LAB REPORT EXPERIMENTS 7 & 8 GAS LAW AND REDOX TITRATION GENERAL CHEMISTRY PDF

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EXPERIMENT 7: REDOX TITRATIONPURPOSEThe purpose of this experiment is to standardize the potassium permanganate solution by titration method.INTRODUCTIONOxidation-reduction processes electrons transfer one species to another. Potassium permanganate has been widely used as an oxidizing agent and requ...

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EXPERIMENT 7: REDOX TITRATION PURPOSE The purpose of this experiment is to standardize the potassium permanganate solution by titration method. INTRODUCTION Oxidation-reduction processes electrons transfer one species to another. Potassium permanganate has been widely used as an oxidizing agent and requires no indicator unless very dilute solutions are used. The oxidizing action of KMnO 4 in the acidic medium can be represented by the following equation: MnO4− + 8H+ + 5e−

Mn2+ + 4H2O

Solution containing MnO 4- ions are purple in colour and the solution containing Mn 2+ ions colourless and hence permanganate solution is decolourised when added 10 a solution of a reducing agent.

CHEMICALS AND APPARATUS

Potassium permanganate solution

Burette

Dried sodium oxalate

Conical flask

1M sulphuric acid

Hot plate Pipette Thermometer

PROCEDURE 1. The burette was rinsed and filled with potassium permanganate solution. the air bubble was removed from the nozzle of the burette by draining some of the solutions.. 2. 0.3 g to 0.4 g of dried sodium oxalate was weighed accurately. The mass of sodium oxalate was noted and transferred to conical flask. 3. 40 ml of 1.0 M sulphuric acid was added to the conical flask and carefully heated to 60ᵒC.

4. The solution was titrated with potassium permanganate until the first persistent pink colour appeared. The temperature was maintained at 55-57ᵒC. 5. The titration was repeated to obtain 2 acceptable results, and the data were then recorded.

OBSERVATIONS:

Trial 1

Trial 2

Trail 3

Mass sodium oxalate

0.338 g

0.338 g

0.338 g

Final burette reading (mL)

29.85

31.05

37.00

Initial burette reading (mL)

3.05

5.00

10.00

Volume of KMnO4 (mL)

26.80

26.05

27.00

DISCUSSION

In this experiment, we are conducting the experiment using potassium permanganate and sodium oxalate to determine the standardizing of potassium permanganate solution. We added sodium oxalate and sulphuric acid before heating it up to 60 ℃ and immediately titrate when the temperature has reached its level. We make sure that the titration is done slowly in order to get the pink solution. The titrant was deep purple at the start of the titration, and the analyte was colorless. When MnO 4 was added to the analyte, the area around the drop turned pink/purple. Titration was repeated until the entire solution was pale pink. The purple MnO 4 that was initially added to the analyte was quickly reduced to clear Mn 2+ and lost its colour. With 0.338 g of sodium oxalate we managed to get 29.85 ml for trial 1, 31.05 ml for trial 2 and 37 ml for trial 3 for the final burette reading. After we obtained the final and initial values of the burette, the volume of KmnO4 was determined. For trial 1, the volume of KMnO4 is 26.80 ml, for trial 2 it is 26.05 ml and for trial 3 it is 27.00 ml. Last but not least, we need to make sure our eyes are perpendicular to the reading to minimise errors. Titrate slowly to ensure that the colour in the conical flask is a pale pink colour. Handle chemicals carefully and clean all equipment with distilled water, including the burette, conical flask, and volumetric flask.

CONCLUSION In conclusion, we determined the standardization of the potassium permanganate solution by the titration method. The actual molarity of potassium permanganate solution is 0.0373 M.

QUESTIONS

1. Write the overall chemical equation for the redox reaction between potassium permanganate and sodium oxalate. 2MnO4- + 5C2O42- + 8H+ MnO4− + 8H+ + 5e− C2O42−

2Mn2+ + 10CO2 + 8H2O Mn2+ + 4H2O

(i)

2CO2 + 2e−

(ii)

2×(i)+5×(ii) 2MnO4- + 5C2O42 + 16H+ + 10e− 2MnO4- +5C2O42+16H+

2Mn2+ + 10CO2 + 10e− + 8H2O

2Mn2+ +10CO2 +8H2O

2. Calculate the actual molarity of potassium permanganate solution Mass of Na2C2O4 = 0.338 g no of mol=

molar mass = 134 g/mol

0.338 g =0.00252 mol 134 g /mol 1 mol of Na2C2O4 = 1 mol of C2O4 0.00252 mol of Na2C2O4 = 0.00252 mol of C2O4 2 mol of MnO4- react with 5 mol C2O42-

