Trial 1 Trial 2 M of NaOH(aq): Average M of NaOH(aq) from part I M of NaOH(aq): Average M of NaOH(aq) from part I Volume of vinegar: 10.00 mL. Volume of vinegar: 10.00 mL. Initial buret reading of NaOH (aq): 0.00 mL. Initial buret reading of NaOH (aq) : 12.35 mL. Final buret reading of NaOh(aq) pink color: 12.35 mL. Final buret reading of NaOh(aq) pink color: 25.00 mL. Volume of NaOH(aq): Volume of NaOH(aq): Calculations: For each trial : NaOH(aq) + CH3COOH(aq) → NaCH3COO(aq) + H₂O(1) a. Moles of NaOH(aq): volume of NaOH(aq) in L. x average M of NaOH(aq) from part I b. Moles of NaOH → moles of CH3COOH: moles of CH3COOH M of CH3COOH = moles of CH3COOH (b) / volume of vinegar in L. Get average M of CH3COOH (aq): (M+ M)/2 Mass% of CH3COOH in the vinegar solution : Molar mass of CH3COOH = 60.01 g/mole, Density of vinegar solution = 1.10 g/mole Moles of CH3COOH (b) →g of CH3COOH : c: moles of CH3COOH x molar mass of CH3COOH 10.00 mL. vinegar x 1.10 g/mL. = 10.1 g vinegar Mass% of CH3COOH = (g of CH3COOH (c) / 10.1 g vinegar) x 100 Get average of mass% of CH3COOH.
Trial 1 Trial 2 M of NaOH(aq): Average M of NaOH(aq) from part I M of NaOH(aq): Average M of NaOH(aq) from part I Volume of vinegar: 10.00 mL. Volume of vinegar: 10.00 mL. Initial buret reading of NaOH (aq): 0.00 mL. Initial buret reading of NaOH (aq) : 12.35 mL. Final buret reading of NaOh(aq) pink color: 12.35 mL. Final buret reading of NaOh(aq) pink color: 25.00 mL. Volume of NaOH(aq): Volume of NaOH(aq): Calculations: For each trial : NaOH(aq) + CH3COOH(aq) → NaCH3COO(aq) + H₂O(1) a. Moles of NaOH(aq): volume of NaOH(aq) in L. x average M of NaOH(aq) from part I b. Moles of NaOH → moles of CH3COOH: moles of CH3COOH M of CH3COOH = moles of CH3COOH (b) / volume of vinegar in L. Get average M of CH3COOH (aq): (M+ M)/2 Mass% of CH3COOH in the vinegar solution : Molar mass of CH3COOH = 60.01 g/mole, Density of vinegar solution = 1.10 g/mole Moles of CH3COOH (b) →g of CH3COOH : c: moles of CH3COOH x molar mass of CH3COOH 10.00 mL. vinegar x 1.10 g/mL. = 10.1 g vinegar Mass% of CH3COOH = (g of CH3COOH (c) / 10.1 g vinegar) x 100 Get average of mass% of CH3COOH.
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter4: Chemical Reactions
Section: Chapter Questions
Problem 4.24QP: Three acid samples are prepared for titration by 0.01 M NaOH: 1 Sample 1 is prepared by dissolving...
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Question
Molarity of NaOH= 0.1018m
![### Titration Data and Calculations
#### Data:
| | **Trial 1** | **Trial 2** |
|-----------------------|---------------------------------------------------------|---------------------------------------------------------|
| **M of NaOH(aq):** | **Average M of NaOH(aq) from part I** | **Average M of NaOH(aq) from part I** |
| **Volume of vinegar:**| 10.00 mL | 10.00 mL |
| **Initial buret reading of NaOH (aq):** | 0.00 mL | 12.35 mL |
| **Final buret reading of NaOH(aq) pink color:** | 12.35 mL | 25.00 mL |
| **Volume of NaOH(aq):** | | |
#### Calculations:
For each trial I:
\[ \text{NaOH}(aq) + \text{CH}_3\text{COOH}(aq) \rightarrow \text{NaCH}_3\text{COO}(aq) + \text{H}_2\text{O}(l) \]
a. **Moles of NaOH(aq):**
- Use the formula `Volume of NaOH(aq) in L × average M of NaOH(aq) from part I`
b. **Moles of NaOH → moles of CH₃COOH:**
- The stoichiometric ratio of NaOH to CH₃COOH is 1:1
\[ \text{M of CH}_3\text{COOH} = \text{moles of CH}_3\text{COOH (b)} / \text{volume of vinegar in L} \]
#### Mass% of CH₃COOH in the Vinegar Solution:
- **Molar mass of CH₃COOH:** 60.01 g/mol
- **Density of vinegar solution:** 1.10 g/mL
Calculate:
- **Volume of vinegar:** 10.00 mL
- **Mass of vinegar:**
\[ 10.00 \text{ mL} \times 1.10 \text{ g/mL} = 10.1 \text{ g vinegar} \]
\[ \text{Mass% of CH}_3\text{COOH} =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffdeaa466-70f4-480d-bd45-ed519d8d81f4%2F56a26360-8364-428e-984a-1f8b0446b2c4%2Fzemth9u_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Titration Data and Calculations
#### Data:
| | **Trial 1** | **Trial 2** |
|-----------------------|---------------------------------------------------------|---------------------------------------------------------|
| **M of NaOH(aq):** | **Average M of NaOH(aq) from part I** | **Average M of NaOH(aq) from part I** |
| **Volume of vinegar:**| 10.00 mL | 10.00 mL |
| **Initial buret reading of NaOH (aq):** | 0.00 mL | 12.35 mL |
| **Final buret reading of NaOH(aq) pink color:** | 12.35 mL | 25.00 mL |
| **Volume of NaOH(aq):** | | |
#### Calculations:
For each trial I:
\[ \text{NaOH}(aq) + \text{CH}_3\text{COOH}(aq) \rightarrow \text{NaCH}_3\text{COO}(aq) + \text{H}_2\text{O}(l) \]
a. **Moles of NaOH(aq):**
- Use the formula `Volume of NaOH(aq) in L × average M of NaOH(aq) from part I`
b. **Moles of NaOH → moles of CH₃COOH:**
- The stoichiometric ratio of NaOH to CH₃COOH is 1:1
\[ \text{M of CH}_3\text{COOH} = \text{moles of CH}_3\text{COOH (b)} / \text{volume of vinegar in L} \]
#### Mass% of CH₃COOH in the Vinegar Solution:
- **Molar mass of CH₃COOH:** 60.01 g/mol
- **Density of vinegar solution:** 1.10 g/mL
Calculate:
- **Volume of vinegar:** 10.00 mL
- **Mass of vinegar:**
\[ 10.00 \text{ mL} \times 1.10 \text{ g/mL} = 10.1 \text{ g vinegar} \]
\[ \text{Mass% of CH}_3\text{COOH} =
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