Volumetric Analysis- Acid-Base Titration Concentration of Acetic Acid in Commercial sample of Vinegar 1. Brand name of vinegar sample: America's Choice 2. Volume of Vinegar sample used: 3.0ml mL 0.0030L %3D 3. Percent (%) of acetic acid on label: 4. Concentration (in Molarity, M) of NaOH solution:_0,28 5 M Trial 1 Trial 2 Trial 3 Initial level of NaOH solution in the burette (V1) 0.8 0-0 1.0 Final Level of NaOH solution in the burette (V2) 48-1 47.4 48.3 Volume (in mL) of NaOH solution used (V2-V1) 47.3ml 47.4ml 47.3ml Average Volume (in mL) of NaOH solution used 47.3 ml Average volume (in L) of NaOH solution used 0.0473 5. Moles of NaOH used in titration (show your calculations) 0.0135 moles of NaOH 0.285 H x O-0473=0.0135 6. Moles of Acetic acid (HC2H3O2) neutralized by NaOH O -0135 moles of HC2H3O2 7. Molarity of acetic acid, HC2H3O2 in vinegar sample (show your calculations) TZ vinerar ontain= Q0135 니.S M 4.5 moles of ctic acid 8 O-003 0.81 g HC2H3O2 8.Grams of Acetic Acid, HC2H3O2 (show your calculations) 0.013 SX 60.052= 0.810 O.819 acelie acid presentin 3ml of Vinegar sample 270 % 9. Percent (m/v) of HC2H3O2 in vinegar sample (show your calculations) 0.81 = 0.27 (glml) 3.0 - 0+27 x10?(g/) = 270 Question 1: Is your experimentally calculated value of percent of HC2H3O2 is high or low compare to the percent of acetic acid value listed on the label of your vinegar sample bottle? Explain, why do you have a different calculated percent concentration?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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$**Volumetric Analysis - Acid-Base Titration** **Concentration of Acetic Acid in Commercial Sample of Vinegar** 1. **Brand name of vinegar sample:** America’s Choice 2. **Volume of Vinegar sample used:** $3.0 \, \text{mL} = 0.0030 \, \text{L}$ 3. **Percent (%) of acetic acid on label:** 5% 4. **Concentration (in Molarity, M) of NaOH solution:** 0.285 M | | Trial 1 | Trial 2 | Trial 3 | |---------------------|---------|---------|---------| | Initial Level of NaOH solution in the burette ($V_1$) | 0.8 | 0.0 | 1.0 | | Final Level of NaOH solution in the burette ($V_2$) | 48.1 | 47.4 | 48.3 | | Volume (in mL) of NaOH solution used ($V_2 - V_1$) | 47.3 | 47.4 | 47.3 | **Average Volume (in mL) of NaOH solution used:** 47.3 mL **Average Volume (in L) of NaOH solution used:** 0.0473 L 5. **Moles of NaOH used in titration (show your calculations):** \[ 0.285 \times 0.0473 = 0.0135 \, \text{moles of NaOH} \] 6. **Moles of Acetic acid ($ \text{HC}_2\text{H}_3\text{O}_2 $) neutralized by NaOH:** $0.0135 \, \text{moles of} \, \text{HC}_2\text{H}_3\text{O}_2$ 7. **Molarity of acetic acid, $\text{HC}_2\text{H}_3\text{O}_2$, in vinegar sample (show your calculations):** \[\left(\frac{0.0135}{0.003}\right) = 4.5 \, \text{M}\] 8$

Transcribed Image Text:**Volumetric Analysis - Acid-Base Titration** **Concentration of Acetic Acid in Commercial Sample of Vinegar** 1. **Brand name of vinegar sample:** America’s Choice 2. **Volume of Vinegar sample used:** $3.0 \, \text{mL} = 0.0030 \, \text{L}$ 3. **Percent (%) of acetic acid on label:** 5% 4. **Concentration (in Molarity, M) of NaOH solution:** 0.285 M | | Trial 1 | Trial 2 | Trial 3 | |---------------------|---------|---------|---------| | Initial Level of NaOH solution in the burette ($V_1$) | 0.8 | 0.0 | 1.0 | | Final Level of NaOH solution in the burette ($V_2$) | 48.1 | 47.4 | 48.3 | | Volume (in mL) of NaOH solution used ($V_2 - V_1$) | 47.3 | 47.4 | 47.3 | **Average Volume (in mL) of NaOH solution used:** 47.3 mL **Average Volume (in L) of NaOH solution used:** 0.0473 L 5. **Moles of NaOH used in titration (show your calculations):** \[ 0.285 \times 0.0473 = 0.0135 \, \text{moles of NaOH} \] 6. **Moles of Acetic acid ($ \text{HC}_2\text{H}_3\text{O}_2 $) neutralized by NaOH:** $0.0135 \, \text{moles of} \, \text{HC}_2\text{H}_3\text{O}_2$ 7. **Molarity of acetic acid, $\text{HC}_2\text{H}_3\text{O}_2$, in vinegar sample (show your calculations):** \[\left(\frac{0.0135}{0.003}\right) = 4.5 \, \text{M}\] 8

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