Preview

Titration Analysis

Satisfactory Essays
Open Document
Open Document
505 Words
Grammar
Grammar
Plagiarism
Plagiarism
Writing
Writing
Score
Score
Titration Analysis
Investigation 8.4- Titration Analysis of Vinegar

Prediction: We predict that the amount concentration of acetic acid in a sample of vinegar is 0.83 mol/L.

VCH3COOH (aq) = 10.00 mL
VNaOH (aq) = ?
CCH3COOH = 0.83 mol/L
CNaOH (aq) = 0.145 mol/L

Materials:

Sodium hydroxide 100 mL
Acetic acid 50 mL
250 mL beaker x 2
Distilled water x 1 bottle
Erlenmeyer flask x 1
Burette x 1
Pipette x 1
Pipette bulb x 1
Volumetric pipette x 1
Phenolphalein

Procedure:

Step 1- Gather materials, put on eyeglasses and apron.
Step 2- Pour 50 mL of acetic acid into one 250 beaker and 100 mL of sodium hydroxide into another 250 mL beaker.
Step 3- Clean burette, volumetric pipette and pipette bulb with distilled water.
Step 4- Use a funnel or just the beaker to pour the sodium hydroxide (not above head) into the volumetric pipette. Use the pipette and the pipette bulb to gather 10 mL of acetic acid and empty into Erlenmeyer flask. Add a few drops of phenolphalein to the flask. Let 9 mL of the sodium hydroxide drain into the flask, placed underneath the volumetric pipette until it turns pink.
Step 5- Record results. Repeat step four to determine more results.
Step 6- Clean up workspace, put away materials and safety tools.

Evidence:



Qualitative/Quantitative: The acetic acid + phenolphalein changed pink quicker, and turned a darker shade of pink when more sodium hydroxide was added to it. For each trial, we added at least 9 mL of sodium hydroxide to the acetic acid + phenolphalein.

Analysis:

CH3COOH (aq) + NaOH (aq) à H2O (l) + NaCH3COO (aq)
V = 10.00 mL V = 11.00 mL
C =? C = 0.145 mol/L

According to the evidence obtained, we concluded that the amount concentration of acetic acid in a sample of vinegar is 0.1595 mol/L.

0.1595 x 5 = 0.80

Evaluation:

% difference: 0.80-0.83 = 0.036 x 100 = 3.61 % 0.83

% yield: 0.80 = 0.9638 x 100 = 96.4 % 0.83

The

You May Also Find These Documents Helpful

  • Better Essays

    Figure 1 also displays the results of the titration of Acetic Acid in vinegar with the NaOH. As seen in the table, it took much more NaOH to titrate the vinegar than it did with the HCL, a difference of almost 9 mL. Furthermore, the Acetic Acid was also determined to have a higher concentration compared to the HCl, with a molarity of 0.649 M. (Figure 2)…

    • 2337 Words
    • 10 Pages
    Better Essays
  • Satisfactory Essays

    CHEM120 W5 Lab Template 1

    • 414 Words
    • 3 Pages

    Step 4: Titrate NaOH into HCl until end point, record initial buret volume and add NaOH (quickly at first then slowly) until the HCl solution turns pink and record the final buret volume of NaOH in buret.…

    • 414 Words
    • 3 Pages
    Satisfactory Essays
  • Good Essays

    6.03 Titration Lab

    • 573 Words
    • 3 Pages

    3.) Place 5mL of vinegar in a 125 mL Erlenmeyer flask. Dilute the vinegar with 25 mL of water and add two drops of phenolphthalein.…

    • 573 Words
    • 3 Pages
    Good Essays
  • Powerful Essays

    Diprotic Acid Lab Report

    • 1283 Words
    • 6 Pages

    Once the pipette has been cleaned, pipette 20mL of the food acid and place it…

    • 1283 Words
    • 6 Pages
    Powerful Essays
  • Good Essays

    7. Using the slider on the right hand side, add NaOH to the HCl in the Erlenmeyer flask (This action is known as titrate). Add the indicator until the color of the indicator turns a light shade of pink.…

    • 1301 Words
    • 6 Pages
    Good Essays
  • Good Essays

    Devry week five chem Ilab

    • 496 Words
    • 2 Pages

    Step 4: Titrate NaOH into HCl until end point, record initial buret volume and add NaOH (quickly at first then slowly) until the HCl solution turns pink and record the final buret volume of NaOH in buret.…

    • 496 Words
    • 2 Pages
    Good Essays
  • Satisfactory Essays

    LAB 20C

    • 561 Words
    • 3 Pages

    2. To titrate an acetic acid solution with 0.50M sodium hydroxide, and determine the molarity and percentage composition of the vinegar.…

    • 561 Words
    • 3 Pages
    Satisfactory Essays
  • Satisfactory Essays

    ○ Commercial Vinegar is 5% acetic acid. Based on your mass from the previous question, how…

    • 536 Words
    • 3 Pages
    Satisfactory Essays
  • Good Essays

    50mL of 0.1 M HCL was obtained in a beaker. A second buret was rinsed with water and then a small amount of HCL, and then filled with HCL. It was then labeled ACID. The initial volume was recorded and then approximately…

