Types of Solutions s1

AP Chemistry Chapter 12 SOLUTIONS 12.1 Types of Solutions Solutions may exist in any of the three phases of matter.

Solute: the substance dissolved. (nail polish – smaller amount) Solvent: the substance that completes the dissolving. (p 500) (nail polish remover – larger amount)

Miscible fluids: fluids that mix or dissolve in all proportions. (water and pop) Immiscible fluids: fluids that do not mix. (p 501) (oil and water)

Concept Check 12.1 Identify the solute (s) and solvent (s) in the following solutions. (p 502) (a) 80 g of Cr and 5 g of Mo (b) 39% N2, 41% Ar and the rest O2 (c) 5 g MgCl2 dissolved in 1000 g of H2O

12.2 Solubility and the Solution Process When a crystal dissolves, particles are dissolving at the same time particles are returning to the crystalline state. When these two processes occur at the same rate, dynamic equilibrium is reached.

Saturated solution: for a given temperature, solute and amount of solvent, as much of the solute that can be dissolved has been.(p 502)

Solubility: the amount that dissolves in a given quantity of water at a given temperature to give a saturated solution. (p 502)

Unsaturated solution: more solute can be dissolved than has been.

Supersaturated solutions: contain more solute than the saturated solution. Usually obtained by raising the temperature of the solvent and then slowly cooling to room temperature. (p 503)

Factors in Explaining Solubility (p 503) 1. The natural tendency of molecules to mix due to their random motions. (food color in water) 2. Intermolecular forces. If strong forces exist between the MC of the solvent and MC of the solute separately, but only weak forces exist between the solvent and solute the two will not mix.

Molecules with similar IMF are soluble in one another. Like dissolves like! (p 505) - Small alcohols are soluble in water because of the OH, but as the MC become larger they become insoluble.

Exercise 12.2 p 506. Which of the following compounds is likely to be more soluble in water: C4H9OH or C4H9SH? Explain.

Ionic Compounds have markedly different solubilities. These differences in solubility can be explained in terms of the different energies of attraction between ions in the crystal and between ions and water. (p 506) - ion-dipole force = between the ions and water. This attraction is called hydration. - Smaller ionic radius (& greater charge) = large hydration energy because the electric field is more concentrated. AP Chemistry Chapter 12 - Lattice energies fight the energy of hydration. If the crystal has a strong lattice energy, it will not dissolve. - Depends on charge of ions & distance between them.

12.3 Effects of Temperature and Pressure Temperature (p 509) - Gases become less soluble as the temperature increases. - Many ionic solids become more soluble. (Ca and Ce decrease when attached to large anions)

Heat may be absorbed (endothermic) or released (exothermic) during the dissolving process. NH4NO3 is used in cold packs and CaCl2 or MgSO4 are used in hot packs.

Pressure (p 510) - no effect on solids or liquids but large effect on gases. - Le Chatelier’s principle: when a system in equilibrium is disturbed by a change of temperature, pressure or concentration, the system shifts in such a way as counteract the change and reestablish equilibrium.

- All gases become more soluble in a liquid at a given temperature when the partial pressure of the gas over the solution is increased or less soluble with pressure decreased! o Demo pop in sealed syringe

Henry’s Law: the solubility of a gas is directly proportional to the partial pressure of the gas above the solution. S1P2 = S2P1. (p 511)

Homework: Read 12.1 – 12.3 Be sure to read 506-508 and use these to assist you in doing the homework. Read through the A Chemist Looks At on p 508. Review 2-5 All Practice 31-40 odds

12.4 Expressing Concentrations

Colligative Properties: properties that depend on the concentration of solute molecules or ions in solution but not on the chemical identity of the solute.

Molarity = moles of solute Liters of solution Practice p68 Dilutions: M1V1 = M2V2 Practice p 69 Normality = molarity X total positive oxidation number of solute Practice p71 Mass % of Solute = mass of solute * 100 Mass of solution Practice p57 How would you prepare 425 g of an aqueous solution containing 2.40% by mass of NaC2H3O2? AP Chemistry Chapter 12 Molality = moles of solute Kilograms of water Practice p 70 What is the molality of a solution containing 5.67g of glucose, C6H12O6 dissolved in 25.2 g of water?

Mole Fraction = moles of substance A Total moles of solution

- Multiplying by 100 give mole percent

What are the mole fractions of glucose and water in a solution containing 5.67 g of C6H12O6 dissolved in 25.2 g of water?

Conversions between concentration units Molality to Mole Fraction: An aqueous solution is 0.120m glucose. What are the mole fractions of each component in the solution?

