Surface tension of liquids

Objective:

1. To understand the basic concept of surface tension in liquids and how it affects the properties of liquids

2. To study the effects of detergent on the surface tension of liquids.

Introduction:

Water molecule consists on a big oxygen atom and two smaller hydrogen atoms. The hydrogen atoms hold the slightly negative charges which making the entire water molecule becomes polar. Eventually, the hydrogen bond is exists in between two neighboring water molecules. Each water molecule experiences a pull from other water molecules from every direction, but water molecules at the surface do not have molecules above the surface of the water to pull at them. These water molecules experience an inward pulling from the water molecules below than them. The difference in force draws the water molecules at the surface creating the surface tension.

Surface tension is the force acting at the angles of 90º to any line on the liquid surface. Surface tension is a property of a liquid surface that caused by the cohesive force between the same liquid molecules. In a liquid, each molecule within the body of the liquid tends to attracted equally at all directions by the cohesive force, so that it experiences no net force. However, those molecules on the surface of liquid have no neighboring molecules above which exhibits stronger attractive force upon on their nearest neighboring molecules on the surface. This results an inward force which pulling the molecules towards the interior of the liquid.

Due to the effect of surface tension, the surface of liquid will form a thin “film” which makes it more difficult to move an object floating on the surface than to move it when it is completely submersed in the liquid. Surface tension is typically measured in the unit of dyne/cm which is the force in the unit of dyne that required to breaks the surface film of water with the length of 1cm. The surface tension in room temperature is 72dyness/cm. It means 72 dynes of forces would be taken to break down a surface film of water 1cm long. The increase in temperature will affects the surface tension of the water dramatically. When the kinetic energy of water molecules increases, it will tend to break down the surface tension of water. Besides, the addition of some solute can influence the surface tension of the liquid but it is depends on the nature of the solutes added. However, some of the solute concentration may not have effect to the surface tension of liquid once the minimum is reached which is surfactant.

Surfactant (surface-active-agents) is a compound in which can lower the surface tension of a liquid or interface tension between two different liquids (or a liquid and a solid). Examples of surfactants are detergent, wetting agent, emulsifiers foaming agents and dispersants. Surfactants usually are amphiphilic, which means they contain both the hydrophobic groups and hydrophilic groups. Due to these properties, the surfactant molecules will migrate to the water surface, where the insoluble component will project out of the water surface and the hydrophilic water soluble component will remain in the water phase. Detergents are the chemicals which consist of hydrophobic (non-polar) hydrocarbon "tails" and a hydrophilic (polar) "head" group. Surfactants can interact with water in a variety of ways which able to disrupt the hydrogen bonding network between the water molecules. Since this will reduce the cohesive force between the molecules, so the surface tension of the water will be altered.

Surface tension is a term that used to describe the cause of the phenomena in which the difference between meniscus of water and mercury. The influence of intermolecular force and surface tension on the interfacial properties of liquid causes the meniscus formation. Stronger intermolecular force between water molecules and glass surface (adhesive force) compared to those between water molecules (cohesive force) cause the water to be drawn up onto the glass wall forming a concave meniscus. For the formation of mercury meniscus, a convex meniscus is formed due to the stronger intermolecular force between mercury atoms (cohesive force) compared to those between mercury atoms and glass surface which prevent mercury from wetting the glass surface.

Apparatus: Petri dish, Toothpick

Materials: Sulfur powder, liquid dish detergent, low fat milk, full cream milk, food colouring

Procedure:

Part A

1. Clean water was poured into a petri dish with depth of 1cm.

2. Powdered sulfur was dusted on the surface of clean water and observation was recorded.

3. The surface of water was touched by a toothpick and observation was recorded.

4. The tip of toothpick was dipped in detergent. The surface of water was touched and toothpick was hold in a place for a while. Observation was recorded.

Part B

1. Low fat milk was poured into a petri dish.

2. Four different coloured food colouring were added. The drops were kept close together in the centre of the petri dish and observation was recorded.

3. The surface of milk was touched with a toothpick and observation was recorded.

4. The tip of a toothpick was dipped in detergent. The surface of milk was touched and toothpick was hold in a place for a while. Observation was recorded.

