Using English to Report

I. Practical Title : Titrimetry And Control Certain (Ph) Bufer Solutions

II. Date And Time : Tuesday, 19 March 2016

III. Objectives Of Experiment :

1. Studying and applying titration techniques to analyze acidic samples

2. Standardize the penitration solution

3. Standardize NaOH solution

4. Describe the titration curve

5. Determine the weak acid equilibrium constant

6. Explain the importance of ph control especially in body physiology system

7. Describe how to maintain ph in a variety of uses

8. Get familiar with some buffer solutions from certain systems and how they work

IV. Praktek Questions

1. What is meant by (a) Acids, (b) Bases, (c) Equivalent Titals, and (d) Indicators

Answer:

·         Acids: compounds that have a sour taste, change the color of the blue litmus to red.

·         Bases: Compounds that have a bitter taste and change the litmus color from red to blue.

·         Equivalent Point: The point that occurs between acidic and basic solutions where the acid solution can react with the amount of the basic solution.

·         Indicator: A substance used as a guide to distinguish acid and base solutions.

2. Explain the difference of end point of titration with the equivalent point.

Answer:

•    Endpoint titration: The point in a titration in which an indicator changes color.

•    Equivalent Point: When the substance in the right titration reacts with the neutralizing agent.

3. A total of 0.774 9 potassium hydrogen citrate in entry into the erlenmeyer and dissolved with distilled water, then in titration with a naoh solution. When used 33.60 ml, what is the molarity of the naoh?

Answer:

Given: KHC6H6O7 + NaOH -> NaKC6H6O 2

Vol NaOH = 33.6 ml = 0.0336

Asked: M NaOH = ...

Answered:

Mol KHC6H6O7 = 0.7742 / 230 = 3,36.10-3

Mol NaOH = Mol KHC6H6O7 = 3,36.10-3 mol

M NaOH = mol / L

= 3.36.10-3 mol / 33,60.10-3 ml

= 0.1 M

4. Explain what is meant by:

•    Acid acid titration curve: An image representing a count of ph by volume liters.

•    Equivalent Point: The point at which acid has reacted perfectly

•    Standardization: The process of determining the concentration of a carefully specified solution.

•    Primary standard solution: concentrated solution.

•    pH: Negative Logarithm H + or expressed negative concentration of H + in laruutan

•    pH Meter: The tool used to measure the pH of the solution

5. Calculate the mass of potassium hydrogenphthalate (khp) to neutralize 25 ml of 0.1 M NaOH and write the reaction equation.

Answer:

Given: V NaOH = 25 ml

 M NaOH = 0.1M

Asked: mass = ...

Answered:

KHC8H4D4 + NaOH NaKC8H4D4 + H2O

0.0025 mol 0.0025 0.0025

Mole NaOH = m.V

= 0.1 x 0.025

= 0.025 mol

Potassium period Hydrogen phthalate = mol x Mr

= 0.0025 x 204

= 0.51gr

6. How to make 50 ml of HCl solution with pH 1 from 1M HCl solution?

Answer:

PH = 1

[H +] = 10-1 m

V Hcl = 50 ml

V1. M1 = v2. M2

V1 .1 = 50. 10-1 = 5ml

V1 = 5ml

V2 = 50 ml

V water = v2-v1 = 45 ml

How to make 5ml HCl solution 1m + 45 ml distilled water

7. What is Bufer's solution? Why is buffer solution important?

Answer:

·         The buffer solution (buffer solution) is a solution which can maintain the price of the ph even in the addition of the acid / base solution into the solution.

·         Because it can maintain the pH of the solution in a given pH region because it contains the acid / base equilibrium salt ion into the solution

8. Define for weak acids and weak bases.

Answer:

·         Weak acid: H + ion is bigger than water so it shifts the water balance to the left as a result (H +) and the water is smaller to that of the weak acid.

·         Weak base: (OH-) and water dapast is ignored because it is so small compared to that of a base

9. Explain by the reaction equation how a solution of sodium ionide (NaCN) with hydrogen sionide (HCN) serves as a buffer solution.

Answer:

HCN + NaOH = NaCN + H2O

HCN H + + CN-

NaCN Na + + CN-

If the acid is added, the H + ion reacts with the CN-forming HCN (the equilibrium shifts left, then the amount of H + in the fixed solution).

