A chemist titrates 210.0 mL of a 0.1242M lidocaine (C₁4H21 NONH) solution with 0.6479M HBr solution at 25 °C. Calculate the pH at equivalence. The PK, of lidocaine is 7.94. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added.
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![**Titration Problem: Calculating pH at Equivalence**
A chemist titrates 210.0 mL of a 0.1242 M lidocaine (\(C_{14}H_{21}NONH\)) solution with 0.6479 M HBr solution at 25°C. Calculate the pH at equivalence. The \(pK_b\) of lidocaine is 7.94.
**Conditions:**
- Volume of lidocaine solution: 210.0 mL
- Molarity of lidocaine solution: 0.1242 M
- Molarity of HBr solution: 0.6479 M
- Temperature: 25°C
- \(pK_b\) of lidocaine: 7.94
**Instructions:**
Round your answer to 2 decimal places.
**Note for advanced students:**
You may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added.
**Explanation:**
The problem involves determining the pH of the resultant solution at the equivalence point during a titration process. At equivalence, the number of moles of the titrant (HBr) equals the number of moles of the analyte (lidocaine). The titration involves the reaction of a weak base (lidocaine) with a strong acid (HBr), and the \(pH\) needs to be calculated after this neutralization reaction is complete.
**Step-by-Step Solution:**
1. Calculate the moles of lidocaine:
\[ \text{moles of lidocaine} = \text{Volume (L)} \times \text{Molarity} = 0.210 \, \text{L} \times 0.1242 \, \text{M} = 0.026082 \, \text{mol} \]
2. Determine the volume of HBr solution needed to reach equivalence:
\[ \text{Volume of HBr} = \frac{\text{moles of lidocaine}}{\text{Molarity of HBr}} = \frac{0.026082 \, \text{mol}}{0.6479 \, \text{M}} = 0.04026 \, \text{L} = 40.26 \, \text{mL} \]
3. Calculate the total volume of the solution at equivalence](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65f958a7-7c6b-44c4-ab5b-ab922054d7b7%2F3efdc1db-2c6b-4376-a160-a0f4ff7bae89%2Fm28rf5_processed.png&w=3840&q=75)
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