Physical Properties of Solutions

1. Physical Properties of Solutions Chapter 13 SolutionStoichiometry

2. A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount 12.1

3. nonelectrolyte weak electrolyte strong electrolyte An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. 4.1

4. A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. 12.1

5. The solution Process 1. The ease of solution depends on two factors a. the change in energy (enthalpy) b. the change in disorder (Entropy) 2. The solution process is favored when there is a decrease in enthalpy an increase in entropy

9. 3. Heat of solution Hsolution a. if a solution gets hotter as a substance dissolves - energy is being released i. -Hsolution designates the release of energy ii. The more negative the Hsolution, the more favorable the solution process

10. 4. Main interactions that affect the solution process strong solvent - solvent attractions does not favor solubility weak solvent-solvent attractions favor solubility weak solute - solute attractions favor solubility 5. three process a. breaking apart the solute particles (endothermic) b. breaking apart solvent solvent particles (endothermic) c. interactions of solute and solvent particles (exothermic)

11. Three types of interactions in the solution process: • solvent-solvent interaction • solute-solute interaction • solvent-solute interaction Requires heat Requires heat Gives off heat DHsoln = DH1 + DH2 + DH3 12.2

12. “like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. • non-polar molecules are soluble in non-polar solvents • CCl4 in C6H6 • polar molecules are soluble in polar solvents • C2H5OH in H2O • ionic compounds are more soluble in polar solvents • NaCl in H2O or NH3 (l) 12.2

13. F. dissolution of solids in liquids the process in which solvent molecules surround and interact with solute ions or molecules is called solvation When the solvent is water - it is calledhydration G Dissolution of liquids in liquids (miscibility) 1. Miscibility - ability of one liquid to dissolve in another miscible - mutually soluble in all proportions

14. d- d+ H2O Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. 4.1

15. H factors that effect solubility 1. Types of solvents and solute a. General rule - like dissolves like- polar dissolves polar - nonpolar dissolves nonpolar b. nonpolar mixed with polar will be immiscible 2. surface area - cause more collisions of the solute and the solvent

16. 3. temperature a. in general - as temperature increase so does solubility for solids b. for gases - an increase in temperature causes a decrease in solubility

17. moles of A XA = sum of moles of all components x 100% mass of solute x 100% = mass of solution mass of solute mass of solute + mass of solvent Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass % by mass = Mole Fraction(X) 12.3

18. moles of solute liters of solution moles of solute m = mass of solvent (kg) M = Concentration Units Continued Molarity(M) Molality(m) 12.3

19. What is the molarity of 200.0 g of H3PO4 in enough water to make 2.00 L M = moles Liters You must first find the number of moles of H3PO4 ( 1 mole H3PO4) = 2.040 moles (200.0 g H3PO4) = 1.02 M (98.02 g H3PO4) 2.00 L

20. Molarity continued How many liters can be prepared of a 2.0 M solution of KOH if 400.0 g of KOH is available Remember Molarity is = moles Liters Change Molarity back into these units 2.0 M = 2.0 moles Liters This gives us units that we can work with

21. Change grams to moles ( ) ( ) 1 mole KOH 400.0 g KOH = 7.142 moles KOH 56.01 g KOH ( ) ) 1 liter ( 7.142 Moles KOH = 3.6 L 2.0 Moles KOH

22. How many gram of NaOH are required to make 5.00 L of a 2.5 Molar solution First change Molar into mole/liter 2.5 Molar = 2.5 moles liters ( ) 40.0 g NaOH ( ) = 500 g 2.5 moles liter ( 5.00 L) 1 mole NaOH

23. Colloids b. Colloids - particles that are intermediate in size. i. Particles are on the order of 1 to 100 nm ii. look cloudy iii. Brownian motion iv. Tyndall Effect v. e.g. mayonnaise, gels, foams, smoke, fog, smog

24. Suspensions a. Suspensions - large particles that will settle if not agitated i. Particles are on the order of 100 nm, which is about 1000 times bigger than atoms

25. solubilitydecreases with increasing temperature solubility increases with increasing temperature Temperature and Solubility Solid solubility and temperature 12.4

26. Temperature and Solubility Gas solubility and temperature solubility usually decreases with increasing temperature 12.4

27. low P high P low c high c Pressure and Solubility of Gases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c is the concentration (M) of the dissolved gas c = kP P is the pressure of the gas over the solution k is a constant (mol/L•atm) that depends only on temperature 12.5

28. 4.5

29. Moles of solute before dilution (i) Moles of solute after dilution (f) = Dilution Add Solvent = MfVf MiVi Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. 4.5

30. MfVf 0.200 x 0.06 Vi = = 4.00 Mi How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? MiVi = MfVf Mi = 4.00 Vi = ? L Mf = 0.200 Vf = 0.06 L = 0.003 L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution 4.5

31. Titrations In a titration a solution of accurately known concentration is added gradually to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 4.7

33. H2SO4 + 2NaOH 2H2O + Na2SO4 M M rx volume acid moles acid moles base volume base base acid coef. 4.50 mol H2SO4 2 mol NaOH 1000 ml soln x x x 1000 mL soln 1 mol H2SO4 1.420 mol NaOH What volume of a 1.420 M NaOH solution is Required to titrate 25.00 mL of a 4.50 M H2SO4 solution? WRITE THE CHEMICAL EQUATION! 25.00 mL = 158 mL 4.7

34. Fractional Distillation Apparatus 12.6

35. The Cleansing Action of Soap 12.8

36. moles of solute M = molarity = liters of solution volume KI moles KI grams KI 1 L 2.80 mol KI 166 g KI x x x 1000 mL 1 L soln 1 mol KI Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. What mass of KI is required to make 500. mL of a 2.80 M KI solution? M KI M KI 500. mL = 232 g KI 4.5