5. The process was repeated by replacing low fat milk with full cream milk and water.

Results: Observation

Part A: Water with sulphur powder on surface

Before Touched by toothpick                 Touched by toothpick

Touched by toothpick with detergent             After 3 minutes

             After 5 minutes

Observation: The sulphur powder on the water surface where touched by toothpick with detergent was sank to the bottom.

Part B: Low fat milk with colouring

Before                                                    Touched by toothpick            

        

Touched by toothpick with detergent                   After 3 minutes

                After 5 minutes

Observation: The food colouring are being pushed away rapidly from the region where touched by the toothpick with detergent.

Part B: Full cream milk with colouring

Touched by toothpick                     Touched by toothpick with detergent

                After 1 minute                                After 3 minutes

               After 5 minutes

Observation: The food colouring are being pushed away slowly from the region where touched by the toothpick with detergent.

Part B: Water with food colouring

                Before                                    Touched by toothpick

Touched by toothpick with detergent               After 3 minutes

               After 5 minutes

Observation: The food colouring are not changed significantly from the region where touched by the toothpick with detergent.

Discussion:

In Part A of the experiment, the sulphur powder was sprinkled evenly on the water surface. The sulphur powder was floated on the surface because the size of sulphur is light and small which the surface tension of water can withstand and maintain them on the surface. When the surface was touched by the toothpick, the sulphur powder starts to float further from the centre where the place touched by the particular toothpick. However, the sulphur powder was started to sink into the bottom of water very quickly once it was touched by the toothpick with detergent. This is because the surface tension of water was disrupted by the detergent since detergent is a surfactant which acts to lower the surface tension of liquid. As a result, the weak surface tension of water no longer able to withstand the sulphur powder on its surface, so the sulphur sank into the bottom of water.

In part B, low fat milk is used as the liquid to show the property of surface tension when exposed to detergent. Low fat milk has a small amount of fat and the water is predominately in it. Due to the densities of colourigs are lower than milk, so they are floated on the surface when the food colourings were added on the surface of milk. When the toothpick was used to touch on the surface of milk, the food colourings were disrupted insignificantly due to the wave generated by the touch. However, the food colourings were dispersed rapidly and became faint in colours when the toothpick with detergent touched on the surface of milk. This is because the surface tension of the milk in the centre was lowered by the surfactant. The stronger surface tensions of the surrounding milk molecules pull the surface of milk away from the weak region where towards the edge of the plate. The detergent decreased the surface tension of milk by dissolving the fat molecules which caused the turbulence. This moment caused the food colourings to swirl, but the swirling of the colors continues for some times before stopping.

Full cream milk was used in the part C of the experiment to identify the difference in surface tension for two different milks. Full cream milk has a larger amount of fat compared to low fat milk. The toothpick without detergent did not affect the distribution of food colourings on the surface of milk. The fat globules in milk were steady and undisturbed. When the surface was touched by the toothpick with detergent, the food colourings started to disperse with a slower speed compared to the dispersion in low fat milk. This may be due to the full cream milk contains less water which limiting the movement of the milk so the colourings spread slower in full cream milk compared to low fat milk. The milk molecules with lower surface tension in the spot were pulled by the milk molecules with higher surface tension in the surrounding. Hence this caused the food colourings moved with the milk molecules streaming away from the detergent dropped. Comparing to the low fat milk, the colourings were scattered more in full cream milk. This may be due to the detergent weakens the milk's bonds more in the full cream milk because it had more fat. Hence, food coloring scattered more in the full cream milk.

In part D of the experiment, food colourings were added on the surface of water. The colourings were sunk to the bottom of water because they have higher density compared to water while there is some amount of colourings were floated on water surface. When the surface was touched by the toothpick, the distribution of the food colourings were not affected much. When the toothpick with detergent was used to touch on the water surface, the floated food colourings were pushed away from the centre where the toothpick had touched. The detergent reduced the surface tension of the water at middle and hence it caused food colourings to spread. This phenomenon was due to the water molecules in the edge with higher surface tension pulled the water molecules with lower surface tension from the centre.

