Fany Fairuz: kumpulan laporan kimia kelas xi ipa

REPORT OF CHEMISTRY

TASKS OF SEMESTER 2

Report 1

The pH Solution of Acid and Base

I. Purpose

To measure the pH of some solution of strong and week acid / base whose concentration is the same using universal indicator

II. Basic Theory

· According to Arrhenus,acid is a substance that releases H⁺ ions in water, while Base is a substance that releases OH ions in water.

· According to Bransted Lowry acid is gives proton, while Base is accepts protons.

· According Lewis, acid is a pair of electrons acceptor, while Base is a pair of electrons electrons donor.

III. Tools and Material

1. Pasteur pipette (1 pcs)

2. Volume pipette (1 pcs)

3. Chemical glass (4 pcs)

4. Universal indicator paper (3 pcs)

5. Water (as needed)

6. Funnel shaped (1 pcs)

7. Measure gourd (1 pcs)

8. NH₃ solution (50 ml)

IV. Procedure

1. Prepare three clean and dry chemical glass.

2. Give each glass a label by number 1,2, and 3.

3. Add 5 ml of 0,1 M NH₃ solution into chemical glass 1.

4. Make NH₃ solution with the concentration of 0,01 M and 0,001 M by dilution.

5. Add 5 ml NH₃ solution into chemical glass 2 and 5 ml NH₃ solution into chemical glass 3.

6. Measure the pH of the three solutions using universal indicator paper.

7. Repeat procedur 1 to 6 using CH₃COOH, NaOH, NH₃, HCl, NaCl.

Calculation

M.V= M_1. V₁

12. V = 1 . 50

V= 4,2 X 2 = 8,4

M₁.V₂= M_3. V₃

1 . V₂= 0,1 . 50

V₂= 5

M₃.V₄= M_5. V₅

0,1. V₄= 0,01 . 50

V₄= 5

M₅.V₆= M_7. V₇

0,01. V₆= 0,001 . 50

V₆= 5

V. Result Experiment

Glass no.	Concentration of acid/base (M)	pH Solution
Glass no.	Concentration of acid/base (M)	HCl	CH₃COOH	NaOH	NH₃	NaCl	HCl
1	0,1	1	3	13	9	7	-
2	0,01	2	3,5	12	8	7	-
3	0,001	3	4	11	7,5	7	-
4	0,0001	-	-	-	-	-	4
5	0,00001	-	-	-	-	-	5
6	0,000001	-	-	-	-	-	6

VI. Question

1. What are the pH differences for acidic solution and base solution?

2. What is the influence of the dilution to the pH of solution of strong acid and base?

3. What is the influence of dilution to the pH of solution of weak acid and base?

Answer:

1. The pH of acidic solution are less than 7, The pH of base solution more than 7

2. When the strong acid dilution with water so the pH is increase because the concentration is decrease, and when the strong base diluted with water the pH is decrease because the concentration of solution is decrease.

3. When the weak acid diluted with water, the pH is increase because the concentration of solution is decrease. And when then weak base diluted with water the pH is decrease because the concentration of solution is decrease.

VII. Conclusion

1. The pH of solution is inversely proportional to the acid strength. It means that the more concentrated the solution of acid, the smaller the pH value is.

2. The pH is proportional to the base strength. It means that more concentrated the solution of base, bigger the pH value.

VIII. Photos

Alat dan bahan Mengukur pH larutan dengan kertas indicator

Gambar 1.3 hasil pengukuran pH

Pengenceran NH₃

IX. Bibliography

· Permana Irfan. 2009. Memahami Kimia. Armico Bandung

· Utami, Budi, Agung Nugroho Catur Saputro, Lina Mahardiani, Sri Yamtinah, Bakti Mulyani. 2006. Simpati. Surakarta:Grahadi.

· Susilowati, Endang. 2009. Chemistry. Solo:Tiga Serangkai.

Report 2

The PH Solution of Acid and Base

I. Purpose :

To measure the pH solution of strong and weak acid / base whose concentration using natural indicator.

