simon

Topics: Base, PH, Acid Pages: 5 (803 words) Published: September 7, 2014
Lab Name:Titration of NaOH
Author: Simon Sun
Lab Partner: Jason Lee
& Fiona Wang
Date: March 10th & 17th

Introduction:
The purpose of this lab is to observe the titration between acid and base, and calculate the morality of unknown solution base on the volume required to reach equivalence point and stoicheiometry. There are two groups of experiments. The first one is the reaction between nitric acid and sodium hydroxide which is a reaction between strong acid and strong base, so, theoretically, the pH value of the the mixture of these two solution at the equivalence point in the titration process should be seven, which means neutral. The second groups is the titration between ammonia and hydrochloric acid. The pH at the equivalence point should be below seven to indicate that the mixture at that point is acid, since hydrochloric acid is a strong acid and ammonia is weak base. After find out the pH of the equivalence point, find out the corresponding volume of the two solutions and calculate the unknown molarity depends on stoicheiometry and the known molarity. In the lab, first step is to prepare 10 milliliter known-molarity solution in the beaker, sodium hydroxide in the first group and ammonia in the second group. Then prepare 30 milliliter unknown-molarity solution (nitric acid in the first group and hydrochloric acid in the second group) in the buret. Check the pH value of the solution by pH-meter in the beaker. Put the beaker under the buret and drop the solution in buret milliliter by milliliter and keep tracking the pH change by pH-meter, until pass through the equivalence point. Record all the data.

Data & Graph
Group 1, Trial I:
Data: 10mL HNO3
V_Nitric Acid (mL)
pH
1.0
2.10
2.0
2.17
3.0
2.28
4.0
2.53
4.5
2.78
5.0
3.69
5.1
6.70
5.2
9.76
5.4
10.55
5.5
10.83
5.7
11.13
6.0
11.36
6.5
11.75
Group 1, Trial II:
Data: 10mL HNO3
V_Nitric Acid (mL)
pH
0.5
2.00
1.0
2.03
2.0
2.10
3.0
2.22
4.0
2.40
4.5
2.63
5.0
3.34
5.2
6.28
5.3
10.04
5.4
10.27
5.5
10.67
5.6
10.87
6.0
11.37
6.5
11.60
7.0
11.75
Equivalence pH(Trial I)=(3.69+10.55)/2=7.12
V_NaOH@equivalence(Trial I)=5.15mL
Molarity_HNO3(Trial I)=5.15*.1/10=.0515M
Equivalence pH(Trial II)=(3.34+10.87)/2= 7.11
V_NaOH@equivalence(Trial II)=5.23mL
Molarity_HNO3(Trial II)=5.23*.1/10=.0523M
Equivalence pH(Group 1)=(7.12+7.11)/2=7.12

Molarity_HNO3(Group 1)=(0.515+0.523)/2=0.52M
Group2, Trial I
Data: 5mL NH3
V_HCl(mL)
pH
0.5
9.90
1.0
9.41
1.5
8.78
1.7
8.16
1.8
3.09
1.9
2.81
2.0
2.59
2.5
3.16
3.0
1.98
3.5
1.88

Group2, Trial II
Data:5mL NH3
V_HCl(mL)
pH
0.5
9.77
1.0
9.28
1.5
8.40
1.7
2.81
1.8
2.52
1.9
2.31
2.0
2.22
2.5
1.95
3.0
1.80
3.5
1.70
Equivalence pH(Trial I)=(8.16+2.81)/2=5.49
V_HCl@equivalence(Trial I)=1.85mL
Molarity_NH3(Trial I)=.2*10/1.85=.54M
Equivalence pH(Trial II)=(8.40+2.52)/2=5.46
V_HCl@equivalence(Trial II)=1.65mL
Molarity_NH3(Trial II)=5*.2/1.65=.61M
Equivalence pH(Group 2)=(5.49+5.46)/2=5.475
Molarity_NH3(Group 2)=.58M

Analysis & Conclusion
Post Lab:
Vinegar is a dilute aqueous solution of acetic acid produced by the fermentation of apple juice (cider vinegar), grapes (wine vinegar), or barley malt (malt vinegar). Federal regulations require that vinegar contain at least 4% acetic acid by mass. If the amount of acetic acid is less than 4%, the acidity level may not be high enough to preven the growth of bacteria in pickled or cannel foods. The amount of acetic acid in vinegar can be determined by microscale titration with a standard solution of sodium hydroxide. 1. write the balanced chemical equation for the reaction acetic acid with sodium hydroxide. H^(+) (aq)+OH^(-)(aq)——>H2O(l)

CH3COOH(aq)+NaOH(aq)——CH3COONa(aq)+H2O(l)

In the microscale titration, the exact number of drops of sodium hydroxide of known molarity (.50 M) needed to react...
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