When added Bases, the OH ion reacts with H + to form H2O (equilibrium shifts right, then HCN decomposes to CN- and H + ions)

The H + ion is bonded by OH-recovered from the decomposition of the ion so that the amount of H + ions remains

10. Mention several buffer solution pairs with the same physiological properties.

Answer:

HC2H3O2 + NaOH -> NaC2H3O2 + H2O

KH2PO4 + NaOH -> K2HPO4 + H2O

V. Theory Of Theory

               An important application and stoichiometry in the laboratory is the analysis of the elements to determine its composition. Measurements that are based on mass in the name of gravimetry, and measurements based on the volume of the solution in the name of volumetry or titration. In this experiment the volumetric analysis technique is applied to the analysis of acidic samples.

               Some types of reactions that can be used for titration are sedimentation, reduction and acid-base reaction, all of which can take place perfectly.

In this experiment an acid-base reaction will be used to standardize the base solution and then be used to analyze the acidic sample. In short the acid-base reaction or neutralization is caused by the transfer of protons (H + ions) from acid to base. A classic example of this type of reaction is the reaction of hydrogen ions with hydracan ion

H + (aq) + OH- (aq) -> H2O (l)

               In this experiment the source of OH-ions is a dilute NaOH solution and the source of H + ions is an acid solution. At first prepare a solution of 0.1 m NaOH then this solution in standardization with acid solution in know concentration. Naoh solution is not available in pure state and the solution may change its concentration as it absorbs CO2. Therefore naoh solution must be standardized before use to sample examples.

               In most acid-base titration. The solution change at the equivalent point is unclear. Therefore, to determine the titration end point in use of the indicator because this substance shows the color change in a particular ph in this experiment in use fenollftalein. This compound is colorless in acidic and pink soluble in base solution.

To calculate the ionization constant of acetic acid through henderson-hasselbalch equation.

PH = pKa + Log

               This equation can be used to calculate the pH value of the buffer solution. It can be used to calculate the pH at each point of the titration curve. The pH value on the curve is seen from the starting pH price before adding nohoh to the point pass. By using the above equation we can calculate the price of Ka. During the titration, the acid-base concentration decreases as the weak acid reacts with added NaOH.

               The quantity of acid and base will be the same at a certain point, the acidity will also occur at ½ equivalent point at the midpoint, the required amount of ½ NaOH reacts perfectly with ½ the amount of weak acid. The quantity of NaOH at the mid point is

  = 13.51 ml

At this time the acid concentration is equal to the base concentration according to the following equation:

           [Acid] = [Base]

Log [Acid] / [Base] = Log 1 = 0

According to the Henderson-Hasselbalch equation

PH = pKa

Then pKa can be determined

               Most physiological processes are very sensitive to changes in pH. For example, the pH of human blood is basically maintained at a pH of 7.2. Only in this pH can blood carry oxygen and carbon dioxide properly. If the pH falls below 7.2 (H + concentration is higher) then the hemoglobin in the blood will not react with oxygen, and when the pH is increased (the concentration of hegoglobin in the blood will not decompose into carbon dioxide in the lungs).

Weak acid, weak base, and Salt

               The buffer solution system is a weak acid solution (or weak base) together with its salt. As for weak acids or weak bases are acids or bases that only ionize slightly. Acetic acid (HC2H3O2) is a weak acid, as shown in the following equation.

HC2H3O2 + H2O = H2O + C2H3O2

               The ammonium hydroxide solution is an example of a weak base, also because only a few percent of these bases reside as nh and oh ions. Acids and bases in gololngkan as strong or weak, depending on the degree of ionization (ionization). Several acids which have high degree of ionisation in 100% aqueous solutions are ionic bases such as NaOH, kOH, and Ca (OH) 2 being ionic in solid state and also completely dissociated in water. On the other hand, large amounts of acids (eg HC2H3O2, HCN, H2CO3, and H3PO4), organic acids (RCOOH) and some organic bases (R-NH2) are only slightly ionized in aqueous solution.

               Salt and weak acids are salts of which one ion is equal to the acidic ion. Salts between other sites may be prepared by allowing the weak acids to react with an appropriate base comprising suitable cations. For example a salt consisting of C3H3O2-ion is a salt of acetate (HC2H3O2). A typical salt, eg sodium acetate (NaC2H3O2) can be formed from the corresponding acid and base.