II. Basic theory

Natural indicators can include kitchen materials, flowers, or fruit - fruit. To be able to be used, such materials must be made form of a solution by means attenuating then examined indicator solution into the acid-base solution. Color changes that occur in each indicator varies naturally. Previously we tested with a solution of NaOH and CH₃ COOH.

III.      Tools and materials
1. Test tube                                     (5 pcs)
2. Second beaker
3. Pipette drops of                          (6 pcs)
4. Pipette drops small                  (1 pcs)
5. Plat drops                                    (2 pcs)
6. spatula                                        (1 pcs)
7. Mortil                                            (1 pcs)
8. NaOH
9. Acid and alkaline solution
10. CH3 COOH
11. Natural Indicator (japanese kambodia, orchid, tapak dara, eporbia, turmeric)
water

IV. Steps :

1. Filling the water into each beaker.

2. Giving the label on each tube reaction, writing A, B, C, D, E on it.

3. Blending of flower ( as natural indicator).

4. Giving the water on flower that have been soft.

Droping CH₃COOH on drop plate 5 coloumb (suitable the sum of indicator).

NaOH water on drop plat 5 coloumb.

6. Taking flower solution using small drop pipette on pipette drop (each of solution drop into one coloumb CH₃COOH and one coloumb NaOH water). Writing the changing colour.

7. Drop A, B, C, D, Esolution (in drop plate).

8. Drop flower solution into A, B, C, D, E solution (in drop plate)

9. Writing the changing of colour.

	Rose	Curcuma	Kol Merah	Bougenvil	Kamboja
CH₃COOH	Pink	yellow	Purple	Pink	Pink
NaOH	Green	Orange	Green	Green	Green

	A	B	C	D	E
Rose	Pink	Yellow	Pink	Yellow	Yellow
Camboja	Pink	Green	Pink	Green	Green
Kol Ungu	Pink	Green	Pink	Green	Green
Kunyit	Yellow	Yellow	Yellow	Yellow	Yellow
Bougenvil	Purple	Yellow	Yellow	Yellow	Yellow

V. Conclusion:
A solution is acidic
B solution is base
C solution is acidic
D solution is base
E solution is base

VI. Photos

Alat dan bahan menghaluskan indicator alam

Perubahan warna pada larutan Perubahan warna pada cuka dan kapur

VII. Bibliography

· Permana Irfan. 2009. Memahami Kimia. Armico Bandung

· Utami, Budi, Agung Nugroho Catur Saputro, Lina Mahardiani, Sri Yamtinah, Bakti Mulyani. 2006. Simpati. Surakarta:Grahadi.

· Sudarmo, Unggul. 2004. Kimia. Surakarta:Seri Made Simple.

· Susilowati, Endang. 2009. Chemistry. Solo:Tiga Serangkai.

Report 3

Determine Concentration of HCl by Using Standard Condensation of NaOH 0,1 M

I. Propose

To determine HCl concentration using by standard solution is NaOH 0,1 M.

II. Based theory

Acid and Base reaction can be used to determine the concentration of unknown acid and base solutions. The determination is conducted by titration. Type titration there is 2 :

a. Acidimetric

Is a base solution with unknown concentration is titrated with acid.

b. Alkali metric

Is an Acid solution with unknown concentration is titrated with base.

At the experiment , we will concentration observation from HCl using by standard solution is NaOH 0,1 M. we do at the experiment titration type Alkalimetric.

III. Tools and Materials

1. Burette 50 ml 1

2. Stand 1

3. Graduated cylinder of 25 ml 1

4. Erlenmeyer flask of 100 ml 1

5. Pasteur pipette 2

6. Test tube 1

7. Test tube shelf 1

8. Chemical glass of 100 ml 3

9. Measure gourd of 100 ml 1

10. Tissue

11. Measure pipette 1

12. Condensed HCl 20 ml

13. PP indicator

14. Aquades

15. NaOH 0,1 M 50 ml

IV. Procedure

1. Tide burette on the stand with klem strong.

2. Pour standard solution is NaOH 0,1 M into the burette used by funnel until 0 scale.

3. Take 25 ml condensed HCl used by measure pipette, pour into the measure gourd of 100 ml then add aquades until 100 ml ( sign boundary ).