    • 716 Words
    • 3 Pages
    Good Essays
  • Good Essays

    Experiment 1 Protocol

    • 402 Words
    • 3 Pages

    13. Repeat Steps 10 - 12 four more times until all the acetic acid has been added to beaker “B”.…

    • 402 Words
    • 3 Pages
    Good Essays
  • Good Essays

    Science Experimeny

    • 1617 Words
    • 7 Pages

    5. Using the graduated cylinder label base, add 5ml of sodium hydroxide (NaOH) to tube 3.…

    • 1617 Words
    • 7 Pages
    Good Essays
  • Satisfactory Essays

    We did three trial for this experiment. We diluted the NaOH in experiment 12 and because of that we also have to dilute the acetic acid to save the amount of NaOH used in every titration. The solution NaOH was diluted to 8.3mL of 6M of NaOH to get 0.1M NaOH and placed in a flask with the oxalic acid and phenolphthalein until it got balanced.The calculated molar concentration of acetic acid was 0.06525M,average of three trials, which had a 3.9% acetic acid. Which this gave a 2.5 % difference compared to the 4.0% concentration reported on the bottle of the rice vinegar.…

    • 279 Words
    • 1 Page
    Satisfactory Essays
  • Good Essays

    Chapter 2 Homework

    • 462 Words
    • 2 Pages

    9 Acetic acid is the active ingredient in vinegar. It consists of 40.00% C, 6.73% H, and 53.28% O. What is the empirical formula of acetic acid?…

    • 462 Words
    • 2 Pages
    Good Essays
  • Good Essays

    Acid Base Titrations

    • 310 Words
    • 2 Pages

    The acid-base titration curves help to find the pKa, Ka, and pH at equivalent point. At the beginning pH for HCl is 1.90 which is lower than the 3.28 for acetic acid; thus, strong acid (HCl) means lower pH and weak acid (acetic acid) means higher pH. Then at the equivalent point for the titration HCl-NaOH the pH is 7, which mans that is neutral, in other words there are enough NaOH mmol to neutralize the HCl mmol present; also, the solution contains only water and NaCl the salt derived from the titration made. On the other hand, at the equivalent point for the titration acetic acid-NaOH the pH is 8.67; consequently is more basic. This is something that it was expected because for a weak monoprotic acid (acetic acid)-strong base (NaOH) titration the pH at equivalence point is always grater than 7 because the anion of the weak acid is a base. Next beyond the equivalent point, the curves for the acetic acid- NaOH and HCl-NaOH titrations are identical because the pH in both cases is determined by the concentration of OH from the excess NaOH., getting all this data the pKa is determined by estimating the pH at the volume point which is 16 divided by two which is 8 (the halfway to the equivalent point), so when the volume is 8, pKa is 4.76, taking the antilog of 4.76 the answer gives a value of 1.74 * 10 ^-5 that is very accurate to the theoretical one. The indicator plays an important role for determining the equivalence point based on the color change. Phenolphthalein is an excellent indicator for the acetic acid-NaOH titration because the pH at equivalent point (8.67) falls within the pH range (8.2-9.8) in which phenolphthalein changes…

    • 310 Words
    • 2 Pages
    Good Essays
  • Powerful Essays

    Iraq

    • 2180 Words
    • 9 Pages

    1. Scope The goal of this analytical procedure is to determine the acetic acid content of two samples of vinegar produced by Kolinska, Ljubljana, Slovenia. These samples are: Sample No. 1: vinegar made from wine, 4% Sample No. 2: vinegar made from alcohol, 9% The smallest quantity of the sample is 100 mL. Interesting results may be obtained by comparing the acidity of both samples of vinegar with factory analysis data. 2. Principle The content of acetic acid in vinegar will be determined by acid – base titration with the standard solution of sodium hydroxide with approximate concentration c(NaOH) = 1 mol/L. This is prepared using the primary standard reagent potassium hydrogen phthalate. The time for one titration is about 10 minutes. 2.1. Acid – base titration An acid – base titration is a procedure used in quantitative chemical analysis to determine the concentration of either an acid or a base. Titration is the slow addition of an acid (or a base) of known concentration from a burette (a narrow graduated cylinder) to a base (or an acid) of unknown concentration in an Erlenmeyer flask. We distinguish between strong and weak acids and also between strong and weak bases. So, four types of acid – base titrations are possible: strong acid – strong base, strong acid – weak base, weak acid – strong base, and weak acid – weak base. The end-point occurs when the stoichiometric amount of base (or acid) has been added to the acid (or the base). The end-point is frequently detected using a visual indicator. An acid – base indicator is a substance, which changes colour with pH. For each type of acid – base titration an appropriate indicator must be chosen. See table 1, below. However, the end-point can also be determined potentiometrically using a pH meter or by a conductometric method. At this point, all the acid has been neutralised and neither excess base nor excess acid is present in the solution. The…

    • 2180 Words
    • 9 Pages
    Powerful Essays