Precaution steps:

1. Do not shake and stir the surface of liquid because it will influence its surface tension.

2. Make sure that all the fans have been switched off before carry out the experiment.

3. Make sure the milk is not expired because it will affect the property of the milk.

4. Do not exhale deeply which may affect the surface tension of liquid when carrying out the experiment.

Relationship between osmosis process and hypertonic, hypotonic and isotonic effects

Objective:

1. To understand the phenomenon in which molecules flow from a location of higher chemical potential to a region of lower potential in which both regions are separated by a semi permeable membrane.

2. To investigate the hypertonic, hypotonic and isotonic effect.

Introduction:

Osmosis is the net diffusion of water molecules from a region of lower solute concentration to a region of higher concentration by passing through a semi permeable membrane. A semi permeable membrane is otherwise known as selectively membrane which only allows some of the molecules to pass through it but not others. However, the membrane is permeable to the solvent. The permeability is depends on the size and property of molecule. In osmosis, the net movement of water molecules will only move in one direction. The water molecules move from the side where water concentration is higher to the other side of the membrane which down the concentration gradient.

Osmosis is a passive process which does not involve any energy input into the system. The osmosis process is spontaneously and naturally. The difference in the concentration of solutions creates an osmotic pressure. Osmotic pressure is the pressure that exists across a semi permeable membrane between two solutions of different concentration. Osmotic pressure of a solution is a colligative property and at given temperature its magnitude depends only on the concentration of the solute but not its identity. Osmotic pressure is calculated by using the van't Hoff's formula below:

Π = cRT

where π = osmotic pressure (atm)

c = concentration of solution (mol/L)

R = gas constant

T = Temperature(K)

The concentration of solution can be categorized into three basic categorizes which have different concentration with respect to a particular concentration of solution. The types of solution are divided into hypertonic, hypotonic, and isotonic. Hypertonic solution is the solution containing a higher solute concentration when compared to a particular solution. The higher the solute concentration, the lower the water potential and osmotic pressure. This is because some water molecules will be attracted to the solute molecule and will no longer be free. In osmosis, the water molecules will flow into the hypertonic solution due to lower water potential. For hypotonic solution, is defined as a solution with lower solute concentration with respect to a solution. Its water potential and osmotic pressure are higher. In hypotonic solution, there are more free water molecules which always diffuse into a region of lower water potential. The solution has same concentration respect to another solution is known as isotonic solution. It has the same water potential and osmotic pressure which the net movements of water molecules are the same for both solutions.

Net movement of the water molecules are from hypotonic (low-concentrated solute) to hypertonic (high-concentrated solute) due to difference of osmotic pressure in two solutions. Due to the thermal agitation, the water molecules will diffuse spontaneously (Δ G < 0) from the region of lower solute concentration towards higher solute concentration. The addition of solute will increase the entropy of water molecules since the solute is being dispersed overall in the solution. The direct osmosis (follow concentration gradient) is entropy generating process (ΔS > 0) which having tendency to equalize the chemical potentials and concentrations of two different solutions which are separated by the semi permeable membrane. The water is acting to dilute the solution with higher solute concentration until the concentration is the same for both sides. However, the water will continue to flow with the similar rate in the both directions.

Apparatus: 50ml beaker, analytical balance

Materials: dialysis tubing, tap water, 30% sucrose solution, 60% sucrose solution, string

Procedure:

1. Three pieces of equal length of dialysis tubing and several lengths of string were obtained.

2. One end of the tubing was folded over and tied closed with the string.

3. To each tube, 5 ml of 30% sucrose solution was added. Then, the bag was squeezed gently to remove excess air and was tied off with string.

4. Some slack was left in the bag as room for expansion, but the air trapped inside was removed. The bags were briefly rinsed in running tap water and then were dried on paper towels.

5. Each bag was weighed and the results were recorded.

6. After weighing, three bags were placed into tap water, 30% sucrose solution and 60% sucrose solution each.

7. The bags were remained undisturbed for 30 minutes.

8. The bags were removed, rinsed, and reweighed. All results were recorded into table.

Results and calculations:

	Tap water	30% Sucrose	60% sucrose
Weight at 0 min	5.7739g	5.9171g	5.9795g
30 min	7.1150g	5.8309g	5.3015g
Change	1.3411g	-0.0862g	-0.6780g

