CHEM 102 CLASS NOTES

CHAPTER 15. The Chemistry of Solutes and Solutions

KEY CONCEPTS

Solutions	Solubility	Solute	Solvent
Saturated solutions	Supersaturated solutions	Solubility gases: Henry's Law	Molarity
Normality	Molality	Mass percentage	Volume & Mass %
Proof	Parts per million (ppm)	Parts per billion (ppb),	Mole fraction
Energy changes of solution process	Solubility of substances	Colligative properties	Vapor pressure
Boiling point	Freezing point	Osmotic pressure	Colligative properties of electrolytic solutions
Colloids	True solutions	Suspensions

Solubility and Intermolecular Forces

The dissolving process of solutes in a solvent depends on the relative strength of three intermolecular attractive forces. The three types of forces between the solute and the solvent are solute-solute and solvent-solvent and solvent-solvent must be considered in the solution process. These interactions could be ion-ion, ion-ion dipole, dipole-dipole and London dispersion forces between the ions, dipoles or non-polar molecules of the solvents or the solute. These intermolecular attractions must be broken before new solute-solvent attractive forces can become effective. Perhaps the bond breaking and bond forming processes take place simultaneously. A solute will dissolve in a solvent if the solute-solvent forces of attraction are great enough to overcome the solute-solute and solvent-solvent forces of attraction. A solute will not dissolve if the solute-solvent forces of attraction are weaker than individual solute and solvent intermolecular attractions. Generally, if all three of the intermolecular forces of attraction are roughly equal, the substances will be soluble in each other.

Non-polar solute - Non-polar solvent:

In all types of non-polar compounds, about the only intermolecular attractions are the very weak induced dipole forces. The weak attractive forces formed by the solute-solvent molecules compensate for breaking those weak bonds in the two pure non-polar substances. An example is solid iodine (I₂) dissolved in liquid bromine (Br₂).

Non-polar solute - Polar solvent:

Non-polar Iodine is not very soluble in water. An intermolecular bond between an induced dipole (I₂) and a polar bond in water is not very strong compared to the hydrogen bonds in water. The water molecules would rather remain hydrogen bonded to each other, then to allow iodine molecule come between them. The water molecules effectively "squeeze" out the non-polar iodine. The intermolecular forces are not roughly equal, therefore, the "unlike" substances are not soluble in each other.

Various gases such as O₂, N₂, H₂, CO₂ are not very soluble because the gases are essentially non-polar. Of course you may say that oxygen must be dissolved in water to sustain fish life -- true, but the solubility is very low. Carbon dioxide is soluble in water such as carbonated beverages -- again this is true but why does it fizz when opened or lose the bubbles on standing? Carbon dioxide is not very soluble in water.

Polar Solute - Polar Solvent:

Polar ammonia molecules dissolve in polar water molecules. These molecules mix readily because both types of molecules engage in hydrogen bonding. Since the intermolecular attractions are roughly equal, the molecules can break away from each other and form new solute (NH₃), solvent (H₂O) hydrogen bonds. A wide variety of solutions are in this category such as sugar in water, alcohol in water, acetic and hydrochloric acids.

Ionic Solute - Polar Solvent:

When an ionic crystal such as NaCl is placed in water, a dissolving reaction will occur. Initially, the positive and negative ions are only attracted to each other. The water molecules are hydrogen bonded to each other. If the crystal is to dissolve, these bonds must be broken.

Negative chloride ions on the surface are attracted by neighboring positive sodium ions and by the partially positive hydrogen atom in the polar water molecule (See the graphic on the right).

Similarly, the positive sodium ions are attracted by both chloride ions and the partially negative oxygen atom in the polar water molecule. (See the graphic on the right).

A "tug-of-war" occurs for the positive and negative ions between the other ions in the crystal and the water molecules. Whether the crystal dissolves is determined by which attractive force is stronger. If the internal ionic forces in the crystal are the strongest, the crystal does not dissolve. This is the situation in reactions where precipitates form. If the attractions for the ions by the polar water molecules are the strongest, the crystal will dissolve. This is the situation in sodium chloride.

Once the ions are released from the crystals, the ions are completely surrounded by water molecules. Note that the proper atom in the water molecule must "point" toward the correct ion. The charge principle and the partial charges in the polar molecule determine the correct orientation.

Enthalpy, Entropy, and Dissolving Solids

Enthalpy of Solution

Typically the dissolving of a solid in water will involve measurable heat either exothermic or endothermic. There is not absolute agreement on whether dissolving itself should be categorized as wholly physical, partly chemical, etc., but intermolecular forces are certainly involved (at the very least).

Regardless of the appropriate "label" for dissolving as a process there is an overall energy term. It is known as the heat of solution, DHsoln. DHsoln could be broken into few steps.

1. solute particles are separated from the solid mass (energy is absorbed, DH1)

2. solvent particles move apart to make space for dissolved solute (energy is absorbed, DH2)

3. solute and solvent particles are attracted to one another (energy is released, DH3)

For most solids dissolving in water, the sum of the first two terms is greater than the third and thus dissolving is frequently endothermic (DHsoln = +) and solubility generally increases with increasing temperature. When heats of solution become very highly positive it is often because the solute and solvent are dissimilar and, in the extreme case, immiscible. The old rule of "like dissolves like" is an approximation, but a good one.

