USING ENGLISH TO REPORT

TITRIMETRY BASIC CHEMICAL PRINCIPLES AND CONTROL PH

BASIC CHEMICAL PRACTICAL REPORT

TITRIMETRY AND CONTROLLING pH

I. TITLE: TITRIMETRY AND PH CONTROL

II. Day / Date: 19 Desenber 2014

III. Objectives: 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. familiar with some buffer solutions of a particular system and how they function

IV. Practice Questions

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

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

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

  (C) Equivalent Point: The point that occurs between acidic and basic solutions in which the acid solution may react with the amount of the basic solution.

  (D) Indicator: A substance used as a guide for distinguishing acid and base solutions.

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

 Titration end point: The point in a titration in which an indicator changes color.

  . Equivalent Titk: 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?

Given: KHC6H6o7 + NaOH NaKC6H6O 2

Vol NaOH = 33.6 ml = 0.0336

Dit: M NaOH?

Mol KHC6H6o7 = 0.7742 / 230 = 3,36.10-3

Mol NaOH = Mol KHC6H6o7 = 3,36.10-3 mol

M NaOH = =

= 0.1 M

4. Explain what is meant by:

(A) acid-base titration curve: The figure representing the count of ph by volume l.

(B) Equivalent Point: The point at which the acid has reacted is perfect

(C) Standardization: The process of determining the concentration of a carefully specified solution.

(D) Primary standard solution: the known solution of concentration.

(E) pH: Negative Logarithm H + or expressed negative concentration of H + in laruutan

(F) pH Meter: The instrument 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.

V NaOH = 25 ml

M NaOH = 0.1M

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 Phosphate = 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?

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. (a) What is Bufer's solution?

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

(B) why is the buffer solution important?

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

8. Define for weak acids and weak bases.

• 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 sionide (NaCN) with hydrogen sionide (HCN) serves as a buffer solution

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. Name some buffer solution pairs that are of similar physiological properties.

HC2H3O2 + NaOH NaC2H3O2 + H2O

KH2PO4 + NaOH K2HPO4 + H2O

V. Theoretical Basis

            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. The classic example of this type of reaction is the reaction of hydrogen ions with hydracane 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.

Acetic acid titration curve with 0.101 M NaOH solution

Figure 9.1 acid-base titration curve between acetic acid solution and naoh solution 0.101 M. Equivalent point after addition of 27.02 ml of NaOH.

            The equivalent point is reached after the addition of naoh 27.02 Ml. From the titration curve in can also data to calculate the ionization of acetic acid constants through the 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 amount of ½ of required 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 = 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 the same as the acidic ion. Salts between other sites may be prepared by allowing the weak acid to react with an appropriate base comprising a suitable cation. 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

            Salt of weak base has the same cation as 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.

            Polypotic acid triclacy is weak evidence strongly 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-that is:

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 changed 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 a weak acid can be calculated as follows:

Ka =

[H +] = Ka

-Log [H +] = -Log Logo

PH = pKa-log

PH = pKa + log

(Harvest, 1991: 235-237)

