Solution Formation A solution is a homogeneous mixture of two or more substances, consisting of ions or molecules. A colloid, although it also appears.

Solution Formation A solution is a homogeneous mixture of two or more substances, consisting of ions or molecules. A colloid, although it also appears to be homogeneous, consists of comparatively large particles of a substance dispersed throughout another substance. In this chapter, we will examine the properties of each of these systems. 2

A solution is a homogenous mixture of 2 or more substances
The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount

Types of Solutions Solutions may exist as gases, liquids, or solids.
The solute is the dissolved substance. In the case of a solution of a gas or solid in a liquid, it is the gas or solid. Otherwise, it is the component of lesser amount. The solvent is the dissolving medium. Generally it is the component of greater amount. 2

Gaseous Solutions Nonreactive gases can mix in all proportions to give a gaseous solution. Fluids that dissolve in each other in all proportions are said to be miscible fluids. If two fluids do not mix, they are said to be immiscible. For example, air is a solution of oxygen, nitrogen, and smaller amounts of other gases. 2

Liquid Solutions Liquid solutions are the most common types of solutions found in the chemistry lab. Many inorganic compounds are soluble in water or other suitable solvents. Rates of chemical reactions increase when the likelihood of molecular collisions increases. This increase in molecular collisions is enhanced when molecules move freely in solution. 2

Solid Solutions Solid solutions of metals are referred to as alloys.
Brass is an alloy composed of copper and zinc. Bronze is an alloy of copper and tin. Pewter is an alloy of zinc and tin. 2

Ways of Expressing Concentration
Concentration expressions are a ratio of the amount of solute to the amount of solvent or solution. The quantity of solute, solvent, or solution can be expressed in volumes or in molar or mass amounts. Thus, there are several ways to express the concentration of a solution. 2

Molarity The molarity of a solution is the moles of solute in a liter of solution. For example, 0.20 mol of ethylene glycol dissolved in enough water to give 2.0 L of solution has a molarity of 2

EXERCISE! Consider separate solutions of NaOH and KCl made by dissolving g of each solute in mL of solution. Calculate the concentration of each solution in units of molarity. 10.0 M NaOH 5.37 M KCl [100.0 g NaOH / g/mol] / [250.0 / 1000] = 10.0 M NaOH [100.0 g KCl / g/mol] / [250.0 / 1000] = 5.37 M KCl Copyright © Cengage Learning. All rights reserved

Mass Percentage of Solute
The mass percentage of solute is defined as: For example, a 3.5% sodium chloride solution contains 3.5 grams NaCl in grams of solution. 2

EXERCISE! What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water? 6.6% [5.5 g / (5.5 g g)] × 100 = 6.6% Copyright © Cengage Learning. All rights reserved

Mole Fraction The mole fraction of a component “A” (A) in a solution is defined as the moles of the component substance divided by the total moles of solution (that is, moles of solute and solvent). For example, 1 mol ethylene glycol in 9 mol water gives a mole fraction for the ethylene glycol of 1/10 = 0.10. 2

A Problem to Consider An aqueous solution is m glucose. What are the mole fractions of each of the components? A m solution contains mol of glucose in 1.00 kg of water. After converting the 1.00 kg H2O into moles, we can calculate the mole fractions. 2

A Problem to Consider An aqueous solution is m glucose. What are the mole fractions of each of the components? 2

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.) 0.0145 8.00 g H3PO4 × (1 mol / g H3PO4) = mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 mol / g H2O) = 5.55 mol H2O Mole Fraction (H3PO4) = mol H3PO4 / [ mol H3PO mol H2O] = Copyright © Cengage Learning. All rights reserved

Molality The molality of a solution is the moles of solute per kilogram of solvent. For example, 0.20 mol of ethylene glycol dissolved in 2.0 x 103 g (= 2.0 kg) of water has a molality of 2

A Problem to Consider What is the molality of a solution containing 5.67 g of glucose, C6H12O6, dissolved in 25.2 g of water? First, convert the mass of glucose to moles. Then, divide it by the kilograms of solvent (water). 2

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.) 0.816 m 8.00 g H3PO4 × (1 mol / g H3PO4) = mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 kg / 1000 g) = kg H2O Molality = mol H3PO4 / kg H2O] = m Copyright © Cengage Learning. All rights reserved

Formation of a Liquid Solution
Separating the solute into its individual components (expanding the solute). Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). Allowing the solute and solvent to interact to form the solution. Copyright © Cengage Learning. All rights reserved