Energy Changes and dissolution

Dissolution is either exothermic or endothermic

Energy is gained or lost during dissolution

Substance DH°soln (kJ/mol)

LiCl -37.0

NaCl 3.9

KCl 17.2

KOH -57.6

Entropy of Solution

When solids dissolve in liquids the entropy which is measure of disorder of the system nearly always increases. This owes mainly to the increased freedom of movement of the solute particles as the forces, which hold the solid together are overcome. Dispersed in the solvent, the molecules and complex ions can translate and rotate as well as vibrate. The energy states for the many possible translational and rotational motions are more closely spaced than those for the vibrational states in the solid. Dispersal of energy is what entropy is all about. So the entropy change for the first step in our dissolving "process" is positive (DS1 = +). Also occurring in the system, the solvent molecules must move about to make space for the solute. In general this results in disruptions of the solvent-solvent forces and more freedom of movement for the solvent molecules, hence additional pathways for energy dispersal. So DS2 is typically positive.

In the final step solvent and solute interactions decrease the free movement of both species to some extent and the entropy may decrease. This leaves two unknowns: what are the relative magnitudes of these three steps which occur in the system AND what about DSsurr?

The question about DSsurr takes us back to the question about the overall DHsoln (actually DHsys) since the flow of heat between the system and the surroundings will determine DSsurr. If heat moves into the surroundings then the entropy there will rise. If the heat flow is in the other direction then DSsurr will be negative. Taking urea as an example, urea certainly dissolves and therefore: .Ssys + .Ssurr >0. But if the solvent-solute interactions are strong enough, DHsys (i.e., DHsoln) might be negative. This would make DSsurr positive but would also tend to make DS3 more negative. If, on the other hand, DHsys were positive, then DSsurr would be negative and so DS1 + DS2 > |DS3 + DSsurr|. As already mentioned, it is possible to find the heat of solution in the lab but determining the entropy of solution (DSsoln) is a more complicated matter. This is true for most entropy determinations, which are typically indirect. In this experiment the thermodynamic requirements for the equilibrium condition (Kc ) can be exploited to determine DSsoln from the readily found heat of solution and the free energy change, DG.

DGsoln = DHsoln -TDSsoln

For a spontaneous process DG should be negative.

DG^osoln = -RT ln Kc

When gases dissolve in water, the hydrated molecules represent a higher degree of order than the random distribution in the gas. DS for the dissolving of a gas is, therefore, negative.

Soluble gases, then, must have a negative DH. Experiment shows that the dissolving of gases is, in fact, an exothermic process.

Solubility and Equilibrium

When crystals are first placed in a solvent, many particles may leave the surface and go into solution. As the number of solute particles in solution increases, some of the dissolved particles return to the surface of the crystal.

Eventually the number of particles leaving the crystal surface equals the number returning to the surface. This point is called solution equilibrium.

At a specific temperature, there is a limit to the amount of solute that will dissolve in a given quantity of solvent. For instance, at 20^oC, a maximum of 64.2 grams of nickel chloride will dissolve in 100 cm³ of water. This quantity is called the solubility of the substance at 20^oC.

Solubility is an equilibrium process leading either to unsaturated, saturated, or super- saturated solutions.

Unsaturated solutions: An unsaturated solution has a lower amount of solute compared to a saturated solution.

Saturated solution. A saturated solution contains the maximum amount of solute that can be dissolved in a solvent at a particular temperature.

Supersaturated solution: A supersaturated solution is a special case of saturated solutions that holds more solute than it would normally hold. Supersaturation results when a solution is cooled and the excess solute is not precipitated out. If you add a tiny crystal of a solute to a super saturated solution precipitation occurs immediately.

Explain the following terms used to describe solutions.
solvent b) solute c) solubility d) saturated solution
e) supersaturated solution f) dilute solutions e) concentrated solutions

a) Solvent: Solvent is the liquid which make up the bulk of a solution. Aqueous solutions are solid, liquid and gaseous solutes dissolved in water. It is important to realize that the definition of solution includes any homogenous phase, gas, liquid and solids. A gas dispersed in a liquid is called an aerosol. There could be solid solutions (solid-solid) of metals that are called alloys such as bronze. Bronze is a homogenous solid solution of copper and tin.

b) Solute: Solute is the other substance in a solution once the solvent is identified. Solvent has the highest concentration in the solution.

c) Solubility: This is a parameter to indicate the quantity of solute that will dissolve in a given quantity of a solvent. Solubility is expressed conveniently as grams of solute per 100 mL of solution.

d) Saturated solution. A saturated solution contains the maximum amount of solute that can be dissolved in a solvent at a particular temperature.

e) Supersaturated solution: A supersaturated solution is a special case of saturated solutions that holds more solute than it would normally hold. Supersaturation results when a solution is cooled and the excess solute is not precipitated out. I f you add a tiny crystal of a solute to a super saturated solution precipitation occurs immediately.

f) Dilute solutions: A dilute solution has a lower amount of solute compared to another solution.

e)Concentrated solutions: A concentrated solution has a higher amount of solute compared to another solution.

Solubility Equilibrium

Consider an ionic solid, MX(s), consisting of M⁺ cations and X^- anions, in equilibrium with its ions in aqueous solution:

MX(s) <===> M⁺(aq) + X^-(aq)

AgCl(s) <===> Ag⁺(aq) + Cl^-(aq)

The equilibrium constant for this process, called the solubility product constant, K_sp, is in K_sp = [M⁺][X^-] or K_sp = [Ag⁺][Cl^-]

Le Chatelier's Principle is used in predicting the effect of temperature on solubility. It has been used to estimate the solubility of a salt in water as a function of temperature. As you learn later the temperature affects solubility product equilibrium and it.

Temperature and Solubility

As you learn later solubility product equilibrium and it is related to temperature. The effect of temperature on the solubility of a solid in water may be predicted using Le Principe du Chatelier if it is known whether the process is endo- or exothermic. If the solution process is endothermic, solubility increases with increasing temperature; if exothermic, solubility decreases. Le Chatelier's Principle permits us to make only qualitative predictions of solubility for a particular solute-solvent system.

