The materials to prepare students for

The materials to prepare students for practical lessons of inorganic chemistry

LESSON № 4.

Theme: Solutions. The methods for expression the composition of solutions. The preparation of solutions by the given concentration

Plan

Solutions. Essence of basic mains: solutions, solvent, permeate. Solubility. Solutions of gas, liquid and solid matters. Water as one of the most important solvents in a biosphere and chemical technology. Role of waters solutions for living organisms. Non-aqueous solvents and solutions.

Process of dissolution as the physical and chemical phenomenon Solubility of solid matters in liquids, factors which influence on solubility. Solubility of gases in liquids, its dependence on: pressure (Henry law), a temperature, concentration of electrolytes in water.

Determinate methods of solution composition. Mass fraction, volume fraction. Molar concentration. Molar concentration of equivalent. Molality, molarity of solution. Mole fraction of solutions.

 

Solutions – a homogeneous mixture of two or more substances in which the molecules or atoms of the substances are completely dispersed at the molecular or ionic level. Consisting of 2 parts – the solvent and solute. Solutions are formed when solvent molecules break apart the solute.

Solvent – the substance that does the dissolving. Usually, it is the larger part and a liquid.

Solute – the substance that is being dissolved. Usually, it is the smaller part and can be a solid, liquid or gas.

Characteristics of a Solution:

1. The solute does not settle out even if left for long periods of time.

2. It is transparent; no particles can be seen in the solvent; it won’t disperse light.

3. It can be different colors.

4. The solute can’t be separated out by filtering; both solvent and solute pass through filter paper.

SOLUBILITY

Solubility – the ability of substance to be dissolved into another substance.

Factors that affect solubility:

Nature of the substances – substances that are similar will form solutions, polar vs. non-polar, “like dissolves like”

Energetics of solution formation

Why solutions form — or don’t

You may recall that in the earlier unit on phase equilibria, we pointed out that aggregations of molecules that are more disordered tend to be the ones that are favored at higher temperature, whereas those that possess the lowest potential energy are favored at lower temperatures. This is a general principle that applies throughout the world of matter; the stable form at any given temperature will always be that which leads to the best balance between low potential energy and high molecular disorder.

To see how these considerations are applied to solutions, think about the individual steps that must be carried out when a solute is dissolved in a solvent:

1  If the solute is a solid or liquid, it must first bedispersed — that is, its molecular units must be pulled apart. This requires energy, and so this step always works against solution formation.

2 The solute must then be introduced into the solvent. Whether this is energetically favorable or unfavorable depends on the nature of the solute and solvent. If the solute is A and the solvent is B, then what is important is the strength of the attractive forces between A-A and B-B molecules, compared to those between A-B pairs; if the latter are greater, then the potential energy will be lower when the substances are mixed and solution formation will be favored.

Endothermic

Exothermic if A-B attractions stronger than A-A + B-B

Endothermic if attractions between like molecules are stronger than those between unlike molecules.

If step 2 releases more energy than is consumed in step 1, this will favor solution formation, and we can generally expect the solute to be soluble in the solvent. Even if the dissolution process is slightly endothermic, there is a third important factor, the entropy increase, that will very often favor the dissolved state.

Entropy of dissolution

As anyone who has shuffled a deck of cards knows, disordered arrangements of objects are statistically more favored simply because there are more ways in which they can be realized. And as the number of objects increases, the more does statistics govern their most likely arrangements. The numbers of objects (molecules) we deal with in Chemistry is so huge that their tendency to become as spread out as possible becomes overwhelming. But in doing so, the thermal energy they carry with them is also spread and dispersed, so the availability of this energy, as measured by the temperature, is also of importance. Chemists use the term “entropy” to denote this aspect of molecular randomness.

A proper understanding of these considerations requires some familiarity with thermodynamics, which most students do not encounter until well into their second semester of Chemistry (see box at left). If you are not there yet, don’t despair; you are hereby granted temporary permission to think of molecular “disorder” and entropy simply in terms of “spread-outedness”.

Thus in the very common case in which a small quantity of solid or liquid dissolves in a much larger volume of solvent, the solute becomes more spread out in space, and the number of equivalent ways in which the solute can be distributed within this volume is greatly increased. This is the same as saying that the entropy of the solute increases.

If the energetics of dissolution are favorable, this increase in entropy means that the conditions for solubility will always be met. Even if the energetics are slightly endothermic the entropy effect can still allow the solution to form, although perhaps limiting the maximum concentration that can be achieved.

In such a case, we may describe the solute as being slightly soluble in a certain solvent. What this means is that a greater volume of solvent will be required to completely dissolve a given mass of solute.

 

Polar and non polar molecules

Polar molecules are those in which electric charge is distributed asymmetrically. The most familiar example is ordinary water, in which the highly electronegative oxygen atom pulls part of the electric charge cloud associated with each O–H bond closer to itself. Although the H₂O molecule is electrically neutral overall, this charge imbalance gives rise to a permanent electric dipole moment.

“Associated” liquids

Chemists use this term to refer to liquids in which the effects of hydrogen bonding dominate the local structure. Water is the most important of these, but ammonia NH₃ and hydrogen cyanide HCN are other common examples.

Thus liquid water consists of an extended network of H₂O molecules linked together by dipole-dipole attractions that we call hydrogen bonds. Because these are much weaker than ordinary chemical bonds, they are continually being disrupted by thermal forces. As a result, the extended structure is highly disordered (in contrast to that of solid ice) and continually changing.

When a solute molecule is introduced into an associated liquid, a certain amount of energy must be expended in order to break the local hydrogen-bond structure and make space for the new molecule. If the solute is itself an ion or a polar molecule, new ion-dipole or dipole-dipole attractions come into play. In favorable cases these may release sufficient potential energy to largely compensate for the energy required to incorporate the solute into the structure.

An extreme example of this occurs when ammonia dissolves in water. Each NH₃ molecule can form three hydrogen bonds, so the resulting solution is even more hydrogen-bonded than is pure water — accounting for the considerable amount of heat released in the process and the extraordinarily large solubility of ammonia in water.

Non polar solutes are sparingly soluble in water

When a non polar solute such as oxygen or hexane is introduced into an associated liquid, we might expect that the energy required to break the hydrogen bonds to make space for the new molecule is not compensated by the formation of new attractive interactions, suggesting that the process will be energetically unfavorable.

We can therefore predict that solutes of these kinds will be only sparingly soluble in water, and this is indeed the case.

The hydrophobic effect

It turns out, however, that this is not an entirely correct explanation for the small solubility of non polar solutes in water. It is now known that the H₂O molecules that surround a non polar intruder and find themselves unable to form energy-lowering polar or hydrogen-bonded interactions with it will rearrange themselves into a configuration that maximizes the hydrogen bonding between the water molecules themselves. In doing so, this creates a cage-like shell around the solute molecule. In terms of the energetics of the process, these new H₂O-H₂O interactions largely compensate for the lack of solute-H₂O interactions.

But this shell of highly organized water molecules exacts its own toll on the solubility by reducing the entropy of the system. Dissolution of a solute normally increases the entropy by spreading the solute molecules (and the thermal energy they contain) through the larger volume of the solvent. But in this case, the H₂O molecules within the highly structured shell surrounding the solute molecule are themselves constrained to this location, and their number is sufficiently great to reduce the entropy by far more than the dissolved solute increases it.

The small solubility of a non polar solute in an associated liquid such as water results more from the negative entropy change rather than from energetic considerations. This phenomenon is known as the hydrophobic effect.

In the next section, we will explore the ways in which these energy-and-entropy considerations come together in various kinds of solutions.

Temperature – increased kinetic energy.

To understand why things dissolve at all, we will look at the solution formation process from a thermodynamic point of view. shows a thermodynamic cycle that represents the formation of a solution from the isolated solute and solvent. From Hess’s law we know that we can add the energies of each step in the cycle to determine the energy of the overall process. Therefore, the energy of solution formation, the enthalpy of solution, equals the sum of the three steps–ΔH_soln = ΔH₁ + ΔH₂ + ΔH₃.

ΔH₁ and ΔH₂ are both positive because it requires energy to pull molecules away from each other. That energy cost is due to the intermolecular forces present within any solute or solvent. The forces acting between molecules such as CH₃Cl are largely van der Waals and dipole-dipole interactions. Some molecules that contain O-H, N-H, or F-H bonds can form hydrogen bonds that are relatively strong intermolecular forces. Ions of opposite charge, such as in a crystal of NaCl, are attracted to each other because of electrostatic forces. Each of those forces increase with decreasing distance. Therefore, it should make sense that it costs energy to pull molecules and ions away from each other. When the expanded form of the solvent and the solute are combined to form a solution, energy is released, causing ΔH₃ to be negative. This makes sense because the solute and solvent can interact through the various types of intermolecular forces.

What determines the enthalpy of solution is, therefore, the difference between the energy required to separate the solvent and solute and the energy released when the separated solvent and solute form a solution. To restate that in simpler terms, solutions will form only when the energy of interaction between the solvent and solute is greater than the sum of the solvent-solvent and solute-solute interactions. That situation can only occur when the solvent and solute have similar properties. For example, if a non-polar molecule, such as oil, is mixed with a polar molecule like water, no solution forms. Water’s solvent-solvent intermolecular interactions are mostly hydrogen bonds and dipole-dipole while oil has only van der Waals. Water can satisfy its hydrogen bonds and become stabilized by dipole-dipole interactions only wheear other water molecules. Therefore, water is destabilized when it forms a solution with oil. That is why such a solution will never form between oil and water. Therefore, the primary rule of solubility is that like dissolves like. Only when the solute and solvent molecules have several common structural features such as their polarities will a solution form.

