Chemistry 102(01) Spring 2002 n Instructor: Dr. Upali Siriwardane n n Office: CTH 311 Phone n Office Hours: n 8:00-9:00 & 11:00-12:00 a.m., M, W, 8:00-10:00 a.m., Tu,Th, F March 27, 2002 (Test 1): Chapter 12 &13. April 26, 2002(Test 2): Chapter 14 & 15. May 15, 2002 (Test 3): Chapter 15 & 17. May 16, 2002 (Comprehensive Test): Chapters 12,13,14,15,17

Chapter 12. Physical Properties of Solutions n solutions n Solubility gases :Henry's Law. n Concentration units. n Energy changes that occur in the solution process n Solubilities of substances in various solvents n Colligative properties. n Colligative properties of electrolyte solutions. n Colloid, true solution and suspension.

Solutions n Homogeneous Mixtures n Types of solutions Mixture of Gases Liquid solutions (L+S,L+L,L+G) Solid solutions (S+S,alloys) Aerosols (L+G) Foam (S+G)

Solution Components n Solute –smaller amount n Solvent –larger amount

n “Like dissolves like.” n Materials with similar polarity are soluble in each other. Dissimilar ones are not. n Miscible n Liquids that are soluble in each other in all proportions such as ethanol and water. n Immiscible n Liquids that are not soluble in each other such as hexane and water.

Solubility of Salts SOLUBILITY RULES SOME SALTS ARE SOLUBLE SOME ARE INSOLUBLE THERE ARE DEGREES OF SOLUBILITY

Amount of Solute A solution that contains as much it can hold is called Saturated solution A solution that contains less than maximum amount is called unsaturated solution A solution that contains more than maximum amount is called supersaturated solution

Factors Affecting Solubility n Miscibility of solute and solvent- ”Like dissolves like” Heat of solution,  H(solution) exothermic -cooling helps endothermic-heating helps n Gases:cooling & pressure helps

Identify Polar and Non-polar groups in Covalent Molecules Acetic acid HC 2 H 3 O 2 CH 3 COOH Hexanol C 6 H 13 OH Hexane C 6 H 14 Propanoic acid C 2 H 5 COOH Like Dissolves Like

Solution Process of Ionic Compounds

Temperature and solubility SO 2 KCl glycine NaBr KNO 3 sucrose Solubility (g/100ml water) Temperature ( o C)

Pressure and solubility of gases n Increasing the pressure of a gas above a liquid increases the concentration of the gas. n This shifts the equilibrium, driving more gas into the liquid.

Henry's law n At constant temperature, the solubility of a gas is directly proportional to the pressure of the gas above the solution. n S = k H p n p= partial pressure of gas above the solution in atm. n S= concentration of gas in the solution in mol/L. n k H = Henry's constant which is characteristic of the gas and the solvent. n S 1 /S 2 =p 1 /p 2

Electrolytes Solutes which dissolves in water to form conducting solution- Electrolytes Solutes which dissolves in water to from non-conducting solution- Non - Electrolytes Solutes which dissolves in water to from weakly conducting solution - Weak - Electrolytes

Concentration Units n a) Molarity (M) n b) Normality (N) n c) Molality (m) d) Mole fraction (a)(a) n e) Mass percent (% weight) n f) Volume percent (% volume) n g) "Proof" n h) ppm and ppb

Molarity(M) n moles of solute n Molarity (M) = n Liters of solution

Calculate the normality of the solution, which is prepared by dissolving 25 g of H 2 SO 4 in water to a final volume of 2L. M.W. (H 2 SO 4 ) = g/mole

Molality (m) n moles of solute n Molality (m) = n kg of solvent

Calculate the molality of C 2 H 5 OH in water solution which is prepared by mixing 75.0 mL of C 2 H 5 OH and 125 g of H 2 O at 20 o C. The density of C 2 H 5 OH is g/mL.

Mole fraction (  a ) n moles of solute (substance)  a = n moles of solute + solvent

Calculate the mole fraction of benzene in a benzene(C 6 H 6 )- chloroform(CHCl 3 ) solution which contains 60 g of benzene and 30 g of chloroform. M.W. = (C 6 H 6 ) M.W. = (CHCl 3 )

When 100. mL of M HCl is diluted to 250. mL, the resulting MOLARITY of the HCl solution is: a) M b) M c) M d) M e) none of these

Mass (w/w) % or Weight % n Mass of solute n Mass (w/w) % = x 100 n Mass of solution

What is the mole fraction of ethanol, C 2 H 5 OH, in a methanol solution that is 40.%(w/w) ethanol, C 2 H 5 OH, by mass? a b c d. 0.54

Calculate the molarity of a solution of water/alcohol containing 35% C 2 H 5 OH by weight. The density of this solution is 1.10 g/mL.

