I.
Title
of Experiment
Acid Base Titration
Acid Base Titration
II.
Date And Time
Monday, April 17, 2017
Monday, April 17, 2017
III.
Research Purposes
1. Knowing the acid-base penetration by titration method
2. Determine the concentration of an acid or base solution by using acid titration
Bases
3. Knowing the equivalence and end points of titration-base
1. Knowing the acid-base penetration by titration method
2. Determine the concentration of an acid or base solution by using acid titration
Bases
3. Knowing the equivalence and end points of titration-base
IV. Basic Theory
Titration is a method to determine the content of a
substance by using another substance that has been determined by its
concentration. Titration
is usually distinguished by the type of reaction involved in the titration
process, for example when it involves an acid-base reaction, it is referred to
as acid-base titration, redox titration for titration involving
oxidation-reduction reactions, a titration of complexometry for titration
involving complex reaction formation and so forth. (Here only
discussed about acid-base titration)
Substances to be determined are referred to as
"titrant" and are usually placed in the Erlenmeyer, whereas the known
substances are referred to as "titers" and are usually placed in a
"burette". Both titer
and titrant are usually solvents.
Titration Acid base principles
Acid base
titration involves both acid and base as titer or titrant. Acid base titration
based on neutralizing reaction. The
acid solution is determined by using an alkaline solution and vice versa.
Titrant is added titer bit by bit until it reaches the
equivalent state (meaning the titrant stoichiometry and the right titer runs
out). This
is called the "equivalent point".
At this point of equivalence the titration process is
stopped, then we record the volume of titer required to reach that state. By
using titrant volume data, volume and concentration of titer we can calculate
titrant level.
How to Know the Equivalent PointThere are two general
ways to determine the equivalence point of acid-base titration.
1. Use pH meter to monitor pH change during titration, then plot between pH
and titrant volume to obtain the titration curve. The
midpoint of the titration curve is the "equivalent point".
2. Use acid-base indicator. The indicator is added to the titrant before the titration process is performed. This indicator will change color when the equivalent point occurs, at which point we stop the titration.
2. Use acid-base indicator. The indicator is added to the titrant before the titration process is performed. This indicator will change color when the equivalent point occurs, at which point we stop the titration.
In general the second way is chosen due to the ease of
observation, no additional tools required, and very practical. The indicator
used in acid-base titration is an indicator whose color is affected by pH. The
addition of the indicator is kept as little as possible and generally is two to
three drops.
Indicator |
Color Change
|
Solvent
|
|
Acid
|
Bases
|
||
Thimol blue
|
Red
|
Yellow
|
Water
|
Yellow Methyl
|
Red
|
Yellow
|
Ethanol 90%
|
Orange Methyl
|
Red
|
Red-orange
|
Water
|
Red Methyl
|
Red
|
Yellow
|
Water
|
Bromtimol blue
|
Yellow
|
Blue
|
Water
|
Phenolphthalein
|
Not colored
|
Purple
|
Ethanol 70%
|
Thimolftalein
|
Not colored
|
Blue
|
Ethanol 90%
|
Acommonly used indicator of acid-base titration is the
phenolphthalein indicator. The
following table is a characteristic of the phenolphthalein indicator.
PH
|
< 0
|
0−8.2
|
8.2−12.0
|
>12.0
|
conditions
|
Very acidic
|
Acid or near Neutral
|
Bases
|
Very Bases
|
Color
|
Orange
|
Not colored
|
Purple Pink
|
Not colored
|
Picture
|
To obtain the accuracy of the titration result, the titration end point is
chosen as close as possible to the equivalent point, this can be done by
selecting the appropriate indicator and in accordance with the titration to be
performed. The circumstance in which titration is stopped by seeing the
indicator color change is referred to as the "end point of the
titration".
Acid base titration is divided into 5 types:
Acid base titration is divided into 5 types:
1.
1.
Strong
acids - Strong bases
2.
Strong
acid - weak base
3.
Weak
acid - Strong base
4.
Strong
acids - Salt of weak acids
5.
Strong
base - Salt of weak base
1.