Mole Fraction to Molality: A solution is 0.150 mole fraction glucose and 0.850 mole fraction water. What is the molality of the glucose?

To convert between molality and molarity, you must know the density of the solution. M  m : An aqueous solution is 0.273m KCl. What is the molar concentration of potassium chloride, KCl? The density of the solution is 1.011*103g/L.

m  M: An aqueous solution is 0.907M Pb(NO3)2. What is the molality of Pb(NO3)2 in this solution? The density of the solution is 1.252 g/L.

Homework: Read 12.4 Q: 41-58.

12.5 Vapor Pressure of a Solution

The vapor pressure of a volatile solvent (ex. NH3) is lowered by the addition of a nonvolatile solute (ex. H2O). (p 518)

This vapor-pressure lowering of a solvent is a colligative property equal to the vapor pressure of the pure solvent minus the vapor pressure of the solution. Vpsolvent - Vpsolution AP Chemistry Chapter 12

Raoult’s law: the partial pressure of solvent, PA, over a solution equals the vapor pressure of the pure o o solve, P A, times the mole fraction of solvent, XA in the solution. (the total X its %) PA = P AXA  observed to hold for dilute solutions

o The amount the vapor-pressure is lowered may be calculated using P = P AXB. (p 520)

EX. Calculate the vapor-pressure lowering of water when 5.67 g of glucose is dissolved in 25.2 g of water at 25oC. The vapor pressure of water at 25oC is 23.8 mmHg. What is the vapor pressure of the solution?

12.6 Boiling Point Elevation and Freezing Point Depression The addition of a nonvolatile solute to a liquid reduces its vapor pressure; therefore, the temperature must be increased in order to achieve a set vapor pressure in order to boil. Hence Bp elevation.

The boiling point elevation, Tb is a colligative property of a solution equal to the boiling point of the solution minus the boiling point of the pure solvent. (p 523) Tb = KbCm Kb = boiling point constant listed on p523 Cm = molal concentration

The freezing point depression, Tf, is a colligative property of a solution equal to the freezing point of the pure solvent minus the freezing point of the solution. Tf = KfCm (Kf = Fp constant p 523)

EX. An aqueous solution (dissolved in H2O) is 0.0222 m glucose. What are the boiling point and the freezing point of this solution?

From the freezing point lowering, you can calculate the molal concentration and from the molality, you can obtain the molecular weight. Important concept to know for AP Test.

 Find molessolute (gsolvent X molality)

 gsolute / moles = MM

EX. A solution is prepared by dissolving 0.131 g of a substance in 25.4 g of water. The molality of the solution is determined by freezing point depression to be 0.056 m. What is the molecular weight of the substance?

EX. Camphor melts at 179.5oC. Its freezing point constant is 40oC/m. A 1.07 mg sample of a compound was dissolved in 78.1 mg of camphor. The solution melted at 176.0oC. What is the molecular weight of the compound? If the empirical formula of the compound is CH, what is the molecular formula?

Homework: Read 12.5 – 12.6 Q: 59 – 68 AP Chemistry Chapter 12 12.7 Osmosis

Osmosis is the phenomenon of solvent flow through a semi-permeable membrane to equalize the solute concentrations on both sides of the membrane. (p 527)

Osmotic pressure,  is a colligative property of a solution equal to the pressure that, when applied to the solution, just stops osmosis. (p527)  = MRT M = molar concentration, R = the gas constant, T = temperature in Kelvin Doesn’t matter which R, the unit of R determines the unit of 

EX. When 0.798 g of starch, (C6H10O5)200 is dissolved in 100.0 mL of water, what is the osmotic pressure in mmHg?

12.8 Colligative properties of ionic solutions

To explain the colligative properties of ionic solutions, you must account for the total concentration of ions, rather than the concentration of the ionic substance.

When salt, NaCl dissolves, we get two ions, Na+ & Cl-. Therefore, the formula for the freezing point depression must be altered slightly. Tf = iKfCm, i = number of ions

EX. Estimate the freezing point of a 0.010m aqueous soln of aluminum sulfate.

The # of ions not always predictable because of the electrical interactions of the ions in solution.

EX. A 0.029m aqueous soln of potassium sulfate has a freezing point of –0.14oC. Calculate i.

12.9 Colloids

A colloid is a dispersion of particles of one substance (the dispersed phase) throughout another substance (the continuous phase). Ex. Fog: water particles in air. (p 532)

Colloids scatter light when a flashlight is shone through them. (p 533)