HC2H3O2 + NaOH NaC2H3O2 + H2O

               Similarly, sodium slanide (NaCN) and calcium cyanide [Ca (CN) 2] are salts of slanidic acid. Potassium Monohydrogen phosphate (K2HPO4), is a hydrogen phosphoric acid salt and KH2PO4 as shown in the following equation:

KH2PO4 + KOH K2HPO4 + H2O

               The salt of a weak base has the same cation as the base. Examples of salts of ammonium hydroxide, NH4OH (NH3 ammonium solution), are ammonium chloride, NH4CL and ammonium sulphate, (NH4) 2 SO4 (Epinur.2012: 61-64).

Important traits that need to be remembered in a weak acid titration curve by a strong base.

 the initial pH is higher than in the strong acid and strong base titration curves-

 There is a rather sharp increment of a suitable ph on a titration-

 Before the point is reached, ph changes occur gradually-

 pH at this point after greater than 7-

 After a point, the curve is titrated on a weak acid by a strong base identical to a strong acid-base curve.-

               The weak polyphic acid titration is strong evidence that polliprotic acid ionizes in the neutralization of phosphoric acid almost all H3PO4 molecules begin to convert to Na2PO4 and finally Na2HPO4 is converted to Na3PO4 ie:

Na3PO4 + OH- H2PO4- + H2O followed by

H2PO4 + OH - PO4-3 + H2O (Sutrisno.1994: 100-101).

For an alkaline solution, the concentration must exceed the concentration of H + in a solution. Such imbalances can be made in two different ways:

First: The base may be a hydroxide, which can only dissociate to produce hydroxide ions. Where M represents cation, usually metal, the most common base is hydroxide-like as it is.

The second line can be done by extracting one ion. Hydrogen from one water molecule, leaving one hydroxide ion:

               The strength of the buffer is not a special one, it is only an expression of two urgent reversible equilibral reactions occurring within the solution of a proton donor and its conjugated proton elvepting. If both have the same concentration.

If we add H + or OH-into the buffer, the result is a small change in the relative concentration ratio of the acid and the anion thereof as well as only a few buffer systems with the addition of a small amount of acid / base is precisely offset by an increase in other components. The number of unchanged buffer components that changes only the ratio (Lehninger.1993: 187).

A solution containing a weak acid plus a salt of the acid or a weak base plus a salt of a strong base. Such a system is referred to as a buffer (buffer) solution because the bit of the addition of strong acid / strong base changes only a small pH.

example:

H + + C2H3O2- - HC2H3O2

               Its pH has not changed significantly. Conversely, if hydrogen ions are added to form more alkaline hydrogen acetate molecules. Standard buffer solutions can be made from weak acids and salts of the weak acid. A convenient equation is used to calculate the pH of such a solution or to calculate the acid-to-salt value of the salt required to obtain a solution with the desired pH pH of a buffer containing weak acids can be calculated as follows:

Ka = (H + [A]) / [A]

[H +] = Ka (H + [A]) / [A])

-Log [H +] = -Log Logo ([HA] / [A])

PH = pKa-log ([HA] / [A])

PH = pKa + log (harvest.1991: 235-237)