Titration
of Strong Acids - Strong Bases
Example:
- Strong acids: HCl
- Strong base: NaOH
- Strong acids: HCl
- Strong base: NaOH
Reaction Equation:
HCl + NaOH → NaCl + H2O
The ionic reaction:
H + + OH- → H2O
Strong Acid Base Acid Titration Curve
HCl + NaOH → NaCl + H2O
The ionic reaction:
H + + OH- → H2O
Strong Acid Base Acid Titration Curve
2.
Titration
of Strong Acids - Weak Bases
Example:
- Strong acids: HCl
- Weak base: NH4OH
Reaction Equation:
HCl + NH4OH → NH4Cl + H2O
The ionic reaction:
H + + NH4OH → H2O + NH4 +
Strong Acid Titration Curve - Weak Base
Example:
- Strong acids: HCl
- Weak base: NH4OH
Reaction Equation:
HCl + NH4OH → NH4Cl + H2O
The ionic reaction:
H + + NH4OH → H2O + NH4 +
Strong Acid Titration Curve - Weak Base
3.
Weak
Acid Titration - Strong Bases
Example:
- Weak acid: CH3COOH
- Strong base: NaOH
Reaction Equation:
CH3COOH + NaOH → NaCH3COO + H2O
The ionic reaction:
H + + OH- → H2O
Weak Acid Titration Curve - Strong Base
Example:
- Weak acid: CH3COOH
- Strong base: NaOH
Reaction Equation:
CH3COOH + NaOH → NaCH3COO + H2O
The ionic reaction:
H + + OH- → H2O
Weak Acid Titration Curve - Strong Base
4.
Titration
of Strong Acid - Salt of Weak Acid
Example:
- Strong acids: HCl
- Salt of weak acid: NH4BO2
Reaction Equation:
HCl + NH4BO2 → HBO2 + NH4Cl
The ionic reaction:
H + + BO2- → HBO2
Example:
- Strong acids: HCl
- Salt of weak acid: NH4BO2
Reaction Equation:
HCl + NH4BO2 → HBO2 + NH4Cl
The ionic reaction:
H + + BO2- → HBO2
5.
Titration
of Strong Bases - Salt of Weak Bases
Example:
- Strong base: NaOH
- Salt of weak base: CH3COONH4
Reaction Equation:
NaOH + CH3COONH4 → CH3COONa + NH4OH
The ionic reaction:
OH- + NH4- → NH4OH
Example:
- Strong base: NaOH
- Salt of weak base: CH3COONH4
Reaction Equation:
NaOH + CH3COONH4 → CH3COONa + NH4OH
The ionic reaction:
OH- + NH4- → NH4OH
General Titration Formulas
At the point equivalent point, the mol-equivalent acid
will be equal to the base mol-equivalent, so we can write as follows:
Mol-equivalent acid = mol-equivalent base
The mol-equivalent is obtained from the multiplication of
Normality with volume then the above formula can we write as:
NxV acid = NxV base
Normality is obtained from the multiplication of molarity
(M) with the amount of H + ions on the acid or the amount of OH ions in the
base, so the above formula becomes:
NxMxV acid = nxVxM basesinformation :
N = Normality
V = Volume
M = Molarity
N = number of H + ions (on acid) or OH - (on base)
IV.
Tools And Materials
Tool:
• Measuring cup
• Erlenmeyer pumpkin
• Beaker
• Buret
• Stative and clamps
• Funnel
• Drop pipette
• Cleaning brush
INGREDIENTS:
• 0.1 M HCl solution
• 0.1 M CH3COOH solution
• NaOH solution
• PP solution
Tool:
• Measuring cup
• Erlenmeyer pumpkin
• Beaker
• Buret
• Stative and clamps
• Funnel
• Drop pipette
• Cleaning brush
INGREDIENTS:
• 0.1 M HCl solution
• 0.1 M CH3COOH solution
• NaOH solution
• PP solution
V.
WAYS
OF WORKING:
1.
Prepare
a stative burette and clamps
2.
Filling
the burette with NaOH solution right up to the zero line with the help of the
funnel
3. Insert
15 ml of 0.1 M HCl into the erlenmeyer flask, then add 3 drops of PP indicator
to the solution
4.
Lay
the erlenmeyer flask just below the burette, then open the burette faucet
slowly so that NaOH can drip into the solution.
5. During
addition of NaOH, wiggle the erlenmeyer flask so that the NaOH can be well
blended and until the earliest color changes occur.
6.