VI. TOOLS AND MATERIALS

Tool

A. Erlenmeyer

B. Drop pipette

C. Balance Sheet

D. Measuring cup

E. Test tube

F. Universal indicator

G. Buret 50 ml

H. 500 ml bottle

I. Funnels

J. Poles

K. Watch glass

L. Stirring bar

material

A. Distilled water

B. Pp indicator

C. NaOH solution

D. Khp 0.1 gr

E. Kitchen vinegar

F. Hcl solution

G. Sodium acetate solution

H. NH4Cl

I. NH4OH

VII. Work procedures

A. Absorption of NaoH 0.1 M Solution

1.6 gr NaoH

• Weighed

• Moved to bottle

• Dissolved with 400 ml of distilled water

• Beat

Observation result

B. Standardization of 0.1 M NaOH solution

Burette 50 ml

• Washed and rinsed with distilled water

• Closed and inserted approximately 5 ml naoh

• To fill buret with naoh s / d 0

• Flow solution

2 Erlenmeyer 250 ml

·       Washed and rinsed

·       DETAILED 25 ml of HCL 0.1 is included on each erlenmeyer

25 ml of distilled water and 3 drops of phenolphthalein indicator

·         Plus into each erlenmeyer

·         Note the initial position of NaOH

·         In a little bit naoh flow on Erlenmeyer 1

·         Recorded final volume on the burette

·         Done 2 times

3 Erlenmeyer

Washed

Filled with 0.14 grams of KHP

Added 10 ml of distilled water, shaken until dissolved

Added 3 drops pp indicator

Note the volume of NaOH used

Observation result

C. Determine the percentage of acetic acid in vinegar

3 Erlenmeyer 250 ml

Washed and rinsed

Dripped 25 ml of vinegar into a seeeeer erlenmeyer

10 ml of distilled water

Added

3 drops pp indicator

Added and titrated with standard solution until red is formed

Calculated percent of mass in each instance

Repeat once more if the results are different> 0.05%

Observation result

POTENSIOMETRY

A set of pH meter tools

Prepared

buffer solution pH 5

Calibrated

5,1 gr KHP

Weighed

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

80 ml Liquid pipette

Entered into a cup glass

A standardized NaOH solution

Entered into the buret

Installed like a picture

Recorded pH

Created a titration curve

Repeated experiment once again starting no 2

Observation result

A. The solution is not a buffer

1. Determination of pH of the solution is not a buffer

3 test tubes

Filled with tube 1 denngan 1 ml distilled water

The 2nd tube is filled with 1 ml of a 0.000001 M HCL solution

Tube 3 is filled with 1 ml of 0.0000 M NaOH solution

Determined and noted pH of the solution with universal indicator

2. Determination of pH of the solution is not a buffer after added acid

3 test tubes

Tube 1 is filled with 1 ml of distilled water

The 2nd tube is filled with 1 ml of a 0.000001 M HCL solution

Tube 3 is filled with 1 ml of 0.0000 M NaOH solution

Dripped HCL 1 M into each tube

PH of the solution is recorded

B. The buffer solution

1. Determination of Ph buffer solution

5 ml of acetic acid HC2H2O2 1 M

Mixed with 5 ml of sodium acetate NaC2H2O2 1 M

PH is recorded with a universal indicator.

2. Determination of pH of buffer solution after acid addition

2 test tubes

Tube 1 is filled with 2 ml of buffer solution

Tube 2 is filled with 2 ml buffer solution

Plus 1 drop of 1 M HCL solution into each tube

The pH of the solution is recorded and compared

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

2 test tubes

•       Tubes 1, 2 filled with 2 ml of buffer solution

•       Added 1 drop of NaOH

·         The pH of the solution is recorded and compared with the buffer solution

VIII. Trial Data

Acid - Base Titrations

A. Standardization with Ha solution

No		Deuteronomy
		1	2	3
		25 ml	25 m	25 ml
			0,1 m
		25 . 10-4
		0,0221 mol	0,022 mol	0,0362 mol
		50 ml	50 ml	50 ml
		29 ml	30 ml	31 ml
		221 ml	220 ml	369 ml
		0,1 m	0,1 m	0,1 m
			0,1 m

B. Standardization with KHP

No		Deutronomy
		1	2	3
		104,75 gr
		105,1 gr
		0,35 gr
		0,0017 mol
		26 .10-4 mol
		50 ml
		24 ml
		26 ml
		0,1 m
			0,1 m

C

NO		deutronomy
		1	2	3
		2 ml	2 ml	2 ml
		1,008 g/m
		0,5 gr	0,5 gr	0,5 gr
		50 ml	50 ml	50 ml
		39,5 ml	39 ml	39,0 ml
		10,5 ml	11 ml	10,1 ml
		0,1 m
		11,1 . 10-4 mol	11,1 . 10-3 mol	10,1 x 10-4 mol
		11,1 . 10-3 mol	11,1 . 10-3 mol	11,01 . 10-3 mol
		10,5 . 10-3 mol	11,1 . 10-3 mol	11,01 . 10-3 mol
		63 . 10-3 gr	64 . 10-3 g	60,6 . 10-3 g
		12,6%	13,2%	12,1%
		12,63%

No	Buret Reading(ml)	Volume  NaOh (ml)	pH
1		10	5
2		20	6
3		30	9
4		35	11
5		40	12
6		45	12
7		46	12
8		47	12
9		48	12
10		49	12