4. Take watery HCl ( step 3 ) used by chemical glass 25 ml, pout into the erlenmeyer flask. Add 3 – 5 drip PP indicator.

5. We do titration watery HCl used by standard solution NaOH 0,1 M carefully. Shaking down enlenmeyer flask! Discontinue titration in th event color changed HCl from transparent color become pink color.

6. Read burette scale and written used standard volume.

7. Repeat procedure 2 – 6 up to thrice.

V. Result Of Observation

Experiment	HCl volume (ml)	NaOH volume (ml)
1	20	17
2	20	17
3	20	17

a. Ma. Va = b. Mb. Vb

1. Ma. 20 = 1. 0,1. 17

Ma =

= 0,0175 º 0,02 M

VI. Conclusion

we earn to determine concentration of HCl by using standard condensation of NaOH 0,1 M.

VII. Photos

Proses titrasi Hasil titrasi

VIII. Bibliography

· Permana Irfan. 2009. Memahami Kimia. Armico Bandung

· Utami, Budi, Agung Nugroho Catur Saputro, Lina Mahardiani, Sri Yamtinah, Bakti Mulyani. 2006. Simpati. Surakarta:Grahadi.

· Sudarmo, Unggul. 2004. Kimia. Surakarta:Seri Made Simple.

· Susilowati, Endang. 2009. Chemistry. Solo:Tiga Serangkai

Report 4

Determine Commerce Vinegar Rate used by Standard Solution is NaOH Solutions 0,1 M

I. Propose

To determine commerce vinegar rate used by standard solution is NaOH solutions 0,1 M.

II. Based theory

Acid and Base reaction can be used to determine the concentration of unknown acid and base solutions. The determination is conducted by titration. Type titration there is 2 :

a. Acidimetric

Is a base solution with unknown concentration is titrated with acid.

b. Alkali metric

Is an Acid solution with unknown concentration is titrated with base.

At the experiment , we will concentration observation from commerce vinegar rate using by standard solution is NaOH 0,1 M. we do at the experiment titration type Acidimetric.

III. Tools and material

1. Burette 50 ml 1

2. Stand 1

3. Graduated cylinder of 25 ml 1

4. Erlenmeyer flask of 100 ml 1

5. Pasteur pipette 2

6. Test tube 1

7. Test tube shlf 1

8. Chemical glass of 100 ml 3

9. Measure gourd of 100 ml 1

10. Tissue

11. Measure pipette 1

12. commerce vinegar rate

13. PP indicator

14. Aquades

15. NaOH 0,1 M 50 ml

IV. Procedure

1. Prepare tools and material we needed.

2. Thinning commerce vinegar rate 10 ml with water up to volume 100 ml used by measure gourd.

a. Measure CH₃COOH counted 100 ml used by measure glass

b. Move CH₃COOH which have been measured into the measure gourd up to volume 100 ml

c. Shake slowly.

3. Pour NaOH 0,1 M into the burette up to 0 scale used by funnel.

4. Pour CH₃COOH thinning result into the Erlenmeyer flask counted 20 ml.

5. Add PP indicator counted 3 drip used by Pasteur pipette into the erlenmeyer flask.

6. We do titrated, at the same time shaken down Erlenmeyer flask up to the changed color become pink color.

7. Discontinue titrated.

8. Repeat procedure 3 – 7 until thrice.

V. Result of observation

Experiment	CH₃COOH volume (ml)	NaOH volume (ml)
1	20	3,5
2	20	3,5
3	20	3,5

VI. Calculations

b. Ma. Va = b. Mb. Vb

16. Ma. 20 = 1. 0,1. 3,5

Ma =

= 0,085 º 0,01M

_·M₁.V₁= M_2.V₂

M₁.10 = 0,01.20

M_{1 =}0,02 M

· M = p.10. kadar

m_m

0,02 = 1,05.10. kadar

1,2 = 10,5. Kadar

Kadar = 1 %

VII. Photos

Proses titrasi

Proses perubahan warna

VIII. Bibliography

· Utami, Budi, Agung Nugroho Catur Saputro, Lina Mahardiani, Sri Yamtinah, Bakti Mulyani. 2006. Simpati. Surakarta:Grahadi.

