1. Chapter 12 Solutions

2. Solutions solute is the dissolved substance ◦ seems to “disappear” ◦ “takes on the state” of the solvent solvent is the substance solute dissolves in ◦ does not appear to change state when both solute and solvent have the same state, the solvent is the component present in the highest percentage solutions in which the solvent is water are called aqueous solutions 2

3. Units of Concentration molarity = moles of solute liters of solution

4. Molarity and Dissociation the molarity of the ionic compound allows you to determine the molarity of the dissolved ions CaCl 2 (aq) = Ca +2 (aq) + 2 Cl -1 (aq) A 1.0 M CaCl 2 (aq) solution contains 1.0 moles of CaCl 2 in each liter of solution ◦ 1 L = 1.0 moles CaCl 2, 2 L = 2.0 moles CaCl 2 Because each CaCl 2 dissociates to give one Ca +2 = 1.0 M Ca +2 ◦ 1 L = 1.0 moles Ca +2, 2 L = 2.0 moles Ca +2 Because each CaCl 2 dissociates to give 2 Cl -1 = 2.0 M Cl -1 ◦ 1 L = 2.0 moles Cl -1, 2 L = 4.0 moles Cl -1 4

5. Using Concentrations as Conversion Factors concentrations show the relationship between the amount of solute and the amount of solvent ◦ 12%(m/m) sugar(aq) means ◦ 5.5%(m/v) Ag in Hg means ◦ 22%(v/v) alcohol(aq) means 5

6. Example A solution is prepared by mixing 15.0 g of Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (% m/m) of the solution. What volume of 10.5% by mass soda contains 78.5 g of sugar? Density of solution is 1.04 g/ml

7. Solution Concentration PPM grams of solute per 1,000,000 g of solution mg of solute per 1 kg of solution 1 liter of water = 1 kg of water ◦ for water solutions we often approximate the kg of the solution as the kg or L of water 7 grams solute grams solution x 10 6 mg solute kg solution mg solute L solution

8. Solution Concentrations Mole Fraction, X A the mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution total of all the mole fractions in a solution = 1 unitless the mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution = mole fraction x 100% 8 mole fraction of A = X A = moles of components A total moles in the solution

9. Example What is the percent by mass of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with 0.500 kg of H 2 O to make 515 mL of solution? What is the mole fraction of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with 0.500 kg of H 2 O to make 515 mL of solution? A water sample is found to contain the pollutant chlorobenzene with a concentration of 15 ppb (by mass). What volume of this water contains 5.00 x 10 2 mg of chlorobenzene? Assume density of 1.00 g/ml

10. Mixing and the Solution Process Entropy formation of a solution does not necessarily lower the potential energy of the system ◦ the difference in attractive forces between atoms of two separate ideal gases vs. two mixed ideal gases is negligible ◦ yet the gases mix spontaneously the gases mix because the energy of the system is lowered through the release of entropy entropy is the measure of energy dispersal throughout the system energy has a spontaneous drive to spread out over as large a volume as it is allowed 10

11. Will It Dissolve? Chemist’s Rule of Thumb – Like Dissolves Like a chemical will dissolve in a solvent if it has a similar structure to the solvent when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other 11

12. Intermolecular Attractions 12

13. Energy changes and the solution process Simply put, three processes affect the energetics of the process: _ Separation of solute particles ΔH 1 ( this is always endothermic) _ Separation of solvent particles ΔH 2 ( this too is always endothermic) _ New interactions between solute and solvent ΔH 3 ( this is always exothermic) The overall enthalpy change associated with these three processes : ΔH soln = ΔH 1 + ΔH 2 + ΔH 3 (Hess’s Law)

14. Intermolecular Forces and the Solution Process Enthalpy of Solution The solute-solvent interactions are greater than the sum of the solute-solute and solvent- solvent interactions. The solute-solvent interactions are less than the sum of the solute-solute and solvent- solvent interactions.