VI. Tools And Materials

A. Tool

o   Erlenmeyer

o   Pipette drops

o   Balance Sheet

o   Measuring cup

o   Reactor tube

o   Universal indicator

o   Buret 50 ml

o   500 ml bottle

o   Helongan

o   Supporting poles

o   Glass watch

o   Mixer rod

B. Material

o   Distilled water

o   pp indicator

o   NaOH solution

o   Khp 0.1 gr

o   Kitchen vinegar

o   HCl solution

o   Sodium acetate solution

o   NH4Cl

o   NH4OH

VII. Work Procedures

A. Preparation of NaoH 0.1 M Solution

1.      1.6 gr NaoH

2.      Moved to bottle

3.      Dissolved with 400 ml of distilled water

Beat

4.      Observation result

B. Standardization of 0.1 M NaOH solution

5.      Buret 50 ml

6.      Washed and rinsed with distilled water

7.      Closed and inserted approximately 5 ml naoh

8.      To fill buret with naoh s / d 0

9.      Flow solution

10.  2 erlenmeyer 250 ml

11.  Washed and rinsed

12.  DETAILED 25 ml of HCL 0.1 is included on each erlenmeyer

13.  25 ml of distilled water and 3 drops of phenolphthalein indicator

14.  Added to each erlenmeyer

15.  Recorded initial NaOH

16.  Flowed naoh naoh on Erlenmeyer 1

17.  Recorded final volume on the burette

18.  Done 2 times

19.  3 pieces of erlenmeyer

20.  Washed

21.  Filled with 0.14 grams of KHP

22.  Added 10 ml of distilled water, shaken until dissolved

23.  Added 3 drops pp indicator

24.  Recorded volume of NaOH used

25.  Observation result

C. Determine the percentage of acetic acid in vinegar

1.      3 erlenmeyer 250 ml

2.      Washed and rinsed

3.      Dabbed 25 ml of vinegar into the seeeeer erlenmeyer

4.      10 ml of distilled water added

5.      3 drops pp indicator

6.      Added and titrated with standard solution until red is formed

7.      Calculated percent of mass in each instance

8.      Repeat once more if the results are different> 0.05%

9.      Observation result

POTENSIOMETRY

1.      A set of pH meter tools

2.      Prepared

3.      Buffer solution of pH 5

4.      Calibrated

5.      5.1 gr KHP

6.      Weighed

7.      Dissolved with distilled water and diluted in a 250 ml measuring flask until the + sign

8.      80 ml liquid pipette

9.      Entered into a cup glass

10.  A standardized NaOH solution

11.  Inserted into the buret

12.  Installed as shown

13.  Noted pH

14.  Created a titration curve

15.  Repeated experiment once again starting no 2

16.  Observation result

A. Non-buffer solution

1. Determination of pH of the solution is not buffer

1.      1 ml distilled water 1 ml HCl 0.0001M 1ml NaOH 0.0001M

2.      Inserted into a tube

3.      Observation result

2. Determination of pH after acid addition

1.      1 ml distilled water 1 ml HCl 0.0001M 1ml NaOH 0.0001M

2.      Inserted into a tube and added1 tets HCl

3.      Observation result

B. Bufer solution

1. Determination of pH of Bufer solution

1)      5ml acetic acid and 5ml sodium acetate 5ml NH4OH and NH4Cl

2)      Mixed in a test tube

3)       Noted

4)      Observation result

2. Determination of pH of Bufer solution after acid addition

1)      2 ml buffer solution 2 ml buffer solution

2)      Inserted into a tube and added

3)      1 drop HCl 1 M

4)      Observation result

3. Determination of pH of Bufer solution after addition of base

1)      2 ml buffer solution 2 ml buffer solution

2)      Inserted into a tube and added

3)      1 drop of 1 M NaOH

4)      Observation result

VIII. Observation Data
Basic Acid Titration
A. Standarization With Hcl Solutions

	Deuteronomy 1	Deuteronomy 2
Volume of HCL solution	25 ml	25 ml
Molarity of HCL solution	0.1 M	0.1 M
Mol HCL used	25.10-4 mol	25.10-4 mol
NAOH mol obtained	0.0221mol	0.0221mol
Early NAOH volume	50ml	50ml
Final NAOH volume	29ml	30 ml
NAOH volume added	221ml	220 ml
Molarity of NAOH solution	0.01 M	0.01 M
Molarity of NAOH solution averaging	0.01M	0.01M

B. Standarization With Khp

	Deuteronomy 1	Deuteronomy 2
Mass weighing bottle contains KHP	10,475 gr	10,475 gr
Mass of weighing bottle after KHP	105,2 gr	105,2 gr
KHP Mass	0,55 gr	0,55 gr
KHP Mole	0,0017 mol	0,0017 mol
NAOH mole is needed	0,026	0,026
Early NaOH volume	50 ml	50 ml
The final NAOH volume	26	26
NAOH volume used	24	24
Molarity of NAOH solution	1 M	1 M

C. Determine% acetic acid in vinegar

	Deuteronomy 1	Deuteronomy 2
Volume of vinegar	2 ml	2 ml
Dosage of vinegar	1,0089 g/m	1,0089 g/m
Mass of vinegar	2,016 g	2,016 g
. Early NAOH vol	50 ml	50 ml
The final NAOH volume	24ml	24ml
NAOH volume used	26 ml	26 ml
Molarity of NAOH solution	0,1 M	0,1 M
NAOH Mole added	0,0026 mol	0,0026 mol
Mol acetic acid	0,0026	0,0026
Acetic acid weight	0,150 g	0,150 g
% Acetic acid mass	7,7	7,7
The average mass of 1ml acetic acid	7,7	7,7

