ACID BASE TITRATION
I. DAY/ DATE OF EXPERIMENT:
We did the experiment on Friday, 4th November 2011.
II. COMPLETED THE EXPERIMENT
We finished the experiment also on Friday, 4th November 2011
III. THE PURPOSE OF THE EXPERIMENT
· To determine the concentration of NaOH by Oxalic Acid solution
· To determine the concentration of HCl by NaOH
IV. BASIC THEORY
Acid-base titration is a technique that is widely used to determine the exact concentration of an acid or alkaline solution. Titration is a measurement of a solution of one reactant needed to react completely with a number of other specific reactants. Acid-base titration is a neutralization reaction. If the solution is called acidic raw acidimetry and if the solution is called alkalimetri raw base. Titration is a quantitative chemical analysis techniques used to determine the concentration of a particular solution, where the determination using a standard solution of known concentration is appropriately. Solution used for the determination of the solution of the unknown concentration is placed in a burette (see figure) and the solution is referred to as the standard solution or titrant or titrator, while the solution of unknown concentration is placed in Erlenmeyer (see picture) and is referred to as the analyte solution. Titrant is added little by little on the circumstances in which the analyte to obtain equivalent titrant reacts with the analyte, which means that all titrant reacts with the analyte out of this situation is referred to as a point equivalent. Equivalent point can be determined by a variety of ways, a common way is to use indicators. The indicator will change color with the addition of as little as possible titrant, in this way then we can immediately stop the process of titration. In other words, the acid-base titration of acid equivalent amount equal to the number of equivalent base.
Phenolphthalein indicator
Phenolphthalein is an organic compound having the formula C20H14O4, solid grain shaped, colorless and soluble in alcohol and organic solvents. Generally, using phenolphthalein as an indicator of acid-base (in acid solution is colorless and colored red). Phenolphthalein is an acid-base indicators are included in kelasftalein. Fenolftalein an indicator of one color. Route pH at 8.3 to 9.8 ie discoloration from colorless to red. This indicator is very suitable as an indicator for the titration of acids with strong bases. The acid can be titrated strong acid or weakacid with Ka price limits are eligible within perfectness. Fenolftalein reaction are compounds containing phenol groups, so that weakly acidic.
Anthocyanin
Anthocyanins are the pigments of red, purple, and blue found in all plants except Fungus. Most of the anthocyanin in the form of glycosides, usually bound one or two units of sugar such as glucose, galactose, ramnosa, and silosa. If monoglikosida, so the part of sugar only bound at position 3, and at position 3 and 5 when a diglikosida danbagian
aglikionnya called anthocyanidins. Most of anthocyanin reddish in acid solution, but the purple and blue with increasing pH which ultimately damaged in strong alkalis solution (Sinaga, 2010). The color resulting from anthocyanin is influenced by medium acidity, anthocyanin in acidic situation pH 1-3 showed a red color while pH caused increase resulting decrease in color intensity (Abbas, 2003). It shows that anthocyanin pigments are not stable against changes in pH and temperature.
Phenolphthalein is an organic compound having the formula C20H14O4, solid grain shaped, colorless and soluble in alcohol and organic solvents. Generally, using phenolphthalein as an indicator of acid-base (in acid solution is colorless and colored red). Phenolphthalein is an acid-base indicators are included in kelasftalein. Fenolftalein an indicator of one color. Route pH at 8.3 to 9.8 ie discoloration from colorless to red. This indicator is very suitable as an indicator for the titration of acids with strong bases. The acid can be titrated strong acid or weakacid with Ka price limits are eligible within perfectness. Fenolftalein reaction are compounds containing phenol groups, so that weakly acidic.
Anthocyanin
Anthocyanins are the pigments of red, purple, and blue found in all plants except Fungus. Most of the anthocyanin in the form of glycosides, usually bound one or two units of sugar such as glucose, galactose, ramnosa, and silosa. If monoglikosida, so the part of sugar only bound at position 3, and at position 3 and 5 when a diglikosida danbagian
aglikionnya called anthocyanidins. Most of anthocyanin reddish in acid solution, but the purple and blue with increasing pH which ultimately damaged in strong alkalis solution (Sinaga, 2010). The color resulting from anthocyanin is influenced by medium acidity, anthocyanin in acidic situation pH 1-3 showed a red color while pH caused increase resulting decrease in color intensity (Abbas, 2003). It shows that anthocyanin pigments are not stable against changes in pH and temperature.
There are 2 kind of standard solution. A standard solution is prepared by dissolving an accurately weighed quantity of a highly pure material and diluting to an accurately known volume of volumetric flask is called primary standard. A solution standardized by titrating a primary standard itself a secondary standard. It will be less accurate than a primary standard due to the error of titration.
A primary standard should fulfill these requirement;
1. It should be 100% pure, although 0.01 to 0.02% impurity is tolerable if it is accurately known.
2. it should be stable to drying temperature, and it should be stable indefinitely at room temperature.