Pressure and Dissolving Gases in Liquids: Henry's Law

Pressure has little effect on solution process of solids and liquids. When the solute is a gas pressure increase the solubility. The amount of gas, which dissolves in a given amount of solvent, is greater at high pressure than it is at low pressure.

The mass of a gas, which will dissolve in a liquid at a given temperature, varies directly with the partial pressure of that gas. This statement is Henry's Law, named in honor of William Henry, the English chemist who first discovered this relationship.

S_g = k_HP_g

where S_g is the solubility

k_H is the Henry’s the Law constant

P_g is partial pressure of gas

Increasing the pressure of a gas above a liquid increases its solubility

Table: Molar Henry's Law Constants for Aqueous Solutions at 25^oC

Gas	Constant (Pa/(mol/litre))	Constant (atm/(mol/litre))
He	282.7 x 10⁺⁶	2865.
O₂	74.68 x 10⁺⁶	756.7
N₂	155. x 10⁺⁶	1600.
CO₂	2.937 x 10⁺⁶	29.76
NH₃	5.69 x 10⁺⁶	56.9

Describe the effect of the following on solubility of a solute in a solvent:

a) Temperature b) Pressure.

Consider following cases: KNO₃(solid solute)/water; mixture of N₂ and O₂(gaseous solutes)/H₂O (Henry's law).

a) Temperature

i) Solid solutes: Solubility of solid solutes in water normally increases with increasing temperature. For example, solubility of KNO₃increases with increasing temperature.

ii) Gaseous solutes: Solubility of gaseous solute normally decreases with increasing temperature. For example, the solubility of N₂ and O₂ in water decreases with temperature.

b) Pressure

i) Solid solutes: The pressure above a solution has little or no effect on the solubility of the solid solute. For example, the pressure has no effect on the solubility of KNO₃in water.

ii)Gaseous solutes: Solubility of gaseous solute normally increases with increasing temperature. Soft drinks are made by dissolving CO₂ in water under high pressure. The effect of solubility of gases in liquids are governed by the Henry's Law:

S_g = k_H p_g

p_g= partial pressure of gas above the solution in atm.

S_g= concentration of gas in the solution in mol/L.

k_H= Henry's constant which is characteristic of the gas and the solvent.

k_H is obtained from the slope of a graph where k_H is plotted against p_g.

Henry's constant, k_N2 for N₂/H₂O solution is greater than k_O2 for O₂/H₂O solution. Therefore, at higher pressures more N₂ dissolves in water (blood) than. This solubility difference at high pressures is the cause of "divers bend" found among deep water divers.

Solution Concentration: Keeping Track of Units

Describe and define the following terms used for solutions:

a) Solvent

b) Solute

c) Molarity (M)

d) Molality (m)

e) Mole fraction (C_a)

f) Mass percent (% weight)

g) Volume percent (% volume)

h) "Proof"

i) ppm and ppb

a) Solvent: Solvent is the liquid which make up the bulk of a solution. Aqueous solutions are solutions of solids, liquids and gaseous solutes in water. It is important to realize that the definition of solution includes any homogenous phase including gases, liquids and solids. A gas dispersed in a liquid is called an aerosol. There could be solid solutions (solid-solid), which are called alloys. There could be solid solutions such as bronze, which is a homogenous solid solution of copper and tin.

b) Solute: Solute is other substance in the solution once the solvent is identified.

c) Molarity(M)

moles of solute

Molarity (M) = ------------------

Liters of solution

d) Molality (m)

moles of solute

Molaity (m) = ------------------------

kg of solvent

e) Mole fraction (C_a)

moles of solute

Mole fraction (C_a) = ------------------------

moles of solute + solvent

f) Mass (w/w) % or Weight %

Mass of solute

Mass (w/w) % = ------------------------ x 100

Mass of solution

g) Volume (v/v) % or Volume %

Volume of solute

Volume (v/v) % = ------------------------ x 100

Volume of solution

h) Proof = Volume % x 2

i) ppm or ppb (w/w or v/v)

Mass (volume) of solute

ppm (w/w or v/v) = ---------------------------- x 10⁶

Mass (volume) of solution

Mass (volume) of solute

ppb (w/w or v/v) = ------------------------------ x 10⁹

Mass (volume) of solution

A solution containing 58.5 g of NaCl in 2206 g of water has a density of 1.108 g/cm³. Calculate the Molarity of the solution.

moles of solute

Molarity(M) = ------------------------

Liters of solution

solute = NaCl; m.w = 58.44 g/mole

moles of NaCl = ?

58.5 ~~g NaCl~~	1 mole NaCl	= 1.00 mole NaCl
	58.44 ~~g NaCl~~	= 1.00 mole NaCl

To get liters of solution:

Total mass of solution = 58.5 g + 2206 g = 2264.5 g

use g/cm³ (density) to convert g -- cm³ and then 1L/1000cm³ to cm³ ---- L

2264.5 ~~g solution~~	1000 ~~cm³ solution~~	1 L solution	= 2.044 L solution
	1.108 ~~g solution~~	1000 ~~cm³ solution~~	= 2.044 L solution

1.00 mole NaCl

Molarity of NaCl solution = ------------------------- = 0.489 M

2.044 L solution

Calculate the molality of C₂H₅OH in water solution which is prepared by mixing 75.0 mL of C₂H₅OH and 125 g of H₂O at 20^oC. The density of C₂H₅OH is 0.789 g/mL.