Summary of Factors Affecting Solubility

Normally, solutes become more soluble in a given solvent at higher temperatures. One way to predict that trend is to use Le Chatelier’s principle. Because ΔH_soln is positive for most solutions, the solution formation reaction is usually endothermic. Therefore, when the temperature is increased, the solubility of the solute should also increase. However, there are solutes that do not follow the normal trend of increasing solubility with increasing temperature. One class of solutes that becomes less soluble with increasing temperature is the gasses. Nearly every gas becomes less soluble with increasing temperature.

Another property of gaseous solutes in summarized by Henry’s law (see ) which predicts that gasses become more soluble when their pressures above a liquid solution are increased. That property of gaseous solutes can be rationalized by using Le Chatelier’s principle. Imagine that you have a glass of water inside of a sealed container filler with nitrogen gas. If the size of that container were suddenly halved, the pressure of nitrogen would suddenly double. To decrease the pressure of nitrogen above the solution (as is required by Le Chatelier’s principle), more nitrogen gas becomes dissolved in the glass of water.

Solids – more solid can dissolve because the solute molecules move apart and stay disconnected.

Gases – less gas can dissolve because more molecules can escape the solution.

Pressure – only affects gases.  Pressure “pushes” the molecules into the solution and prevent them from escaping.

Factors the affect the rate of solubility:

Surface area – increased access of solvent molecules to solute molecules.

Stirring and temperature – increased kinetic energy increases collisions between solvent and solute.

Solubility curve – graph of the relationship between temperature and solubility depicting saturation points.

Other important terms:

electrolyte = a substance that dissolves in water to give a solution that conducts electric current

nonelectrolyte = a substance that dissolves in water to give a solution that does NOT conduct an electric current

colloid = microscopic particles that are too large to be dissolve into solutions and instead suspended

suspension = a heterogeneous mixture in which the particles are so large that they settle out unless the mixture is constantly stirred or agitated

Tyndall effect = when light is reflected and scattered by the suspended colloid particles.

Miscible = molecules “mix” together

Physical and chemical characteristics of a water

Water (H₂O) is composed of two atoms of hydrogen and one of oxygen. Each hydrogen atom is linked to the oxygen atom by a single covalent bond. Because oxygen is more electronegative than hydrogen, there is a separation of charge within the molecule. The electron distribution in oxygen-hydrogen bonds may therefore be described as polar or asymmetrical. If water molecules were linear, then the bond polarities would cancel each other out, and water would be nonpolar. However, water molecules have a bent geometry with a bond angle of 104.5°

Molecules such as water, which have an unbalanced distribution of charge, are called dipoles. Such molecules have opposite charges on two points. When molec­ular dipoles are subjected to an electric field, they orient themselves in the direc­tion opposite to that of the field.

Water’s properties are directly related to its molecular structure.

One consequence of the large difference in electronegativity of hydrogen and oxygen is that the hydrogens of one water molecule are attracted to the unshared pairs of electrons of another water molecule. This noncovalent relationship is called a hydrogen bond. In addition to hydrogen bonds, three other types of noncovalent interactions play important roles in determining the capacity of water to interact with other types of molecules. These are electrostatic interactions, van der Waal’s forces, and hydrophobic interac­tions. Because biological reactions take place in a water medium, an understand­ing of noncovalent bonding is important.

Covalent bonds between hydrogen (electropositive atoms) and oxygen are polar. For example, each of the two hydrogens in wa­ter molecules will be weakly attracted to oxygen atoms in other nearby water mol­ecules. The resulting intermolecular “bonds” act as a bridge between adjacent molecules. Although considerably weaker than ionic and covalent bonds, hydrogen bonds are stronger than most other types of noncovalent bonds.

Electrostatic interactions occur between oppositely charged atoms or groups.  An important aspect of all electrostatic interac­tions in aqueous solution is the hydration of ions that occurs. Because water mol­ecules are polar, they are attracted to charged ions. Shells of water molecules, referred to as solvation spheres, cluster around both positive and negative ions. As ions become hydrated, the attractive force between them is reduced, and the charged species dissolves in the water. Water, sometimes called the universal solvent.

Melting point  of water – 0 ^oC; boiling point – 100 ^oC.

Water plays an important role in the thermal regulation of liv­ing organisms. Water’s high heat capacity coupled with the high water content found in most organisms (between 50% and 95%, depending on species) con­tributes to the maintenance of an organism’s internal temperature. The evapora­tion of water is used as a cooling mechanism, since it permits large losses of heat. For example, an adult human may eliminate daily as much as 1200 g of water in ex­pired air, sweat, and urine. The associated heat loss may amount to approximately 20% of the total heat generated by metabolic processes.

Water is a remarkable solvent. Water’s ability to dissolve a large variety of ionic and polar substances is determined by its dipolar structure and its capacity to form hydrogen bonds. Salts such as sodium chloride (NaCI) are held together by ionic (or electrostatic) forces. They dissolve easily in water because dipolar water molecules are attracted to the Na⁺ and Cl^– ions.

Organic molecules with ionizable groups and many neutral organic molecules with polar functional groups also dissolve in water. Their solubility is due primar­ily to the hydrogen bonding capacity of water. Nonpolar compounds are not soluble in water. Because they lack polar functional groups, such molecules cannot form hydrogen bonds.

Liquid water molecules have a limited capacity to ionize to form a hydrogen ion (H⁺) and a hydroxide ion  (OH^–). (H⁺ does not actually exist in aqueous solution. In water a proton combines with a water molecule to form the hydrated hydrogen ion, H₃O⁺, commonly called a hydronium ion. For convenience, however, the hydrated proton is usually represented as H⁺.)

The state and distribution of water in the organism.

There are two water compartments in the body:

1. Intracellular water

2. Extracellular water

Extracellular fluid is divided into:

1. interstitial fluid

2. plasma

Distribution of water in an adult man, weighing 70 kg

Biological role of water:

1. Water is an essential constituent of cell structures and provides the media in which the chemical reactions of the body take place and substances are transported.

2. It has a high specific heat for which, it can absorb or gives off heat without any appreciable change in temperature.

3. It has a very high latent heat. Thus, it provides a mechanism for the regulation of heat loss by sensible or insensible perspiration on the skin surface.

4. The fluidity of blood is because of water

5. Water is the most suitable solvent in human body

6. Dielectric constant : Oppositely charged particles can coexist in water. Therefore, it is a good  ionizing medium. This increases the chemical reactions.

7. Lubricating action: Water acts as lubricant in the body to prevent friction in joints, pleura, conjunctiva and peritoneum

A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

Solutions can be classified on the basis of their state: solid, liquid, or gas.

TYPES OF SOLUTION

The substances making up the solutions are called components

The components of a binary solution are solute and solvent.

Solvent is a component which is present in excess, in other words a solvent is a substance in which dissolution takes place. Solvent doesn’t change its physical state during reaction of dissolution. Solute is a component which is present in lesser quantity. Or solute is a substance that dissolves. In a solution, the particles are of molecular size (about 1000 pm) and the different components cannot be separated by any of the physical methods such as filtration, setting ,centrifugation, etc.)

1. Depending upon the total components present in the solution:

a)  Binary solution (two components)

b)  Ternary solution (three components)

c) Quaternary solution (four components)…..etc.

2. Depending upon the ability of the dissolution some quantity of the solute in the solvent:

a) Saturated solution

b) Not saturated solution

An unsaturated solution is one in which the concentration of solute is less than its concentration in a saturated solution. (Additional solute can be dissolved in an unsaturated solution, until the solution becomes saturated.)

A supersaturated solution is one in which the concentration of solute is greater than its concentration in a saturated solution. A supersaturated solu­tion is unstable and its solute tends eventually to crystallize out of solution, much as a supercooled liquid tends eventually to crystallize. Solubility equilib­rium is not possible in a supersaturated solution. If more solute is added, crys­tallization occurs, usually rapidly, as solute leaves the solution to crystallize on the surfaces of the crystals of added solute. (The situation is very much like the rapid freezing of a supercooled liquid brought about by adding a seed crystal.)

The solubility of a solute in a given solvent is defined as the concentration of the saturated solution. At 25°C, urea (see above) can be dissolved in water until a total of 19 mol has been added for each liter of solution formed. Some solutes are infinitely soluble in a given solute (this means that solute and solvent will mix in all proportions), while others have solubility’s so low that they are not measurable by direct methods. Although there is probably no such thing as complete insolubility, the term insoluble is commonly applied to a substance whose solubility is extremely low.

 

The change in solubility with change in temperature can be used to create solutions with more solute dissolved than is predicted by the solubility of the substance. For example, the solubility of glucose at 25 °C is 91 g glucose per 100 mL of water, and the solubility of glucose at 50 °C is 244 g glucose per 100 mL of water. Therefore, if we add 100 g of glucose to 100 mL water at 25 °C, 91 g dissolve. Nine grams of solid remain on the bottom, and the solution is saturated at this temperature. If we then heat the mixture to 50 °C, the remaining 9 grams of glucose will dissolve. At the new temperature, the solubility limit in 100 mL of water is 244 g glucose. With only 100 g of glucose dissolved, our system is now unsaturated.

If we now slowly cool the mixture back to 25 °C, 9 g of glucose should precipitate from solution. Sometimes this happens immediately, but sometimes it takes a while for the glucose molecules to find their positions in a solid structure. In the time between the cooling of the solution and the formation of glucose crystals, the system has a higher amount of dissolved glucose (100 grams) than is predicted by the solubility limit at 25 °C (91 grams). Because the solution contains more dissolved solute than is predicted by the solubility limit, we say the solution is supersaturated.