1. It should be 100% pure, although 0.01 to 0.02% impurity is tolerable if it is accurately known.
2. it should be stable to drying temperature, and it should be stable indefinitely at room temperature.
The primary standard is always dried before weighing.*
3. It should be readily available and fairly inexpensive.
4. Although not necessary, it should have a high formula weight. This is so that relatively large amount of it will have to be weighed to get enough to titrate. The relative error in weighing a greater amount of material will be smaller than that for a small amount.
5. If it is to be used in titration, it should possess the properties required for a titration listed above. in particular, the equilibrium of the reaction should be far to the right so that a very sharp end point will be obtained.
4. Although not necessary, it should have a high formula weight. This is so that relatively large amount of it will have to be weighed to get enough to titrate. The relative error in weighing a greater amount of material will be smaller than that for a small amount.
5. If it is to be used in titration, it should possess the properties required for a titration listed above. in particular, the equilibrium of the reaction should be far to the right so that a very sharp end point will be obtained.
V. EXPERIMENTAL METHODS
A. Tools and Materials:
1. Stative and clamps
2. Burette
3. Erlenmeyer flask 250 ml
4. Funnel
5. Pipette mumps 25 ml
6. Pipette drops
7. Beaker 100 ml
8. Measuring cup
9. NaOH 0.1 m
10. C2H2O4 0.1 m
11. HCl 0.1 m
12. Phenolptalein
13. Aquades
14. Plant extracts
15. Pipette filler
16. Mortar and alu
VI. EXPERIMENT RESULT
1. Determining of NaOH solution concentration with 5 ml of C2H2O4 0,05 M using PP indicator.
Titration
|
NaOH needed (ml)
|
First
|
9,7
|
Second
|
9,4
|
Third
|
9,6
|
2. Determining of HCl concentration with NaOH using PP indicator.
Titration
|
NaOH needed (ml)
|
First
|
9,6
|
Second
|
9,8
|
Third
|
9,25
|
3. Determining of HCl concentration with NaOH using plant extract indicator.
Titration
|
NaOH needed (ml)
|
First
|
9,5
|
Second
|
9,3
|
Third
|
9,2
|
VII. DATA ANALYSIS
1. Determining of NaOH solution concentration with C2H2O4 using PP indicator.
Titration
|
NaOH needed (ml)
|
NaOH Molarity
|
First
|
9,7
|
0,025
|
Second
|
9,4
|
0,0265
|
Third
|
9,6
|
0,026
|
Reaction:
2NaOH + C2H2O4 → Na2C2O4 + 2H2O
2. Determining of HCl concentration with NaOH using PP indicator.
Titration
|
NaOH needed (ml)
|
HCl Molarity
|
First
|
9,6
|
0,049
|
Second
|
9,8
|
0,050
|
Third
|
9,25
|
0,048
|
Reaction :
HCl + NaOH → NaCl + H2O
3. Determining of HCl concentration with NaOH using plant extract indicator.
Titration
|
NaOH needed (ml)
|
HCl Molarity
|
First
|
9,5
|
0,049
|
Second
|
9,3
|
0,048
|
Third
|
9,2
|
0,047
|
Reaction :
HCl + NaOH → NaCl + H2O
VIII. EXPLANATION
1. Determining of NaOH solution concentration with C2H2O4 using PP indicator.
Reaction happened is
2NaOH + C2H2O4 → Na2C2O4 + 2H2O
Titration
|
NaOH needed (ml)
|
NaOH Molarity
|
First
|
9,7
|
0,025
|
Second
|
9,4
|
0,0265
|
Third
|
9,6
|
0,026
|
NaOH molarity can be gotten like in the above data is obtained from calculation with diluting formula, that is M1.V1 = M2.V2 which solution 1 is NaOH, solution 2 is C2H2O4. For more complete, can be seen in calculation attachment pages.
2. Determining of HCl concentration with NaOH using PP indicator.
Reaction happened is
HCl + NaOH → NaCl + H2O
Titration
|
NaOH needed (ml)
|
HCl Molarity
|
First
|
9,6
|
0,049
|
Second
|
9,8
|
0,050
|
Third
|
9,25
|
0,048
|
HCl Molarity can be gotten like in the above data is obtained from calculation with diluting formula, that is that is M1.V1 = M2.V2 which solution 1 is NaOH, solution 2 is HCl. Here used average of NaOH Molarity from the previous experiment (First Experiment). For more complete, can be seen in calculation attachment pages.
3. Determining of HCl concentration with NaOH using plant extract indicator.
Reaction happened is
HCl + NaOH → NaCl + H2O
Titration
|
NaOH needed (ml)
|
HCl Molarity
|
First
|
9,5
|
0,049
|
Second
|
9,3
|
0,048
|
Third
|
9,2
|
0,047
|
Here plant extract indicator used is “bunga sepatu”
HCl Molarity can be gotten like in the above data is obtained from calculation with diluting formula, that is that is M1.V1 = M2.V2 which solution 1 is NaOH, solution 2 is HCl. Here used average of NaOH Molarity from the previous experiment (First Experiment). For more complete, can be seen in calculation attachment pages.
