Academy of Art University The Determination of Acetic Acid in Vinegar Lab 8 Hi I have HW in chemistry class, they are 5 Qs and I need the full answers to

Academy of Art University The Determination of Acetic Acid in Vinegar Lab 8 Hi

I have HW in chemistry class, they are 5 Qs and I need the full answers to give them to my teacher, Just watch the videos and answers the Qs

QUESTION 1
Describe the appearance of the solution at the end of reaction.
QUESTION 2
Concentration of standardized NaOH = M
Volume of vinegar sample = mL
Answer with the correct number of significant figures.

QUESTION 3
Trial 1Complete the table below:
Final buret reading mL
Initial buret reading mL
Volume titrated mL

Concentration of acetic acid in vinegar = M

QUESTION 4
Trial 2Complete the table below:
Final buret reading mL
Initial buret reading mL
Volume titrated mL

Concentration of acetic acid in vinegar = M
QUESTION 5
Trial 3Complete the table below:
Final buret reading mL
Initial buret reading mL
Volume titrated mL

Concentration of acetic acid in vinegar = M

QUESTION 6
Average molarity of the three trials: M
Standard deviation: %
Given the average molarity of acetic acid in your vinegar sample, what is the percent acetic acid in your sample? %
What is the % error? Generally, vinegar contains 5% of acetic acid. %

QUESTION 7

see the file for this one CHEM 118 Summer 2020
Experiment 8
THE DETERMINATION OF ACETIC
ACID IN VINEGAR
INTRODUCTION
A
technique commonly used in chemistry laboratories is one known as a titration. In general
this involves the addition of a certain measured amount of one solution, such as an acid,
to a measured amount of a second solution, such as a base, until the basic solution has
been just neutralized. Alternatively, the base may be added to the acid, as will be done in
this experiment. In either case, the generalized neutralization reaction that results is:
ACID + BASE
→
SALT + WATER
A substance is called an acid if it is a proton donating substance. In the presence of water, an acid
releases the H+ ions:
H(CH3CO2)
→
H+
+
acetic acid
CH3COOacetate ion
A base is a substance that can accept these H+ ions in the presence of water.
+ H+
NaOH
→
Na+
+ OH-
→
H2O
+
Na+
Sodium Hydroxide
A neutralization reaction occurs if both the acid and base are present in the water solution in
equivalent amounts.
H(CH3CO2)
acid
+
NaOH
base
→
H2O
water
+
Na (CH3CO2)
salt
An amount of either compound that is less than the equivalent amount would leave the solution
either acidic or basic, as the case may be.
There are several methods for ascertaining whether a solution has been neutralized, such
as using litmus paper, indicator dyes, and pH meters. In this experiment you will determine the
amount of acetic acid in a sample of vinegar by titrating with a standardized base, and using
phenolphthalein as an indicator.
EXPERIMENTAL
PROCEDURE:
Obtain a 50 mL buret from the stockroom. Using a repipetter set for 4.00 mL, obtain a
sample of vinegar. Record the volume on the data sheet. Record volume to three
significant figures!
Clean the buret, rinse with distilled water, and then rinse with the NaOH. Fill the buret
with NaOH and read and record the volume to the nearest 0.02 mL. Be sure to record the molarity
of the base used on your data report form!!
Add 3 drops phenolphthalein to the vinegar sample. Place the sample under the buret
and slowly add the base from the buret, while gently swirling the flask to mix
As you get closer to the neutralization point of the titration a faint pink color will appear,
and then it will disappear within a few seconds. When this happens, add the base very slowly
until the pink color persists for at least 30 seconds. Record the volume to the nearest 0.02 mL.
Repeat the above procedure for two other samples.
CALCULATIONS
Calculate the molarity of acetic acid in each sample. Remember you need
to know the concentration of the standardized base, the volume of base you titrated, and the
volume of acid sample you used. Determine the average molarity of the three titrations and its
standard deviation.
Standard deviation (S) =
2
!∑ X –
( ∑ x)2
n
n-1
To show how this equation works, suppose we measure the length of several objects and obtain
lengths of 50.1, 49.4, 51.2, and 49.9 cm. What is the standard deviation of our sample? Using the
above equation, we first must add up the measurements, as well as their squares:
x
x2
50.1
2510.0
49.4
2440.4
51.2
2621.4
49.9
2490.0
Sx = 200.6
Sx2 = 10061.8
Note that n = 4, since there have been four measurements. Now, substituting into the above
equation, we get:
Standard deviation (S) =
!10061.8 –
(200.6)2
4
3
= 0.753
Next, express the molarity of your acid as a percent (wt/vol). Finally, do an error
analysis to determine the accuracy of your average value compared to the expected value for
percent acetic acid.
% Error =
|observed value –expected value|
expected value
´ 100
Question: For the reaction: N2 (g) + O2 (g)
2NO(g) Kc = 4.10 x 10-4 at 2000°C. A
reaction mixture initially contains 0.30 mol of N2 (g) and 0.34 mol of O2 (g) in a 2.0 L container.
What is the concentration of NO (g) at equilibrium?

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