Observe
the color changes that occur in HCl solution.
7.
Recording
the amount of NaOH used is the difference between the final volume and the
initial volume of NaOH.
8.
Determining
the concentration of NaOH used with the formula V1.M1 = V2.M2
9. Perform
1-7 activities once again and calculate the average amount of NaOH used to know
the equivalent point
10. Repeat activities 1-9 for 0.1 M CH3COOH solution
VI.
Discussion
A. Table Of Observation Results
A. Table Of Observation Results
|
No
|
Solution
|
Trial to-
|
Volume of NaOH Used (mL)
|
Color of solution
|
|
|
1
|
HCl
+ PP + NaOH
|
1
|
28
drops
= 1,4 ml
|
Purple to pink-pink an
|
|
|
2
|
HCl
+ PP + NaOH
|
2
|
33
drops =
1,65 ml
|
Young purple
|
|
|
3
|
HCl
+ PP + NaOH
|
3
|
20
drops =
1 ml
|
Old purple
|
|
|
4
|
HCl
+ PP + NaOH
|
4
|
28
drops =
1,4 ml
|
Young purple
|
|
|
5
|
HCl
+ PP + NaOH
|
5
|
34
drops =
1,7 ml
|
Pink
|
|
|
6
|
HCl
+ PP + NaOH
|
6
|
33
drops =
1,65 ml
|
Pink
|
|
|
7
|
HCl
+ PP + NaOH
|
7
|
67
drops = 3,35 ml
|
Purple Pink
|
|
|
8
|
HCl
+ PP + NaOH
|
8
|
27
drops =
1,35 ml
|
Purple Pink
|
|
|
No
|
Solution
|
Trial to-
|
Volume of NaOH
Used (mL)
|
Color of solution
|
|
1
|
CH3COOH + PP + NaOH
|
1
|
58 drops
= 2,9 ml
|
Pink
|
|
2
|
CH3COOH + PP + NaOH
|
2
|
59 drops =
2,95 ml
|
Pink
|
|
3
|
CH3COOH + PP + NaOH
|
3
|
49 drops =
2,45 ml
|
Young purple
|
|
4
|
CH3COOH + PP + NaOH
|
4
|
52 drops =
2,6 ml
|
Pink
|
|
5
|
CH3COOH + PP + NaOH
|
5
|
41 drops =
2,05 ml
|
Pink purple
|
|
6
|
CH3COOH + PP + NaOH
|
6
|
55 drops =
2,75 ml
|
Pink purple
|
|
7
|
CH3COOH + PP + NaOH
|
7
|
34 drops =
1,7 ml
|
Pink purple
|
|
8
|
CH3COOH + PP + NaOH
|
8
|
33 drops =
1,65 ml
|
Pink purple
|
VI.
Discussion
Titration is
an analytical way of measuring the amount of solution needed to react
constantly with a substance present with another solution. In this experiment
we determined the molarity of NaOH by using a titration process between 15 ml
0.1 M HCl solution with NaOH solution. 15
ml of HCl solution was put into Erlenmeyer flask and then added 3 drops of PP
indicator, then dropped with NaOH solution already provided in buret dropwise
until equivalent or finished reacting. Similarly,
titration between CH3COOH solution of 15 ml 0.1 M with NaOH solution. 15
ml of CH3COOH solution was put into Erlenmeyer flask and then added 3 drops of
PP indicator, then dropped with NaOH solution which had been provided in buret
dropwise until equivalent or finished reacting.
The
equivalent point can be known with the help of PP solution, the color range is
not colored to purple, ie if colorless means acidic and if purple is a base. If
the solution is equivalent then, the solution will experience the earliest
discoloration, and the color is very young and bright when that titration is
stopped. When
the solution shows the earliest discoloration is the so-called end point of the
titration.
Experiment 1 uses HCl
Titration strong acid + strong
base
In our titration experiments on 15 ml of HCl solution
titrated with NaOH yielded the following reaction equations;
HCl + NaOH à NaCl + H2O
Titration 1
In the first experiment, the first step was HCl 15 ml 0.1
M inserted into Erlenmeyer glass, then added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is 1.4 ml and the color is purple pink
1.4 M = 1.5
Titration 2
HCl 15 ml 0.1 M was inserted into Erlenmeyer glass, then added 3 drops of
penoftaline. 50
ml of NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
is obtained at 1.65 ml and the color is purple.
