Chapter 12 Properties of Solutions

Liquids 2 Properties of Liquids A. Viscosity B. Surface Tension

Viscosity Viscosity is the resistance of a liquid to flow. The stronger the intermolecular forces, the higher the viscosity. Some Properties of Liquids

Viscosity Some Properties of Liquids

Surface Tension

Surface tension is the amount of energy required to increase the surface area of a liquid. Resistance to increased surface area or the tendency of a liquid to maintain minimum surface area. Cohesive forces bind molecules to each other. Adhesive forces bind molecules to a surface. Ex. miniscus Some Properties of Liquids

Surface Tension Meniscus is the shape of the liquid surface. –If adhesive forces are greater than cohesive forces, the liquid surface is attracted to its container more than the bulk molecules. Therefore, the meniscus is U-shaped (e.g. water in glass). –If cohesive forces are greater than adhesive forces, the meniscus is curved downwards. Some Properties of Liquids

Definitions Solvent – what does the dissolving –The one that is greater quantity Solute – what gets dissolved – can be liquid, solid or gas –The one that is lesser in quantity

A solution is always homogeneous!

Factors Affecting Solubility Temperature Pressure Solvent-Solvent Interactions

Rule of Thumb LIKE DISSOLVES LIKE! Polar solutes dissolve in polar solvents. Non-polar solutes dissolve in non-polar solvents. Compounds with metals (Na, K, Li, etc…) are ionic and therefore polar.

Solution Concentration –Molarity (M) –% Mass, % Volume, %Mass/Vol. –Mole Fraction –Molality (m) –Normality (will be skipped)

Molarity (M) Molarity = moles of solute Liter of solution Moles=Molarity x Liter of soln. Moles=grams molar mass

Sample Problem Calculate the molarity of a solution containing 3.65 grams of HCl in enough water to make mL of solution. Ans:0.200 M

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

Mole Fraction (  ) To Solve for the Mole Fraction of A :  A =Mole A Mole A + Mole B

Sample Problem 1.00 gram of CH 3 CH 2 OH (ethanol) is mixed with grams of water. Calculate the mole fraction of ethanol in this solution.

3 Types of % Concentrations –Percent by Mass –Percent by Volume –Percent by Mass-Volume

Percent by Mass % Mass = mass of solutex100 total mass of solution Sample Problem: Find the concentration of 20 grams of sugar in enough water to make 350 grams of solution. Answer: 5.7%

Things to Note: Identity of the substance (molecular formula) is not taken into account since only the masses are needed in the equation.

Percent by Volume % Volume = volume of solute x 100 Total Vol. of solution

Sample Problem 10.0 mL of benzene is added to 40.0 mL of carbon tetrachloride. Find the concentration of the solution. Ans: 10 mL / 50 mL) x 100 = 20%

Percent by Mass-Volume % Mass-Volume =mass of solutex 100 Total Vol. of Solution

Answers to Problem Set o C 2. a. molar mass = 181 b. Tantalum c x a b c

Molality (m) Molality =moles kg. solvent

Usefulness of Molality Chemists find that molality (m) is a useful concept when they must deal with the effect of a solute on boiling point and freezing point of a solution.

Problem A solution is made containing 25.5 g phenol (C 6 H 5 OH) in 495 g ethanol (CH 3 CH 2 OH) Calculate: A. mole fraction of phenol B. mass % of phenol C. molality of phenol

Colligative Properties Depend upon the number of particles in solution rather than the identity of the compound Therefore if solute does not ionize, the concentration of the solution does not change and there is no change in the calculation of the mole fraction (  soln )

Colligative Properties If solute ionizes, then the number of particles the solute breaks down into should be factored in. Example: NaCl Na + +Cl - »1 mole 1 mole1 mole

Colligative Properties If solute is Na 2 SO 4 Then number of particles is 3 Na 2 SO 4 = 2Na + +SO 4 2-

Colligative Properties Boiling Point Elevation Freezing Point Depression Osmotic Pressure Vapor Pressure lowering

Freezing Point Depression Presence of solute in solution decreases the freezing point of solution  T = k f m solute where  T = change in Temperature in o C k f = molal freezing point depression constant ( o C/molal) m = molality

Effect of Ionizing substances The presence of solutes that ionize have to be factored into the equation such that if :  T = k f m solute, Now:  T = ik f m solute where i = the # of ions produced.

Boiling Point Elevation Non-volatile solute elevates the boiling point of solvent Presence of non-volatile solute decreases vapor pressure, thus solution has to be heated to a higher temperature to boil.

Boiling Point Elevation  T = k b m solute Where:  T = change in Temp. (in o C) k b = molal boiling point depression constant ( o C/molal) m = molality

Sample Problem A solution is prepared by dissolving 4.9 grams sucrose (C 12 H 22 O 11 = Mol. Mass: g/mol) in 175 grams of water. Calculate the boiling point of this solution. Calculate the freezing point of this solution. Sucrose is a non- electrolyte. K b of water is 0.51 o C/molal. K f of water is 1.86 o C/molal.