Steps in the Dissolving Process
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Steps in the Dissolving Process
Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent. Step 3 usually releases energy. Steps 1 and 2 are endothermic, and step 3 is often exothermic. Copyright © Cengage Learning. All rights reserved

Three types of interactions in the solution process:
solvent-solvent interaction solute-solute interaction solvent-solute interaction DHsoln = DH1 + DH2 + DH3

Enthalpy (Heat) of Solution
Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔHsoln = ΔH1 + ΔH2 + ΔH3 ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released). Copyright © Cengage Learning. All rights reserved

Enthalpy (Heat) of Solution

CONCEPT CHECK! Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role. Oil is a mixture of nonpolar molecules that interact through London dispersion forces, which depend on molecule size. ΔH1 will be relatively large for the large oil molecules. The term ΔH3 will be small, since interactions between the nonpolar solute molecules and the polar water molecules will be negligible. However, ΔH2 will be large and positive because it takes considerable energy to overcome the hydrogen bonding forces among the water molecules to expand the solvent. Thus ΔHsoln will be large and positive because of the ΔH1 and ΔH2 terms. Since a large amount of energy would have to be expended to form an oil-water solution, this process does not occur to any appreciable extent. Copyright © Cengage Learning. All rights reserved

The Energy Terms for Various Types of Solutes and Solvents
ΔHsoln Outcome Polar solute, polar solvent Large Large, negative Small Solution forms Nonpolar solute, polar solvent Large, positive No solution forms Nonpolar solute, nonpolar solvent Polar solute, nonpolar solvent Copyright © Cengage Learning. All rights reserved

Temperature Change Heat can be evolved or absorbed when an ionic compound dissolves in water. This heat of solution can be quite noticeable. When NaOH dissolves in water, it gets very warm (the solution process is exothermic). On the other hand, when ammonium nitrate dissolves in water, it becomes very cold (the solution process is endothermic). 2

Solubility and the Solution Process
The amount of a substance that will dissolve in a solvent is referred to as its solubility. Many factors affect solubility, such as temperature and, in some cases, pressure. There is a limit as to how much of a given solute will dissolve at a given temperature. A saturated solution is one holding as much solute as is allowed at a stated temperature. 2

A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate.

“like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. non-polar molecules are soluble in non-polar solvents CCl4 in C6H6 polar molecules are soluble in polar solvents C2H5OH in H2O ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (l)

Molecular Solutions H O
Polar molecules interact well with polar solvents such as water. The dipole-dipole interactions of water with a polar solvent can be easily explained as electrostatic attraction. d+ d- polar solute H O 2

Molecular Solutions Nonpolar solutes interact with nonpolar solvents primarily due to London forces. Heptane, C7H16, and octane, C8H18, are both nonpolar components of gasoline and are completely miscible liquids. However, for water to mix with gasoline, hydrogen bonds must be broken and replaced with weaker London forces between water and the gasoline. Therefore gasoline and water are nearly immiscible. 2

Ionic Solutions Polar solvents, such as water, also interact well with ionic solutes. Since ionic compounds are the extreme in polarity, we can illustrate the electrostatic attractions of water for cations and anions. + - H O d- d+ 2

Effects of Temperature and Pressure on Solubility
The solubility of solutes is very temperature dependent. For gases dissolved in liquids, as temperature increases, solubility decreases. On the other hand, for most solids dissolved in liquids, solubility increases as temperature increases. 2

Summeary - In General One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed. Processes that require large amounts of energy tend not to occur. Overall, remember that “like dissolves like”. Copyright © Cengage Learning. All rights reserved

Pressure Effects C = concentration of dissolved gas k = constant
Little effect on solubility of solids or liquids Henry’s Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas in direct contact with the liquid. Henry’s law: C = kP C = concentration of dissolved gas k = constant P = partial pressure of gas solute above the solution Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.

Pressure and Solubility of Gases
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c is the concentration (M) of the dissolved gas c = kP P is the pressure of the gas over the solution k is a constant for each gas (mol/L•atm) that depends only on temperature low P high P low c high c

Temperature Effects (for Aqueous Solutions)
Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature. Predicting temperature dependence of solubility is very difficult. Solubility of a gas in solvent typically decreases with increasing temperature. Copyright © Cengage Learning. All rights reserved

The Solubilities of Several Solids as a Function of Temperature

Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO3 contaminated with 10 g NaCl. Fractional crystallization: Dissolve sample in 100 mL of water at 600C Cool solution to 00C All NaCl will stay in solution (s = 34.2g/100g) 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g

The Solubilities of Several Gases in Water
solubility usually decreases with increasing temperature Copyright © Cengage Learning. All rights reserved