moles of solute Moles ofC₂H₅OH

Molality(m) = ----------------------- = ------------------------------

kg of solvent kg of solvent

m.w. (C₂H₅OH) = 46.08 g/mole d(C₂H₅OH) = 0.789 g/mL

To get moles of C₂H₅OH

75.0 ~~mL C₂H₅OH~~	0.789 ~~g C₂H₅OH~~	1 kg H₂O	= 1.284 mole C₂H₅OH
	1 ~~mL C₂H₅OH~~	46.08 ~~g C₂H₅OH~~	= 1.284 mole C₂H₅OH

To get kg of H₂O

125 ~~g H₂O~~	1 kg H₂O	= 0.125 kg H₂O
	1000 ~~g H₂O~~	= 0.125 kg H₂O

1.284 mole C₂H₅OH

Molality(m) = ------------------------ = 10.27 m

0.125 kg H₂O

Calculate the molarity of a solution of water/alcohol containing 35% C₂H₅OH by weight. The density of this solution is 1.10 g/mL.

moles of solute Moles ofC₂H₅OH

Molarity(M) = --------------------- = ---------------------

L of solution volume(L) of the solution

m.w (C₂H₅OH) = 46.08 g/mole d(C₂H₅OH) = 1.10 g/mL
To get moles of C₂H₅OH: ( 35% C₂H₅OH solution has 35g C₂H₅OH + 65g H₂O)

35 ~~g C₂H₅OH~~	1 mole C₂H₅OH	= 0.7595 mol C₂H₅OH
	46.08 ~~g C₂H₅OH~~	= 0.7595 mol C₂H₅OH

To get liters of solution:

100 ~~g solution~~	1 ~~mL solution~~	1 L solution	= 0.0901 L
	1.10 ~~g solution~~	1000 ~~mL solution~~

0.7595 mole C₂H₅OH

Molarity (M) = ------------------------------- = 8.36 M

0.0901 L

Calculate the mole fraction of benzene in a benzene(C₆H₆)-chloroform(CHCl₃) solution which contains 60 g of benzene and 30 g of chloroform.

For a mixture containing

moles(n) of a n_a

Mola Fraction(c_a)= --------------------------- = --------------

moles(n) of a + moles(n) of b n_a + n_b

a = C₆H₆

b = CHCl₃

n_C6H6

Mola Fraction(c_a) = ------------------

n_C6H6 + n_CHCl3

m.w (C₆H₆) = 78.12 g/mole m.w (CHCl₃) = 119.37 g/mole

To get moles of C₆H₆:

60 ~~g C₆H₆~~	1 mole C₆H₆	= 0.768 mole C₆H₆
	78.12 ~~g C₆H₆~~	= 0.768 mole C₆H₆

To get moles of CHCl₃:

30 ~~g CHCl₃~~	1 mole CHCl₃	= 0.251 mole CHCl₃
	119.37 ~~g CHCl₃~~	= 0.251 mole CHCl₃

c_{C6H6 =}	0.768	0.768	= 0.754
	0.768 + 0.251	1.019

c_{CHCl3 =}	0.251	0.251	= 0.246
	0.251 + 0.768	1.019

c_C6H6= 0.754

c_CHCl3= 0.246

Note: c_C6H6+ c_CHCl3= 1

0.754 + 0.246 = 1

Vapor Pressures, Boiling Points, and Freezing Points of Solutions

Osmotic Pressure of Solutions

What is the effect of concentration of solute (non-volatile) on the following properties of solvents:

a) Vapor pressure (Raoult's Law); ( P_solution= C_solvent P^o_solvent)

b) Boiling point ( DT_b = K_b m_solute)

c) Freezing point ( DT_f = K_fm_solute)

d) Osmotic pressure ( P = MRT )

Vapor pressure, boiling point, freezing point and osmotic pressure are called colligative properties since they are related to number of solute particles in a solution. Number of non-dissociating solute particles in a solution is expressed as c( mole fraction), m(molality), or M(molarity).

a) Vapor pressure (Raoult's Law); P_solution = C_solventP^o_solvent)

The vapor pressure of a liquid is decreased or lowered by dissolving a solid solute in water. The vapor pressure a solution is given by the equation

Vapor pressure of a solution is governed by Raoult's law.

P_solution = c_{solvent P}^o_solvent

Note: c_solvent= (1-c_solute) making an equivalent form to the equation:

P_solution = (1-c_solute)P^o_solvent

Psolution = vapor pressure over the solution due to solvent molecules

c_solvent = mole fraction solvent in the solution

P^o_solvent =vapor pressure of the pure solvent

vapor pressure over a solution increases with the increasing c_solvent.

b) Freezing point

The freezing (melting or fusion) point of a liquid is decreased by dissolving a solid solute in water. The boiling point decrease or the depression of a solution is given by the equation:

DT_f = K_f m_solute

DT_f = boiling point elevation.

K_f = molal freezing point depression constant-molal means concentration is in given as molality(m).

m_solute= concentration of solute expressed as molality(m).

c) Boiling point

The boiling point of a liquid is increased by dissolving a solid solute in water. The boiling point increase or the elevation of a solution is given by the equation:

DT_b = K_{b msolute}

DT_b = boiling point elevation.

K_b = molal boiling point elevation constant-molal means concentration is in given as molality(m).

m_solute= concentration of solute expressed as molality(m).

d) Osmotic pressure ( P = MRT )

Osmotic pressure is a colligative property which depends on the number and size of solute particles. For example, two solutions, water and salt water, separated by a semipermeable membrane, a membrane that allows only water molecles to pass through, the pressure exerted by the pure water to push more water into salt solution is called the osmatic pressure. Osmotic pressure of a solution can be calculated using the equation:

P = MRT

P = Osmatic pressure of the solution

M= Molarity of the solute in the solution

R = Ideal gas constant (R= 0.08026 or 62.4)

T= Temperature of the solution in Kelvin

To calculate the colligative properties of a solutions containing ionic compounds or salts that can dissocite into ions, for example-NaCl --- Na⁺ and Cl^-, these equations,
P_solution = (1-c_solute)P^o_solvent , DT = K_b m_solute, DT = K_f m_solute, P = MRT ,
should be multiplied by a factor called Vant Hoff factor (i) to include the effect of extra particles or more ions generated through the complete dissociation of NaCl. The Vant Hoff factor or NaCl is equal to 2 (NaCl ---1 Na⁺ + 1Cl^-). The Vant Hoff factor or CaCl₂ is equal to 3(CaCl₂ --- 1 Ca²⁺ + 2 Cl^-). For weak electrolytes which dissociate in completely the Vant Hoff factor is always less than the what is expected from complete dissociation.