Percentage of weight of dialysis tubing (in tap water) increase

= 1.3411g/5.7739g x 100%

= 23.22%

Percentage of weight of dialysis tubing (in 30% sucrose solution) decrease

= 0.0862g/ 5.8309g x 100%

= 1.48%

Percentage of weight of dialysis tubing (in 60% sucrose solution) decrease

= 0.6780g/5.9795g x 100%

= 11.34%

Discussion:

In this experiment, three dialysis tubing were filled with 5 ml of 30% sucrose solution. These dialysis tubing were separately put into the three beaker which containing tap water, 30% sucrose solution and 60% sucrose solution respectively. The dialysis tubing is semi permeable which allows some molecules that are small enough to pass through it. In this case, the small molecules are referred to water molecules. The contents in the dialysis tubing are not flow out through the small pores before place them into beakers with 30% sucrose solution. This may be due to the cohesion force between the water molecules in which try to pull the outermost water molecules in dialysis tubing inwardly.

When one of the dialysis tubing containing 30% sucrose is placed into the beaker with tap water, this tap water is said to be hypotonic with respect to 30% sucrose solution. Due to the high water potential in tap water, the water molecules will tend to move from the tap water into the dialysis tubing. The osmotic pressure of tap water is higher than the osmotic pressure of sucrose solution. The water molecules are forced to move out from the tap water to 30% sucrose solution and move to 30% sucrose solution by diffusing the membrane of dialysis tubing. The reason is because the higher osmotic pressure in the tap water forced the water molecules to diffuse into sucrose solution. Eventually, the rate of water diffuse into the tubing is higher than the rate of water diffuse out from the tubing until equilibrium is reached. The increase in the water caused the weight of dialysis increases as well.

The 60% sucrose solution is considered as hypertonic solution when compared to the 30% sucrose solution in it. The 60% sucrose solution has higher solute concentration and lower water potential due to lack of free water molecules. This is because more water molecules are attracted to the sucrose molecules in the solution and hence the particular water molecules are not free to move. The low osmotic pressure and low water potential in 60% sucrose solution causes the water molecules in 30% sucrose solution to move into it by osmosis. Spontaneously, a net movement of water tends to minimize the difference in concentration between two solutions. The diffusing rate of water molecules move out from the dialysis tubing is higher than the water molecules move into the particular tubing. As a result, the volume of tubing has been decreased as well as its weight. The net movement will become zero after the same concentration of solution is reached for both sides of dialysis membrane.

Isotonic solution is a solution has the same concentration with respect to another solution. In this experiment, one of the 30% sucrose solutions was placed into a beaker containing 30% sucrose solution. So, the solution in the beaker is called isotonic since it has the same concentration with the solution in dialysis tubing. The water potentials of both solutions are the same as the two solutions have the same amount of water molecules free to move in the solutions. Thus, the rates of water molecules diffuse in and out of dialysis tubing are the same since there is no difference in osmotic pressure for both solutions. The net movement of water molecules is zero as the water potentials between two solutions has reached equilibrium. Theoretically, the weight and volume of dialysis has no change. However, the weight of dialysis tubing was reduced in practical. This might be due to the content in beaker was contaminated by impurity which caused the water molecules to move out. The second reason might be due to there is some water left inside the tubing which diluted its content, so the concentration has been reduced and the solution is beaker is considered as hypertonic with respect to the sucrose solution in dialysis tubing. The decrease in concentration might be small but it is enough to dilute the contents which caused the water potential increased and hence diffuse out to the solution (hypertonic) in the beaker.

Precaution steps:

1. The dialysis tubing should not be too tight and turgid in order to leave some slack in the tubing as space for expansion when water flow into it.

2. The exterior of dialysis tubing should be roughly dried before weighing to prevent the weight of water is being counted.

3. The interior of dialysis tubing should be dried in order to prevent dilution of sucrose solution.

Determination of the activation energy for the reaction of bromide and bromate ions in acid solution.

Objectives

1. To understand the chemistry of activation energy.

2. To determine the activation energy for the reaction of bromide and bromate ions in acid solution.

Introduction

Activation energy is defined as the minimum energy barrier that must be overcome for a chemical reaction to take place. It is usually denoted as E_a, and given in unit of kiloJoule, kJ/mol. For a chemical reaction, an appreciable number of molecules with the energy equal to or greater than activation energy should be exist in the system. In order for a reaction to occur, the reactant particles must collide according to the collision theory. However, not all collision are able cause the reaction to happen, only a certain collisions in the system can cause chemical reaction, which is called effective collision. The effective collisions of molecules must collide with the correct orientation and sufficient energy to overcome the activation energy barrier. The energy is needed to break the existing bonds and form the new bonds of the molecules which resulting in the formation of products.