15.0.1 = 1.65 M2
1.5 = 1.65 M2
15.0.1 = 1.65 M2
1.5 = 1.65 M2
Titration 3
HCl 15 ml 0.1 M was inserted into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of titration is reached, and the titration
volume is 1 ml and the color is purple.
15.0,1 = 1
M2
1.5 = 1 M2
1.5 = 1 M2
Titration 4
HCl 15 ml 0.1 M was inserted into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is 1.4 ml and the color is purple pink.
15.0.1 = 1.4
M2
1.5 = 1.4 M2
Titration 5
1.5 = 1.4 M2
Titration 5
HCl 15 ml 0.1 M was inserted into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is obtained at 1.7 ml and the color is pink / pink.
15.0.1 = 1.7
M2
1.5 = 1.7 M2
Titration 6
1.5 = 1.7 M2
Titration 6
HCl 15 ml 0.1 M was inserted into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is found at 1.65 ml and the color is pink / pink.
15.0.1 =
1.65 M2
1.5 = 1.65 M2
Titration 7
1.5 = 1.65 M2
Titration 7
HCl 15 ml 0.1 M was inserted into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml of NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the
end point of the titration is reached, and the titration volume is 3.35 ml
and the color is purple pink.
15.0.1 =
3.35 M2
1.5 = 3.35 M2
1.5 = 3.35 M2
Titration 8
HCl 15 ml 0.1 M was inserted into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is obtained at 1.35 ml and the color is purple pink.
15.0.1 =
1.35 M2
1.5 = 1.35 M2
1.5 = 1.35 M2
The molarity of NaOH is:
So the molarity of NaOH is 0.99
Experiment 2 uses
CH3COOH
Titration of weak
acid + strong base
In our titration experiments on 15 ml of CH3COOH solution
titrated with NaOH yielded the following reaction equations;
Titration 1
CH3COOH 15 ml 0.1 M was put into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator changes
color or the end point of the titration is reached, and the titration volume is
2.9 ml and the color is pink.
2.9 M = 1.5
Titration 2
CH3COOH 15 ml 0.1 M was put into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is 2.95 ml and the color is pink.
15.0.1 =
2.95 M2
1.5 = 2.95 M2
1.5 = 2.95 M2
Titration 3
CH3COOH 15 ml 0.1 M was put into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is 2.45 ml and the color is purple.
15.0.1 =
2.45 M2
1.5 = 2.45 M2
1.5 = 2.45 M2
Titration 4
CH3COOH 15 ml 0.1 M was put into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml of NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
is 2.6 ml and the color is pink.
15.0.1 = 2.6
M2
1.5 = 2.6 M2
1.5 = 2.6 M2
Titration 5
CH3COOH 15 ml 0.1 M was put into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml of NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is 2.05 ml and the color is purple pink.
15.0.1 =
2.05 M2
1.5 = 2.05 M 2
Titration 6
1.5 = 2.05 M 2
Titration 6
CH3COOH 15 ml 0.1 M was put into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml of NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is 2.75 ml and the color is purple pink.
15.0.1 =
2.75 M2
1.5 = 2.75 M2
1.5 = 2.75 M2
Titration 7
CH3COOH 15 ml 0.1 M was put into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml of NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is obtained at 1.7 ml and the color is purple pink.
15.0.1 =
1.7M2
1.5 = 1.7 M2
1.5 = 1.7 M2
Titration 8
CH3COOH 15 ml 0.1 M was put into Erlenmeyer glass, then
added 3 drops of penoftaline. 50
ml NaOH is fed into the burette, then let drip dropwise until the indicator
changes color or the end point of the titration is reached, and the titration
volume is obtained at 1.65 ml and the color is purple pink.
15.0.1 =
1.65 M2
1.5 = 1.65 M2
The molarity of NaOH is:
So the molarity of NaOH is 0.662 M
1.5 = 1.65 M2
The molarity of NaOH is:
So the molarity of NaOH is 0.662 M
In this experiment we performed titration eight times
each, a strong acid titration with a strong base eight times and a weak acid
titration with a strong base of guard eight times. In
the eight experiments on the titration of HCl with NaOH there are some that
fail where the color change is too old as well as in the titration of CH3COOH
with NaOH. However,
there are also some who succeed.