The forms of the colligative property equations for dissociating solutes:

P_solution = (1-i c_solute) P^o_solvent

DT = i K_b m_solute,

DT = i K_f m_solute,

P = i MRT

i = Vant Hoff Factor

moles of ions in the solution

i = --------------------------------

moles of solute dissolved

The vapor pressure above a glucose-water solution at 25^oC is 23.8 torr. What is the mole fraction of glucose (non-dissociating solute) in the solution. The vapor pressure of water at 25^oC is 30.5 torr.

This is a problem to use Raoult's law [P_solution = (1- c_solute)P^o_solvent] and calculate the mole fraction of solute, glucose (m_solute= ?) given the vapor pressure of the solution (P_solution= 23.8 torr) and the pure solvent(P^o_solvent= 30.5 torr).

P_solution= (1- c_solute)P^o_solvent

23.8 torr = (1- c_solute) 30.5 torr

(1-c_solute) = 23.5/30.5

(1- c_solute) = 0.7803

_-c_solute = 0.7803 -1

_-c_solute= -0.2196

c_solute= 0.2196

Mole fraction of glucose = 0.2196

A 2.25g sample of a compound is dissolved in 125 g of benzene. The freezing point of the solution is 1.02^oC. What is the molecular weight of the compound.
K_f for benzene = 5.12 ^oC/m, freezing point = 5.5^oC.

This is a problem using the equation, DT_f =K_f m_solute to calculate the molality of a solute (m_solute= ?) given the freezing point (1.02^oC) of a solution, freezing point of the pure solvent (5.5^oC) and K_f (5.12 ^oC/m), molal freezing point depression constant.

First we have to calculate DT_f:

Note that DT is expressed as an absolute value: DT =5.5 -1.02 = 4.48

DT_f = K_f m_solute

4.48 = 5.12 x m_solute

m_solute = 4.48/5.12

m_solute = 0.875 mol/kg

Once the mole fraction is calculated, moles of solute can be can be calculated because

moles of compound Moles ofcompound

Molality(m) = -------------------------- = -----------------------

kg of solvent kg of H₂O

Moles of compound = Molality(m) x 1 kg of H₂O

Molality = 0.875 mol/kg of H₂O

Moles of compound = 0.875 moles

Grams of the compound in 1 kg water: 125 g of water has 2.25 g of the compound.
Therefore, grams of the compound in 1 kg water = (2.25/125) x 1000 = 18 g

Grams of the compound in 1 kg water = 18 g

Once the number of moles and the grams of the of the compound in 1kg of water is calculated, molecular weight of the compound can be can be calculated from the because

grams of the compound in 1 kg water 18g

Molecular weight(grams/mole) = ----------------------------------------------- = --------------

moles of the compound in 1kg water 0.875 moles

Molecular weight of the compound =20.57 g/mole

Calculate the osmotic pressure in torr of a 0.500 M solution of NaCl in water at 25^oC. Assume a 100% dissociation of NaCl.

To calculate the osmotic pressure of a solution containing salts that can be dissociated into ions, for example: NaCl --- Na⁺ and Cl^-, these equations, P = MRT , should be multiplied by a factor called Vant Hoff factor (i) to include the effect of the more particles or more ions generated through the complete dissociation of NaCl. The Vant Hoff factor or NaCl is equal to 2 (NaCl ---1 Na⁺ + 1Cl^-), 1 + 1 = 2.

P = i MRT

P = Osmatic pressure of the solution = ?

M= Molarity of the solute in the solution = 0.500 M

R = Ideal gas constant = 62.4 L-torr/mol K

T= Temperature of the solution in Kelvin = 25^oC +273.15 = 298.15 K

i = (NaCl ---1 Na⁺ + 1Cl^-), 1 + 1 = 2

P = 2 x 0.500 M x 62.4 L-torr/mol K x 298.15 K

P = 18605 torr

The osmotic pressure of the solution = 18605 torr

Predict the type of behavior (ideal, negative, positive) based on vapor pressure of the following pairs of volatile liquids and explain it in terms of intermolecular attractions:
a) Acetone((CH₃)₂CO)/water (H₂O) ; b) Ethanol(C₂H₅OH)/hexane (C₆H₁₄);
c) Benzene (C₆H₆)/toluene(CH₃C₆H₅).

a) Acetone((CH₃)₂CO)/water (H₂O).
Acetone and water will form strong intermolecular forces of attraction because of their capacity to form hydrogen bonds. This solution, therefore, shows negative deviation from the Raoults law.

