Chapter 11: Properties of Solutions

Properties of Solutions: An Overview

Solution Composition

• A solution is a homogeneous mixture whose components include:

  • Solute: The substance that dissolves in the solvent.

  • Solvent: The substance in which the solute dissolves, typically present in a larger quantity.

• Soluble (Miscible):

  • The solute dissolves in the solvent as a "whole" unit.

  • It does NOT ionize.

• Insoluble (Immiscible):

  • The solute and solvent do NOT mix.

  • A solution does NOT form.

• Ionization:

  • The solute dissolves in the solvent in "pieces," forming ions.

Solubility Terminology

• A solution is a homogeneous mixture of two or more substances.

• Solutions can be composed of substances in the same phases or different phases.

• One constituent is usually the solvent, and the other is the solute.

• The ability of a solute to dissolve in a solvent, or its solubility, depends on:

  • Nature's tendency toward mixing (thermodynamics).

  • The types of intermolecular attractive forces between components.

• Solvents and solutes must be compatible; otherwise, a solution will not form.

  • This typically means they should have similar polarities, following the principle: "LIKE DISSOLVES LIKE".

Various Types of Solutions

Practice Problem Example:

• Decide whether liquid hexane (C6H{14}) or liquid methanol (CH3OH) is the more appropriate solvent for grease (C{20}H_{42}) and potassium iodide (KI).

Classification of Solutions

• Unsaturated Solution:

  • Contains just below the maximum quantity of solute that can dissolve at a given temperature in a specific amount of solvent.

  • The rate of dissolution is equal to the rate of deposition.

  • In a truly unsaturated solution, more solute can still be dissolved.

• Saturated Solution:

  • Contains the maximum quantity of solute that dissolves at that temperature for a given quantity of solvent.

  • No more solute will dissolve at that temperature.

• Supersaturated Solution:

  • Contains more solute than is normally possible for a given quantity of solvent at a specific temperature.

  • These solutions are unstable and will lose all the solute above the saturation point when disturbed (e.g., by adding a seed crystal or scratching the container).

Other Factors Affecting Solubility

• Temperature

• Pressure (specifically for gases)

• Volume of solvent to amount of solute (ratio)

• Solubility is generally given in grams of solute that will dissolve per 100 grams of water (solvent).

Solute Solubility Based on Temperature and Pressure

Effect of Temperature

• For most solids: Usually, an increase in temperature results in an increase in solubility.

  • This occurs when the dissolution process (ΔH{solution}) is endothermic (absorbs heat, ΔH{solution} > 0).

  • Example: More sugar can be dissolved in warm water when making candy because sugar is more soluble in warmer solvents.

• For all gases: Their solubility will decrease as the temperature increases.

  • The dissolution of gases is generally an exothermic process (ΔH_{solution} < 0) because there is no significant need to overcome solute-solute attractions of gas particles; instead, the primary energy change is the interaction with the solvent.

  • Examples: Warm soda pop fizzes more than cold soda because less CO2(g) remains in solution. CO2(g) is less soluble in warm solvents.

Effect of Pressure (Henry's Law)

• The larger the partial pressure of a gas in contact with a liquid, the more soluble the gas is in the liquid.

• Henry's Law illustrates this direct relationship:

  • Formula: C{gas} = kH P_{gas} where:

    • C_{gas} is the concentration of the dissolved gas (often in Molarity).

    • k_H is Henry's Law constant, specific to the gas and solvent at a given temperature.

    • P_{gas} is the partial pressure of the gaseous solute above the solution.

Practice Problem: Henry's Law

• Problem: If 2.64 g of CO2 dissolves in 500.0 mL water at 25.0 ext{ } ^oC and at 3.50 atm CO2 pressure, what is the kH (Henry's law constant) for CO2?

• Solution:

  1. Convert grams of CO2 to moles: 2.64 ext{ g } CO2 imes rac{1 ext{ mol}}{44.01 ext{ g}} = 0.0600 ext{ mol } CO_2

  2. Calculate the concentration (Molarity) of CO2: C{gas} = rac{0.0600 ext{ mol}}{0.500 ext{ L}} = 0.120 ext{ M}

  3. Apply Henry's Law to find kH: C{gas} = kH P{gas}
     0.120 ext{ M} = kH imes 3.50 ext{ atm} kH = rac{0.120 ext{ M}}{3.50 ext{ atm}} = 0.0343 ext{ M/atm}

The Energies of Solution Formation (Concept Only)

Steps Involved in the Formation of a Liquid Solution

Formation of a solution involves three main steps, each with an associated enthalpy change:

1. Expand the solute (ΔH_{solute}):

   • Requires energy to separate the solute into its individual components (molecules or ions).

   • This step is endothermic (positive value, ΔH_{solute} > 0).