The activation energy of a reaction can be measured by using Arrhenius equation as shown in the equation below:

k = Ae^-Ea/RT

where k = rate constant

T = absolute temperature

E_{a =} energy of activation

R = gas constant

The pre-exponential term, A is the property of particular reaction related to the collision frequency of the reactive species and thus is temperature dependent. However, according to the equation, the dependence of k on temperature is dominated by the strong exponential term, so the dependence of A on temperature is usually ignored as a first approximation. By taking logarithms of both sides,

Log₁₀ k = -E_a / 2.303RT + log ₁₀ A

= -E_a /2.303RT + constant

So, when a reaction has a rate constant that obeys Arrhenius equation, a plot of log₁₀ k versus 1/T gives a straight line. The gradient of the straight line is –E_a / 2.303R while the interception of the straight line on the y-axis of the graph can be used to determine the values of log₁₀ A.

Now, the rate of reaction is higher when the time taken for a fixed amount of reaction to complete is shorter. This makes the time taken, t to complete a fixed amount of reaction is inversely proportional to the rate constant, k.

T α 1/k

or t = constant/k

By taking logarithms of both sides,

Log₁₀ t = - log₁₀ k + constant

= E_a / 2.303RT + constant

A plot of log₁₀ t versus 1/T gives a straight line as well and the slope of the graph is E_a / 2.303R. Thus, if t is measured at several temperatures then the energy of activation can be found.

In this experiment, the above method is applied to the reaction of bromide and bromate ions in an acid solution which occurs slowly at room temperature.

KBrO₃ + 5 KBr + 3 H₂SO₄ --> 3 K₂SO₄ + 3 Br₂ + 3 H₂O

or BrO₃^- + 5 Br^- + 6H⁺ --> 3 Br₂ + 3 H₂O

The time required for a fixed amount of the reaction to be completed, t is found by adding a fixed amount of phenol and some methyl red indicator to the reaction mixture. The bromine produced in the first reaction reacts very rapidly with the phenol to form tribromophenol.

C₆H₅OH + 3 Br₂ -->  C₆H₂Br₃OH + 3 HBr

When all the phenol has reacted, the bromine continuously produced in the first reaction will then react with the methyl red indicator and bleaches its colour.

Methyl red + Br₂ --> colourless compound

Apparatus : 1 dm^-3 beaker, 3 100cm³ beakers, 2 boiling tubes, 1 5cm³ pipette, 1 10-cm³ pipette, thermometer (0 - 110°C), stopwatch

Material : 0.01 mole dm^-3 aqueous phenol solution, bromide/bromate solution (0.0833 mole dm^-3 potassium bromide and 0.0167 mole dm^-3 potassium bromate, equivalent to 0.05 mole dm^-3 bromine), 0.3 mole dm^-3 sulphuric acid, methyl red indicator.

Procedures

1. 10 cm³ of phenol solution and 10 cm³ of bromide/bromate solution were pipette into one boiling tube.

2. Four drops of methyl red indicator were added to the mixture.

3. 5 cm³ of sulphuric acid was pipette into another boiling tube.

4. The two boiling tubes were immersed in the water bath of (75 ± 1) °C.

5. The contents of the two tubes were mixed by pouring rapidly from one tube to the other twice and the stopwatch was started at the same time.

6. The boiling tube containing the reaction mixture was kept immersed in the water.

7. The time required for the red colour of the methyl red indicator to disappear was determined.

8. The whole experiment was repeated at 65, 55, 45, 35, 25 and 15 °C.

9. Ice was used to achieve the lowest temperature.

Results & Calculations

(Assume R = 8.314 J K^-1 mol^-1)

From Graph 1,

Slope of the graph = E_a / 2.303R

E_a / 2.303R = (2.94 – 2.20) / (3.35×10^-3 – 3.10×10^-3)

E_a /2.303R = 0.74 / (2.50×10^-3)

E_a = 2960 x 2.303R

E_a = 2960 x 2.303 x 8.314

E_a = 56676 J/ mol

Energy of activation, E_a = 56676 J /mol

Discussion

The reaction between bromide and bromate ions in acid solution is a slow chemical reaction at room temperature. This may be due to the high activation energy of the reaction, which required 56.676 kJ of energy in order for a reaction to take place. According to collision theory, high activation energy will cause the product more difficult to form since it is not sufficient energy for that molecule collide without enough energy. Any reaction cannot occur if the colliding molecules do not have the energy equal or higher than its activation energy. There are also other factors that can reduce the effectiveness of collisions of molecules such as present or absent of catalyst.

