Solutions

Various Types of Solutions Example State of Solution State of Solute State of Solvent Air, natural gas Gas Vodka, antifreeze Liquid Brass Solid Carbonated water (soda) Seawater, sugar solution Hydrogen in platinum

Solution Composition

Molarity

Exercise You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.

Exercise You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?

Exercise Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity. 7

Mass Percent

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?

Mole Fraction

Exercise A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.)

Molality

Exercise A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)

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.

Steps in the Dissolving Process

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.

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).

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. 19

The Energy Terms for Various Types of Solutes and Solvents Hsoln 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

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”.

Structural Effects: Polarity Pressure Effects: Henry’s law Temperature Effects: Affecting aqueous solutions

Pressure Effects 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.

A Gaseous Solute

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.

The Solubilities of Several Solids as a Function of Temperature

The Solubilities of Several Gases in Water

An Aqueous Solution and Pure Water in a Closed Environment

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

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.

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 Hsoln T 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 33

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. 34

Colligative Properties Depend only on the number, not on the identity, of the solute particles in an ideal solution: Boiling-point elevation Freezing-point depression Osmotic pressure

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

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

Changes in Boiling Point and Freezing Point of Water

Exercise A solution was prepared by dissolving 25.00 g glucose in 200.0 g water. The molar mass of glucose is 180.16 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.

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 -0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

Exercise A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of –0.246°C. Will the cell explode, shrivel, or do nothing?

Osmosis – flow of solvent into the solution through a semipermeable membrane. = MRT = osmotic pressure (atm) M = molarity of the solution R = gas law constant T = temperature (Kelvin)

Exercise When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound.

van’t Hoff Factor, i The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as:

Ion Pairing At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle.

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

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.

Modified Equations

A suspension of tiny particles in some medium. Tyndall effect – scattering of light by particles. Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm.

Types of Colloids

Coagulation Destruction of a colloid. Usually accomplished either by heating or by adding an electrolyte.