This failure is due to several factors:
1.
Less careful eye
when paying attention to the color changes that occur, which may be the initial
color changes have occurred but because it is not paid attention carefully so
that the penetration is continued and the resulting color is too concentrated
and conspicuous
2.
Less careful when
carrying out the titration proces
3.
Less precisely the
preparation of 0.1 M HCl and CH3COOH in the weighing process.
4.
Less precise in the
calculation of droplets of NaOH solution that allows excessive penetrations so
that the resulting color more concentrated
5.
At the time of almost reaching the equivalent point
of buret faucet flow
VI.
Conclusion
1.
The equivalent
point is the point where the acid concentration is equal to the base
concentration (depleted reacting) or the point at which the amount of base
added is equal to the amount of neutralized acid accompanied by the indicator
color change.
2.
The end point of
the titration is the state where the titration is stopped by looking at the
indicator color change
3.
The PP indicator
needs to be added to the solution because it knows the color change that occurs
at the equivalence point
4.
The reaction
equations for each experiment:
A. Strong acid + strong base
HCl + NaOH NaCl + H2O
B. Weak acid + strong base
VII.Bibliography
Www.isolabgmbh.com/product.asp%3...rup%3D19
Www.indigo.com/glass/gphglass/buret.html
Www.daym.gov.tr/index2.php%3Fad%...kat%3D57www.analisateknisia.blogspot.com/20 11 ... ive.html www.oldprint.blogspot.com/20 11/12 / pe. ..int.html
Www.indigo.com/glass/gphglass/buret.html
Www.daym.gov.tr/index2.php%3Fad%...kat%3D57www.analisateknisia.blogspot.com/20 11 ... ive.html www.oldprint.blogspot.com/20 11/12 / pe. ..int.html
Http://kimia.upi.edu/utama/bahanajar/kuliah_web/20011/SRIYANI(050679)/latihan.html
Http://kimia.upi.edu/utama/bahanajar/kuliah_web/2011/Sri%20Ratisah%20054828/materi.HTMhttp://akhitochan.wordpress.com/201 1/02/13 / titration-acid-base /
Http://kimia.upi.edu/utama/bahanajar/kuliah_web/2011/Sri%20Ratisah%20054828/materi.HTMhttp://akhitochan.wordpress.com/201 1/02/13 / titration-acid-base /
VIII. Attachment
What does the addition of a solution of acid-base reactions pp in this?
BalasHapusPhenolphthalein
Hapuscan you give example redox titration for titration involving oxidation-reduction reactions?
BalasHapusExamples of well-known redox titration are iodimetry, iodometry, permanganometry using titrant potassium permanganate for the determination of Fe2 + and oxalate, Potassium dichromate is used for iron (II) and Cu (I) determination titrations in CuCl. Bromate is used as a titrant for the determination of phenol, and iodide (as I2 which is titrated with thiosulfate), and Cerium (IV) which can be used for titration titrant redox determination of ferocyanide and nitrite.
HapusHow to recognize a solution acidic or alkaline?
BalasHapusBy testing using indicators of both natural and artificial indicators such as litmus paper and phenolptalein
Hapusaccording to you, what is the best indicator can we used to determine pH of a solution?
BalasHapusPhenolphthalein, Because by using these indicators we can easily distinguish acid and base solutions, if In acidic and neutral solutions, phenolphthalein is colorless. Whereas when put into an alkaline solution, the color will turn red.
HapusWhy you used PP indicator to determined pH of a solution? and in your experiment do you have a mistake?why that happened?
BalasHapusBecause by using pp we more easily distinguish acid and base solutions, the function of pp is as an indicator used to facilitate distinguish acid and base solutions. There is an error because in this process we include more pp of the rules to speed up the titration process
HapusHow to determine the concentration of acid or base solution by using acid titration?
BalasHapusHow to calculate it using the formula: M1 * Valence 1 * Volume 1 = M2 * Valence 2 * Volume 2
HapusHow can we know if the solution is at the equivalence point?
BalasHapusBy looking at the color change at the time of titration when it has turned red to easily indicate the solution has reached the equivalent point
HapusWhat is the equivalent point?
BalasHapusThe equivalent point is 6
Hapus