2. Expand the solvent (ΔH_{solvent}):

   • Requires energy to overcome intermolecular forces in the solvent to create space for the solute particles.

   • This step is also endothermic (positive value, ΔH_{solvent} > 0).

3. Allow the solute and solvent to interact to form the solution (ΔH_{mixture}):

   • This step releases energy as new interactions form between solute and solvent particles.

   • This step is exothermic (negative value, ΔH_{mixture} < 0).

Enthalpy (Heat) of Solution (ΔH_{solution})

• The enthalpy change associated with the overall formation of the solution is the sum of the ΔH values for these three steps:

  • Formula: ΔH{solution} = ΔH{solute} + ΔH{solvent} + ΔH{mixture}

• ΔH_{solution} can have:

  • A positive sign when energy is absorbed (endothermic process, resulting in a colder solution).

  • A negative sign when energy is released (exothermic process, resulting in a warmer solution).

Driving Forces for Solution Formation

• When the sum of endothermic terms (ΔH{solute} + ΔH{solvent}) is approximately equal in magnitude to the exothermic term (|ΔH_{mixture}|):

  • ΔH_{solution} is approximately zero.

  • The increasing entropy (disorder) upon mixing is the primary driving force for the formation of the solution, while the overall energy of the system remains nearly constant.

• When the sum of the endothermic terms is smaller in magnitude than the exothermic term (|ΔH{solute} + ΔH{solvent}| < |ΔH_{mixture}|):

  • ΔH_{solution} is negative, and the solution process is exothermic.

  • In this case, both the tendency toward lower energy (exothermic ΔH_{solution}) and the tendency toward greater entropy drive the formation of a solution.

• When the sum of the endothermic terms is greater in magnitude than the exothermic term (|ΔH{solute} + ΔH{solvent}| > |ΔH_{mixture}|):

  • ΔH_{solution} is positive, and the solution process is endothermic.

  • As long as ΔH_{solution} is not excessively large, the tendency toward greater entropy still drives the formation of a solution.

  • However, if ΔH_{solution} is too large (meaning too much energy is required to form the solution), a solution will not form.

Example Problems on Solubility

• Which of the following is soluble in water: Vitamin C or Vitamin K_3?

  • (Requires knowledge of molecular structures and polarity to determine.)

• Rank the following compounds according to increasing solubility in water:

  • CH3–CH2–CH2–CH3 (butane)

  • CH3–CH2–O–CH2–CH3 (diethyl ether)

  • CH3–CH2–OH (ethanol)

  • CH_3–OH (methanol)

  • (Requires understanding of hydrogen bonding and polarity influencing solubility in water).

Concentration Units

• Molarity (M):

  • Moles of solute per liter of solution (mol/L).

  • Formula: M = rac{ ext{moles of solute}}{ ext{liters of solution}}

  • Example: A 1.0 ext{ M } CaCl2(aq) solution contains 1.0 mole of CaCl2 in each liter of solution.

    • CaCl_2(aq)
      ightarrow Ca^{2+}(aq) + 2 Cl^{-}(aq)

    • This means 1.0 ext{ M } CaCl_2 produces 1.0 ext{ M } Ca^{2+} ions and 2.0 ext{ M } Cl^{-} ions, totaling 3.0 ext{ M} of ions in solution.

• Mass Percent (weight percent):

  • Formula: ext{Mass percent} = rac{ ext{mass of solute}}{ ext{mass of solution}} imes 100 ext{%}

• Molality (m):

  • Moles of solute per kilogram of solvent (mol/kg).

  • Formula: m = rac{ ext{moles of solute}}{ ext{kilograms of solvent}}

• Percent Concentration (% concentration):

  • Weight/Volume (w/v)%: 4.0 ext{ (w/v)\%} means 4.0 grams of solute in 100.0 mL of solution.

  • Volume/Volume (v/v)%: 8.5 ext{ (v/v)\%} means 8.5 mL of solute in 100.0 mL of solution.

  • Weight/Weight (w/w)%: 12.0 ext{ (w/w)\%} means 12.0 grams of solute in 100.0 grams of solution.

Practice Problem: Molarity, Mass Percent, and Molality

• Problem: A solution is prepared by mixing 1.00 g ethanol (C2H5OH) with 100.0 g water to give a final volume of 101 mL. Calculate the molarity, mass percent, and molality of ethanol in this solution.

• Molarity Solution:

  1. Molar mass of C2H5OH = (2 imes 12.01) + (6 imes 1.008) + (1 imes 16.00) = 46.068 ext{ g/mol}

  2. Moles of C2H5OH = rac{1.00 ext{ g}}{46.068 ext{ g/mol}} = 0.0217 ext{ mol}

  3. Volume of solution = 101 ext{ mL} = 0.101 ext{ L}

  4. Molarity = rac{0.0217 ext{ mol}}{0.101 ext{ L}} = 0.215 ext{ M} (Not explicitly calculated in transcript, but derived).