Sample Problem Calculate the boiling point and freezing point of 0.50 m FeCl 3 solution. Assume complete dissociation. K b of water is 0.51 o C/molal. K f of water is 1.86 o C/molal.

Sample Problem A solution of an unknown nonvolatile compound was prepared by dissolving g of the substance in 40.0 g of CCl 4. The boiling point of the resulting solution was o C higher than that of the pure solvent. Calculate the molar mass of the solute. K b for CCl 4 is Normal boiling point of CCl 4 is 76.8 o C

Lab Experiment Determine molality (m) using the equation  T = k f m solute Determine molecular mass of sulfur using the equation molality (m) = gram mol. mass Kg. solvent

Osmosis Osmosis – direction of flow is from less concentrated to more concentrated - shrivelling Reverse Osmosis - from more concentrated to less concentrated - bloating

Equation Osmotic Pressure  = MRT where M = Molarity R = ideal gas constant ( L-atm/K-mol) T = Temperature in Kelvin  = osmotic pressure in atm

Effect of Ionizing substances Again the presence of solutes that ionize has to be considered:  = MRT, Now:  = iMRT where i = the # of ions produced.

Sample Problem A grams sample of a non- dissociating protein is dissolved in water to make 25.0 mL of solution. The osmotic pressure of the solution is 0.56 torr at 25 o C. What is the molar mass of the protein?

Solution Formation Energies are involved in solution formation to: A. Break the solute apart (endothermic) B. Break the intermolecular forces of the solvent (endothermic =  H is positive) C. Allow solute and solvent molecules to interact (exothermic =  H is negative)

Therefore the enthalpy of the solution  H solution =  H A +  H B +  H C If  H soln is (+) = means  H C is small If  H soln is (-) = means  H C is large

Factors Affecting Solubility A. Structure Effects –Hydrophilic – loves water –Hydrophobic – hates water B. Pressure Effects – more important for gases C. Temperature effects

IDEAL SOLUTION When Raoult’s Law is closely followed Very little interaction between molecules (i.e., bonding between solute and solvent) If only LDF were present, Raoult’s Law closely followed.

Temperature Effect Increasing temperature ONLY increases the RATE of solubility and not solubility itself. In some cases, higher Temp. decreases solubility. Experimentation needed.

Raoult’s Law States that the partial pressure exerted by solvent vapor above the solution (P A ), equals the product of the mole fraction of the solvent in the solution (  A ) times the vapor pressure of the pure solvent (P o ) P A = (  A )( P o ) Partial pressure = vapor pressure above soln.

Vapor Pressure The total vapor pressure of the solution is the sum of the partial pressures of each volatile component. P tot (above soln) = P solvent(above soln) + P solute(above soln) Partial Pressure of A(solv.) in soln: P solvent =  solvent P o Partial Pressure of B(solute) in soln: P solute =  solute P o

Case 1- all volatile components (Ideal Solution) The total vapor pressure over the solution is the sum of the partial pressures of each volatile component P tot(above soln) = P solv. (above soln) + P solute (above soln) = (  solvent in soln )( P o ) + (  solute in soln )( P o )

Case 2 – non-volatile solute (Non-Ideal Solution) P tot above soln = P solvent (above soln) + P solute (above soln) Since pure pressure of a non-volatile solute is ZERO, the equation simplifies to: P tot above soln = (  A in soln )( P o )

Sample Consider a sample containing 3.0 mole of benzene (C 6 H 6 ) and 5 moles of toluene (C 7 H 8 ). The vapor pressure of pure benzene is 88 torr. For pure toluene, it is 34 torr. A. What is the mole fraction of benzene? B. What is the mole fraction of toluene? C. What are the partial pressures of benzene and toluene above the solution. D. What is the total vapor pressure of the solution?

Problem A. Calculate the vapor pressure of water prepared by dissolving 35.0 grams of glycerin (C 3 H 8 O 3 ) in 125 grams of water at 343 K. The vapor pressure of pure water is in Appendix B.) B. Calculate the mass of ethylene glycol (C 2 H 6 O 2 ) that muct be added to 1.00 kg of ethanol (C 2 H 5 OH) to reduce its vapor pressure by 10.0 torr at 35 o C. The vapor pressure of pure ethanol at 35 oC is 1.00 x 10 2 torr.

Problem At 20 o C, the vapor pressure of pure benzene is 75 torr and that of pure toluene (C 7 H 8 ) is 22 torr. Assume that benzene and toluene form an ideal solution. A. What is the composition in mole fractions of a solution that has a vapor pressure of 35 torr at 20 o C? B. What is the mole fraction of benzene in the vapor above the solution described in part (a)?

Mole fraction in vapor C in vapor = P volatile component P total

HOMEWORK Do Problems: and 13.61

Henry’s Law States that the amount of gas dissolved is proportional to the amount of gas above the solution

Pressure Effect Henry’s Law:S = Pk –Where k is the constant characteristic of a particular solution –S is the solubility of dissolved gas

END of CHAPTER 12