Liquid/Vapor Equilibrium
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Colligative Properties of Solutions
The colligative properties of solutions are those properties that depend on solute concentration. Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. These properties include: vapor pressure reduction freezing point depression boiling point elevation osmosis 2

Vapor Pressure Lowering: Addition of a Solute

Vapor Pressures of Solutions
Nonvolatile solute lowers the vapor pressure of a solvent. Raoult’s Law: Psoln = observed vapor pressure of solution solv = mole fraction of solvent = vapor pressure of pure solvent Copyright © Cengage Learning. All rights reserved

A Solution Obeying Raoult’s Law

Nonideal Solutions Liquid-liquid solutions where both components are volatile. Modified Raoult’s Law: Nonideal solutions behave ideally as the mole fractions approach 0 and 1. Copyright © Cengage Learning. All rights reserved

Vapor Pressure for a Solution of Two Volatile Liquids

Summary of the Behavior of Various Types of Solutions
Interactive Forces Between Solute (A) and Solvent (B) Particles ΔHsoln ΔT for Solution Formation Deviation from Raoult’s Law Example A  A, B  B  A  B Zero None (ideal solution) Benzene-toluene A  A, B  B < A  B Negative (exothermic) Positive Negative Acetone-water A  A, B  B > A  B Positive (endothermic) Ethanol-hexane Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult’s Law), show a positive deviation, or show a negative deviation? Hexane (C6H14) and chloroform (CHCl3) Ethyl alcohol (C2H5OH) and water Hexane (C6H14) and octane (C8H18) a) Positive deviation; Hexane is non-polar, chloroform is polar. b) Negative deviation; Both are polar, and the ethyl alcohol molecules can form stronger hydrogen bonding with the water molecules than it can with other alcohol molecules. c) Ideal; Both are non-polar with similar molar masses. Copyright © Cengage Learning. All rights reserved

Fractional Distillation Apparatus

Boiling Point Elevation
The normal boiling point of a liquid is the temperature at which its vapor pressure equals 1 atm. Because vapor pressure is reduced in the presence of a nonvolatile solute, a greater temperature must be reached to achieve boiling. The boiling point elevation, DTb is a colligative property equal to the boiling point of the solution minus the boiling point of the pure solvent. 2

Boiling-Point Elevation
Nonvolatile solute elevates the boiling point of the solvent. ΔT = Kbmsolute ΔT = boiling-point elevation Kb = molal boiling-point elevation constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved

Boiling Point Elevation: Liquid/Vapor Equilibrium

Boiling Point Elevation: Addition of a Solute

Freezing-Point Depression
When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. ΔT = Kfmsolute ΔT = freezing-point depression Kf = molal freezing-point depression constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved

Freezing Point Depression: Solid/Liquid Equilibrium

Freezing Point Depression: Addition of a Solute

Freezing Point Depression: Solid/Solution Equilibrium

A Problem to Consider An aqueous solution is m in glucose. What are the boiling point and freezing point for this solution? Kb and Kf for water as oC/m and 1.86 oC/m, respectively. Therefore, The boiling point of the solution is oC and the freezing point is –0.041oC. 2

DTf = Kf m Kf water = 1.86 0C/m DTf = Kf m
What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is g. DTf = Kf m Kf water = C/m = 3.202 kg solvent 478 g x 1 mol 62.01 g m = moles of solute mass of solvent (kg) = 2.41 m DTf = Kf m = C/m x 2.41 m = C DTf = T f – Tf Tf = T f – DTf = C – C = C

EXERCISE! A solution was prepared by dissolving g of glucose in g water. The molar mass of glucose is g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution °C The change in temperature is ΔT = Kbmsolute. Kb is 0.51 °C·kg/mol. To solve formsolute, use the equation m = moles of solute/kg of solvent. Moles of solute = (25.00 g glucose)(1 mol / g glucose) = mol glucose Kg of solvent = (200.0 g)(1 kg / 1000 g) = kg water msolute = ( mol glucose) / ( kg water) = mol/kg ΔT = (0.51 °C·kg/mol)( mol/kg) = 0.35 °C. The boiling point of the resulting solution is °C °C = °C. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

EXERCISE! You take 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be °C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture. 72.8% sucrose and 27.2% sodium chloride; mole fraction of the sucrose is 0.313 The solution is 72.8% sucrose and 27.2% sodium chloride. The mole fraction of the sucrose is To solve this problem, the students must assume that i = 2 for NaCl. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

EXERCISE! A plant cell has a natural concentration of m. You immerse it in an aqueous solution with a freezing point of –0.246°C. Will the cell explode, shrivel, or do nothing? The cell will explode (or at least expand). The concentration of the solution is m. Thus, the cell has a higher concentration, and water will enter the cell. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