D. POTENSIOMETRY

No	Volume of NaOH (ml)	pH
1	10	4
2	20	5
3	30	5
4	40	5
5	45	5
6	46	5
7	48	6
8	50	6
9	55	6
10	60	6

IX. Discussion
            At this practicum entitled trimetri and pH control we do a lot of standardization. A solution needs to be standardized as well as NAOH as it is a primary solution. That is a solution that already has a fixed concentration. Standardization for primary standsar solutions must have the following conditions:
A. Easily obtainable in pure form
B. Must be stable
C. The substance is easy to dry, not dipemboskopis not absorb water vapor
D. Masaa is a large equivalent

A. Standardize with HCL solution
In this experiment the first step was to wash a burette size of 50ml using distilled water, then weighing 1.6 g NaOH and transferred into the bottle. NaOH solution already made. Prepare 3 erlenmeyer and insert 25ml of distilled water and 3 drops of pp indicator, then place the already loaded erlenmeyer under buret, slowly drop the NaOH solution present inside the burette into the erlenmeyer containing the HCl solution, until the erlenmeyer solution turns pink.
In this experiment the initial volume of NaOH is 50 ml, so the molarity of the NaOH solution can be calculated:
M = = 0.0001 x 1000 l = 0.1 M
In calculations obtained molarity of NaOH is 0.1 M, mean molarity of NaOH is theoretically the same as NaOH in practice. Based on the theory it is said that the equivalence point will be achieved when the mole of the titrant is equal to the mole of the titrated substance, this means that mole of NaOH is equal to mole of HCl,
Mol HCL = M HCl x V HCl
= 0.1 M x 0.025 l
= 0.0025 mol
Mol NaOH = M NaOH x V HCl
= 0.02 x 0.031
= 0.0025 mol

From the above calculation it can be seen that the HCl solution changes color when mole HCl = mol NaOH = 0.0025 mol
From the experiments that have been done turns out the difference in molarity of NaOH gained too much between practice and theory. We experiment with 3 repetitions. The second repetition uses a volume of 31 ml NaOH and the HCl solution in the erlenmeyer changes color to pale pink. And the molarity of NaOH obtained are:
M = = 0.0001 x 1000 l = 0.1 M
Then the average molarity of NaOH can be calculated:
M NaOH average = = = 0.1 M

B. Standardize with KHP
            The next experiment was to standardize using 0.35 g of potassium hydrogen phthalate (khp), which was added with 3 drops of phenolphthelain indicator (pp) and then titrated with NaOH solution until a pink color was performed with 2 repetitions
Deuteronomy 1
• mass khp = 0.55 gr
• mol khp = gr / mr
• mole of NaOH required 0.0026 M
• the volume of NaOH used = 0.1 M
• molarity of NaOH solution = 0.1 M
Repeat 2
• mass khp = 0.55 gr
• mol khp = 0.0017 mol
• mole of NaOH required 0.00026 M
• the volume of NaOH used = 34ml
• molarity of NaOH solution = 0.1 M

C. Determine% acetic acid in vinegar
            In this experiment three erlenmeyer 250 ml were used. Which added 20 ml of distilled water into it and then drop 3 drops of pp indicator and then titrated with standard NaOH solution until formed pink color, this experiment also done as much as 2 times

Deuteronomy 1
• Volume of vinegar = 1 ml
• Vinegar density = 1.008 g / m
• Mass vinegar = 1.008 g
• Initial NaOH volume = 100 ml
• The final NaOH volume = 44.5 ml
• Used NaOH volume = 55.5 ml
• Molarity of NaOH solution = 0.1 M
• Mol NaOH added = 5.55 mol
• Mole acetic acid = 5.55 mol
•% of the mass of acetic acid in the sample
Acetic acid mass x 100% / vinegar mass = 0.325%
Deuteronomy II
• Volume of vinegar = 1 ml
• Vinegar density = 1.008 g / m
• Mass vinegar = 1.008 g
• Initial NaOH volume = 50 ml
• Used NaOH volume = 48.6 ml
• Molarity of NaOH solution = 0.1 M
• Mol NaOH added = 48.6 mol
• Mol acetic acid = 4.86 mol
• Acetic acid weight = 291.6
•% of the mass of acetic acid in the sample
Acetic acid mass x 100% / vinegar mass = 0.325%

D. Potentiometry
The next experiment was potentiometry which in this experiment used a set of pH and calibration with buffer solution of pH 5. Then a solution of 5.1 g of khp with distilled water and dilute in a 250 ml measuring flask until the mark is tarred. Furthermore, 0.1 M NaOH solution is introduced into the burette. Record the pH readable on the pH scale before addition of NaOH and after addition of 10, 20, 30, 40, 45, 46, 48, 50, 55 and 60 ml NaOH solution.