· Permana Irfan. 2009. Memahami Kimia. Armico Bandung

· Sudarmo, Unggul. 2004. Kimia. Surakarta:Seri Made Simple.

· Susilowati, Endang. 2009. Chemistry. Solo:Tiga Serangkai

Report 5

Precipitation Reaction

I. Propose

To learn about precipitation reaction of some substances.

II. Based theory

Solubility product of an ionic compound which is difficult to dissolve may provide information about the solubility of these compounds in water. The greater the price Ksp of a substance, the more easily soluble compound. Price Ksp of a substance can be used to predict whether precipitation of a substance occurs when two solutions containing ions of compounds difficult to dissolve the precipitate to mix A_MB_N make prediction if a solution containing ions Aⁿ⁺ and B^m-mixed use concept of the product A_MB_N ion (Qsp) following this:

If Qsp> Ksp precipitate it will happen.
If Qsp = Ksp, then it will happen A_MB_N saturated solution.
If Qsp <Ksp, there will be a saturated solution or precipitate A_MB_N.

III. Tools and Materials

1. Test tube

2. Graduate cylinder of 10 ml

3. Pasteur pipette

4. 0,05 M BaCl₂ solution

5. 0,05 M PbCl₂solution

6. 0,05 M CaCl₂solution

7. 0,05 M NaOHsolution

8. 0,05 M Na₂SO₄solution

9. 0,05 M Na₂CO₃solution

10. 0,05 M NaCrO₄solution

IV. Procedure

1. Based on Ksp predict the precipitate formation of metal compounds of Ba, Pb, and Ca that are reacted with a reactant according to observation teble.

2. Take three test tube. Then add 3 ml of 0,005 M BaCl₂ solution into test tube 1, 3 ml of 0,05 M PbCl₂solution into test tube II, and 3 ml 0,05 M CaCl₂solution into test tube III.

3. Add 3 ml 0,05 M NaOHsolution into each tube. Make notes about your observation on the observation table. Give (+) mark when precipitate formed and(-) when there is on precipitate.

4. Repet procedure 1 and 2 by substitute NaOH with Na₂CO_3,Na₂SO₄, and NaCrO_4.

V. Result of observation

0,05 M Metal ion	0,05 M ion Reactan	Ksp	Prediction	The result of the test
Ba²⁺		5x10^-3	-	+
Pb²⁺	OH-	1,43x10^-20	+	+
Ca²⁺		6,5x10^-9	+	-
Ba²⁺		2,58x10^-9	+	+
Pb²⁺	CO3^2-	7,4x10^-14	+	+
Ag²⁺		10	+	+
Ba²⁺		1,5x10^-9	+	+
Pb²⁺	SO4^2-	1,6x10^-8	+	=
Ag²⁺		2x10^-9	+	-
Ba²⁺		22x10^-10	+	+
Pb²⁺	CrO4^2-	2,8x10^-13	+	+
Ca²⁺		1,9x10^-12	+	-

VI. Question

1. Which solution result in precipitate?

2. Does the experiment match your prediction?what does not match?when anything does not match, give the explanation.

Answere

1. All solutions have sediment.

2. There is, occurs in solution

VII. Conclusion

From the observation BaCl₂ solution of 0.05 M, 0.05 M PbCl₂, 0.05 M CaCl₂ if added with a solution of 0.05 M NaOH, 0.05 M Na₂SO₄, 0.05 M Na2CO3, 0.05 M NaCrO₄ it will deposition occurs because Qsp> Ksp precipitate A_MB_N it will happen.