 

Solubility of Glucose

 

Pressure and Temperature Effects

If you have not yet studied thermodynamics or you do not know what ΔG or ΔS stands for, then please skip to the next heading in which the following discussion on temperature and pressure effects on solubility is summarized without the thermodynamics. The following discussion is a slightly more advanced treatment of the same phenomena.

The creation of disorder during the solution formation process is its essential driving force. In fact, most compounds that are soluble in water have positive enthalpies of solution. The only reason why those solutions form is due to the positive entropy of solution, ΔS_soln. As shows, both the solvent and the (solid or liquid) solute become less ordered upon solution formation. Therefore, from the equation ΔG = ΔH – TΔS we should predict that the solubility of every compound should increase with increasing temperature. That prediction turns out to be correct for nearly every solvent and solute. However, there are some exceptions, such as sodium sulfate in water that actually become less soluble at higher temperatures. That is usually due to their negative entropies of solution. The reason why some solutions have a negative entropy of solution is beyond the scope of this SparkNote. If you wish to pursue this topic further, search for the hydrophobic effect.

Using the idea of the entropy of solution, we can predict other properties of solutions. For example, we should predict that all gasses should bcome less soluble in water with increasing temperature because they have a negative entropy of solution. Gasses have a negative entropy of solution in water because they are confined to a smaller volume when dissolved as compared to their volumes as gasses. As we should predict, all gasses become less soluble in water with increasing temperature.

EFFECT OF TEMPERATURE ON SOLUBILITY:

The solubility of solutes is dependent on temperature. When a solid dissolves in a liquid, a change in the physical state of the solid analogous to melting takes place. Heat is required to break the bonds holding the molecules in the solid together. At the same time, heat is given off during the formation of new solute — solvent bonds.

CASE I: Decrease in solubility with temperature:

If the heat given off in the dissolving process is greater than the heat required to break apart the solid, the net dissolving reaction is exothermic (energy given off). The addition of more heat (increases temperature) inhibits the dissolving reaction since excess heat is already being produced by the reaction. This situation is not very common where an increase in temperature produces a decrease in solubility.

CASE II: Increase in solubility with temperature:

If the heat given off in the dissolving reaction is less than the heat required to break apart the solid, the net dissolving reaction is endothermic (energy required). The addition of more heat facilitates the dissolving reaction by providing energy to break bonds in the solid. This is the most common situation where an increase in temperature produces an increase in solubility for solids.

The use of first-aid instant cold packs is an application of this solubility principle. A salt such as ammonium nitrate is dissolved in water after a sharp blow breaks the containers for each. The dissolving reaction is endothermic – requires heat. Therefore the heat is drawn from the surroundings, the pack feels cold.

Solubility of Gases vs. Temperature:

The variation of solubility for a gas with temperature can be determined by examining the graphic on the below.

As the temperature increases, the solubility of a gas decrease as shown by the downward trend in the graph .

More gas is present in a solution with a lower temperature compared to a solution with a higher temperature.

The reason for this gas solubility relationship with temperature is very similar to the reason that vapor pressure increases with temperature. Increased temperature causes an increase in kinetic energy. The higher kinetic energy causes more motion in molecules which break intermolecular bonds and escape from solution.

This gas solubility relationship can be remembered if you think about what happens to a “soda pop” as it stands around for awhile at room temperature. The taste is very “flat” since more of the “tangy” carbon dioxide bubbles have escaped. Boiled water also tastes “flat” because all of the oxygen gas has been removed by heating.

To make gasses more soluble in water, we could think about trying to decrease the magnitude of their negative entropies of solution. One way to accomplish that is to make the gas above the solvent more ordered by increasing the pressure of the gas. In fact, William Henry discovered this property of gaseous solutes around the turn of the nineteenth century. Because he discovered that behavior first, the law that describes the increase in the solubility of a gas with increasing pressure is called Henry’s law. Henry’s law is given below in , C stands for the concentration of the dissolved gas and P represents the partial pressure of the gas above the solution.

Solubility of gases increases with pressure: Henry’s Law

The pressure of a gas is a measure of its “escaping tendency” from a phase. So it stands to reason that raising the pressure of a gas in contact with a solvent will cause a larger fraction of it to “escape” into the solvent phase.

The direct-proportionality of gas solubility to pressure was discovered by William Henry (1775-1836) and is known as Henry’s Law. It is usually written as

p = k_H c

in which p is the partial pressure of the gas above the liquid, c is the concentration of gas dissolved in the liquid, and k_H is the Henry’s law constant. The latter can be expressed in various units, and in some instances is defined in different ways, so be very careful to note these units when using published values.

In the table below, k_H is given as (partial pressure of gas in atm) ÷ (concentration in liquid, mol L^–1)

More on Henry’s law constants

·         Conversion factor tables

·         On-line converter (EPA)

·         Table of k_H values (pdf, 107 p)

 

Problem Example 1

Some vendors of bottled waters sell pressurized “oxygenated water” that is [falsely] purported to enhance health and athletic performance by supplying more oxygen to the body.

(a) How many moles of O₂ will be in equilibrium with one liter of water at 25° C when the partial pressure of O₂ above the water is 2.0 atm?

(b) How many mL of air (21% O₂ v/v) must you inhale in order to introduce an equivalent quantity of O₂ into the lungs (where it might actually do some good?)

Solution:

(a) Solving Henry’s law for the concentration, we get

c = p / k_H = (2.0 atm) ÷ (769 L atm mol^–1) = 0.0026 mol L^–1

(b) At 25° C, .0026 mol of O₂ occupies (22.4 L) × (.0026 mol) × (298/273) =
.063 L. The equivalent volume of air would be (.063 L) (.21) = .303 L. Given that the average tidal volume of the human lung is around 400 mL, this means that taking one extra breath would take in more O₂ than is present in 1 L of “oxygenated water”.

Solutions of liquids in liquids

Whereas all gases will mix to form solutions regardless of the proportions, liquids are much more fussy. Some liquids, such as ethyl alcohol and water, are miscible in all proportions. Others, like the proverbial oil and water , are not; each liquid has only a limited solubility in the other, and once either of these limits is exceeded, the mixture separates into two phases.

The reason for this variability is apparent from the table. Mixing of two liquids can be exothermic, endothermic, or without thermal effect, depending on the particular substances. Whatever the case, the energy factors are not usually very large, but neither is the increase in randomness; the two factors are frequently sufficiently balanced to produce limited miscibility.

The range of possibilities is shown here in terms of the mole fractions X of two liquids A and B. If A and B are only slightly miscible, they separate into two layers according to their relative densities. Note that when one takes into account trace levels, no two liquids are totally immiscible.

A useful general rule is that liquids are completely miscible when their intermolecular forces are very similar iature; “like dissolves like”. Thus water is miscible with other liquids that can engage in hydrogen bonding, whereas a hydrocarbon liquid in which London or dispersion forces are the only significant intermolecular effect will only be completely miscible with similar kinds of liquids.

Substances such as the alcohols, CH₃(CH₂)_nOH, which are hydrogen-bonding (and thus hydrophilic) at one end and hydrophobic at the other, tend to be at least partially miscible with both kinds of solvents. If n is large, the hydrocarbon properties dominate and the alcohol has only a limited solubility in water. Very small values of n allow the –OH group to dominate, so miscibility in water increases and becomes unlimited in ethanol (n = 1) and methanol (n = 0), but miscibility with hydrocarbons decreases owing to the energy required to break alcohol-alcohol hydrogen bonds when the non polar liquid is added.

These considerations have become quite important in the development of alternative automotive fuels based on mixing these alcohols with gasoline. At ordinary temperatures the increased entropy of the mixture is great enough that the unfavorable energy factor is entirely overcome, and the mixture is completely miscible. At low temperatures, the entropy factor becomes less predominant, and the fuel mixture may separate into two phases, presenting severe problems to the fuel filter and carburetor.

Solutions of molecular solids in liquids

The stronger intermolecular forces in solids require more input of energy in order to disperse the molecular units into a liquid solution, but there is also a considerable increase in entropy that can more than compensate if the intermolecular forces are not too strong, and if the solvent has no strong hydrogen bonds that must be broken in order to introduce the solute into the liquid.

For example, at 25° C and 1 atm pressure, 20 g of iodine crystals will dissolve in 100 ml of ethyl alcohol, but the same quantity of water will dissolve only 0.30 g of iodine.

As the molecular weight of the solid increases, the intermolecular forces holding the solid together also increase, and solubilities tend to fall off; thus the solid linear hydrocarbons CH₃(CH₂)_nCH₃ (n > 20) show diminishing solubilities in hydrocarbon liquids.

Solutions of ionic solids in liquids

Since the coulombic forces that bind ions and highly polar molecules into solids are quite strong, we might expect these solids to be insoluble in just about any solvent. Ionic solids are insoluble in most non-aqueous solvents, but the high solubility of some (including NaCl) in water suggests the need for some further explanation.

 

The key factor here turns out to be the interaction of the ions with the solvent. The electrically-charged ions exert a strong coulombic attraction on the end of the water molecule that has the opposite partial charge.

As a consequence, ions in solution are alwayshydrated; that is, they are quite tightly bound to water molecules through ion-dipole interaction. The number of water molecules contained in theprimary hydration shell varies with the radius and charge of the ion.

Hydration shells around some ions in a sodium chloride solution. The average time an ion spends in a shell is about 2-4 nanoseconds. But this is about two orders of magnitude longer than the lifetime of an individual H2O–H2O hydrogen bond.