No.	Volume NaOH(ml)	pH
1.	10 ml	4
2.	20 ml	5
3.	30 ml	5
4.	40 ml	5
5.	45 ml	5
6.	46 ml	5
7.	48 ml	6
8.	50 ml	6
9.	55 ml	6
10.	60 ml	6

From this experiment has been obtained the results so that the results can be made graphya as follows.
E. Determination of non-buffer solution
            This experiment was performed using 3 tubes, tube 1 was filled with 1 ml of distilled water, tube 2 with 1 ml of HCL 0.0001 M solution. And tube 3 with 1 ml of 0.0001 M NaOH solution. Then formed pH of the solution with universal indicator. Based on experimental results obtained pH water = 5 pH NaOH = 10 and pH HCL = 4
Furthermore, to determine the pH of the solution instead of the buffer after added acid. Take three clean reaction tubes and then in tube 1 fill with 1ml distilled water. Tube 2 with 1 ml of HCL solution 0.0001 Ms tube 3 with 1 ml of 0.0001 M NaOH solution Added 1 drop of HCL 1 M. into each tube so that based on the experiment obtained pH water = 1 pH NaOH = 1 and PH HCL = 1.
At the initial pH of distilled water has a pH of 5 and after it is dropped with HCl to 1, the pH of the distilled water should be 7, but we have experimented with the distilled water in the weak acid region.
F. Bufer solution
            The first step we made was mixing 5 ml of 1M HC2H2O2 acetic acid with 5 ml sodium acetate NaC2H2O2 solution pH of the solution was 4 then the weak acid solution and the salt was added with 1 drop of 1M HCl solution. And the pH recorded in the universal indicator is 1 thus the acidity of this object is acidic.
            Then the base pair is weak and the salt is a weak base NH4OH 5ml 1M and the salt NH4Cl 5ml 1M. When mixed the pH is 10 and when the mixture is dropped with NaOH solution turns the pH up ie 11 represents the basic properties of the solution.
Thus, the conclusion of the buffered solution experiments that acetic acid (HC2H2O2) and sodium acetate (NaHC2H2O2) solutions are buffered solutions, since they can maintain the pH from the addition of a few acids and basic bases in our experimental pH, which means that there is a mistake in the experiment.

X. DISCUSSION
A. Standardization of HCl solution
            In this titration process, standard HCl 0.1 M titrated produces a pink color, this occurs because the indicator used is pp which is a weak acid form. In this case, the weak acid is colorless and the ions are bright pink. The use of excess ions can shift the equilibrium position towards the left and change the indicator to be colorless. The addition of hydrogensia ions removes hydrogen ions from the equilibrium that leads to the right to turn the indicator into pink.
B. Standardization by using KHP
            In this experiment will be determined the actual molarity of NaOH by using KHP, NaOH itself will be hygroscopic and quickly absorb CO2 so that the concentration will change at any time.
The primary satandar used is Potassium Hydrogen Phthalate (KHP), because KHP is a very good standard for alkaline solutions. So, if you want to make acid or base solution with the desired concentration, then to know the actual concentration is done standardization with the main standard, that is KHP.
C. Standardization of acetic acid percentage in vinegar
            In this experiment we will determine the percentage of acetic acid therein by using a standard NaOH solution according to the equation
NaOH + HC2H2O2 -> NaHC2H2O2 + H2O
We found that the required volume was 44.5 ml, and after we performed the calculation the mass of acetic acid in vinegar was 6,55 x 10-3 gr from 2.016 gr vinegar 2 ml so the percentage was 0.325% and this result is reasonable for vinegar acid.

