Question

Please help, I need 3 to 4 sentences each for questions 1-4.

Chemical reaction equations Lab 1. KHC8H404(aq) + NaOH(aq) + H20 (1) + NaKC8H404(aq) Lab 2. HCH3CO2(aq) + NaOH(aq) → H20(1) + NaCH3CO2 (aq) (Eq 1) (Eq 2) Results Brand of vinegar analyzed in Lab 2: Great Value Amount of acetic acid in the vinegar claimed on the label: 5 % acidity Name of indicator used: Thymol Blue Experimental concentration (molarity) of NaOH from Lab 1: 0.4705 mol/L Experimental concentration (molarity) of acetic acid in the vinegar from Lab 2: 0.8630 mol/L Experimental % by mass of acetic acid in the vinegar from Lab 2: 5.13 % Discussion (1) Write the Results (above) into a short paragraph reporting the main conclusions of the two Labs. Use a scientific writing style and keep the focus to a concise factual statement that could be submitted to a supervisor. (Paragraph of 3 – 4 sentences.) (type your answer here....)

When answering, please use a scientific writing style as it states in the discussion section.

Answer 1

Answer to part 1:

The main purpose of the experiments in two labs was to determine and verify the concentration of acetic acid in the vinegar. At first in the Lab 1, the NaOH solution is standardized by titration with a known concentration of potassium phthalate. For the titration thymol blue was used as an indicator. After knowing the strength of NaOH, the NaOH was used as a titrant for the titration of the vinegar with NaOH where thymol blue is an indicator. The titration rreveals the presence of 5.13 % acetic acid in the vinegar which is very close to the value specified in the label of 5 %.

Answer to part 2:

The amount of acetic acid claimed on the label is 5 % and the experimentally obtained amount is 5.13 %. Keeping in mind that in vinegar the acetic acid is produced by fermentation, with time the amount of acetic acid would increase slightly with time. Thus, the experimentally determined amount of 5.13 % acetic acid is matching very close to the 5 % amount claimed on the label, and the company's claim is valid.

Answer to part 3

The 5.13 % acetic acid is equivalent to
(5.13 x 10)/1000 = 51.39
in 1000 mL = 0.85428 M AcOH, which is slightly different by 0.0087 M than the experimentally determined value of 0.8630 M. This amount of difference can happen due to different values (slight changes) of equivalence point from the titration. Here the average the data from average values for the molarity of the experimental concentration of AcOH is omitted because the data considered is upto 4 decimal points while the data should be more than 4 decimal points. The titrant would fail for 0.15 mL titrant test because the 0.15 mL titrant equals to
(0.15 x 0.4705)/1000 = 0.0000705M
NaOH which is less than the value of 0.0087 M.

Answer to part 4:

In the titration of AcOH and NaOH, at the equivalence point salt of a weak acid and NaOH a strong base is present (P^H > 7.0), additionally the P^H of AcO- would be > 7.0 as it is conjugate base of a strong weak acid. Below P^H 8.0 the color of the solution would be yellow and above 9.6 the color of the solution would be blue as the P^H range for thymol blue is 8.0-9.6. At the end point, the color would be green which is equimolar mixtures of blue and yellow. The equivalence point occurs when the color is green.