Sample Lab Report
March 28, 2013
Ms. Au-Yeung
Honors Chemistry
Title:
Learning about Acids and Bases through Titration
Introduction/Purpose:
The purpose of this experiment was to observe how titration neutralizes a solution made of an
acid and a base. The pH scale lets one know if a substance is acidic or basic. Neutral water is 7 on the pH
scale. Anything less than 7 on the pH scale is acidic. An acid is a chemical that dissolves in water to
create more H+ ions than there are in neutral water. A base is a chemical that dissolves in water to
create fewer H+ ions than there are in neutral water or, equivalently, more OH- ions. When equal
amounts of an acid and a base are combined they create neutral water.
In the experiment juice was titrated to find the amount of vitamin C in the sample. Titration is a
type of neutralization that is used to determine the pH in a solution. The chemical formula for vitamin C
is C6H8C6. The Acid in the experiment was the vitamin C. Vitamin C can also be called ascorbic acid. The
base in the experiment was the KIO3 that was slowly poured in to the acid mixture. When the base was
poured in to the acid mixture the solution turned a different color to indicate it had been titrated. Below
are the two chemical equations that show how the ions react with each other for the base and ascorbic
acid to neutralize each other.
IO3- + 5I + 6H+  3I2+ + 3H2O
I2 + C6H8O6  C6H6O6 + 2H + + 2I-
The two equations above show how the ions react with each other in the experiment to neutralize the
base in acid. As shown in the first equation the ion of KIO3 – IO3 accepts a proton when it is neutralized.
One of the products is Iodine. The Iodine is then plugged in to the second equation and it reacts with the
vitamin C. The reaction between the iodine and the vitamin C is what causes the acid mixture to change
a color in the lab.
Hypothesis:
The lemonade that was used to perform this lab contained 240mL per serving size. 10% of the
240mL of lemonade is the daily amount of vitamin C that the lemonade contains. A human is supposed
to consume 60mg of vitamin C per day. Refer to the figure below to see how many milliliters will of
lemonade would be needed to be consumed for a human to get his/her daily amount of Vitamin C.
60mg Vitamin C * .10 = 6mg Vitamin C
Every 240mL serving of lemonade = 6mg Vitamin C
6mg * 10 = 60mg so 240mL * 10 = 2400mL
2400mL serving of lemonade = 60mg of vitamin C
As shown above through a few calculations above 2400mL of lemonade will need to be consumed by a
human for them to get the 60mg of vitamin C they are supposed to consume a day.
Methods:
There were three trials done for this experiment. For each experiment 25 mL of Country Time
lemonade was measured and poured in to a beaker. Then 5 mL of HCl, 6-7 mL of KI and 15-20 drops of
starch were added to the lemonade. Then KIO3 was poured in to the buret, and measured. Then slowly
KIO3 was released from the buret in to the lemonade mixture in the beaker. Occasionally the lemonade
mixture was stirred lightly to make sure the base was completely mixing with it, when the mixture
turned a different color that indicated that it was titrated. The color the lemonade turned was navy
blue.
Results:
Trial
Base (mL)
Indicator
1
0.5 KIO3
5mL KI, 15-20 drops of starch
2
1.0 KIO3
5mL KI, 15-20 drops of starch
3
0.6 KIO3
10mL KI, 15-20 drops of starch
Above is the table the lab group used to keep track of how much base and indicator was used.
The trials vary. Below is the one of the calculations performed to find how much vitamin C was in one of
the trials above. The calculation shown below is on Trial #2.
1mL KIO3 = .001 L * .01/L * 3 mole I2/1 mole KIO3 * 1 mole C6H8O6/ 1 mole I2 = .0003 mole C6H8O6
.0003 C6H8O6 * 176.1232g C6H8O6/1 mole = .005283696g C6H8O6
Stoichiometry was performed on the amount of KIO3 used to titrate the juice sample to discover
how much Vitamin C was in it. The final amount was .00528369g of vitamin C in the acid of Trial #2.
Discussion and Conclusion:
The data from the table is confusing. It does not make sense how on Trial 3 even though 10mL
of KI were used the amount of KIO3 used to titrate that substance was still in the same range as the
other two trials. However, the KI is not an acid, so maybe it just contributed to there being more
indicator then there should be in the lemonade mixture. The results from the lab were not close to the
hypothesis made. Below are the calculations showing how much milliliters of lemonade should be
consumed to reach the daily amount of vitamin C a day, which is 60mg.
25 mL lemonade = .005283696gVitamin C
.005283696g * 1000 = 5mg Vitamin C
5mg * 12 = 60mg Vitamin C
25mL lemonade * 12 = 300mL
300mL lemonade = 60mg Vitamin C
As shown above only 300mL of lemonade need to be consumed to reach the daily amount of vitamin C
humans should consume. However, this is wrong because the serving size of the lemonade is 240mL,
which is only 60mL less than the given amount from the calculations given above. The nutritional facts
on the lemonade bottle said that the amount of vitamin C in the bottle was only 10% of how much
vitamin C should be consumed. The amount of vitamin C that is said to be consumed by the calculations
above (5mg) is only one less than the amount of vitamin C that is said to be consumed by the nutritional
facts on the lemonade bottle (6mg). There must have been an error in the lab. Maybe too much KIO3
was used to titrate the acid mixture, and that affected the results. However, the hypothesis is the
correct amount of milliliters that needs to be consumed for a human to get their daily dosage of vitamin
C. It is not based off of a lab; it is based solely off of the numbers given by facts. In conclusion 2400mL of
lemonade still needs to be consumed to get the daily amount of vitamin C.