pH?
5. What would happen if a buffer were formed with a different \( \text{HNO}_3 \) concentration?

**Tip:** When strong acids are in excess, the pH is determined almost entirely by the concentration of \( \text{H}^+ \).

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 200.0 mL of HNO3. The Kb of NH3 is 1.8 × 10^-5.

Step-by-step solution to find the pH of a 0.10 M NH3 solution titrated with 0.10 M HNO3. Learn how to calculate pH after adding 200.0 mL of HNO3, covering acid-base chemistry, stoichiometry, and titration calculations.

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