Osmotic Pressure (p) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (p) is the pressure required to stop osmosis. more concentrated dilute

Chemistry In Action: Desalination

Osmotic pressure is a colligative property of a solution equal to the pressure that, when applied to the solution, just stops osmosis. 2

Osmotic Pressure (p) p = MRT High P Low P
M is the molarity of the solution R is the gas constant T is the temperature (in K)

EXERCISE! When 33.4 mg of a compound is dissolved in mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound. 111 g/mol The molar mass is 111 g/mol. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

A cell in an: isotonic solution hypotonic hypertonic

Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P1 = X1 P 1 Boiling-Point Elevation DTb = Kb m Freezing-Point Depression DTf = Kf m Osmotic Pressure (p) p = MRT

Colligative Properties of Ionic Solutions
The colligative properties of solutions depend on the total concentration of solute particles. Consequently, ionic solutes that dissociate in solution provide higher effective solute concentration than nonelectrolytes. For example, when NaCl dissolves, each formula unit provides two solute particles. 2

Colligative Properties of Electrolyte Solutions
0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution 0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be nonelectrolytes 1 NaCl 2 CaCl2 3

Examples The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). NaCl i = 2 KNO3 i = 2 Na3PO4 i = 4 Copyright © Cengage Learning. All rights reserved

Ion Pairing Ion pairing is most important in concentrated solutions. As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs. Ion pairing occurs to some extent in all electrolyte solutions. Ion pairing is most important for highly charged ions. Copyright © Cengage Learning. All rights reserved

Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation DTb = i Kb m Freezing-Point Depression DTf = i Kf m Osmotic Pressure (p) p = iMRT

A Problem to Consider Estimate the freezing point of a m aqueous solution of aluminum sulfate, Al2(SO4)3. Assume the value of i is based on the formula. When aluminum sulfate dissolves in water, it dissociates into five ions. Therefore, you assume i = 5. 2

A Problem to Consider Estimate the freezing point of a m aqueous solution of aluminum sulfate, Al2(SO4)3. Assume the value of i is based on the formula. The freezing point depression is The estimated freezing point is –0.093oC. 2

A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. Colloid versus solution collodial particles are much larger than solute molecules collodial suspension is not as homogeneous as a solution

Colloids A colloid is a dispersion of particles of one substance (the dispersed phase) throughout another substance or solution (the continuous phase). A colloid differs from a true solution in that the dispersed particles are larger than normal molecules. The particles range from 1 x 103 pm to about 2 x 105 pm. 2

The Tyndall Effect The scattering of light by colloidal-size particles is known as the Tyndall effect. For example, a ray of sunshine passing against a dark background shows up many fine dust particles by light scattering. 2

Types of Colloids Colloids are characterized according to the state of the dispersed phase and the state of the continuous phase. A sol consists of solid particles dispersed throughout a liquid. An aerosol consists of liquid droplets or solid particles dispersed throughout a gas. An emulsion consists of liquid droplets dispersed throughout another liquid. 2

Hydrophilic and Hydrophobic Colloids
Colloids in which the continuous phase is water are divided into two major classes. A hydrophilic colloid is a colloid in which there is a strong attraction between the dispersed phase and the continuous phase (water). A hydrophobic colloid is a colloid in which there is a lack of attraction of the dispersed phase for the continuous phase (water). 2

Coagulation Coagulation is the process by which the dispersed phase of a colloid is made to aggregate and thereby separate from the continuous phase. Soil suspended in river water coagulates when it meets the concentrated ionic solution of the ocean. The Mississippi Delta was formed this way. 2

Association Colloids A micelle is a colloidal-sized particle formed by the association of molecules, each of which has a hydrophobic end and a hydrophilic end. A colloid in which the dispersed phase consists of micelles is called an association colloid. Ordinary soap in water provides an example of an association colloid. 2

The Cleansing Action of Soap

WORKED EXAMPLES

Worked Example 12.2

Worked Example 12.4

Worked Example 12.7

Worked Example 12.8

Worked Example 12.9

Worked Example 12.10

Worked Example 12.11

Solution Formation A solution is a homogeneous mixture of two or more substances, consisting of ions or molecules. A colloid, although it also appears.

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Solution Formation A solution is a homogeneous mixture of two or more substances, consisting of ions or molecules. A colloid, although it also appears.

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Presentation on theme: "Solution Formation A solution is a homogeneous mixture of two or more substances, consisting of ions or molecules. A colloid, although it also appears."— Presentation transcript:

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