Buffer Control Trial

NO	Solution	PH (acidity)
		Early	After Addition of Chloride	After Addition of Nitroxide
A	The solution is not a buffer
	1. water	5	1
	2. sodium hydroxide	6	1
	3. hydrochloric acid	4	1
B	Buffer solution
	1. a mixture of acetic acid and sodium acetate	4	1	4
	2. A mixture of ammonium hydroxide and ammonium chloride	10	4	11

IX. Discussion

            In this experiment, we did some experiments:

A. Standardize with HCL

            In this experiment, the first step we took was we weighed 1.6 g NaOH and removed the bottle. Then dissolved with 400 ml of distilled water, stirred until dissolved.

In this experiment with 3 repetitions, we get the following results:

Mol NaOH obtained = 0.0221` mol, 0.022 mol, 0.03621 mol

Initial NaOH volume = 50 ml, 50 ml, 50 ml

The final NaOH volume = 29 ml, 30 ml, 31 ml

Used HCL mol = 25. 10-4

Molarity of NaOH solution = 0.1 M, 0.1 M, 0.1 M

Average molarity of NaOH = = 0.1 M

B. Standardization of KHP

            First of all we prepare the tools and materials needed then assemble them. Then we weigh the erlenmeyer flask and also weigh the KHP of 0.4 gr, then KHP is added to the erlenmeyer with 25 ml of water added. After we have added 3 drops of pp indicator, then we mined KHP with NaOH and the results we get are:

KHP mass = 0.35 gr

Initial NaOH volume = 50 ml

The final NaOH volume = 24 ml

Used NaOH volume = 26 ml

From this data, we can determine the concentration of NaOH:

Mol KHP - = - = 1.96 x 10-3

C. Determine the percentage of acetic acid in vinegar

            To perform this experiment, we use the same tool with standardized experiments with KHP. The difference is only on the materials used. 1 drop of vinegar acid and 10 ml of distilled water after that plus 3 drops of pp indicator, then titration with NaOH solution.

The results of our observations, obtained the results as follows:

Volume vinegar = 2 ml

Initial NaOH volume = 50 ml

Vinegar density = 1.008 g / mr

The final NaOH volume = 39.5

Used NaOH volume = 10.5

NaOH concentration

From the data we get, we can find the percentage of acetic acid that is:

Mass of vinegar = ρ. V

= 1.008 gr / mr. 2 ml

= 2.016 gr

(Ulannel 1)

V. M = mol of acetic acid

10.5. 0.1 = mole of acetic acid

1.05. 10-3 = mole of acetic acid

60. 1.05. 10-3 = acetic acid weight

63. 10-3 = acetic acid weight

Acetic acid period

= Mole. Mr

= 10.5 x 10-3. 60

= 63. 10-3

% Acetic acid (repeat 1) = x 100%

= X 100%

= 13, 22%

(Deuteronomy 2)

V. M = mole of acetic acid

N.0,1 = mole of acetic acid

M. 10-3 moles of acetic acid

Acetic acid mass

= Mole. Mr

= 1,1 x 10-3 .60

= 6.6 x 10-3

% Acetic acid = = x 100%

= X 100%

= 13, 22%

(Deuteronomy 3)

V. M = mole of acetic acid

10.1. 0.1 = mole of acetic acid

1.01. 10-3 = mole of acetic acid

Mass of acetic acid = mol x mr

                                    = 1.01. 10-3 x 60

                                    = 60.6. 10-3

% Acetic acid = = x 100%

= X 100%

= 12, 1%

POTENSIOMETRY

            Prior to doing this experiment, we prepared a pH meter tool and calibrated a pH 5 buffer solution, carefully weighed 5.1 g of potassium hydrogen phthalate (KHP) then we dissolved it with water and we diluted it in a 250 ml measuring flask until the mark was tarred. Then we make a standardized NaOH (about 0.1 m) and put in a burette. We note the pH of each NaOH addition. From our experiments we got results as following:

10 ml = 5

20 ml = 6

30 ml = 9

35 ml = 11

40 ml = 12

45 ml = 12

46 ml = 12

47 ml = 12

48 ml = 12

49 ml = 12

Control buffer

            In this experiment we made an observation to determine the pH of the solution instead of buffer and pH bufer.