Lattice and hydration energies

The dissolution of an ionic solid MX in water can be thought of as a sequence of two steps:

(1) MX(s) → M⁺(g) + X^–(g)     ΔH >0   (lattice energy)

(2) M⁺(g) + X^–(g) + H₂O → M⁺(aq) + X^–(aq)  ΔH < 0  (hydration energy)

·         The first reaction is always endothermic; it takes a lot of work to break up an ionic crystal lattice.

·         The hydration step is always exothermic as H₂O molecules are attracted into the electrostatic field of the ion.

·         The heat (enthalpy) of solution is the sum of the lattice and hydration energies, and can have either sign.

As often happens for a quantity that is the sum of two large terms having opposite signs, the overall dissolution process can come out as either endothermic or exothermic, and examples of both kinds are common.

Two common examples illustrate the contrast between exothermic and endothermic heats of solution of ionic solids:

 

Hydration entropy can make a difference!

Hydration shells around some ions in a sodium chloride solution. The average time an ion spends in a shell is about 2-4 nanoseconds. But this is about two orders of magnitude longer than the lifetime of an individual H₂O–H₂O hydrogen bond. The balance between the lattice energy and hydration energy is a major factor in determining the solubility of an ionic crystal in water, but there is another factor to consider as well. We generally assume that there is a rather large increase in the entropy when a solid is dispersed into the liquid phase. However, in the case of ionic solids, each ion ends up surrounded by a shell of oriented water molecules. These water molecules, being constrained within the hydration shell, are unable to participate in the spreading of thermal energy throughout the solution, and reduce the entropy. In some cases this effect predominates so that dissolution of the salt leads to a net decrease in entropy. Recall that any process in which the the entropy diminishes becomes less probable as the temperature increases; this explains why the solubilities of some salts decrease with temperature.

 

Gas Pressure and Solubility:

Liquids and solids exhibit practically no change of solubility with changes in pressure. Gases as might be expected, increase in solubility with an increase in pressure. Henry’s Law states that: The solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.

If the pressure is increased, the gas molecules are “forced” into the solution since this will best relieve the pressure that has been applied.

The number of gas molecules is decreased. The number of gas molecules dissolved in solution has increased as shown in the graphic on the left.

Carbonated beverages provide the best example of this phenomena. All carbonated beverages are bottled under pressure to increase the carbon dioxide dissolved in solution. When the bottle is opened, the pressure above the solution decreases. As a result, the solution effervesces and some of the carbon dioxide bubbles off.

3. Depending upon the physical states of the solute and solvent, the solution can be classified into the following nine type:

Out of the nine types of solutions, namely solid in liquid, liquid in liquid and gas in liquid are very common. In all these types of solutions, liquid  acts as solvent.

4. According to the nature of solvent the solutions can be classified

such as:

a) aqueous solution – the solution in which water is a solvent;

b) non- aqueous solution in which water is not the solvent (ether, benzene…)

The basic rule for solubility is “like dissolves like”

5. Depending upon component’s solubility in liquid solutions (which are themselves liquids), these mixtures may be classified into the following three types:

1) The two components are completely miscible (ethyl alcohol in water)

2) The two components are almost immiscible (oil and water, benzene and water)

3) The two components are partially miscible (ether and water)

6. The binary solutions may be classified into two types:

1) Ideal solutions. Such solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecular forces. In these solutions, the intermolecular interactions between the components (A-B) are of same magnitude as the intermolecular interactions in pure components ( A-A and B-B). Ideal solutions obeys Raoult’s law.

2) Non-ideal solutions

NONPOLAR SOLVENTS. First, consider carbon tetrachloride, a nonpolar solvent. (CCl₄ has no dipole moment because it is tetrahedral.) Those solutes that have high solubilities in CCl₄ are also nonpolar, such as iodine (I₂) and sulfur (S₈). Ionpolar substances, the only intermolecular attractions are compar­atively weak London forces. Thus, when a solution is formed from two nonpolar substances, solute-solute and solvent-solvent forces are eas­ily replaced by solute-solvent forces. (Here, “easily” means that there is no great energy deficit.) Wheonpolar solute molecules are inserted among nonpolar solvent molecules, neither experiences a great change in environment, and so the solution is readily formed. On the other hand, nonpolar liquids tend to be poor solvents for polar and ionic compounds. Lattice energies in these com­pounds tend to be higher than ionpolar solutes, and because of the lack of polarity of the solvent molecules, solute-solvent forces are too weak to overcome solute-solute forces. All other factors being equal, polar compounds tend to be less soluble ionpolar liquids than are nonpolar compounds, and ionic com­pounds tend to be still less soluble.

POLAR SOLVENTS. Now, consider the solvent properties of a polar solvent, water. Water is a good solvent for polar solutes, such as hydrogen chloride (HC1), and for ionic solutes, such as sodium chloride (NaCI). In these cases, the establishment of strong solute-solvent attractions provides the energy needed to overcome strong attractive forces in the solute. Water is an especially good solvent for many ionic compounds because the ion-dipole forces established in the solution are strong (recall that water is a highly polar molecule), and many H₂O molecules cluster around each ion. However, water is usually a poor sol­vent for nonpolar substances. This is true because the solvent-solvent (dipole-dipole) attractions in water are so strong that they cannot be overcome by weaker solute-solvent interactions, such as those provided by a nonpolar solute. (The formation of weak solute-solvent attractions does not release enough en­ergy to compensate for that required to separate the solvent dipoles.)

Although the like-dissolves-like rule is a handy generalization, it tends toward oversimplification. For example, not all ionic compounds are highly sol­uble in water. Although barium sulfate, BaSO₄, is an ionic compound, its sol­ubility in water is very low. Apparently the 2 + and 2 – ions in BaSO₄ cause the lattice energy in BaSO₄ to be so high that the solubility of this compound in water is low. Evidently we must look more closely at the specific interactions in solute, solvent, and solution. Another case in point: The ethanol molecule (C₂H₅OH; dipole moment, 1.70 D) is less polar than ethyl chloride (C₂H₅Cl; 2.05 D), yet ethanol is miscible with water in all proportions (the term infinitely soluble is often used), but the solubility of ethyl chloride in water is extremely low. Here, again, considering only polarity of solvent and solute leads to an error; specific strong interactions between ethanol and water (described below) account for their mutual solubility.

Gas solution is not possible to prepare a heterogeneous mixture of two gases because all gases mix uniformly with each other in all proportions. Gaseous solutions have the structure that is typical of all gases. (The molecules are spaced far apart and are in rapid, random motion, colliding frequently with each other and with the walls of the container.) The only difference between a gaseous solution and a pure gas is that in a solutioot all the molecules are alike. Air, the gaseous solution with which we come in closest contact, is composed primarily of N₂ (78 % by volume), O₂ (21 %), and Ar (1 %), with smaller concentrations of CO₂, H₂O, Ne, He, and dozens of other substances at very low levels.

Liquid solutions have the internal structure that is typical of pure liquids: closely spaced particles arranged with little order. Unlike a pure liquid, how­ever, a liquid solution is composed of different particles. Much of this chapter is devoted to the properties of liquid solutions, and special emphasis is given to aqueous solutions, in which the major component is water.

Two kinds of solid solutions are common. The first, the substitutional solid solution, exhibits a crystal lattice that has structural regularity but in which there is a random occupancy of the lattice points by different species. For ex­ample, mixtures of potassium chloride and potassium bromide ranging in com­position from pure KC1 to pure KBr can be crystallized out of aqueous solution. For each intermediate composition, a single crystalline phase with the rock salt structure is formed, but it has Cl^– and Br^– ions randomly distributed at the anion sites. The radii of Cl^– and Br^– are sufficiently similar to allow one ion to substitute for the other. Similarly, I^– ions can be incorporated into the rock salt structure of AgBr by substitution for Br^– ions, but in this case a maximum of only about 70 % of the Br^– ions can be replaced.

In the second type of solid solution, the interstitial solid solution, foreign atoms, ions, or molecules occupy the nooks and crannies, the interstices, in the host lattice. One example of this is austenite, in which carbon atoms occupy some of the interstices in a face-centered cubic array of iron atoms. Most of the transition elements can form such solid solutions with small atoms such as H, B, C, and N.

 In many solutions, one component is present in considerable excess over the others. This component is called the solvent, and other components are called solutes. For example, in the case of a solution prepared by dissolving 1 g each of table salt and table sugar in 100 g of water, we refer to the water as the solvent and the salt and sugar as solutes. As a practical matter, it is often useful to consider the solvent as the component throughout which the particles of the solute(s) are randomly dispersed.

The terms concentrated and dilute are commonly used to give a qualitative indication of the concentration of the solute in a solution; “concentrated” im­plies a relatively high concentration of solute, “dilute” a relatively low one.

The composition of a solution is described quantitatively by specifying the concentrations of its components. Commonly used concentration units include mole fraction, mole percent, molarity, molality, percent by mass, and normality. Definitions and examples of each of these units follow, except for normality.

Example

A purchased standard solution of sodium hydroxide had a concentration of 1.0 mol/dm³. How would you prepare 100 cm³ of a 0.1 mol/dm³ solution to do a titration of an acid?

• The required concentration is 1/10th of the original solution.

• To make 1dm³ (1000 cm³) of the diluted solution you would take 100 cm³ of the original solution and mix with 900 cm³ of water.

• The total volume is 1dm³ but only 1/10th as much sodium hydroxide in this diluted solution, so the concentration is 1/10th, 0.1 mol/dm³.

• To make only 100 cm³ of the diluted solution you would dilute 10cm³ by mixing it with 90 cm³ of water.

• How to do this in practice is described at the end of Example 14.3.2 below and a variety of accurate/’less inaccurate’ apparatus is illustrated above.