Negative deviation: Lower vapor pressure than the sum of the individual vapor pressures.

b) Ethanol (C₂H₅OH)/hexane (C₆H₁₄);
Ethanol and hexane molecules have different bonding. Ethanol have polar O-H bond and the hexane have non polar covalent C-H bonds. Because of the unlike nature of molecules, they repel each other and lead to an increase in the vapor pressure of the solution called positive deviation from the Raoults law.
Positive deviation: Higher vapor pressure than the sum of the individual vapor pressures.

c) Benzene (C₆H₆)/toluene CH₃C₆H₅.
Benzene and toluene are two similar molecules both having non polar covalent C-H bonds. Because of their like nature, molecules do not repel or attract each other and the vapor pressure of the solution is the sum of the vapor pressures of the two individual components. The no change in the pressure of the solution is called ideal behavior according to the Raoults law

In the diagram below shows the indicated plot of vapor pressure versus mole fraction of one component of a solution made from two volatile liquids. Categorize the behavior of this solution as based on Raoult's law as ideal, positive deviation or negative deviation.

a) shows ideal behavior (b) shows positive behavior (c) shows negative behavior

Define the Van't Hoff factor (i). Which of the following solutions will show the highest osmotic pressure:
a) 0.2 M Na₃PO₄ b) 0.2 M C₆H₁₂O₆ (glucose) c) 0.3 M Al₂(SO₄)₃ d) 0.3 M CaCl₂
e) 0.3 M NaCl

Vant Hoff factor ( i ) is defined as:
                  moles of ions in the solution
           i = -------------------------
                    moles of solute dissolved

To calculate the osmotic pressure of a solution containing salts that can be dissociated into ions, for example: NaCl --- Na⁺ and Cl^-, the equation,
P = MRT ,
should be multiplied by a factor called Vant Hoff factor ( i ) to include the effect of the more particles or more ions generated through the complete dissociation of NaCl. The Vant Hoff factor or NaCl is equal to 2NaCl ---1 Na⁺ + 1Cl^-). The Vant Hoff factor or CaCl2 is equal to 3(CaCl₂ --- 1 Ca²⁺ + 2 Cl^-). For weak electrolytes which dissociate in completely the Vant Hoff factor is always than the one expected from complete dissociation.
The forms of the colligative property equations for dissociating solutes:

P= i MRT
i = Vant Hoff Factor

a) 0.2 M Na₃PO₄

P= i MRT

i = Vant Hoff Factor = 4; (Na₃PO₄ dissociates in water into three Na⁺ and one PO₄^3- ions giving
total of 4 particles).
M = 0.2; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K
P= 4 x 0.2 x 62.4 x 298.15 = 14,884 torr

b) 0.2 M C₆H₁₂O₆ (glucose)

P = i MRT

i = Vant Hoff Factor = 1; (C₆H₁₂O₆does not break up in the solution. It's a covalent compound)
M = 0.2; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K
P = 1 x 0.2 x 62.4 x 298.15 = 3,721 torr

c) 0.3 M Al₂(SO₄)₃

P = i MRT

i = Vant Hoff Factor = 5 ; Al₂(SO₄)₃ dissociates in water into two Al³⁺and three SO₄^2- ions giving
total of 5 particles.
M = 0.3; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K

P = 5 x 0.3 x 62.4 x 298.15 = 27,907 torr

0.3 M Al₂(SO₄)₃ shows the highest osmotic pressure.

d) 0.3 M CaCl₂

P = i MRT

i = Vant Hoff Factor = 3 ; CaCl₂ dissociates in water into one Ca²⁺ and two Cl ^- ions giving
total of 3 particles
M = 0.2; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K
P = 3 x 0.3 x 62.4 x 298.15 = 16,744 torr

e) 0.3 M NaCl

P = i MRT
i = Vant Hoff Factor = 2 ; NaCl dissociates in water into one Na⁺and one Cl ^-ions giving
total of 2 particles
M = 0.2; R = 62.4 L torr/mol K; T = 273.15 + 25.0 = 298.15 K
P = 3 x 0.3 x 62.4 x 298.15 = 16,744 torr

Colloids and Suspensions

Describe the following terms:

a) True solutions b) Colloids (Tyndall effect) c) Suspensions.

a) True solutions
Normally light passes through true solutions or they transparent to visible light. Because of small solute particle sizes they do not scatter light.. True solutions have solute particles with diameters less than 1 x10³ pm. Colloids, on the other hand have solute particles with diameters in the range 1 x10³ to 1 x10⁵ pm. Suspensions have larger particle diameters which are greater than 1 x10⁵ pm.

b) Colloids (Tyndall effect)
Colloids have solute particle diameters in the rage 1 x10³ to 1 x10⁵ pm. Colloids scatter light and the solution become opaque and relative degree of opacity depends on the sizes and amount of the particles. The scattering of light by colloidal particles is called Tyndell effect which has been used to distinguish between true solutions and colloids. This effect will increase with increasing solute particle diameter of a colloid. For example, milk is opaue because of the higher diameters of the solute particles such as proteins in the milk.

c) Suspensions.
Suspensions have particle diameters greater than 1 x10⁵ pm. The particles in a suspension are lage enough to be affected by gravity and settle to the bottom of the mixture with time. For example, muddy water.

Surfactants

Surfactants constitute the most important group of detergent components. Generally, these are water-soluble surface-active agents comprised of a hydrophobic portion, usually a long alkyl chain, attached to hydrophilic or water solubility enhancing functional groups. The hydrophilic end, which is either polar or ionic, dissolves readily in water. The hydrophobic, or non-polar, end, however, does not dissolve in water. In fact, the hydrophobic, or "water-hating" end will move as far away from water as possible. This phenomenon is called the hydrophobic effect. Because one end of a surfactant resists water and the other end embraces it, surfactants have very unique characteristics.

Emulsifiers

A surfactant is also called an emulsifier. Emulsifiers do help oil and water remain in stable emulsions. Aggregates of oil and emulsifiers form micelles and stay dispersed in water solution. Examples of emulsifiers include sodium dodecyl sulfate.