2 mol x 0.00252 =0.001008 mol MnO 4−¿ 5 mol

∴ no. of mol KMnO4 = 0.001008 mol Molarity =

=

No . of mol volume of solution(L) 0.001008mol 0.027 L

= 0.0373 mol/L KMnO4 = 0.0373 M

3. Explain why no indicator is needed in this experiment This is because without a pH change, the endpoint of the reaction can be determined using only by the colour change. The end point means that the standard solution has been completely oxidized. In this case, the solution will turn permanently from white to pale pink. Since there is no more standard solution to oxidize, the pale pink color is caused by the presence of very little additional KMnO4. Consequently, the presence of KMnO4 indicates that the standard solution has been completely oxidized. As a result, KMnO4 serves as a self-indicator. 4. Explain why heating is necessary in the standardization of potassium permanganate If the temperature is too low, the interaction between the oxalate and the potassium permanganate will move too slow. Above 60 degrees Celsius, oxalate acid begins to decompose, so it’s important to stay in this range.

REFERENCE Editor. (2018, March 15). Preparation and standardization of potassium permanganate. Labmonk. https://labmonk.com/preparation-and-standardization-of-potassium-permanganate Mott, V. (n.d.). Redox Titrations | Introduction to Chemistry. Introduction To Chemistry. https://courses.lumenlearning.com/introchem/chapter/redox-titrations/

EXPERIMENT 8: GAS LAW PURPOSE 1.

To verify the Graham’s law by measuring the distances travelled during the same period of time by two different gases of known molecular mass.

2.

To determine the molar mass of a volatile liquid by measuring mass of vapour of the liquid is needed to fill a flask of known volume at a particular temperature and pressure.

INTRODUCTION A. Graham’s Law The rate of diffusion of different gaseous substance is inversely proportional to the square root of its molar mass. This relationship is referred to as Graham’s law, after the Scottish chemist Thomas Graham. The ratio of the diffusion rates of two gases can be written as:

Where, 𝑅𝑅𝑅𝑅 1 = the rate of diffusion for the first gas 𝑅𝑅𝑅𝑅2 = the rate of diffusion for the second gas. M1 = the molar mass of gas 1 M2 = the molar mass of gas 2.

In this experiment, the relative rate of diffusion of hydrogen chloride will be determined by measuring the distance travelled by two gases in same time period. For a given time period, a lighter gas should be able to diffuse further than heavier gas. A white ring of ammonium chloride, NH4Cl (s) will form in the tube when two gases meet and react to diffuse each other.

HCl (g) + NH3 (g) ⟶ NH4Cl (s) The white ring position along the tube can determine the gases that diffuse the further.

B. Molar Mass of Volatile Liquid The ideal gas law; PV= nRT directly related to the quantity (mol) of gas in the sample. Only one possible quantity of ideal gas can be present in the container for a given fixed volume at particular temperature and pressure: n = PV/RT Where, P=pressure, V=volume, n= number of moles, R= ideal gas constant (8.21×10−2L atm mol-1 K-1), T = absolute temperature (in Kelvin)

The molar mass, M of the gas sample can be calculated since it is represents the number of grams, g, of volatile substance per mol.

Number of moles = n = In this experiment, a small amount of easily volatized liquid will be placed in a flask of known volume. Number of moles of gas can be calculated by refer to the temperature of boiling water bath and atmospheric pressure from the volume of flask used when heated in a boiling water bath at atmospheric pressure. Molar mass can be calculated by refer to the mass of liquid required to fill the flask with vapour when it is in water bath.

APPARATUS:

Stopper,

Medicine

dropper,

Tweezers/

crucible

tongs,

Stock

watch, Marker pen, long ruler (1M), Retort stand, 400/500 mL beaker, 100 /125 mL conical flask, 500mL graduated cylinder, Bunsen burner, Pin (to make a hole), Thermometer, Cotton, Glass tube (50 cm length, 10mm internal diameter) and aluminium foil

CHEMICAL: NH 3 (conc.), HCl (conc.), Acetone (to dry the glass tube), Unknown volatile liquid-acetone (or other compounds recommended by lecturer), Boiling chips