VIII. Photos

Measuring 0.05 M Na2CO3 solution

IX. Bibliography

· Utami, Budi, Agung Nugroho Catur Saputro, Lina Mahardiani, Sri Yamtinah, Bakti Mulyani. 2006. Simpati. Surakarta:Grahadi.

· Permana Irfan. 2009. Memahami Kimia. Armico Bandung

· Sudarmo, Unggul. 2004. Kimia. Surakarta:Seri Made Simple.

· Susilowati, Endang. 2009. Chemistry. Solo:Tiga Serangkai

Report 6

Differences between solution, colloid, and suspension

I. Propose

To learn the differences between solution, colloid, and suspension.

II. Based theory

When a substance is mixed with other substances, there will be spread evenly from one substance into another substance called dispersion system. Example: water + flour, water is the dispersing medium and flour are dispersed substance. Disperse divided into 3 groups: the solution, colloid, suspension. At first glance the difference between suspense (often is a with suspense rough) with a solution (often called a true solution) will be evident from the homogeneity, but it would be difficult to distinguish between solutions or colloids with roughly suspense.

III. Tools and materials

1. Chemical glass of 100 ml (8 pcs)

2. Funnel (3 pcs)

3. Filter paper (7 pcs)

4. Spatula (1 pc)

5. Erlenmeyer (7pcs)

6. Sugar

7. Flour

8. Milk

9. Urea

10. Soap

11. Coconut milk

12. Sulphur powder

13. Jelly

14. Aquades

IV. Procedure

1. Pour about 80 ml of water into chemical glass.

2. Add one gram of sugar into the glass and stir it about one minute.

3. Let the solution for five minutes and make notes on what happens.

4. Filter the mixture using filter paper and make notes on what happens.

5. Repeat procedure 1 to 4 using Flour milk, Urea, Soap, Coconut milk, Sulphur powder, Jelly.

Note : stirrer and funnel should be rinsed and dried before used it stir and to filter different mixtures.

V. Observation result

Mixture of water with	Mixture characteristic
Mixture of water with	Sugar	flour	milk	urea	soap	sulphur	Coconut milk	jelly
Dissolved/not	√	‐	√	√	√	‐	√	‐
Stable/not	√	‐	√	√	√	‐	√	‐
Clear/turbid	√	‐	‐	√	‐	‐	‐	‐
Precipitated/not	‐	√	‐	‐	‐	√	‐	√
Clear/turbid filtrate	√	√	‐	√	‐	√	‐	√

Description of sign (√) chose the first option.
( - ) chose the second option.

VI. Question

1. Classify the mixtures in this experiment into solution, colloid system, and suspension.

2. Explain the differences between solution, colloid, and suspension.

Answer

1. Solution : urea, sugar.

Colloid system : milk, jelly, soap, coconut milk.

Suspension : sulphur, flour.

2. Solution :

· Homogen

· Disperse molekuler

· <10^-7 cm atau <1 nm

· Tidak dapat disaring

· Stabil

· Jernih

· Satu fase

Colloid system :

· Tampak heterogen

· Disperse padatan

· 10^-7 - 10⁵cm atau 1 nm – 100nm

· Dapat disaring dengan penyaring ultra

· Pada umumnya stabil

· Tidak jernih

· Dua fase

Suspensi :

· Heterogen

· Disperse padatan

· >10^-5 cm atau >100 nm

· Dapat disaring dengan penyaring biasa

· Tidak stabil

· Tidak jenih

· Dua fase

VII. Conclusion

Dari percobaan yang dilakukan dapat disimpulkan bahwa

VIII. Photos

Hasil penyaringan

Proses penyaringan larutan

IX. Bibliography

· Utami, Budi, Agung Nugroho Catur Saputro, Lina Mahardiani, Sri Yamtinah, Bakti Mulyani. 2006. Simpati. Surakarta:Grahadi.