Detsu9 Detsu8 All surfactants possess the common property of lowering surface tension when added to water in small amounts, as illustrated in the plot below. The characteristic discontinuity in the plots of surface tension against surfactant concentration can be experimentally determined. The corresponding surfactant concentration at this discontinuity corresponds to the critical micelle concentration (CMC). At surfactant concentrations below the CMC, the surfactant molecules are loosely integrated into the water structure (monomer see figure below). In the region of the CMC, the surfactant-water structure is changed in such a way that the surfactant molecules begin to build up their own structures (micelles in the interior and monolayers at the surface).

When a small concentration of surfactant is added to water, the hydrophobic end will immediately rise to the surface. The surfactant is then stable. The polar end is happily immersed in the polar solvent while the non-polar end rises above the surface. This configuration is called a monolayer. If more surfactant is added and there is not enough room for all of the hydrophobic ends to stick out of the water, a bilayer will form. However, if the concentration of surfactant is large enough, even the bilayer configuration will not be stable. At this point, a micelle forms, in which a group of hydrophilic ends surround their hydrophobic "tails" and shield them from the water. The micelle, therefore, consists of a hydrophilic shell and a hydrophobic core.

A very important effect of surfactants in cleaning products is the wetting effect. Because of the reduced surface tension, the water can be more evenly distributed over the surface and this improves the cleaning process. The emulsifying effect of surfactants is important for both cleansing and washing of textiles. Due to the hydrophobic and hydrophilic parts, surfactants can sorb to non-polar and polar materials at the same time. During cleansing and washing, the non-polar materials are kept in emulsions in the aqueous solution and removed by rinsing. By varying the hydrophobic and hydrophilic part of a surfactant, a number of properties may be adjusted, e.g. wetting effect, emulsifying effect, dispersive effect, foaming ability and foaming control.

Soap the most common surfactant

One of the organic chemical reactions known to ancient man was the preparation of soaps through a reaction called saponification. Natural soaps are sodium or potassium salts of fatty acids, originally made by boiling lard or other animal fat together with lye or potash (potassium hydroxide). Hydrolysis of the fats and oils occurs, yielding glycerol and crude soap.

In the industrial manufacture of soap, tallow (fat from animals such as cattle and sheep) or vegetable fat is heated with sodium hydroxide. Once the saponification reaction is complete, sodium chloride is added to precipitate the soap. The water layer is drawn off the top of the mixture and the glycerol is recovered using vacuum distillation.

The crude soap obtained from the saponification reaction contains sodium chloride, sodium hydroxide, and glycerol. These impurities are removed by boiling the crude soap curds in water and re-precipitating the soap with salt. After the purification process is repeated several times, the soap may be used as an inexpensive industrial cleanser. Sand or pumice may be added to produce a scouring soap. Other treatments may result in laundry, cosmetic, liquid, and other soaps. Soap is an anionic surfactant.

Detergent

The term "detergent" is used for a surfactant material which cleans (or is used for cleaning), but in this definition soap is not included. Even so, this is still a wide definition, because, of course, it can refer to the active ingredient, or the solid, liquid, paste or powder compounded from this active matter.

Surfactants are grouped according to their ionic properties in water:

Anionic surfactants have a negative charge
Nonionic surfactants have no charge
Cationic surfactants have a positive charge
Amphoteric surfactants have positive or negative charge dependent on pH A list of surfactants which are commonly used in biochemistry is presented in the table below.

Structures of common surfactants:

	Chemical or Trade Name Sodium dodecylsulfate (SDS)
	Sodium cholate
	Sodium deoxycholate (DOC)
	N-Lauroylsarcosine Sodium salt
	Lauryldimethylamine-oxide (LDAO

	Cetyltrimethylammoniumbromide (CTAB)
	Bis(2-ethylhexyl)sulfosuccinate Sodium salt

Water: Natural, Clean, and Otherwise

Fresh Water

Fresh water is an essential ingredient of modern life. Thought it’s often available as the result of natural processes (natural water), there are times when it must be extracted from impure water, typically salt water. In some middle eastern countries where rain water is scarce, desalinated sea water through reverse osmosis is the main source of drinking water.

Water Purification

Any extraction process that purifies water must separate water molecules from contaminating liquids, solids, or gases.

Desalination: Removing Salt from Sea water

Distillation

One way to purify water is by distillation. Distillation is a general technique for

separating various chemicals from one another. The chemicals are heated to form

a vapor and that vapor is condensed to form a new mixture of chemicals. Because

the various chemicals have different tendencies to form vapors at a particular

temperature, the newly formed mixture has a different balance of the chemicals

from the original mixture. In some cases, the condensed liquid contains primarily

a single chemical—all of the other chemicals are left behind in the original liquid.

Freezing

Freezing salt water to form pure ice works best in cold climates where low temperatures are available directly. Active refrigeration can also freeze salt water

to obtain fresh water, but it’s expensive. Because of water’s latent heat of melting,

you must remove a large amount of heat from salt water to freeze it. Although

refrigerated water desalination plants have been built, they have proven to be

less economical than distillation plants.

Reverse Osmosis

Another way to desalinate water is by reverse osmosis, a process resembling filtration

except that it takes place at the molecular scale. In effect, salt water is converted

to fresh water by filtering the impurities out of it with an incredibly fine filter. This filter is a semipermeable membrane. It is a surface that only allows certain molecules to pass through it.

However, because it operates at such a tiny size scale, reverse osmosis encounters

some peculiar pressure effects that you don’t see with larger filters where gravity or suction is the main force. The way to stop osmosis from diluting concentrated salt water is to increase the pressure of the salt water side. Increasing that pressure tends to drive water molecules back out of the salt water and through the semipermeable membrane to pure water side. The two fluids will still reach a phase equilibrium, but the salt water will retain a higher concentration of immobile molecules than the fresh water. Maintaining a large difference in concentrations between the two fluids require a huge pressure difference of many tens or hundreds of atmospheres.