PROCEDURE A. Graham’s Law 1. 50 cm length of glass tubes (10 mm inner diameter), which were completely dried, were procured. Two cotton swabs were procured and prepared. 2. 25 drops of concentrated HCl were placed on one cotton swab using dropper pipettes and 25 drops of concentrated NH3 solution were placed on another cotton swab. 3. The moistened ends of the cotton swabs were immediately and simultaneously inserted into the opposite ends of the tube using tweezers. The time taken for the appearance of the faint white cloud (ring) of ammonium chloride was noted. 4. After a few minutes, a white ring formed where the gases HCl and NH3 meet inside the tube, forming the white compound NH4Cl (ammonium chloride). The point on the tube where the white ring forms was marked. The distance travelled by each gas was measured. The data were recorded in data sheet 8A. 5. Using tweezers, cotton swabs were removed and dipped into a beaker of tap water. Two different beakers half filled with tap water were used to dispose the cotton swabs. the cotton swabs were disposed properly. 6. The tubing was rinsed with water. It may be dried by rinsing it with acetone.

B. Molar Mass of Volatile Liquid 1. A boiling water bath was set up with a 400-mL beaker containing 250-mL water (or enough water to immerse the flask). 2. 125mL conical flask was taken and placed in boiling chips. The opening of the flask was tightly covered with a small square of aluminium foil. A straight pin was used to poke a small hole in the foil lid. 3. The empty, sealed flask was weighed along with the boiling chips. 4. The foil cap was removed. 2 mL of the liquid to be tested was added to the flask and the foil was replaced.

5. The flask was clamped with a single burette clamp. The flask was transferred to the boiling water bath, immersed and heated.

6. The liquid flowing back in the flask was noted. The flask was held tilted slightly; it will be easier to notice when the liquid disappears. 7. The liquid was heated until it was no longer visible and no vapour was emitted from the pinhole. Continue heating for 30 seconds beyond this time. The flask was removed, placed on a hot surface or tile, the clamp removed, and waited for the flask to cool to room temperature. 8. The flask was dried. The flask, cap and condensed vapour were weighed. 9. The contents of the flask were disposed of in a waste bottle or as directed. The flask was filled with tap water (to overflowing). The water was poured into a 500-mL graduated cylinder, and the volume was measured and recorded. 10. The barometric pressure (assumed to be 760 torr or equal to 1 atm) was measured and recorded. The data were recorded in Data Sheet 8B. DATA A. Graham’s Law Observation of NH4Cl appearance: White cloudy ring formed in glass tube

Trial 1

Trial 2

Start time (s)

0

0

Finish time(first visible) (s)

721 s

724 s

Distance travelled by NH3 (cm)

62.0

62.5

Distance travelled by HCl (cm)

38.0

38.2

Ammonia diffusion rate (cm/sec)

0.0860

0.0863

HCl diffusion rate (cm/sec)

0.0527

0.0528

B. Unknown liquid number or letter: C3H60 Mass of flask , foil, boiling chips and condensed vapour

73.5870 g

Mass of flask, boiling chips and foil

73. 2610 g

Mass of condensed vapour (mass of vapour)

0.326 g

Temperature of vapour

65 °C

Barometric pressure (pressure of vapour)

760 mmHg (torr)

Volume of flask ( volume of vapour)

152 mL

Mass of condensed vapour Mass of flask, foil, boiling chips and condensed vapour = 73.5870 g Mass of flask, boiling chips and foil = 73. 2610 g Mass of condensed vapour: 73.5870 g - 73. 2610 g = 0.326 g So, the mass of condensed vapour is 0.326 g

QUESTIONS 1. Experiment A a) Calculate the rate of diffusion for NH3 and HCl 

From first trial, Rate of diffusion of NH3: Distance travelled by NH3 = 62.0 cm Time required

= 721 sec

Rate of diffusion

= 62.0 cm / 721 sec = 0.0860 cm/sec



From the second trial Rate of diffusion of NH3: Distance travelled by NH3 = 62.5 cm Time required

= 724 sec

Rate of diffusion

= 62.5 cm / 724 sec

= 0.0863 cm/sec



From first trial, Rate of diffusion of HCl: Distance travelled by HCl = 38.0 cm Time required

= 721 sec

Rate of diffusion

= 38.0 cm / 721 sec = 0.0527 cm/sec



From the second trial, Rate of diffusion of HCl:

Distance travelled by HCl = 38.2 cm Time required

= 724 sec

Rate of diffusion

= 38.2 cm / 724 sec = 0.0528 cm/sec

b)

Calculate the ratio of the rate of diffusion of NH3 to the rate of diffusion of HCl. Trial 1 rate of diffusion NH 3 rate of diffusion HCl