We prepared 3 tubes: tube 1 filled with 1 ml of distilled water, tube 2 filled with 1 ml of HCL 0,00001 M solution and 3rd tube filled with 1 ml of NaOH solution 0,00001 M. both of the solution added acid, Each tube added 1 drop HCL 1 M then the results we can that is:

Tube 1 = initial pH 5 after added hydrochloric acid pH = 1

Tube 2 = initial pH of 6 after added hydrochloric acid pH = 1

Tube 3 = initial pH 4 after added hydrochloric acid pH = 1

            In the experiments of the buffer solution, the first 2 tubes ie tube 1 filled 5 ml of HC2H2O2 1 M acetic acid with 5 ml of sodium acetate NaC2H2O 1 M. tube 2 filled 5 ml NH4OH with 5 ml NH4CL 1 M. Then we added 2 ml of buffer solution and 1 Drops HCL 1 M on tube 1. and in tube 2 we add 2 ml of buffer solution, 1 tets NaOH 1 M. from this treatment we get the following data:

X. Discussion

            In this experiment, we were titrated and determined pH. The solution we titration is HCL with NaOH, KHP with NaOH and also determines the percentage of acetic acid in vinegar. At the time of pH determination with burette readings, non buffer and buffer solutions in our experiments there was an error in determining the percentage of acetic acid. Because according to theory percentage of 3 samples can not be more than 0.05% while we get is more than 0.05%

% Acetic acid = 12.6% = 13.2%: 12.1%

= 0.07%> 0.05%

            This is because we are less careful when experimenting and the lack of tools used

            On burette readings, we did not calculate the pH when adding NaOH to 50, 51, 52, 55 and 60. should be searched but we were not looking for. This is caused by the lack of pH meter so we can not determine the pH.

.

XI Question Post-Practice

1) Is the result of standardization of nohoh solution using hcl and khp solution giving the same result? If not, give your comment

Answer: No, because Hcl is a strong acid while Khp is a weak acid, so the results are different.

2) Comment on the results of the acetic acid analysis in the open sample you are working on

Answer: From the data we got, from 3 experiments that is: 12.6%, 13.2%, and 12.1% with a ratio of 0.07% not in accordance with the theory and should be done repetition but not we do.

3) In order to fit the second and third instances fast, what can be done?

Replied: By estimating from the first example on the volume of how NaOH is used there is a change in the color of the solution.

4) In order for the endpoint of the titration to approach the equivalent point, how and how is it observed for this titration?

Answer: By doing a more thorough observation when we are jaundiced with naoh and the color is not too thick. Observations should be done on white paper so that the color change is easier to observe.

5) Of all the procedures, why is the indicator so important in the titration? Give a brief explanation.

Answer: Because with the indicator, more can do the titration faster and also to make it easier in titration because we know the color that happened when we do titration.

6) if the phthalate in part B is excessive with naoh, is the error in the weight of KHC8H404 on the part of Batau acetic acid in vinegar yielding a positive or negative result. Explain

Answer: The positive result because the weight of acetic acid on vinegar will increase so the percentage of aseetat acid in vinegar will become seedikit

7) Complete the following reaction equation

KHC8H404 + NAOH = KHC8H404 + H20

CONCLUSION

A) In determining the titration, we must make very careful observations because the right titration point is the point near the equivalent point of the titles in the substance in which the titration may change color. Change the color not too fast.

B) We can determine the percentage of acetic acid in vinegar by titration of acetic acid in the example.

C) At the time of the minute, the indicator is needed because with the indicator we can do the titration quickly, and we can know the color that occurs when we do the titration.

BIBLIOGRAPHY

Epinur dan wiwik ernawati. 2012. penuntun pratikumkimia dasar.  Jambi: Universitas Jambi

Keenan. 1990. kimia untuk universitas. Jakarta: Erlangga.

Lehninger. 1993. kimia untuk universitas. Jakarta: Erlangga.

Sutrisno. 1994. kimia dasar. Bandung:Ganessa

Pettsuci, ralp. 1987. kimia dasar. Jakarta: Erlangga