According to Table 1, we can observed that the higher the temperature of the reactant species, the shorter the time taken for the disappearance of red colour of methyl red indicator. This is shows that shorter time taken in the reaction once the reaction is faster and higher rate of reaction. Although the activation energy for the reaction to occur remains unchanged at all the temperature, but the rate of reaction increased as the temperature increased. This means that the rate of reaction is depends on the temperature of the reactant. This has also been proven in Arrhenius equation where rate constant, log₁₀ k is proportionally to 1/T with the slope of the graph, -E_a/R and a constant of log₁₀ A.

Log₁₀ k = -E_a/2.303RT + log₁₀ A

The higher temperature caused the value of the -E_a/2.303RT closer to the value of 0. The constant log₁₀ A will then minus off the value of -E_a/2.303RT and results in a larger value. The larger value will caused the value of the rate constant, k to become larger as well. Larger value of k will then results in faster reaction. Hence, this is proven that the higher the temperature of reactants, the reaction will proceed faster with the higher rate of reaction.

Besides, the rate of reaction roughly doubles for every 10 °C increase in temperature. This is because increase in temperature increase the kinetic energy of the molecules, the molecules with higher kinetic energy can move faster. With the higher speed, the molecules will collide more frequently results the larger amount of successful collision. With the higher kinetic energy of the molecules, the molecules can overcome the activation energy barrier during the collision and hence the reaction can be take places. As a result, the rate of the reaction could be increased if more heat energy is provided to the reactants.

Fixed amount of phenol and methyl red indicator were added to the mixture contents for the different temperature. This is because phenol can provides an intermediate state before the bromine molecules produced in the reaction between bromate and bromide ions in acid solution which is able to bleach the methyl red immediately. In other words, phenol is used to observe the time taken for the bromine molecules to react completely with phenol before bleach the methyl red indicator. The purpose of adding methyl red into the solution is to provide a colour which can easily to be observed. When the sulphuric acid was poured to the bromate/bromide ions solution, the methyl red indicator turns to pink colour. The bleaching effect from the bromine molecules caused the methyl red indicator to turn colourless after all the phenol is used up. This shows a colour changes which the time taken should be stopped. In order to compare the time taken for the bleaching of methyl red colour, the amount of phenol used in the repeated experiment must be equal.

The reason of choosing phenol as the reactant is due to it can form an intermediate state because phenol can react with bromine molecules quickly to produce tribromophenol and hydrogen bromide. This is happens before the bromine molecules react with the methyl red.

The –OH group in the phenol is an activating group in the benzene ring which can donates electron into the benzene ring to stabilize it. So, the product can be easily to form in the reaction. –OH group is ortho-para activating group in the benzene ring. This means that the incoming substituent will go into the ortho position or the para position, but hardly go into the meta position. This is due to the high energy is required for the product with a substituent at meta position to form in the reaction. The bromine molecules undergo substitution reaction in this reaction by substituting three hydrogen atoms from the benzene ring with three bromine atoms to maintain the aromaticity of the ring in phenol.

The reaction between bromate and bromide ions in acid solution is a redox reaction.

BrO₃^- + 5 Br^- + 6H⁺ --> Br₂ + 3H₂O

The potassium and sulphate ions act as spectator ions in this experiment and they are not participate in any of the redox reaction or changing of their state. The bromide ions undergo oxidation by donating one of its electrons to the bromine atom in the bromate ions. The bromine atom in the bromate ions then undergoes reduction by receiving electron from the bromide ions. The hydrogen ions and oxygen atom in bromate ions does not involve in increase or decrease in oxidation number but they were involved in changing the state from the ions in aqueous solution to the water molecule in liquid state.