IX. CONCLUSION
· The average of NaOH Molarity obtained from First Experiment data is 0.026 M.
· The average of HCl Molarity obtained from Second Experiment data is 0,049 M.
· The average of HCl Molarity obtained from Third Experiment data is 0,048 M.
QUESTION AND ANSWER
1. Why in titration of NaOH with oxalic acid using phenolptalein as indicator?
2. What is the differences between equivalent point and final point?
3. On solution above, which serves as primary standard solution, secondary standard solution, and tertiary standard solution?
1. In this titration we use the indicator phenolptalein(PP) because oxalic acid is classified as a very weak acid in unionized condition but if in alkaline environment, PP will be ionized more and he will give a bright color and the color changes are easier to be observed
2. The equivalence point is the point in a titration when the amount of added standard reagent is chemically equal to the amount of analyte
(the point where the number of moles of base equal the number of moles of acid).
The end point is the point in a titration when a physical change occurring immediate after the equivalence point
The end point is the point in a titration when a physical change occurring immediate after the equivalence point
3. primary standard solution : H2C2O4 , because primary standard solution is a solution that has been known its concentration carefully
secondary standard solution : NaOH, because secondary standard solution ia a solution standardized by titrating a primary standard itself
tertiary standard solution : HCl,
REFERENCES
§ http://www.wikipedia.org
§ http://www.newton.dep.anl.gov/askasci/chem03/chem03996.htm
§ http://robbaniryo.com/ilmu-kimia/larutann-standar-primer-dan-sekunder/
ATTACHMENT
CALCULATION
1st Experiment
2NaOH + H2C2O4 ® Na2C204 + 2H2O
Finding Molarity of NaOH
a. First titration
M NaOH = ?
V NaOH = 9,7 ml
V H2C2O4 = 5 ml
M H2C2O4 = 0,05 ml
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M H2C2O4. V H2C2O4
1. x. 9,7 = 2. 0,05. 5
9,7 x = 10. 0,05
x = 0,05154 M
The normality of NaOH = 0,051 M
The molarity of NaOH = 0,025 N
b. Second titration
M NaOH = ?
V NaOH = 9,4 ml
V H2C2O4 = 5 ml
M H2C2O4 = 0,05 ml
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M H2C2O4. V H2C2O4
1. x. 9,4 = 2. 0,05. 5
9,4 x = 10. 0,05
x = 0,0531 M
The normality of NaOH = 0,053 M
The molarity of NaOH = 0,0265 N
c. Third titration
M NaOH = ?
V NaOH = 9,6 ml
V H2C2O4 = 5 ml
M H2C2O4 = 0,05 ml
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M H2C2O4. V H2C2O4
1. x. 9,6 = 2. 0,05. 5
9,6 x = 10. 0,05
x = 0,05208 M
The normality of NaOH = 0,052 M
The molarity of NaOH = 0,026 N
2nd Experiment
NaOH + HCl ® NaCl + H2O
Finding molarity of HCl using PP indicator
Data : VHCl = 5 ml
NNaOH = 0,052 N
MNaOH = 0,026 M
a. First titration
Got 9,6 ml of NaOH
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M HCl. V HCl
1. 0,026. 9,6 = 1. X. 5
5x = 0,249
x = 0,049
The molarity of HCl = 0,049 M
The normality of HCl = 0,049 N
b. Second titration
Got 9,8 ml of NaOH
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M HCl. V HCl
1. 0,026. 9,8 = 1. X. 5
5x = 0,25
x = 0,05 M
The molarity of HCl = 0,05 M
The normality of HCl = 0,05 N
c. Third titration
Got 9,25 ml of NaOH
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M HCl. V HCl
1. 0,026. 9,25 = 1. X. 5
5x = 0,24
x = 0,048
The molarity of HCl = 0,048 M
The normality of HCl = 0,048 N
The molarity of HCl is found : = 0,049 M
3rd Experiment
NaOH + HCl ® NaCl + H2O
Finding molarity of HCl using Plant extract indicator
Data : VHCl = 5 ml (V2)
M NaOH = 0,052 M(M1)
a. First titration
Got 9,5 ml of NaOH ( V1)
M1. V1 = M2. V2
M NaOH. V NaOH = M HCl. V HCl
0,026. 9,5 = x. 5
5x = 0,247
x = 0,049 M
The molarity of HCl = 0,049 M
The normality of HCl = 0,049 N
b. Second titration
Got 9,3 ml of NaOH ( V1)
M1. V1 = M2. V2
M NaOH. V NaOH = M HCl. V HCl
0,026. 9,3 = x. 5
5x = 0,2418
x = 0,048 M
The molarity of HCl = 0,048 M
The normality of HCl = 0,048 N
c. Third titration
Got 9,2 ml of NaOH ( V1)
M1. V1 = M2. V2
M NaOH. V NaOH = M HCl. V HCl
0,052. 9,2 = x. 5
5x = 0,239
x = 0,047 M
The molarity of HCl = 0,047 M
The normality of HCl = 0,047 N
