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Acetic Acid Lab

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by

hannah yi

on 30 October 2014

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Transcript of Acetic Acid Lab

Prelab
Challenge
Determine the concentration of acetic acid in a sample of household Vinegar
Hypothesis
If sodium hydroxide is added, then the pH of acetic acid solution will rise initially and rise drastically and then slow down again. Concentration of acetic acid in vinegar won't be very high since it is used in food.

Prelab (cont.)
Independent Variables
NaOH (base with known solution)
Dependent Variables
pH and equivalence point of acetic solution and NaOH is added in an excess amount
Controls
Concentration of acetic acid
0.1mol NaOH
Procedure Outline
1) Put 10 mL of Vinegar and 50 mL distilled water
2) Place pH probe into beaker and on clamp
3) Fill the burette with 50 mL of titrant
4) Place the beaker under the burette
5) Turn the stopcock of the burette slowly and let a small amount of the titrant out at a time.
6) Record the volume every .15pH increases
7) Add titrant drop by drop as you near the endpoint
8) Stop when the pH of acetic acid remains constant
9) Calculate the concentration of the acetic acid

Procedure diagram
Acetic Acid Lab
Data:
Analysis
Analysis
Calculation
1) Equivalence point
(10.5mL+7mL)/2=8.75mL NaOH
2) Find moles of NaOH
8.75mL NaOH *1L/1000mL*o.1oomolNaOH= 8.75*10^-4
3) Calculate moles of CH3COOH
Mole ratio is 1:1
8.75*10^-4molNaOH*1molCH3COOH/1molNaOH=8.75*10^-4
4) Find concentration of CH3COOH in Vinegar
mol/vol=8.75*10^-4mol/.010L=[.0875]

Conclusion:
Error Analysis
Background
Titration of a strong acid with a strong base. We had to find the equivalence point for this we took two points one where the graph started to become straight( 11.5 ml) and the point where the straight line came to become a curve( 12.1 ml). Then we add the two points and then divide it by 2. The equivalence point is 11.8ml.
From the equivalence point we got 11.8 ml of CH3COOH. We got the the moles of CH3COOH which was 11.8 * 10^-4 mol NaOH.
CH3COOH+ NaOH ----> CH3COONa + H2O is the balanced fromula. We can easily determine the mols of NaOH used as the reaction is in the ratio of 1:1 therefore 11.8*10^-4 mols of NaOH.
Post Lab Questions:
Q1. Determine the equivalence post.
Q2. Determine the moles of CH3COOH.
Q3.Write a balanced formula for the reaction. Determine the moles of NaOH used.
Safety
NaOH is hazardous and corrosive:
-causes serious skin burns, scars or blindness
-contact with eyes or skin should be avoided
-goggles and aprons must be worn at all times
to avoid contact with NaOH

Balance Equation
CH3COOH(aq)+NaOH(aq)-->CH3COONa(aq)+H2O
The ratio is 1:1

Causes of Error:
The air bubbles in the burette can cause error in true reading of NaOH used
Water left after washing glassware can decrease the concentration
Not stirring properly can cause errors
The pH did not stabilize fully every time we took a reading



Analyte:
Acetic acid (ethanoic acid=vinegar acid) is a weak organic acid (pH 4.76) that has sour taste and strong taste. It is colourless and soluble in water. Due to its properties, it is commonly used in making perfumes, inks, pesticides, and even in medicine because it cures coughs and an antibiotic.
Titrant:
Sodium hydroxide, also known as caustic soda, is a white inorganic compound and it is hazardous and corrosive. Contact with eyes and skins should be avoided.
Finding the concentration of Acetic Acid in Vinegar
CH3COOH(aq)+NaOH(aq)-->CH3COONa(aq)+H2O
Bibliography
"Properties of Vinegar." Properties of Vinegar. N.p., n.d. Web. 30 Oct. 2014 <http://www.enzyme-facts.com/properties-of-vinegar.html>

"The Germ-Killing Characteristics of Vinegar." EHow. Demand Media, 23 Apr. 2011. Web. 30 Oct. 2014.
<http://www.ehow.com/info_8285896_germkilling-characteristics-vinegar.html>

122L, Chem, General Chemistry Laboratory, and Revision 1.4. The Titration of Acetic Acid in Vinegar (n.d.): n. pag. Web.
<http://infohost.nmt.edu/~jaltig/Vinegar.pdf>

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