Example

Given a stock solution of sodium chloride of 2.0 mol/dm³, how would you prepare 250cm³ of a 0.5 mol/dm³ solution?

• The required 0.5 mol/dm³ concentration is 1/4 of the original concentration of 2.0 mol/dm³.

• To make 1dm³ (1000 cm³) of a 0.5 mol/dm3 solution you would take 250 cm³ of the stock solution and add 750 cm³ of water.

• Therefore to make only 250 cm³ of solution you would mix 1/4 of the above quantities i.e. mix 62.5 cm³ of the stock solution plus 187.5 cm³ of pure water.

• This can be done, but rather inaccurately, using measuring cylinders and stirring to mix the two liquids in a beaker.

• It can be done much more accurately by using a burette or a pipette to measure out the stock solution directly into a 250 cm³ graduated-volumetric flask.

• Topping up the flask to the calibration mark (meniscus should rest on it). Then putting on the stopper and thoroughly mixing it by carefully shaking the flask holding the stopper on at the same time!

Example

In the analytical laboratory of a pharmaceutical company a laboratory assistant was asked to make 250 cm³ of a 2.0 x 10^-2 mol dm^-3 (0.02M) solution of paracetamol (C₈H₉NO₂).

o    (a) How much paracetamol should the laboratory assistant weigh out to make up the solution? Atomic masses: C = 12, H = 1, N = 14, O = 16

§   method (i): M_r(paracetamol) = (8 x 12) + (9 x 1) + (1 x 14) + (2 x 16) = 151

§   1000 cm³ of 1.0 molar solutioeeds 151g

§   1000 cm³ of 2.0 x 10^-2 molar solutioeeds 151 x 2.0 x 10^-2/1 = 3.02g

§   (this is just scaling down the ratio from 151g : 1.0 molar)

§   Therefore to make 250 cm³ of the solution you need 3.02 x 250/1000 = 0.755 g

o    method (ii): M_r(paracetamol) = 151

§   moles = molarity x volume in dm³

§   mol paracetamol required = 2.0 x 10^-2 x 250/1000 = 5.0 x 10^-3 (0.005)

§   mass = mol x M_r = 5.0 x 10^-3 x 151 = 0.755 g

(b) Using the 2.0 x 10^-2 molar stock solution, what volume of it should be added to a 100cm³ volumetric flask to make 100 cm³ of a 5.0 x 10^-3 mol dm^-3 (0.005M) solution?

§     The ratio of the two molarities is stock/diluted = 2.0 x 10^-2/5.0 x 10^-3 = 4.0 or a dilution factor of 1/4 (0.02/0.005).

§     Therefore 25 cm³ (¹/₄ of 100) of the 2.0 x 10^-2 molar solution is added to the 100 cm³ volumetric flask prior to making it up to 100 cm³ with pure water to give the 5.0 x 10^-3 mol dm^-3 (0.005M) solution.

Example

You are given a stock solution of concentrated ammonia with a concentration of 17.9 mol dm^-3 (conc. ammonia! ~18M!)

(a) What volume of the conc. ammonia is needed to make up 1dm³ of 1.0 molar ammonia solution?

§   Method (i) using simple ratio argument.

§   The conc. ammonia must be diluted by a factor of 1.0/17.9 to give a 1.0 molar solution.

§   Therefore you need 1.0/17.9 x 1000 cm³ = 55.9 cm³ of the conc. ammonia.

§   If the 55.9 cm³ of conc. ammonia is diluted to 1000 cm³ (1 dm³) you will have a 1.0 mol dm^-3 (1M) solution.

§   Method (ii) using molar concentration equation – a much better method that suits any kind of dilution calculation involving molarity.

§   molarity = mol / volume (dm³), therefore mol = molarity x volume in dm³

§   Therefore you need 1.0 x 1.0 = 1.0 moles of ammonia to make 1 dm³ of 1.5M dilute ammonia.

§   Volume = mol / molarity

§   Volume of conc. ammonia needed = 1.0 / 17.9 = 0.0559 dm³ (55.9 cm³) of the conc. ammonia is required,

§   and, if this is diluted to 1 dm³, it will give you a 1.0 mol dm^-3 dilute ammonia solution.

(b) What volume of conc. ammonia is needed to make 5 dm³ of a 1.5 molar solution?

§     molarity = mol / volume (dm³), therefore mol = molarity x volume in dm³

§     Therefore you need 1.5 x 5 = 7.5 moles of ammonia to make 5 dm³ of 1.5M dilute ammonia.

§     Volume (of conc. ammonia needed) = mol / molarity

§     Volume of conc. ammonia needed = 7.5 / 17.9 = 0.419 dm³ (419 cm³) of the conc. ammonia is required,

§     and, if this is diluted to 5 dm³, it will give you a 1.5 mol dm^-3 dilute ammonia solution.

14.4 Water of crystallisation in a crystallised salt

Example 14.4.1: Calculate the % of water in hydrated magnesium sulphate MgSO₄.7H₂O salt crystals

• Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1

• relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246

• 7 x 18 = 126 is the mass of water

• so % water = 126 x 100 / 246 = 51.2%

Example. The % water of crystallisation and the formula of the salt are calculated as follows:

• Suppose 6.25g of blue hydrated copper(II) sulphate, CuSO₄.xH₂O, (x unknown) was gently heated in a crucible until the mass remaining was 4.00g. This is the white anhydrous copper(II) sulphate.

• The mass of anhydrous salt = 4.00g, mass of water (of crystallisation) driven off = 6.25-4.00 = 2.25g

• The % water of crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%

• [ Ar’s Cu=64, S=32, O=16, H=1 ]

• The mass ratio of CuSO₄ : H₂O is 4.00 : 2.25

• To convert from mass ratio to mole ratio, you divide by the molecular mass of each ‘species’

• CuSO₄ = 64 + 32 + (4×18) = 160 and H₂O = 1+1+16 = 18

• The mole ratio of CuSO₄ : H₂O is 4.00/160 : 2.25/18

• which is 0.025 : 0.125 or 1 : 5, so the formula of the hydrated salt is CuSO₄.5H₂O

14.5 Calculation of quantities required for a chemical reaction, % yield and atom economy

Suppose you want to make 50g of the blue hydrated copper(II) sulphate crystals

All of section 14.5 is based on this quantity of 50g of the familiar blue crystals. The preparation is briefly described in the GCSE Acids, Bases and Salts Notes.

o        Your reactants are dilute sulphuric acid and the black solid copper(II) oxide.

o        You can use copper(II) carbonate, but this is not a pure simple compound and the predictive nature of the calculations will not be as good.

o        copper(II) oxide + sulphuric acid ==> copper(II) sulphate + water

o        (i) CuO_(s) + H₂SO_4(aq) ==> CuSO_4(aq) + H₂O_(l)

o        on crystallisation you get the blue hydrated crystals of formula CuSO₄.5H₂O

o        so strictly speaking, after evaporation-crystallisation the equation is

o        (ii) CuO_(s) + H₂SO_4(aq) + 4H₂O_(l) ==> CuSO₄.5H₂O_(s)

o        How much copper(II) oxide is needed?

o        A ‘non-moles’ calculation first of all, involving a reacting mass calculation.

o        The crucial change overall is CuO ==> CuSO₄.5H₂O

o        (Note: In reacting mass calculations you can often ignore other reactant/product masses)

o        Atomic masses: Cu = 64, S = 32, H = 1, O = 16

o        Formula masses are for: CuO = 64 + 16 = 80, CuSO4 = 64 + 32 + (4 x 16) = 160

o        and CuSO₄.5H₂O = 64 + 32 + (4 x 16) + [7 x (1+1+16)] = 250

o        The crucial reacting mass ratio is: 80 ==> 250 since formula ratio is 1:1 in the equation.

o        Therefore, theoretically, to make 50g of the crystals (¹/₅th of 250),

o        you need ¹/₅th of 80g of copper(II) oxide,

o        and 80/5 = 16g of copper(II) oxide is required.

o        However, in reality, things are not so simple because the method involves adding excess copper(II) oxide to the dilute sulphuric acid. (see salt preparation method (b))

o        So in practice you would need to take more of the CuO to get anything like 50g of the salt crystals.

There is another way to calculate the quantities required based on the acid.

How much dilute sulphuric acid (of concentration 1 mol dm^-3) is required?

§    Mol = mass in g / formula mass,

§    so moles of CuSO₄.5H₂O required = 50/250 = 0.2 mol (see basics of moles)

§    Therefore 0.2 mol of H₂SO₄ is required (¹/₅th mol), since the mole ratio CuO : H₂SO₄ is 1 : 1 in the equation.

§    1dm³ (1000 cm³) of a 1 mol dm^-3 solution of contains 1 mole (by definition – see molarity page)

§    Therefore 1/5th of 1dm³ is required, so 200 cm³ of 1 mol dm^-3 dilute sulphuric acid is required,

§    or 100 cm³ of 2M dilute sulphuric acid. (2M is old fashioned notation for 2 mol dm^-3 as seen on many laboratory bottles!)

§    You would then add copper(II) oxide in small amounts until no more dissolves in the warm-hot acid and the excess black powder is filtered off. There is no need to weigh out an exact amount of copper oxide.

Suppose after carrying out the preparation you finally crystallise 34g of pure the blue crystals of CuSO₄.5H₂O and weigh the product when dry.

o    What is the ‘atom economy’ of the preparation? (you need to refer to equations (i)/(ii)  at the start of section 14.5

§   Atom economy = useful theoretical products x 100/mass of all reactants

§   based on equation (i) Atom economy = mass CuSO₄ x 100 / (mass CuO + H₂SO₄)

§   = 160 x 100 / (80 + 98) = 16000/178 = 89.9%

§   based on equation (ii) the atom economy is 100% if you include water as a ‘reactant’, can you see why?