• Mass Percent Solution:
  ext{Mass percent } C2H5OH = rac{ ext{mass of } C2H5OH}{ ext{mass of solution}} imes 100 ext{%}
  = rac{1.00 ext{ g } C2H5OH}{100.0 ext{ g } H2O + 1.00 ext{ g } C2H5OH} imes 100 ext{%} = rac{1.00 ext{ g}}{101.0 ext{ g}} imes 100 ext{%} = 0.990 ext{% } C2H_5OH

• Molality Solution:
  ext{Molality of } C2H5OH = rac{ ext{moles of } C2H5OH}{ ext{kilogram of } H2O} ext{Moles of } C2H5OH = 0.0217 ext{ mol} (from molarity calculation) ext{Kilogram of } H2O = 100.0 ext{ g } H_2O imes rac{1 ext{ kg}}{1000 ext{ g}} = 0.1000 ext{ kg}
  ext{Molality} = rac{0.0217 ext{ mol}}{0.1000 ext{ kg}} = 0.217 ext{ m}

Practice Problem: Sulfuric Acid Solution

• Problem: The electrolyte in automobile lead storage batteries is a 3.75 ext{ M} sulfuric acid (H2SO4) solution that has a density of 1.230 ext{ g/mL}. Calculate the mass percent and molality of the sulfuric acid.

• Answer Strategy:

  1. Assume 1.00 ext{ L} of solution for convenience.

  2. From molarity, find moles of H2SO4 in 1.00 ext{ L}.

  3. Convert moles of H2SO4 to grams of H2SO4 (solute).

  4. Use solution density to find the total mass of 1.00 ext{ L} of solution.

  5. Subtract mass of solute from total mass of solution to find mass of solvent (water).

  6. Calculate mass percent.

  7. Calculate molality.

• Partial Solution provided: We know 1.00 ext{ L} of this solution contains 1230 g of the mixture of sulfuric acid and water.

  • Since the solution is 3.75 ext{ M}, we know that 3.75 moles of H2SO4 is present per liter of solution.

  • The number of grams of H2SO4 present is:
    3.75 ext{ mol } H2SO4 imes rac{98.08 ext{ g } H2SO4}{1 ext{ mol } H2SO4} = 363.0 ext{ g } H2SO4

  • Mass of water in 1.00 ext{ L} of solution:
    ext{Mass of solution} - ext{Mass of solute} = 1230 ext{ g} - 363.0 ext{ g} = 867.0 ext{ g } H_2O

  • Now, one can calculate mass percent and molality:

    • Mass Percent: rac{363.0 ext{ g } H2SO4}{1230 ext{ g solution}} imes 100 ext{%} = 29.5 ext{% } H2SO4

    • Molality: rac{3.75 ext{ mol } H2SO4}{0.8670 ext{ kg } H_2O} = 4.32 ext{ m}

Colligative Properties of Solutions

• Definition: Adding a solute to a solvent modifies the properties of the pure solvent. These changes are referred to as colligative properties.

• Dependency: Colligative properties depend ONLY on the NUMBER of solute particles relative to solvent particles in solution and not on the TYPE/KIND of solute particles.

• Effects on Pure Solvent Properties:

  • Raoult's Law: Vapor pressure lowering.

  • Melting point depression: Temperature decreases.

  • Boiling point elevation: Temperature increases.

  • Osmotic pressure.

The Vapor Pressures of Solutions (Raoult's Law)

• The presence of a nonvolatile solute (a solute that does not readily evaporate) lowers the vapor pressure of a solvent.

• This occurs because the solute particles inhibit the escape of solvent molecules from the liquid surface into the gas phase.

• Illustration: If an aqueous solution and pure water are in a closed environment, over time, water molecules will transfer from the pure water side to the solution side until equilibrium, indicating the solution has a lower vapor pressure.

Boiling-Point Elevation

• A nonvolatile solute elevates the boiling point of the solvent.

• The magnitude of the boiling-point elevation depends on the concentration of the solute.

• The change in boiling point (ΔT_b) can be represented by the following formula:

  • Formula: ΔTb = kb imes m imes i where:

    • ΔTb = Change in boiling point (T{solution} - T_{solvent}).

    • k_b = Boiling-point elevation constant (a characteristic property of the solvent).

    • m = Molality of the solute (moles of solute per kg of solvent).

    • i = van't Hoff factor (number of particles a solute dissociates into).

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.

• The change in freezing point (ΔT_f) can be represented by the following formula:

  • Formula: ΔTf = kf imes m imes i where:

    • ΔTf = Change in freezing point (T{solvent} - T_{solution}).