Volume (v/v) % or Volume % n Volume of solute n Volume (v/v) % = x 100 n Volume of solution n Proof = Volume % x 2

A solution of hydrogen peroxide is 30.0% H 2 O 2 by mass and has a density of 1.11 g/cm 3. The MOLARITY of the solution is: a) 7.94 M b) 8.82 M c) 9.79 M d) e) none of these M.W. = (H 2 O 2 )

ppm & ppb (w/w or v/v) n Mass (volume) of solute n ppm = x 10 6 n Mass (volume) of solution n n Mass (volume) of solute n ppb = x 10 9 n Mass (volume) of solution

ppm and ppb conversions n 1 ppm = (1g/ 1x 10 6 g) 1x 10 6 n = (1/1000 g) n 1x 10 6 /1000g n = mg/ 1x 10 3 g n = mg/ L n 1 ppb = (1g/ 1x 10 9 g) 1x 10 9 n = (1/ g) n 1x 10 9 / g =  g/ 1x 10 3 g =  g/ L

Effect of solutes on Solution Properties n Solution Properties –a) Vapor Pressure –b) Freezing Point –c) Boiling Point –d) Osmotic Pressure n There are two types of solutes –a) Volatile solutes (covalent) –b) nonvolatile solutes (ionic)

Colligative properties n “Bulk” properties that change when you add a solute to make a solution. Based on how much you add but not – what the solute is. Effect of electrolytes is based on number of ions produced. n Colligative properties vapor pressure lowering freezing point depression boiling point elevation osmotic pressure

Colligative Property A property which depends only on the concentration or number of solute particles not on the nature of solutes

Vapor Pressure n Raoult's Law Nonvolatile solutes P solution =  solvent P o (solvent) P solution =  solute )P o (solvent) –vapor pressure lowering Volatile solutes P solution =  a  P o (solute) +  b P o (solvent)

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

At a given temperature the vapor pressures of pure liquid benzene and toluene are 745 torr and 290 torr,respectively. A solution prepared by mixing benzene and toluene obeys Raoult's law. At this temperature the vapor pressure over a solution in which the mole fraction of benzene is equal to is

Benzene and toluene form an ideal solution. At 298K, what is the mole fraction of benzene in the liquid that is in equilibrium with a vapor that has equal partial pressures of benzene and toluene? At 298K, the vapor pressures of pure benzene and pure toluene are 95 and 28 torr, respectively. a) 0.50 b) 0.77 c) 0.23 d) 0.30

Boiling Point Elevation (  T)  T b = K b m solute  K b = molal boiling point elevation constant n molal means concentration is given in molality(m). n m solute = concentration of solute expressed as molality(m).

What is the boiling point of a m aqueous solution of glucose? (K b for H 2 O is o C/m)  T b = K b m solute

Freezing Point Depression (  T)  T f = K f m solute  – K f = molal Freezing Point Depression constant –molal means concentration is given in molality(m). –m solute = concentration of solute expressed as molality(m).

What is the freezing point of a m aqueous solution of glucose? (K f for H 2 O is 1.86 o C/m)  T f = K f m solute

A 2.25g sample of a compound is dissolved in 125 g of benzene. The freezing point of the solution is 1.02 o C. What is the molecular weight of the compound? K f for benzene = 5.12 o C/m, freezing point = 5.5 o C.

Osmotic Pressure n Process of solvent moving through a semi-permeable membrane is called n Osmosis n The pressure created by moving solvent is called n Osmotic Pressure

Osmotic Pressure (   = MRT  = Osmotic pressure of the solution n M= Molarity of the solute in the solution n R = Ideal gas constant n T= Temperature of the solution in Kelvin

Calculate the osmotic pressure in atm at 20 o C of an aqueous solution containing 5.0 g of sucrose (C 12 H 22 O 11 ), in mL solution. M.W.(C 12 H 22 O 11 )=  = MRT R = L-atm/mol K = 62.4 L-torr/mol K Calculation

Ionic vs. covalent substances n Ionic substances have a greater effect per mole than covalent. n 1 mol/kg of water for glucose= 1 molal n 1 mol/kg of water for NaCl = 2 molal ions n 1 mol/kg of water for CaCl 2 = 3 molal ions n Effects are based on the number of particles!

Colligative Properties of Electrolytes n Number of solute particles in the solution depends on dissociation into ions expressed as Van’t Hoff facotor(i) n Van’t Hoff facotor (i) moles of particles in solution moles of solutes dissolved

Colligative Properties of Electrolytes n Vapor Pressure P solution = (1- i  solute )P o (solvent) n Boiling Point Elevation  T = i K b m solute n Freezing Point Depression  T = i K f m solute n Osmotic Pressure  = i MRT n i =Van’t Hoff factor

Calculate the osmotic pressure in torr of a M solution of NaCl in water at 25 o C. Assume a 100% dissociation of NaCl. Calculation

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/water(CH 3 ) 2 CO/H 2 O b) Ethanol(C 2 H 5 OH)/hexane(C 6 H 14 ) c) Benzene (C 6 H 6 )/toluene CH 3 C 6 H 5. Ideal, Negative, Positive Behavior

Acetone/water(CH 3 ) 2 CO/H 2 O

Ethanol(C 2 H 5 OH)/hexane(C 6 H 14 )

Benzene (C 6 H 6 )/toluene CH 3 C 6 H 5

Define the Van't Hoff factor (i). Which of the following solutions will show the highest osmotic pressure: a) 0.2 M Na 3 PO 4 b) 0.2 M C 6 H 12 O 6 (glucose) c) 0.3 M Al 2 (SO 4 ) 3 d) 0.3 M CaCl 2 e) 0.3 M NaCl

5.00 M.W. = = 128 g/mol

a) True solutions (diameter less than 1 x10 3 pm) b) Colloids (Tyndall effect) (range 1 x10 3 to 1 x10 5 pm) c) Suspensions. (greater than 1 x10 5 pm) Types of Solutions

Tyndall Effect