§   What is the % yield? i.e. comparing what you actually get with the maximum possible, i.e. a ‘reality check’!

§   % yield = mass of product obtained x 100 / theoretical mass from the equation

§   % yield = 34 x 100 / 50 =  68%

Heat effect of a dissolution

THE MECHANISM OF DISSOLVING. Consider what happens when we add a solid solute to a liquid solvent. Imme­diately after the addition, the solid-state structure begins to disintegrate, as, little by little, solvent molecules chip away at the surface of the crystal lattice, prying out solute particles, surrounding them, and finally dispersing them throughout the body of the solution. The ease with which all this takes place depends primarily on the relative strengths of three attractive forces: (1) the forces between the particles of the solute before it has dissolved {solute-solute forces), (2) the forces between solvent particles before dissolution has taken place {solvent-solvent forces), and (3) the forces that are formed between solute and solvent particles during the dissolving process {solute-solvent forces). As dissolving takes place, solute-solute and solvent-solvent forces are replaced by solute-solvent forces. In general, high solubility of a solid in a liquid is favored by weak solute-solute forces (measured in solids by lattice energy) and by strong solute-solvent forces (measured by solvation energy; see below). A gen­eralization that is old but still very useful is “like dissolves like,” which means that a solvent will dissolve a solute if they have similar properties. More spe­cifically, polar solvents tend to dissolve polar and ionic solutes, and nonpolar solvents tend to dissolve nonpolar solutes. Although this rule is not perfect, it is very useful as a rough guideline for predicting solubilities. Let us see how it can be justified in terms of the structural changes that take place during dissolution.

PROCESSES OF DISSOLUTION AND DISSOLVING CAPACITY OF WATER

The water well solves many matters, its dissolving capacity is determined by high inductivity (= 78). The moleculas of water enter with ions of salts ion – dipole interplay, derivating hydrated ions.

The hydration of ions results in their stabilization in solution. Each negatively ionized atom attracts positive poles of dipoles of water and aims to hold them about itself. The positively ionized atoms hydrated even more strongly, as the charge density for them is higher.

The dissolubility of organic matters in water depends on availability of polar groups (having large affinity to water) in moleculas; such groups are called hydrophylic, are capable to interact with water (formation of hydrogen bindings).

In common cases dissolubility of matters are subject to the rule of thumb “similar solves in similar”. For example, the hydrocarbon a naphthalene well solves in gasoline (mixture of hydrocarbon), it is worse – in alcohol (names hydrocarbon radicals), almost water soluble.

The dissolubility of solid matters depends on temperature. For the majority of salts she increases with temperature rise.

The process of dissolution is conditioned by interplay of fragments of solved matter and solvent. The dissolution of solid matters in water and molecular dissociation on ions can be presented as follows: the dipoles of water, falling in an electrical field of polar moleculas, will orient around of polar groups or ions, which one are on a surface of a chip. Attracting to themselves of a molecula or ions, the dipoles of water relax, and then lacerate intermolecular or ionic bonds. For example, the water reduces strength of an ionic bond in a chip of sodium salt, between ions Na⁺ and Cl^–. The separate molecules or ions under influencing of heat motion of all fragments of solution are displaced (diffuse), uniformly being arranged between molecules of a solvent. At dissolution often descends not only gap of bonds in solute, but also the associates of molecules of a solvent are blasted. In solution there are new associates as well from molecules of a solvent, as from fragments of solute.

The processes of dissolution are accompanied by heat effects, which depend on energy consumption on transfer of solid matter in a liquid state (that is its dissolution), and from a secured heat at interplay of fragments of solved matter and solvent (solvation). If quantity of energy expended for a gap of bonds  in a chip, is more than a heat generated of a solvation, dissolution – endothermal process. At minor heat of dissociation on ions and large heat of a solvation – exothermal process.

At formation of solutions the polarity of moleculas of a solvent matters, than their electrical dipole moment more, the above their capacity to a solvation, the is more excreted of energy and better matter solves. Capacity of matters will be dissolved in solvent liquids, as depends on polarity of their moleculas.

The polar matters well solve in polar solvents, (for example, sodium salt in water). To the contrary, the matter with low-polarity by moleculas, for example, fatty acids, solve in solvents, molecula which one poorly polar (ethers) or not polar (benzole) better.

Molarity, volumes and the concentration of solutions

Revise section 7. moles and mass before proceeding in this section 11 and eventually you may need to be familiar with the use of the apparatus illustrated above, some of which gives great accuracy when dealing with solutions and some do not.

It is very useful to be know exactly how much of a dissolved substance is present in a solution of particular concentration or volume of a solution. So we need a standard way of comparing the concentrations of solutions.

The concentration of an aqueous solution is usually expressed in terms of moles of dissolved substance per cubic decimetre, that is mol dm^-3 and this is called molarity, sometimes denoted in shorthand as M.

Note: 1dm³ = 1 litre = 1000ml = 1000 cm³, so dividing cm³/1000 gives dm³, which is handy to know since most volumetric laboratory apparatus is calibrated in cm³ (or ml), but solution concentrations are usually quoted in molarity, that is mol/dm³ (mol/litre).

Equal volumes of solution of the same molar concentration contain the same number of moles of solute i.e. the same number of particles as given by the chemical formula.

You need to be able to calculate the number of moles or mass of substance in an aqueous solution of given volume and concentration: the concentration of an aqueous solution given the amount of substance and volume of water, for this you use the equation:

o    (1a) molarity (concentration) of Z = moles of Z / volume in dm³

you need to be able to rearrange this equation  therefore

§   (1b) moles = molarity (concentration) x volume in dm3 and

§   (1c)  volume in dm3 = moles / molarity (concentration)

You may also need to know that …

§   (2) molarity x formula mass of solute = solute concentration in g/dm³ and dividing this by 1000 gives the concentration in g/cm³, and

§   (3) concentration in g/dm³ / formula mass = molarity in mol/dm³ both equations (2) and (3) result from equations (1) and (4), work it out for yourself.

§    (4) moles Z = mass Z / formula mass of Z and (5) 1 mole = formula mass in grams

Example

What mass of sodium hydroxide (NaOH) is needed to make up 500 cm³ (0.500 dm³) of a 0.500 mol dm^-3 (0.5M) solution? [A_r‘s: Na = 23, O = 16, H = 1]

o    1 mole of NaOH = 23 + 16 + 1 = 40g

o    molarity = moles / volume, so mol needed = molarity x volume in dm³

o    mol NaOH needed = 0.500 x 0.500 = 0.250 mol NaOH

o    therefore mass = mol x formula mass

o    = 0.25 x 40 = 10g NaOH required

Example

(a) How many moles of H₂SO₄ are there in 250cm³ of a 0.800 mol dm^-3 (0.8M) sulphuric acid solution?

(b) What mass of acid is in this solution? [A_r‘s: H = 1, S = 32, O = 16]

(a) molarity = moles / volume in dm³, rearranging equation for the sulfuric acid

mol H₂SO₄ = molarity H₂SO₄ x volume of H₂SO₄ in dm³

mol H₂SO₄ = 0.800 x 250/1000 = 0.200 mol H₂SO₄

(b) mass = moles x formula mass

formula mass of H₂SO₄ = 2 + 32 + (4×16) = 98

0.2 mol H₂SO₄ x 98 = 19.6g of H₂SO₄

 

Example

5.95g of potassium bromide was dissolved in 400cm³ of water. Calculate its molarity. [A_r‘s: K = 39, Br = 80]

moles = mass / formula mass, (KBr = 39 + 80 = 119)

mol KBr = 5.95/119 = 0.050 mol

400 cm³ = 400/1000 = 0.400 dm³

molarity = moles of solute / volume of solution

molarity of KBr solution = 0.050/0.400 = 0.125M

 

Example

What is the concentration of sodium chloride (NaCl) in g/dm³ and g/cm³ in a 1.50 molar solution?

At. masses: Na = 23, Cl = 35.5, formula mass NaCl = 23 + 35.5 = 58.5

since mass = mol x formula mass, for 1 dm³

concentration = 1.5 x 58.5 = 87.8 g/dm³, and

concentration = 87.75 / 1000 = 0.0878 g/cm³

Example

A solution of calcium sulphate (CaSO₄) contained 0.500g dissolved in 2.00 dm³ of water.

Calculate the concentration in (a) g/dm³, (b) g/cm³ and (c) mol/dm³.

(a) concentration = 0.500/2.00 = 0.250 g/dm³, then since 1dm³ = 1000 cm³

(b) concentration = 0.250/1000 = 0.00025 g/cm³

(c) At. masses: Ca = 40, S = 32, O = 64, f. mass CaSO₄ = 40 + 32 + (4 x 16) = 136

moles CaSO₄ = 0.5 / 136 = 0.00368 mol

concentration CaSO₄ = 0.00368 / 2 = 0.00184 mol/dm³

 

The methods for expression the composition of solutions. The preparation of solutions by the given concentration

 

For scientific or technical applications, a qualitative account of concentration is almost never sufficient; therefore quantitative measures are needed to describe concentration. There are a number of different ways to quantitatively express concentration; the most common are listed below. They are based on mass, volume, or both. Depending on what they are based on it is not always trivial to convert one measure to the other, because knowledge of the density might be needed to do so. At times this information may not be available, particularly if the temperature varies.