0.0860 cm / sec 0.0527 cm / sec

=

= 1.632

Trial 2 rate of diffusion NH 3 rate of diffusion HCl

0.0863 cm/ sec

= 0.0528 cm / sec = 1.634

c) Using the molecular masses of NH3 and HCl, calculate the theoretical ratio of the rates of diffusion of these gases. rate1 rate2

=

√

M HCl M NH 3

=

√

1+ 35.46 14+ 1(3)

= 1.46

d) Calculate the % error in your experimentally determined value for the ratio of the rates of diffusion of NH 3 and HCl. Use the theoretical ratio calculated in (c) as the accepted value for the ratio. theoretical ratio−experimental ratio % error = absolute value of [ ] × 100% theoreticalratio Trial 1 1.632 −1.46 1.46

Trial 2 × 100%

= 11.8 %

1.634 −1.46 1.46

× 100%

= 11.9 %

2. Experiment B Calculated the molecular weight of the unknown liquid. Show your calculations, and include units of the different quantities in your calculations. Answer: Use pV=nRT (P = 1 atm, V= 0.152, R= 0.08206 LatmK-1mol-1, T= 65 + 273.15 = 338.15 K) n = pV/ RT (1 atm× 0.152 L) atm L number of moles = ×338.15 K ) (0.08206 mol K = 0.00548 The molar mass = mass/n Mass = 0.326 g Molar mass = (0.3051 gram ÷ 0.00548 mole) = 55.68 g/mol So, the molecular weight of unknown liquid is 55.68 g/mol

DISCUSSION From experiment 1, the reaction which taking place is HCl + NH 3 → NH4Cl. The amount of time needed to form the ring is determined by the dimensions of the tube, the amount of solution applied to the wads, and the temperature of the room. Since hydrogen chloride diffuses slower than ammonia, the ring usually forms near the hydrochloric acid end of the tube. This is due to the fact that hydrogen chloride has nearly twice the molecular weight of ammonia and that the rate of diffusion is inversely proportional to the square root of the gas's molecular mass. The glass tube's purpose is to remove air currents and see if the gas molecules can move on their own. In the first trial, the distance travelled by the NH 3 is 62 cm, which took 721 seconds. Thus, the diffusion rate of the NH 3 is 0.0860 cm/sec. In the second trial, the distance travelled by NH 3 is 62.5 cm, which took 724 seconds to complete. Thus, the diffusion rate of NH3 is 0.0863 cm/sec. For HCl, the distance travelled in the first experiment is 38.0 cm, which took 721 seconds. Thus, the diffusion rate is 0.0527 cm/sec. In the second experiment, the distance travelled by HCl is 38.2 cm, which took 724 seconds. Thus, the diffusion rate of HCl is 0.0528 cm/sec. The percentage error for trial 1 is 11.8% and for trial 2 is 11.9%. From Experiment 2, these relationships can be combined into a single equation that describes the relationship between temperature, pressure, volume, and quantity (number of moles) of all gases under ordinary conditions. The equation is the ideal gas law: pV = nRT. Based on the equation of the ideal gas law, we could find the number of moles, which is 0.00548. The unknown liquid is acetone (C3H6O). The weight of acetone is 0.326 g, so we calculated the molecular weight of C3H6O, which is 55.58 g/mol. There are several errors that could occur during the experiment, causing the results to differ slightly. To avoid this, we must ensure that our eyes are perpendicular to the scale in order to accurately calculate the amount of solution. After that, we must wash all of the equipment with distilled water before using it. Some safety precautions should also be taken. Since concentrated HCl and concentrated NH 3 solutions are both harsh on the skin, use rubber gloves and a face mask when handling them.

CONCLUSION In conclusion, Graham's law was verified in this experiment by measuring the distances travelled in the same time by two different gases of known molar mass. The molar mass of a volatile liquid can be determined by measuring the mass of vapour of the liquid required to fill a known volume at a given temperature and pressure. The molar mass of a volatile liquid of acetone (C3H6O) obtained from this experiment was 55.68 g/mol.

REFERENCE O. (n.d.-b). The Ideal Gas Law | Physics. Temperature, Kinetic Theory, and the Gas Laws. https://courses.lumenlearning.com/physics/chapter/13-3-the-ideal-gas-law/ What is the ideal gas law? (article). (n.d.). Khan Academy. https://www.khanacademy.org/science/physics/thermodynamics/temp-kinetic-theory-idealgas-law/a/what-is-the-ideal-gas-law...