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?

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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 $\text{H}^+$.