Units of concentration – particularly the most popular one, molarity ‑ require knowledge of a substance’s volume, which unlike mass is variable depending on ambient temperature and pressure. In fact (partial) molar volume can even be a function of concentration itself. This is why volumes are not necessarily completely additive when two liquids are added and mixed. Volume-based measures for concentration are therefore not to be recommended for non-dilute solutions or problems where relatively large differences in temperature are encountered (e.g. for phase diagrams). Unless otherwise stated, all the following measurements of volume are assumed to be at a standard state temperature and pressure (for example 0 degrees Celsius at 1 atmosphere or 101.325 kPa). The measurement of mass does not require such restrictions. Mass can be determined at a precision of < 0.2 mg on a routine basis with an analytical balance and more precise instruments exist. Both solids and liquids are easily quantified by weighing. The volume of a liquid is usually determined by calibrated glassware such as burettes and volumetric flasks. For very small volumes precision syringes are available. The use of graduated beakers and cylinders is not recommended as their indication of volume is mostly for decorative rather than quantitative purposes. The volume of solids, particularly of powders, is often difficult to measure, which is why mass is the more usual measure. For gases the opposite is true: the volume of a gas can be measured in a gas burette, if care is taken to control the pressure, but the mass is not easy to measure due to buoyancy effects.

The concentration of a solution may be defined as the amount of solute present

In the given quantity of the solution.

1.Mass percentage or volume percentage

The mass percentage of a component in a given solution is the mass of the com ponent  per 100 g of the solution.

Volume-volume percentage (sometimes referred to as percent volume per volume and abbreviated as % v/v) describes the volume of the solute in mL per 100 mL of the resulting solution. This is most useful when a liquid – liquid solution is being prepared, although it is used for mixtures of gases as well. For example, a 40% v/v ethanol solution contains 40 mL ethanol per 100 mL total volume. The percentages are only additive in the case of mixtures of ideal gases.

Work 2. The preparation of Sodium chloride and the determination of mass percentage of the prepared solution.

Materials: NaCl (solid), an electronic scales, a measured cylinder, a set of hydrometers, measuring pipettes, the beakers (50, 100 ml), the cylinder (100 ml), the volumetric flask (250 ml), distilled water, 10% H₂SO₄ solution.

On an electronic scales to weigh 3 g; 5 g; 8 g; 10 g; 12g of Sodium chloride on the paper basket. To measure 100 ml of water. To carry the salt in the beaker with the capacity 200 ml, to pour out 200 ml of water in the beaker. To mix the solution using a glass stick. To pour out the solution in the cylinder with capacity 100 ml 4/5 part or its volume.

To define the density of solution using a hydrometer. To calculate a mass percentage of Sodium chloride in the prepared solution and to compare it with a data handbooks table. To calculate the relative error of the experience. For the prepared solution to calculate Molarity, Molality and Titre.

Work 3.The preparation of sodium chloride solution with a given mass percentage and the total volume of the solution.

To do the calculation for the preparation of following variants: 200 ml of 1% , 100 ml of 1.5, 250 ml of 2%, 100 ml of 5%, 200 ml of 2% solution of Sodium chloride. (or risk) on the volumetric flask.

On an electronic scales to weigh Sodium chloride according to its mass percentage on the paper basket. Carefully to pour off salt into a volumetric flask, through the glass funnel, and to watch the funnel with a small amount of distilled water. To fill a volumetric flask which contains the salt with water of 2/3 parts of the flask’s volume. Carefully to mix the solution till the almost dissolving of salt. To add water till the mark

Work 4. To prepare the physiological solution of sodium chlorite from the hypertonic solution of 10% sodium chloride.

In the cylinder with the capacity of 100ml to pour of 10% sodium chloride solution till 4/5 part of the cylinder’s volume. To determine the density of the hypertonic solution by hydrometer.

To calculate the mass and the volume of the hypertonic solution and water, which is used for preparation of the physiological solution. Measure the calculated amount of the hypertonic solution, to move it in a beaker and to dilute it with the calculated amount of water. To mix the solution. To define the density of the prepared solution using a hydrometer. To compare it with a data handbooks table and to calculate the error.

2. Molarity

It is the number of moles of the solute dissolved per litre of the solution. It’s represented as M or  C_M

 (М) C_M = Moles of solute / Volume of solution in litres 

or

(М) C_M  = Mass of component A/ Molar mass of A *Volume of solution in litres

The unit of molarity is mol/L, 1L = 1000 ml

Molarity (in units of mol/L, molar, or M) or molar concentration denotes the number of moles of a given substance per liter of solution. A capital letter M is used to abbreviate units of mol/L. For instance:

The actual formula for molarity is:

Such a solution may be described as “0.50 molar.” It must be emphasized a 0.5 molar solution contains 0.5 moles of solute in 1.0 liter of solution. This is not equivalent to 1.0 liter of solvent. A 0.5 mol/L solution will contain either slightly more or slightly less than 1 liter of solvent because the process of dissolution causes volume of liquid to increase or decrease.

Following the SI system of units, the National Institute of Standards and Technology, the United States authority on measurement, considers the term molarity and the unit symbol M to be obsolete, and suggests instead the amount- of – substance concentration (c) with the units mol/m³ or other units used alongside the SI such as mol/L. This recommendation has not been universally implemented in academia or chemistry research yet.

Preparation of a solution of known molarity involves adding an accurately weighed amount of solute to a volumetric flask, adding some solvent to dissolve it, then adding more solvent to fill to the volume mark.

When discussing molarity of minute concentrations, such as in pharmacological research, molarity is expressed in units of millimolar (mmol/L, mM, 1 thousandth of a molar), micromolar (μmol/L, μM, 1 millionth of a molar) or nanomolar (nmol/L, nM, 1 billionth of a molar).

Although molarity is by far the most commonly used measure of concentration, particularly for dilute aqueous solutions, it does suffer from a number of disadvantages. Masses can be determined with great precision as balances are often very precise. Determining volume is ofteot as precise. In addition, due to a thermal expansion, molarity of a solution changes with temperature without adding or removing any mass. For non-dilute solutions another problem is the molar volume of a substance is itself a function of concentration so volume is not strictly additive.

Dilution calculation example:

How do you prepare 100ml of 0.40M MgSO4 from a stock solution of 2.0M MgSO4?

Answer:

There are two solutions involved in this problem. Notice that you are given two concentrations, but only one volume. Solution #1 is the one for which you have only concentration – the solution that is already sitting on the shelf. Solution #2 is the one for which you have both concentration and volume – the solution that you are going to prepare.

At least until you are comfortable with this type of problem, it may be helpful to write out what numbers go with what letters in our equation.

M1 = 2.0M MgSO₄ ; V1 = unknown M2 = 0.40M MgSO₄ ; V2 = 100ml

A solution contains 5.7 grams of potassium nitrate dissolved in enough water to make 233 mL of solution. What is its molarity ?

Answer: The formula weight of KNO₃ is 101.103 g/mol.

3. Molality

It is the number of moles of the solute dissolved per 1000 g (or 1 kg) of the solvent. It’s denoted by m or C_m

(m) C_m = Moles of solute/Weight of solvent in kg

or

(m) C_m= Moles of solute * 1000/Weight of solvent in gram

The unit of Molality is m or mol/kg

Molality is considered better for expressing the concentration as compared to molarity because the molarity changes with temperature because of expansion of the liquid with the temperature

Molality (mol/kg, molal, or m) denotes the number of moles of solute per kilogram of solvent (not solution). For instance: adding 1.0 mole of solute to 2.0 kilograms of solvent constitutes a solution with a molality of 0.50 mol/kg. Such a solution may be described as “0.50 molal”. The term molal solution is used as a shorthand for a “one molal solution”, i.e. a solution which contains one mole of the solute per 1000 grams of the solvent.

Following the SI system of units, the National Institute of Standards and Technology, the United States authority on measurement, considers the unit symbol m to be obsolete, and suggests instead the term ‘molality of substance B’ (m_B) with units mol/kg or a related unit of the SI. This recommendation has not been universally implemented in academia yet.

Note: molality is sometimes represented by the symbol (m), while molarity by the symbol (M). The two symbols are not to be confused, and should not be used as symbols for units. The SI unit for molality is mol/kg. (The unit m – not italicized – means meter.)

Like other mass-based measures, the determination of molality only requires a good scale, because masses of both solvent and solute can be obtained by weighing, and molality is independent of physical conditions like temperature and pressure, providing advantages over molarity. In a dilute aqueous solution near room temperature and standard atmospheric pressure, molarity and molality will be very similar in value. This is because 1 kg of water roughly corresponds to a volume of 1 L at these conditions, and because the solution is dilute, the addition of the solute makes a negligible impact on the volume of the solution.

However, in all other conditions, this is usually not the case.

For example What is the molality of dentist’s amalgam? Recall that dentist’s amalgam is 70% mercury and 30 % copper, by mass.

Answer: Assume 100grams of amalgam and two significant figures, as in the previous example. By mass, mercury is the major component, so it must be the solvent. 100 grams of amalgam contains 70 grams of mercury, which is 0.070 kg of mercury. It also contains 30 grams of copper, which is 0.47 moles of copper. This gives

Note that we have used the mass of the solvent rather than the mass of the mixture.

 4. Normality

It is the number of  gram equivalents of the solute dissolved per litre of the solution. It’s denoted by N or 

(N) Cn = Number of gram equivalents of solute/Volume of solution in litres

or

(N) Cn = Number of gram equivalents of solute *1000/Volume                                                        of solution in ml

Number of gram equivalents of solute = Mass of solute /       Equivalent mass of solute

 Relationship between Normality and Molarity of Solutions

Normality = Molarity  *  Molar mass/Equivalent mass

Normality highlights the chemical nature of salts: in solution, salts dissociate into distinct reactive species (ions such as H⁺, Fe³⁺, or Cl^–). Normality accounts for any discrepancy between the concentrations of the various ionic species in a solution. For example, in a salt such as MgCl₂, there are two moles of Cl^– for every mole of Mg²⁺, so the concentration of Cl^– is said to be 2 N (read: “two normal”). Further examples are given below. It may also refer to the concentration of a solute in any solution. The normality of a solution is the number of gram equivalent weight of a solute per liter of its solution. The definition of normality depends on the exact reaction intended.

