Rutgers University Inorganic and Organic Titration of Fruit Juice Chemistry Lab Introductory Video https://www.youtube.com/watch?v=_NMkxjpemeg&feature=yout
Rutgers University Inorganic and Organic Titration of Fruit Juice Chemistry Lab Introductory Video https://www.youtube.com/watch?v=_NMkxjpemeg&feature=youtu.be
Experimental Video: https://www.youtube.com/watch?v=j8QsCqILKlM
Document attached
Using the data provided, perform the calculations need to fill in the missing values on the report sheet. Do not use the data given in the videos.
Submit the lab as an attachment to this assignment (Word document only).
Professor’s Instructions Name:
CH 111 Introduction to Inorganic and Organic Chemistry
Exp. 625 Titration of Fruit Juice
INSTRUCTIONS
1.
2.
3.
4.
5.
Print out these instructions and the report sheet.
Read the Background/Introduction section of the lab manual and watch the intro video.
Watch the videos attached under the instructions.
Do the calculations as shown on the next page.
Submit both the report sheet and any supplementary questions as either a .doc or .pdf file and submit them as
an attached file under the lab assignment. No labs will be accepted by e-mail.
In this titration orange juice is used as the acid in place of HCl. That is because most fruit juices are acidic, with citric acid
(H3C6H5O7) being one of the more common acids, especially and not surprisingly in citrus fruits. One big difference
between citric acid and HCl is that citric acid is triprotic, meaning it has three protons (H+) to donate, so as you can see in
the balanced chemical equation below, it would take 3 times as many moles of NaOH to neutralize the H3C6H5O7. This
means the molar ratio of H3C6H5O7:NaOH is 1:3, and it is reflected in the calculations shown below. For this experiment,
we assume that all the acid present in the fruit juice is citric acid to actually make the calculations possible.
H3C6H5O7 (aq) + 3 NaOH (aq)
Na3C6H5O7(aq) + H2O (aq)
(Citric Acid)
Report Sheet
Data
Molarity of NaOH: 0.1065 M
Volume of Fruit Juice: 20.0 mL
Trial
Final buret reading, mL
Initial buret reading, mL
Volume of NaOH (titrant)
1
0.00
16.80
2
16.80
33.12
Results and Calculations
Trial
1
Calculate the number of mL of NaOH required for the titration
Calculate the number of moles of NaOH
Calculate the number of moles of H3C6H5O7 titrated
Calculate the mass of H3C6H5O7 present in the juice
Calculate the mass of H3C6H5O7 present per mL of juice
Calculate the average mass of present per mL of juice
2
Supplementary Questions
None
Calculations
Volume NaOH used in titration = Volume NaOH final – Volume NaOH initial
Moles NaOH = volume NaOH used in Liters (convert!) x Molarity NaOH
To find Moles H3C6H5O7 – Since the molar ratio of H3C6H5O7H and NaOH from the balanced equation is 1:3, you need 3
moles of NaOH to neutralize 1 mole of H3C6H5O7. Or you need 1/3 the moles of H3C6H5O7 to neutralize the moles of
NaOH. So
.
Moles of H3C6H5O7 = (moles NaOH)/3
Mass of H3C6H5O7 = (moles of H3C6H5O7) x (Molar Mass H3C6H5O7)
Mass H3C6H5O7 in 1.0 mL of juice (g/mL) =
Mass ??3 ??6 ??5 ??7
??????. ??????????
Name:
CH 111 Introduction to Inorganic and Organic Chemistry
Exp. 625 Titration of Fruit Juice
INSTRUCTIONS
1.
2.
3.
4.
5.
Print out these instructions and the report sheet.
Read the Background/Introduction section of the lab manual and watch the intro video.
Watch the videos attached under the instructions.
Do the calculations as shown on the next page.
Submit both the report sheet and any supplementary questions as either a .doc or .pdf file and submit them as
an attached file under the lab assignment. No labs will be accepted by e-mail.
In this titration orange juice is used as the acid in place of HCl. That is because most fruit juices are acidic, with citric acid
(H3C6H5O7) being one of the more common acids, especially and not surprisingly in citrus fruits. One big difference
between citric acid and HCl is that citric acid is triprotic, meaning it has three protons (H+) to donate, so as you can see in
the balanced chemical equation below, it would take 3 times as many moles of NaOH to neutralize the H3C6H5O7. This
means the molar ratio of H3C6H5O7:NaOH is 1:3, and it is reflected in the calculations shown below. For this experiment,
we assume that all the acid present in the fruit juice is citric acid to actually make the calculations possible.
H3C6H5O7 (aq) + 3 NaOH (aq)
Na3C6H5O7(aq) + H2O (aq)
(Citric Acid)
Report Sheet
Data
Molarity of NaOH: 0.1065 M
Volume of Fruit Juice: 20.0 mL
Trial
Final buret reading, mL
Initial buret reading, mL
Volume of NaOH (titrant)
1
0.00
16.80
2
16.80
33.12
Results and Calculations
Trial
1
Calculate the number of mL of NaOH required for the titration
Calculate the number of moles of NaOH
Calculate the number of moles of H3C6H5O7 titrated
Calculate the mass of H3C6H5O7 present in the juice
Calculate the mass of H3C6H5O7 present per mL of juice
Calculate the average mass of present per mL of juice
2
Supplementary Questions
None
Calculations
Volume NaOH used in titration = Volume NaOH final – Volume NaOH initial
Moles NaOH = volume NaOH used in Liters (convert!) x Molarity NaOH
To find Moles H3C6H5O7 – Since the molar ratio of H3C6H5O7H and NaOH from the balanced equation is 1:3, you need 3
moles of NaOH to neutralize 1 mole of H3C6H5O7. Or you need 1/3 the moles of H3C6H5O7 to neutralize the moles of
NaOH. So
.
Moles of H3C6H5O7 = (moles NaOH)/3
Mass of H3C6H5O7 = (moles of H3C6H5O7) x (Molar Mass H3C6H5O7)
Mass H3C6H5O7 in 1.0 mL of juice (g/mL) =
Mass ??3 ??6 ??5 ??7
??????. ??????????
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