Municipal or Utility Water Treatment

Most municipal water found in a city or community today has been treated extensively. Specific water treatment methods and steps taken by municipalities to meet local, state, national, or international standards vary but are categorized below.

Coarse filtration

A coarse screen, usually 50 to 100 mesh (305 to 140 microns), at the intake point of a surface water supply, removes large particulate matter to protect downstream equipment from clogging, fouling, or being damaged.

Addition of alum or lime

Addition of alum or lime produces aggregate suspended solids (microbes, organic matter, toxic contaminants) which is then removed by sedimentation in a settling basin. The sedimentation process is aided by the application of alum (aluminum sulfate) to the raw lake water this material when added to water containing normal debris such a mud, particles and color causes these materials to clump and settle to the bottom of sedimentation tanks where it removed. The clear water at the very top of these units is skimmed off leaving the alum and debris behind.

Settling or Clarification
Settling or clarification is generally a multi-step process to reduce turbidity and suspended matter. Steps include the addition of chemical coagulants or pH-adjustment chemicals that react to form floc. The floc settles by gravity in settling tanks or is removed as the water percolates through a gravity filter. The clarification process effectively removes particles larger than 25 microns. The clarification process is not 100% efficient; therefore, water treated through clarification may still contain some suspended materials.

Sand filtration

Sand filtration is a "physical and sometimes a chemical process for separating suspended and colloidal impurities from water by passage through a bed of granular sand material. The purpose of this filtration is to remove any particulate matter left over after flocculation and settling. Water fills the pores of the filtering sand medium, and the impurities are adsorbed on the surface of the grains or trapped in the openings." The key to this process is the relative grain size of the filter medium.

Aeration

Aeration consists of breaking the incoming water into small droplets (spray) into the air, drawing fresh air through that spray, collecting the water into a storage tank. The air drawn though the spray must be vented outside the house -- remember, it can have odors, toxic substances. Although this system necessitates another pump to repressurize your supply, you are not adding any chemicals to your water, which makes it attractive. This system is low maintenance and no chemicals to purchase. Initial cost may be higher, and space requirements may be greater.

Chlorination or Ozonation: Disinfection

Disinfection is one of the most important steps to municipal water treatment. Usually, chlorine gas is fed into the supply after the water has been clarified and/or softened. The chlorine kills bacteria. In order to maintain the "kill potential", an excess of chlorine is fed into the supply to maintain a residual. The chlorine level must be constantly monitored to assure that no harmful levels of chloramines or chlorinated hydrocarbons develop.

Ozonation
A chemical-free disinfection is known as ozonation. Ozonation relies on oxygen to ensure that our purified water remains free of any possible microbiological contamination. The ozonation process involves taking basic molecular oxygen (O₂) and passing this oxygen through a special chamber in which it is exposed to a high voltage electrical charge. (This type of ozone generation is called cold-plasma discharge.) The electricity causes the oxygen molecule to split and recombine in a higher-energy form known as ozone (O₃). This ozone is then continuously circulated through the purified water. Ozone is a very powerful disinfectant and is capable of oxidizing a very broad range of contaminants. In fact, ozone is highly effective against many types of impurities and organisms, such as cryptosporidium, that are utterly impervious to chlorination.

Other Chemical Treatment

pH adjustment. Certain chemicals, membranes, ion exchange resins and other materials are sensitive to specific pH conditions. An example of this is to prevent acid corrosion in boiler feed water by adjusting the pH so it is between 8.3 to 9.0.
Dispersants. Dispersants are added when scaling may be expected due to concentration of specific ions in the stream. Dispersants disrupt the scale formation, preventing growth of precipitate crystals.
Sequestering (chelating) agents. Sequestering agents are used to prevent the negative effects of hardness, preventing the deposition of Ca, Mg, Fe, Mn and Al.
Oxidizing agents. Oxidizing agents have two distinct functions: as a biocide, or to neutralize reducing agents.
Ion Exchangers: Strong base anion resin can be used for removing anionic metal complexes from acidic waters like ZnCl₄^2- in spent pickle solution or Cr(VI) in rinse water after chromating (Tan). Weakly acidic exchangers (-COO^-) have shown good separation performance for Zn in plating waste (Uy) and for nickel (Halle), however weakly acidic resins have no widespread applications in the plating industry.
Potassium permanganate. Potassium permanganate (KMnO₄) is a strong oxidizing agent used in many bleaching applications. It will oxidize most organic compounds and is often used to oxidize ferrous iron to ferric for precipitation and filtration.
Reducing agents. Reducing agents, like sodium metabisulfite (Na₂S₂O₅), are added to neutralize oxidizing agents such as chlorine or ozone. In membrane and ion exchange systems, they prevent the degradation of certain membranes or resins, which are sensitive to oxidizing agents.

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Pressure has little effect on solution process of solids and liquids. When the solute is a gas pressure increase the solubility. The amount of gas, which dissolves in a given amount of solvent, is greater at high pressure than it is at low pressure.

Table: Molar Henry's Law Constants for Aqueous Solutions at 25oC

Gas

Constant (Pa/(mol/litre))

Constant (atm/(mol/litre))

He

282.7 x 10+6

2865.

74.68 x 10+6

756.7

N2

155. x 10+6

1600.

CO2

2.937 x 10+6

29.76

NH3

5.69 x 10+6

56.9

Table: Molar Henry's Law Constants for Aqueous Solutions at 25^oC

282.7 x 10⁺⁶

74.68 x 10⁺⁶

N₂

155. x 10⁺⁶

CO₂

2.937 x 10⁺⁶

NH₃

5.69 x 10⁺⁶