15. Relative Interactions and Solution Formation Solute-to-Solvent> Solute-to-Solute + Solvent-to-Solvent Solution Forms Solute-to-Solvent= Solute-to-Solute + Solvent-to-Solvent Solution Forms Solute-to-Solvent< Solute-to-Solute + Solvent-to-Solvent Solution May or May Not Form when the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the entropy 15

16. Heats of Hydration for aqueous ionic solutions, the energy added to overcome the attractions between water molecules and the energy released in forming attractions between the water molecules and ions is combined into a term called the heat of hydration ◦ attractive forces in water = H-bonds ◦ attractive forces between ion and water = ion-dipole ◦ H hydration = heat released when 1 mole of gaseous ions dissolves in water 16

17. Ion-Dipole Interactions when ions dissolve in water they become hydrated each ion is surrounded by water molecules 17

18. Solubility Limit a solution that has the maximum amount of solute dissolved in it is said to be saturated ◦ depends on the amount of solvent ◦ depends on the temperature  and pressure of gases a solution that has less solute than saturation is said to be unsaturated a solution that has more solute than saturation is said to be supersaturated 18

19. Example Example: The solubility of NaNO 3 in water at 50 o C is 110g/100g of water. In a laboratory, a student use 50.0 g of NaNO 3 with 200 g of water at the same temperature ◦ How many grams of NaNO 3 will dissovle? ◦ Is the solution saturated or unsaturated? ◦ What is the mass, in grams, of any solid NaNO 3 on the bottom of the container?

20. Temperature Dependence of Solubility of Solids in Water Solubility depends on temperature most solids increases as temperature increases. ◦ Hot tea dissolves more sugar than does cold tea because the solubility of sugar is much greater in higher temperature When a saturated solution is carefully cooled, it becomes a supersaturated solution because it contains more solute than the solubility allowssolubility is generally given in grams of solute that will dissolve in 100 g of water for most solids, the solubility of the solid increases as the temperature increases ◦ when  H solution is endothermic 20

21. Solubility Curve solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line)

22. Temperature Dependence of Solubility of Gases in Water solubility is generally given in moles of solute that will dissolve in 1 Liter of solution generally lower solubility than ionic or polar covalent solids because most are nonpolar molecules for all gases, the solubility of the gas decreases as the temperature increases ◦ the  H solution is exothermic because you do not need to overcome solute-solute attractions the solubility of gases in water increases with increasing mass as the attraction between the gas and the solvent molecule is mainly dispersion forces ◦ Larger molecules have stronger dispersion forces. 22

23. Henry’s Law the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid. at higher pressures, more gas molecules dissolve in the liquid.

24. Henry’s Law Solubility = k ·P where k is the Henry’s law constant for that gas in that solvent at that temperature P is the partial pressure of the gas above the liquid. Example: Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO 2 in water at this temperature is 3.1 x 10 –2 mol/L-atm

25. Thirsty Solutions 25 When equilibrium is established, the liquid level in the solution beaker is higher than the solution level in the pure solvent beaker – the thirsty solution grabs and holds solvent vapor more effectively Beakers with equal liquid levels of pure solvent and a solution are place in a bell jar. Solvent molecules evaporate from each one and fill the bell jar, establishing an equilibrium with the liquids in the beakers.

26. Raoult’s Law the vapor pressure of a volatile solvent above a solution is equal to its mole fraction of its normal vapor pressure, P° P solvent in solution =  solvent ∙ P° ◦ since the mole fraction is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent 26

27. Raoult’s Law for Volatile Solute when both the solvent and the solute can evaporate, both molecules will be found in the vapor phase the total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent ◦ for an ideal solution P total = P solute + P solvent the solvent decreases the solute vapor pressure in the same way the solute decreased the solvent’s P solute =  solute ∙ P° solute and P solvent =  solvent ∙ P° solvent 27

28. Ideal vs. Nonideal Solution in ideal solutions, the made solute- solvent interactions are equal to the sum of the broken solute-solute and solvent-solvent interactions ◦ ideal solutions follow Raoult’s Law effectively, the solute is diluting the solvent if the solute-solvent interactions are stronger or weaker than the broken interactions the solution is nonideal when the solute-solvent interactions are stronger than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be less than predicted by Raoult’s Law ◦ because the vapor pressures of the solute and solvent are lower than ideal when the solute-solvent interactions are weaker than the solute-solute + solvent-solvent, the total vapor pressure of the solution will be larger than predicted by Raoult’s Law 28 Vapor Pressure of a Nonideal Solution

29. Example – Calculate the vapor pressure of water in a solution prepared by mixing 99.5 g of C 12 H 22 O 11 with 300.0 mL of H 2 O. P H2O = 23.8 atm