· Sudarmo, Unggul. 2004. Kimia. Surakarta:Seri Made Simple.

· Susilowati, Endang. 2009. Chemistry. Solo:Tiga Serangkai

Report 7

The Making of Colloids

I. Propose

To learn the colloids making b.y dispersion and condensation.

II. Based theory

The size of colloidal particles is between suspension particles and solution particles. Therefore, colloidial particles can be madeby refine suspension particles until they are of colloid size and aggregating the solution particles. The colloid made by refine suspension particles is called dispersion , while the colloid made by aggregating solution particles is called condentasion.

There are three method of dispersion : mechanical method, peptization method, and Bredig’s method.

1. Chemical method

The mechanicl method insolves reducing coarse solid particles into colloidal size particles.

2. Peptization method

Peptization is a dipersion of a precipitate to form a colloidal system by adding a peptizing agent. Common peptizing agents include electrolytes especially those containing the same type of ions , or certains solvents.

3. Bredig’s method

Bredig’s are method is solid to prepre metal sold such us Ag, Au, and Pt. The metal will be transformed into colloidal particles is used is the electrodes. Two emetal electrodes are dipped into the dispersing medium ( cold distilled water) where both and positioned close to each other. An electic are is then struckbetween the two electrodes. The intense heat produces caused the metal to evaporate. The vapor would then condense in a cold dispersing medium to fform colloidal particles.

III. Tools and materials

1. Mortat and Pestle (1 pcs)

2. Chemical glass of 100 ml (2 pcs)

3. Test tube (3 pcs)

4. Tabe clip (1 pc)

5. Sulphur powder (1 teas poonffull)

6. Sugar (1 teas poonfull)

7. Pudding powder (1 teas poonfull)

8. Kerosene (4 ml)

9. 5 % detergent solution (2ml)

10. Water (as needed)

IV. Procedure

1. The making of sulphur sol.

Add 1 teas poonfull of sugar and 1 teas poonfull of sulphur into the mortar. Crush the mixture until refined. Take 1 teas poonfull of the mixture to be mixed with 1 teas poonfull of sugar. Do this activity of 4 times. Pour the last mixture into the chemical glass and add 50 ml of water and stir it.

2. The making of pudding sol/gel.

Pour the water to the test tube until one-third. Add 1 teas poonfull of sugar powder and stir. Heat the solution until boiled. Freeze the mixture until the pudding gel made.

3. The making oil in water emulsion.

Pour 5 ml of water and 25 ml kerosene in the last tube. Shake the mixture for a moment. Pour 5ml of water, 2 ml of kerosene, and 2 ml of detergent solution into another test tube. Shake the test tube.

V. Observation result

Mixture Component	Observation Result
Mixture Component	Mixed	Not Mixed
1. Sulphur powder and water		√
2. Pudding powder and water
b. Before transferred	√
c. After transferred		√
3. Water and Kerosene		√
4. Water, kerosene and detergent solution	√

VI. Question

1. Explain how to make colloids by dispersion?

2. How does detergent emulsity oil in water?

3. Explain the differences between sol and gel?

Answer

1. The making of colloids by dispersion is conducted by reducing coars particles into colloidal size particles which will then be dispersed into dispering medium.

2. Detergent ( emulgator) has polar and nonpolar groups. The nonpolar part will interct with oil or surround the oil particles. The polar part will interact with water. If the polar part ionizes and becomes negatively charged, the particles of oil will become negatively charged. The negatively charge causes the oil particles to repel and not combine. In other words, the emulsion becomes stable.

3. Sol is the colloid system with solid dispersed phase and liquid dispersion medium. And gel is a half rigid (between liquid and solid) colloid system. Gel will be formed when sol dispersed phase of a sol absorbs it dispersion medium so a reather solid colloid system will be created.

VII. Conclusion

The making of colloid by dispersion is conduted by reducing coarse particles into colloidal size particles which will then be dispersed into the dispersing medium. And to making colloid from and oil (emulsion) is needed detergent is emuglator.

VIII. Photos