For example, hydrochloric acid (HCl) is a monoprotic acid and thus has 1 mol = 1 gram equivalent. One liter of 1 M aqueous solution of HCl acid contains 36.5 grams HCl. It is called 1 N (one normal) solution of HCl. It is given by the following formula:

In contrast, for sulfuric acid, which is diprotic acid, 2 N is usually 1 M, but may be defined as 2 M if pH < 2, where the once-deprotonated species, hydrogen sulfate, does not deprotonate.

5. Mole fraction

It is the ratio of number of moles of  one component to the total number of moles (solute and solven) present in the solution. It’s denoted by X. Let suppose that solution contains n_a  moles of solute and n_b  moles of the solvent. Then

 

For example: Dentist’s amalgam is 70% mercury and 30% copper, by mass. What is the mole fraction of copper in dentist’s amalgam?

Answer: For the purpose of assigning significant figures, we will assume that the percentages are accurate to at least 1%. Assume 100 grams of amalgam, which contains 70 grams of mercury and 30 grams of copper, each value having at least two significant figures. Convert each to moles and find X. Note that X does not have units.

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In measure analysis for the characteristic the composition of solution will use molar mass of an equivalent (equivalent mass)

Molar mass of an equivalent of element is the mass of the element which combines with or displaces 1.008parts by mass of hydrogen or 8 part by mass of oxygen or 35.5 parts by mass of chlorine:

E = f_equiv · M_B

The factor of equivalence (f_equiv) – number, which is demonstrated which part of matter (equivalent) can react with one atom of Hydrogen, or one electron in reduction reactions.

For example:

1. Concentrated hydrochloric acid is 31% hydrochloric acid and 69% water, by mass. If the density of concentrated hydrochloric acid is 1.16g/mL, what is its molarity?

Answer: Assume 100 grams of concentrated hydrochloric acid; this has 31 grams of HCl and 69 grams of water. The formula weight of hydrochloric acid is 36.46 g/mol, so 31 grams is 0.85moles. Find the volume of the solution.

Find the molarity.

2. An aqueous solution of sulfuric acid has a molarity of 18 mol/L. if its density of 1.84 g/mL, what is the mole fraction of water in the solution?

Answer: Assume one liter of solution, which has 18 moles of sulfuric acid. The formula weight of sulfuric acid is 98.08 g/mol, so 18 moles has a mass of 1800 grams (1756 with two significant figures). Use the density formula to find the mass of the solution

If 1800 grams of the solution is sulfuric acid, 40 grams must be water. The formula weight of water is 18.015 g/mol, so 40 grams of water is 2 moles. The mole fraction of water is

An aqueous solution of hydrogen peroxide, H₂O₂, is 16.9 mol/L. if it has a density of 1.196 g/mL, what is its molality?

Answer: Assume one liter of solution; this has 16.9 moles of hydrogen peroxide. The formula weight of hydrogen peroxide is 34.015 g/mol, so the liter of solution contains 575 grams of hydrogen peroxide. Use the density formula to find the mass of one liter of solution:

3. If the mixture has 575 grams of hydrogen peroxide in 1196 grams of solution, it must have 621 grams of water (or 0.621 kg of solvent). The molality is

4. 25mL of 6M sodium hydroxide are diluted to 1.3 L. What is the resulting concentration?

Answer:

5. A student wishes to make 0.5 L of a 3.0 M sulfuric acid solution using concentrated sulfuric acid, 18 M. How should she do this?

Answer: Very Carefully! Diluting concentrated acids is dangerous. The process releases so much thermal energy that the water used in the dilution will boil, which may cause boiling hot acid burns. Ouch!

The amount of acid to use can be calculated by the equation above.

The student needs to carefully add 80 mL of concentrated acid to roughly 400 mL of water then add enough water to make a total of 500 mL of solution. Always add the acid to the water. It takes more energy to boil a lot of water than to boil a little acid. Besides, if something has to splash out, you want it to be the water not the acid.

Disperse systems are called systems, which consist of two phases, one of which is scattered or dispersed in other.

The disperse phase – phase which is scattered (dispersed) in medium. The disperse medium – phase in which dispersion done. For colloids or temporary suspensions the phrase dispersed material or the word dispersants describes the material in suspension, analogous to the solute of a solution. The phrase dispersing medium is used for the material of similar function to a solvent in solutions.

Classification disperse systems

-By stat of dispersed phase and dispersed medium

-By size of dispersed phase

Colloidal solutions are disperse systems, which have dispersed phase particle, which size between 10^-9 to 10^-7m or 1 nm to 100 nm.

As with true solutions, it is a bit of a stretch to consider solids as a dispersing medium or gases as forming a large enough particle to be a colloid, but most texts list some such. A sol is a liquid or solid with a solid dispersed through it, such as milk or gelatin. Foams are liquids or solids with a gas dispersed into them. Emulsions are liquids or solids with liquids dispersed through them, such as butter or gold-tinted glass. Aerosols are colloids with a gas as the dispersing medium and either a solid or liquid dispersant. Fine dust or smoke in the air are good examples of colloidal solid in a gas. Fog and mist are exampes of colloidal liquid in a gas.

Liquid dispersion media with solid or liquid dispersants are the most often considered. Homogenized whole milk is a good example of a liquid dispersed into a liquid. The cream does not break down into molecular sized materials to spread through the milk, but collects in small micelles of oily material and proteins with the more ionic or hydrophilic portions on the outside of the globule and the more fatty, or oily, or non-polar, or hydrophobic portions inside the ball-shaped little particle. Blood carries liquid lipids (fats) in small bundles called lipoproteins with specific proteins making a small package with the fat.

Proteins are in a size range to be considered in colloidal suspension in water. Broth or the independent proteins of blood or the casein (an unattached protein) in milk are colloidal. There are many proteins in the cellular fluids of living things that are in colloidal suspension.

Colloidal dispersants in water stay in suspension by having a layer of charge on the outside of the particle that is attractive to one end of water molecules. The common charge of the particles and the water solvation layer keep the particles dispersed. A Cottrel precipitator collects the smoke particles from air by a high voltage charge and collection device. Boiling an egg will denature and coagulate the protein in it. Proteins can be fractionally ‘salted out’ of blood by adding specific amounts of sodium chloride to make the proteins coagulate. The salt adds ions to the liquid that interfere with the dispersion of the colloidal particles.

Colloids with liquid as a dispersing agent have the following properties:

PROPERTIES OF COLLOIDS

1. The particles of dispersant are the between about 500 nm to 1 nm in diameter.

2.  The mixture does not separate on standing in a standard gravity condition. (One ‘g.’)

3. The mixture does not separate by common fiber filter, but might be filterable by materials with a smaller mesh.

4.The mixture is not necessarily completely homogeneous, but usually close to being so.

5. The mixture may appear cloudy or almost totally transparent, but if you shine a light beam through it, the pathway of the light is visible from any angle. This scattering of light is called the Tyndall effect

6. There usually is not a definite, sharp saturation point at which no more dispersant can be taken by the dispersing agent.

7. The dispersant can be coagulated, or separated by clumping the dispersant particles with heat or an increase in the concentration of ionic particles in solution into the mixture.

8. There is usually only small effect of any of the colligative properties due to the dispersant.

Many food ingredients are completely immiscible and so will form separate phases within the food. However the sizes of these phases can be very small, so to the naked eye the food will appear homogeneous. The techniques and principles of colloid science are suited to dealing with the properties of fine particles and their applications to foods will be explored in this section. A colloidal particle is many times larger than an individual molecule but many times smaller than anything that can be seen with the naked eye. Some examples of colloidal particles in foods are listed in the adjacent table (Table is reproduced from Fennema text). A colloidal particle has at least one length dimension in the range (approximately) of tens of nanometers and tens of micrometers. Although colloidal particles are small, each contains a huge number of molecules. To get an idea of the number of molecules needed to form a colloidal particle: a 10 nm droplet of water would contain about 10⁵ molecules, a 10 μm about 10¹⁴ molecules. To stretch the analogy, if a water molecule was a person, a 10 nm droplet would be have as many people as a city the size of Bethlehem (PA), while a 10 μm droplet 10,000 times the population of the planet.

Particles must be dispersed in a second phase. (Note that the particles are the dispersed phase and the phase they are dispersed in is the continuous phase). Depending what the phases are (solid, liquid, gas) it is possible to generate different types of colloidal system. Their names and some examples are given in the following table: Continuous phase

 

References:

1. The abstract of the lecture.

2. intranet.tdmu.edu.ua/auth.php

3. Atkins P. W. Physical chemistry / P.W. Atkins. – New York, 1994. – P.299‑307.

4. Cotton F. A. Chemical Applications of Group Theory / F. A. Cotton. ‑ John Wiley & Sons : New York, 1990.

5. Girolami G. S. Synthesis and Technique in Inorganic Chemistry / G. S. Girolami, T. B. Rauchfuss, R. J. Angelici. ‑ University Science Books : Mill Valley, CA, 1999.

6. Russell J. B. General chemistry / J B. Russell. New York.1992. – P. 550‑599.

7. Lawrence D. D. Analytical chemistry / D. D. Lawrence. –New York, 1992. – P. 218–224.

 

Prepared by PhD Falfushynska H.