    • k_f = Freezing-point depression constant (a characteristic property of the solvent).

    • m = Molality of the solute.

    • i = van't Hoff factor.

Van't Hoff Factor (i)

• 
  

• Definition: For ionic compounds, they produce multiple solute particles for each formula unit dissolved.

• 
  

• The van't Hoff factor (i) is the ratio of moles of solute particles to moles of formula units dissolved.

  • Formula: i = rac{ ext{moles of solute particles produced}}{ ext{moles of formula units dissolved}}

• 
  

• Theoretical vs. Measured Values: Measured van't Hoff factors are often lower than theoretically expected due to ion pairing in solution, where some ions associate rather than completely dissociate.

• 
  

• Examples:
  

  Compound	Theoretical Value of i
  Glycol	1
  NaCl	2
  CaCl_2	3
  Osmotic Pressure (Concept Only)

  • Osmosis: The flow of solvent into a solution through a semipermeable membrane.

  • Semipermeable membrane: A membrane that permits solvent molecules to pass through but not solute molecules.

  • Osmotic Pressure: The result of increased hydrostatic pressure on the solution side of the semipermeable membrane compared to the pure solvent side.

    • It is caused by the difference in solvent levels observed at equilibrium when osmosis has occurred.

  • Process: In an osmosis cell, solvent (e.g., water) flows from the pure-solvent side through the semipermeable membrane to the solution side. This net transfer continues until the hydrostatic pressure built up on the solution side equalizes the rate of solvent flow in both directions across the membrane. At equilibrium, the pressure of the excess fluid is equal to the osmotic pressure of the solution.

  Practice Problem: Boiling Point and Freezing Point

  • Problem Statement: If 85.00 g of ethylene glycol (C2H6O_2) is dissolved in 250.0 grams of water, what is the boiling point and freezing point of the solution?

  • Useful Information: kb for water is 0.512 ext{ } ^oC/m. kf for water is 1.86 ext{ } ^oC/m. Molar mass of ethylene glycol (C2H6O_2) is 62.07 ext{ g/mol}. Ethylene glycol is a non-electrolyte, so i=1.

  • Solution Steps:

    1. Calculate moles of ethylene glycol:
       ext{moles} = 85.00 ext{ g } imes rac{1 ext{ mol}}{62.07 ext{ g}} = 1.369 ext{ mol} (Using 62.1 ext{ g/mol} as per transcript calculation: 1.368 ext{ mol})

    2. Calculate molality (m):
       ext{Molality } m = rac{1.369 ext{ mol}}{0.250 ext{ kg } H_2O} = 5.476 ext{ m}
       (Using 1.368 ext{ mol} from transcript: 5.475 ext{ m})

    3. Calculate boiling point elevation (ΔTb):
       ΔTb = kb imes m imes i ΔTb = 0.512 ext{ } ^oC/m imes 5.476 ext{ m} imes 1
       ΔT_b = 2.804 ext{ } ^oC
       (Using transcript's 5.475 ext{ m} gives 2.80 ext{ } ^oC)

    4. Calculate the new boiling point of the solution (T{b, solution}):
       ΔTb = T{b, solution} - T{b, solvent}
       2.804 ext{ } ^oC = T{b, solution} - 100.0 ext{ } ^oC (Boiling point of pure water is 100.0 ext{ } ^oC) T{b, solution} = 100.0 ext{ } ^oC + 2.804 ext{ } ^oC = 102.804 ext{ } ^oC
       The solution's boiling point is 102.80 ext{ } ^oC.

    5. Calculate freezing point depression (ΔTf):
       ΔTf = kf imes m imes i ΔTf = 1.86 ext{ } ^oC/m imes 5.476 ext{ m} imes 1
       ΔT_f = 10.186 ext{ } ^oC
       (Using transcript's 5.475 ext{ m} gives 10.18 ext{ } ^oC)

    6. Calculate the new freezing point of the solution (T{f, solution}):
       ΔTf = T{f, solvent} - T{f, solution}
       10.186 ext{ } ^oC = 0.0 ext{ } ^oC - T{f, solution} (Freezing point of pure water is 0.0 ext{ } ^oC) -T{f, solution} = 10.186 ext{ } ^oC
       T_{f, solution} = -10.186 ext{ } ^oC
       The solution's freezing point is -10.18 ext{ } ^oC.

  Additional Practice Problems

  • How much NaCl must be dissolved in 4.00 kg of water to lower the freezing point to -10.00 ext{ } ^oC?

    • (Note: NaCl is an electrolyte, so i=2 in ideal solutions).

  • Why does the addition of a solute to a solvent raise the boiling point of the solution?

    • (Relates to vapor pressure lowering; a lower vapor pressure requires a higher temperature to reach the external atmospheric pressure, thus elevating the boiling point).