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Write a Short Lab Report for a conducted experiment on Voltaic Cell.A screen shots were provided of the Lab Manual. For your , procedure, part of the lab report process use,95-102 as pages. Short lab...

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Sample Chemistry 218 Laboratory Report
Short Report Format
Title: Acid Base Titrations — The Percentage of Acetic Acid in Vinega
A. Student, #990001
Lab Partner: M. Y. Partne
Purpose: To use a quantitative titration method to determine the percentage of acetic acid in
vinegar. Also to learn the concept of neutralization, moles, dilution, and stoichiometry as it applies
to titration data. This was accomplished by first determining what volume of standard hydrochloric
acid (of known concentration) will neutralize a known volume of sodium hydroxide of unknown
concentration, using titration with the aid of a pH indicator (see Equation 1). Once the concentration
of sodium hydroxide was standardized, it was used to determine the concentration of acetic acid in a
diluted unknown vinegar solution (see Equation 2).
A minor objective of the experiment was to learn how to properly clean, operate and
ead a burette.
Equation 1: HCI (ag) + NaOH (aq) =<—— NaCl (aq) + H,0 () + heat
Equation2: CH,COOH (ag) + NaOH (ag) ==== CH,COO" Na* (aq) + H,O (I) + heat
Procedure: The experiment was ca
ied out as outlined in Experiment R3 of the Chemistry 218
Laboratory Manual, pp. xx-yy. Note: 1.60 g of NaOH was dissolved in ~400 mL water in Part A step 1.
Otherwise no changes or modification were made.
Observations: Part A. During Titration of XXXXXXXXXXM HCI with Unknown [NaOH]:

1. Cresol red pH indicator in HCl initially reddish-orange.
2. After addition of NaOH, at the endpoint, solution was pink.
Part B. During Titration of CH3COOH with Standardized NaOH:

1. Phenolphthalein pH indicator in CH;COOH initially was clear and colourless.
Mass Percentage of Acetic Acid in Vinega
Given: Density of vinegar = 1.005 g/mL = 1.005 g/mL x 1000 mL/L = 1005 g/L
Therefore, 60.06 g/L CHsCOOH/1005 g/L) x 100% = 5.57% CH3COOH in vinega
Answers to Questions:
1. Write the net ionic equation for:
a. HCl + NaOH
HCI (aq) + NaOH (aq) =~— NaCl (aq) + H,0 (I)
H* (aq) + CI" (aq) + Na* + OH (aq) =— Na" (aq) + CI" (aq) + H,0 (I)
H* (ag) + OH (aq) === H,0 ()
. Acetic acid + NaOH
CH;COOH (aq) + NaOH (aq) === CH,COO" Na* (aq) + H,O (I)
CH4;COOH (aq) + Na*(aq) + OH" (ag) === CH;COO" Na* (aq) + H,O0 (I)
CH,COOH (aq) + OH (aq) =< CH;COO" (aq) + H,0 (I)
Conclusion: The percentage of acetic acid in vinegar was determined to be 5.57% (w/w).
The concentration of the sodium hydroxide solution used to titrate the acetic acid was determined
to be XXXXXXXXXXM. The titration method proved to be easy to perform and very precise in measuring
concentrations of acids and bases. It is worthy of note that e
or was introduced into the percentage
of acetic acid measurement because of the use of uncali
ated pipettes and burettes.
Answered 2 days After Apr 18, 2024


Dr Shweta answered on Apr 21 2024
6 Votes
Voltaic cell Experiment
1. Purpose of the experiment: The objective of this experiment is to construct multiple voltaic cells and quantify the electromotive force (emf), represented as Ecell, generated by each cell.
2. Introduction and Theory:
When zinc metal is introduced into a copper (II) sulfate solution, the zinc metal undergoes dissolution while metallic copper is precipitated. During this reaction, each zinc atom loses two electrons to become a zinc ion, which means that the zinc is undergoing oxidation. On the other hand, each copper (II) ion gains two electrons to become a copper atom, indicating that the copper (II) ions are being reduced to copper metal.
Zn (s) + CuSO4 (aq) Cu (s) + ZnSO4 (aq)
Zinc metal copper (II) sulfate copper metal zinc sulfate
To express these two processes, we can write the following half-equations:
Zn (s) Zn2+ (aq) + 2e- (aq)
Cu2+ (aq) + 2e- (aq) Cu (s)
Consequently, the entire procedure may be depicted as:
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
In the aforementioned reaction, the electrons are immediately transported from the zinc metal to the copper (II) ions. In a basic voltaic cell, the species responsible for giving and receiving electrons are placed in separate containers called half-cells. In each half-cell, the metal involved is partially...

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