D. Potentiometry
The first is to calibrate the buffer solution pH 3 then weigh thoroughly 5.1 g KHP, then dissolve it with distilled water and diluted, in a 250 ml measuring flask until the marks are listed. Then create a standardized NaOH (about 0.1M) and insert NAOH standardized into the burette. Then each addition of NaOH is recorded pH. The pH obtained from the experiment is:

1.      10 ml   = 4

2.      20 ml   = 5

3.      30 ml   = 5

4.      40 ml   = 5

5.      45 ml   = 5

6.      46 ml   = 5

7.      48 ml   = 6

8.      50 ml   = 6

9.      55 ml   = 6

10.  60 ml   = 6

E. Non-buffer solution
In this experiment the distilled water, HCl, and NaOH were measured pH and the distilled water pH was found to be 5, HCl 4 and NaOH 6. The distilled water pH should have been 7 but in this case the error may be due to distilled water contaminated with weak acid. So also with NaOH should the pH of NaOH close to 10 or even more because NaOH is a strong base. The wrong acid-containing drops used for NaOH may be the cause.

F. The buffer solution
The buffer solution is a solution which can maintain pH even with a small amount of acid or base. Bufer consists of a weak acid or a weak base with its salts and acids with its conjugate base and a base with conjugate acid.
5 ml of 1M acetic acid HC2H2O2 with 5 ml sodium acetate NaC2H2O2 solution pH of the solution was 4 then the weak acid solution and the salt was added with 1 drop of 1M HCl solution. And the pH recorded in the universal indicator is 4.
The weak base NH4OH 5ml 1M and the salt NH4Cl 5ml 1M. When mixed the pH is 10 and when the mixture is dropped with NaOH solution it turns up pH ie 11.
Thus, the conclusion of the buffered solution experiments that acetic acid (HC2H2O2) and sodium acetate (NaHC2H2O2) solutions are buffered solutions, since they can maintain the pH from the addition of a few acids and basic bases in our experimental pH, which means that there is a mistake in the experiment.

XI. POST-FACTORY QUESTIONS
1. Is the result of standardization of NaOH solution using HCL solution with KHP gives the same result?
Answer:
No, because HCL is a strong acid while KHP is a weak acid so the results are different.
2. The concentration of acetic acid analysis results in the vinegar sample you are working on
Answer:
Analysis of acetic acid in our example, the greater the mass% of acetic acid the more NaOH solution is needed to standardize
3. In order for titration for the second and third instances to run quickly, what action do you take?
Answer:
Continue to shake the Erlenmeyer and enlarge the NaOH solution coming out of the faucet but keep it to the point that the outflow NaOH does not exceed the end point of the titration.
4. In order for the endpoint of the titration to approach the equivalent point, how?
Answer:
Slow down / dilute the NaOH solution out of the biuret and keep the Erlenmeyer wiggling when the solution has turned into purple then the titration is stopped and when the color changes we must be careful and prepare for the NaOH that comes out according to the equivalent point
5. Why are indicators so important in titration?
Answer:
Because the indicator can do the titration quickly and easily visible change of color
6. If the apparatus in part B is excessive with NaOH, if the error in KHC8H4O4 in part B or acetic acid on vinegar produces a positive or negative result?
Answer:
Positive because the weight of acetic acid on vinegar will increase so that% acetic acid in vinegar will be less.
7. Adjust the following reaction equation:
Answer:
KHC8H4OH + NaOH -> NaKC8H4O4 + H2O

XII. CONCLUSION
1. Titration is a meter based measure
2. The standard solution is a solution whose concentration is known precisely.
3. In the titration there is an equivalent point and end point of titration.
4. Based on the calculation of molarity then for each experiment obtained as follows:
A. Standardize HCl solution
            HCl = 0.1 M
            NaOH = 0.1 M
B. Standardize KHP
            KHP = 0.0688 M
            NaOH = 0.0661M
C. Concentrate the percentage of acetic acid in cukA
            % Acetic acid = 0.325%
5. If the titration will reach the equivalent point or end point of the titration there will be a pink change due to the indicator used by phenolphthalein.
6. The buffer solution is very useful in the physiology of the human body, especially in the control of blood pH.

XIII. BIBLIOGRAPHY
Hendayana. 1994. Basic 2 Chemistry. Jakarta: New Script
Rival. 1995. Basic chemistry . Jakarta . Erland
Ryan. 2001. Basic chemistry . Jakarta: Erland
Sudarmono, superior. 2005. Chemistry . Surakarta: Erland
Grateful, s. 1999. Basic Chemistry 1. Bandung: ITB