An analytical chemist is titrating 88.9 mL of a 0.2700M solution of ammonia (NH3) with a 0.4900M solution of HIO3. The pK, of ammonia is 4.74. Calculate the pH of the base solution after the chemist has added 54.8 mL of the HIO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO3 solution added. Round your answer to 2 decimal places.

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**Titration of Ammonia Solution with HIO₃ Solution**

*Problem Statement:*
An analytical chemist is titrating 88.9 mL of a 0.2700 M solution of ammonia (NH₃) with a 0.4900 M solution of HIO₃. The pKᵦ of ammonia is 4.74. Calculate the pH of the base solution after the chemist has added 54.8 mL of the HIO₃ solution to it.

*Note for advanced students:* You may assume the final volume equals the initial volume of the solution plus the volume of HIO₃ solution added.

*Round your answer to 2 decimal places.*

---

In this problem, the analytical chemist is performing a titration of a basic solution, ammonia (NH₃), with an acidic solution, iodic acid (HIO₃). The goal is to calculate the pH of the solution after a certain amount of HIO₃ solution has been added.

### Step-by-Step Solution:

1. **Calculate the amount of \( \text{NH}_3 \) in moles**:
   \[
   \text{Moles of } \text{NH}_3 = \text{Volume} \times \text{Concentration} = 0.0889 \, \text{L} \times 0.2700 \, \text{M} = 0.024003 \, \text{moles}
   \]

2. **Calculate the amount of \( \text{HIO}_3 \) in moles**:
   \[
   \text{Moles of } \text{HIO}_3 = \text{Volume} \times \text{Concentration} = 0.0548 \, \text{L} \times 0.4900 \, \text{M} = 0.026852 \, \text{moles}
   \]

3. **Determine the limiting reactant**:
   Ammonia (\( \text{NH}_3 \)) is the limiting reactant because it is present in a smaller amount.

4. **Calculate the amount of \( \text{NH}_3 \) and \( \text{HIO}_3 \) left or reacted**:
   Since \( \text{NH}_3 \) is the limiting reactant,
Transcribed Image Text:**Titration of Ammonia Solution with HIO₃ Solution** *Problem Statement:* An analytical chemist is titrating 88.9 mL of a 0.2700 M solution of ammonia (NH₃) with a 0.4900 M solution of HIO₃. The pKᵦ of ammonia is 4.74. Calculate the pH of the base solution after the chemist has added 54.8 mL of the HIO₃ solution to it. *Note for advanced students:* You may assume the final volume equals the initial volume of the solution plus the volume of HIO₃ solution added. *Round your answer to 2 decimal places.* --- In this problem, the analytical chemist is performing a titration of a basic solution, ammonia (NH₃), with an acidic solution, iodic acid (HIO₃). The goal is to calculate the pH of the solution after a certain amount of HIO₃ solution has been added. ### Step-by-Step Solution: 1. **Calculate the amount of \( \text{NH}_3 \) in moles**: \[ \text{Moles of } \text{NH}_3 = \text{Volume} \times \text{Concentration} = 0.0889 \, \text{L} \times 0.2700 \, \text{M} = 0.024003 \, \text{moles} \] 2. **Calculate the amount of \( \text{HIO}_3 \) in moles**: \[ \text{Moles of } \text{HIO}_3 = \text{Volume} \times \text{Concentration} = 0.0548 \, \text{L} \times 0.4900 \, \text{M} = 0.026852 \, \text{moles} \] 3. **Determine the limiting reactant**: Ammonia (\( \text{NH}_3 \)) is the limiting reactant because it is present in a smaller amount. 4. **Calculate the amount of \( \text{NH}_3 \) and \( \text{HIO}_3 \) left or reacted**: Since \( \text{NH}_3 \) is the limiting reactant,
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