A. Standardization of Sodium Hydroxide Solution Balanced chemical equation:
Data: Trial 1 Trial 2 Trial 3 Mass of oxalic acid 0.1503g 0.1511g 0.1504g Initial burette reading 0.00ml 0.00ml 0.00ml Final burette reading 14.1ml 11.3ml 12.9ml Calculations (show work): Volume NaOH added moles of oxalic acid moles of NaOH added Molarity of NaOH Average molarity
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The Correct Answer and Explanation is:
A. Standardization of Sodium Hydroxide Solution Balanced chemical equation: Data: Trial 1 Trial 2 Trial 3 Mass of oxalic acid 0.1503g 0.1511g 0.1504g Initial burette reading 0.00ml 0.00ml 0.00ml Final burette reading 14.1ml 11.3ml 12.9ml Calculations (show work): Volume NaOH added moles of oxalic acid moles of NaOH added Molarity of NaOH Average molaritygenerate the correct answer and 300 words explanation without using the em dashes
ChatGPT said:
To calculate the molarity of NaOH using standardization with oxalic acid, we follow these steps for each trial. The balanced chemical equation is:
H₂C₂O₄ + 2 NaOH → Na₂C₂O₄ + 2 H₂O
Step 1: Volume of NaOH added
This is calculated from the difference in burette readings (Final – Initial):
- Trial 1: 14.1 mL
- Trial 2: 11.3 mL
- Trial 3: 12.9 mL
Convert to liters (1 mL = 0.001 L):
- Trial 1: 0.0141 L
- Trial 2: 0.0113 L
- Trial 3: 0.0129 L
Step 2: Moles of oxalic acid
Molar mass of oxalic acid (H₂C₂O₄):
= 2(1.008) + 2(12.01) + 4(16.00) = 90.03 g/mol
Moles = mass / molar mass:
- Trial 1: 0.1503 g / 90.03 g/mol = 0.00167 mol
- Trial 2: 0.1511 g / 90.03 g/mol = 0.001678 mol
- Trial 3: 0.1504 g / 90.03 g/mol = 0.001671 mol
Step 3: Moles of NaOH added
From the balanced equation, 1 mol oxalic acid reacts with 2 mol NaOH:
- Trial 1: 0.00167 × 2 = 0.00334 mol
- Trial 2: 0.001678 × 2 = 0.003356 mol
- Trial 3: 0.001671 × 2 = 0.003342 mol
Step 4: Molarity of NaOH
Molarity = moles / volume (in L)
- Trial 1: 0.00334 mol / 0.0141 L = 0.237 mol/L
- Trial 2: 0.003356 mol / 0.0113 L = 0.297 mol/L
- Trial 3: 0.003342 mol / 0.0129 L = 0.259 mol/L
Step 5: Average molarity
= (0.237 + 0.297 + 0.259) / 3 = 0.264 mol/L
Final Answers:
| Calculation | Trial 1 | Trial 2 | Trial 3 |
|---|---|---|---|
| Volume NaOH added (L) | 0.0141 | 0.0113 | 0.0129 |
| Moles of oxalic acid | 0.00167 | 0.001678 | 0.001671 |
| Moles of NaOH added | 0.00334 | 0.003356 | 0.003342 |
| Molarity of NaOH (mol/L) | 0.237 | 0.297 | 0.259 |
Average Molarity: 0.264 mol/L
This process standardizes NaOH solution concentration through stoichiometric relationships, where oxalic acid acts as a primary standard due to its known purity and stability.
