4. Chapter 5: The Nature and Properties of Solutions

5. Solution: A homogenous mixture that looks the same throughout. Solute: Substance being dissolved. Present in smaller amount. Solvent: Substance doing the dissolving. Present in greater amount.

7. Solute Solute - KMnO 4 Solvent Solvent - H 2 O Demo!! Substance doing the dissolving. Present in greater amount. Substance being dissolved. Present in smaller amount.

8. Solvation: Solvation: solute particles are separated and pulled into solution solute particles are surrounded by solvent particles The process of dissolving.

9. Conducting Electricity When a solute (solid or liquid) is placed in a solvent 1 of two things can happen. 1. The solute can sink to the bottom and be insoluble (slightly soluble). 2. The solute can dissolve in the solvent and do MANY things.

10. Conducting Electricity If a compound dissolves in the solvent it can either break apart and create positive and negative ions. Or it can stay together and NOT form ions. Whether or not a substance breaks apart depends on what type of compound it is. Na + Cl - Na + Cl - NaCl (Salt)

11. Molecular Compounds (Non-Metal + Non-Metal) Molecular Compounds do not break up and form ions. Thus, cannot conduct electricity. Sugar(s) Sugar(aq)

12. Molecular Acids (Weak Acids only) Weak molecular acids only break up slightly, only a few ions are created. Thus, poor conductors of electricity. Acetic Acid (Vinegar) (CH 3 COOH) H+H+H+H+ CH3COO -

13. Ionic Compounds (Metal + Non-Metal) Ionic Compounds break apart when they dissolve and produce lots of ions. Thus are excellent conductors of electricity. Salt (NaCl) Na + Cl - Na + Cl - Na + Cl - Na +

15. Conducting Electricity Strong Electrolyte Non- Electrolyte solute exists as ions only - + salt - + sugar solute exists as molecules only - + acetic acid Weak Electrolyte solute exists as ions and molecules DISSOCIATIONIONIZATION

16. Dissociation (Ionic): separation of an ionic solid into aqueous ions separation of ions that already existed before being added to water NaCl(s)  Na + (aq) + Cl – (aq) Dissociation Equation Back

17. Ionization (Molecular Acids) Ionization (Molecular Acids) breaking apart of some polar molecules into aqueous ions the production of new ions, specifically hydrogen/hydronium ions HNO 3 (aq) + H 2 O(l)  H 3 O + (aq) + NO 3 – (aq) Ionization Equation

18. Molecular Solvation (Molecular) Molecular Solvation (Molecular) Polar molecules stay intact. C 12 H 22 O 11 (s)  C 12 H 22 O 11 (aq) Dissociation Equation

19. Molecular Solvation NONPOLAR POLAR “Like Dissolves Like”

21. Substances in Water Ionic Substances: Solubility table allows you to predict if substance will dissociate 100% or not (s or aq) Molecular Compounds: require the understanding of intermolecular forces to predict entities in water (s/l/g or aq) Polar substances dissolve in water (like dissolves like) Hydrogen bonding makes them even more soluble Elements: Most will remain the same (s,l, or g) Exception: Group 1 elements like sodium will dissolve in water to form basic solutions (aqueous ions)

22. Universal Solvent Water is often referred to as the “universal solvent” due to its ability to dissolve many things. However, there is no actual universal solvent. How would you store such a liquid?

23. Acids Empirical Definitions (observable properties): Acids: taste sour Electrolytes (Weak Acids Low…..Strong Acids High) React with metals to produce hydrogen gas Turn blue litmus red Neutralize bases

24. Acids Properties of both ionic and molecular compounds. Strong acids (6) are strong electrolytes (complete ionization). 1. HCl 2.HBr 3.HNO 3 4.H 2 SO 4 5.____________ 6. _____________ Strong Acids are Strong electrolytes (Complete dissociation).

27. Weak Acids are Weak electrolytes (Partial dissociation).

29. Bases Empirical Definitions (observable properties): Bases: taste bitter (coffee) Electrolytes (Weak Bases Low…..Strong Bases High) Feel greasy/slippery Turn Red litmus blue Neutralize acids

30. Bases Properties of both ionic and molecular compounds. Strong bases are strong electrolytes (complete dissociation). Strong bases are any compounds containing OH (hydroxide). -NaOH -Ba(OH) 2 -Ca(OH) 2 -Li 2 (OH) -Mg(OH) -KOH

31. Definitions Arrhenius - In aqueous solution… Arrhenius - In aqueous solution… HCl + H 2 O  H 3 O + + Cl – AcidsAcids form hydronium ions (H 3 O + ) H HHHH H Cl OO – + acid

32. Definitions Arrhenius - In aqueous solution… Arrhenius - In aqueous solution… BasesBases form hydroxide ions (OH - ) NH 3 + H 2 O  NH 4 + + OH - H H H H H H N NO O – + H H H H base

36. Qualitative vs. Quantitative Qualitative: what is present in a solution Is the solution acidic/basic/ionic etc.. Quantitative: how much is present in a solution What is the concentration of the solution

37. Concentration The amount of solute in a solution. Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists

38. Concentration SAWS Water Quality Report - June 2000

39. % by Volume Vinegar is usually labeled as 5% acetic acid by volume This means there is 5mL of acetic acid in every 100mL of vinegar solution

40. % Weight per volume Hydrogen Peroxide used as antiseptic is 3% W/V This means there are 3g of hydrogen peroxide for every 100mL of solution

41. % weight by weight This form is used when making a solution of two solids like in the production of a metal alloy such as sterling silver Merely a flesh wound

42. ppm Parts per million is often used for very dilute solutions and is very important in environmental studies This is a special case of the W/W ratio

43. Molarity/amount concentration

44. molar mass (g/mol) 6.02  10 23 (particles/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molarity (mol/L)

45. Example: Find the molarity of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 = 3.2 mol/L MgCl 2 0.25 L water X mol MgCl 2

46. Example: How many grams of NaCl are req’d to make a 1.54mol/L solution using 0.500 L of water? 0.500 L water 1.54 mol NaCl 1 L water = 0.77 mol X mol NaCl

47. 58.44 g NaCl 1 mol NaCl X g NaCl 0.77 mol NaCl = 45 g

48. Concentration of Ions Balanced dissociation or ionization equations for soluble ionic substances and strong acids allows us to determine the amount concentration of ions or compounds in solution

49. Concentration of Ions Use the mole ratio from the balanced equation to set up the need over got conversion factor. ex: What is the amount concentration of aluminum ions and sulfate ions in a 0.40 mol/L solution of aluminum sulfate? Al 2 (SO 4 ) 3(aq)  2Al 3+ (aq) + 3SO 4 2- (aq) C = 0.40 mol/L N G NGNG ( )( ) 2121 C Al3+(aq) = (0.40 mol/L)(2/1) = 0.80 mol/L C = 0.80 mol/L C = ?

50. Concentration of Ions Al 2 (SO 4 ) 3(aq)  2Al 3+ (aq) + 3SO 4 2- (aq) C = 0.40 mol/L N G NGNG ( )( ) 3131 C SO4 2 -(aq) = (0.40 mol/L)(3/1) = 1.20 mol/L C = ? C=1.20 mol/L

52. Definitions: Standard Solution: solutions of accurate concentration made via dilution or dissolving a solid Stock Solution: an initial, usually concentrated, solution from which solutions are taken for dilution

53. Dilution Preparation of a standard solution by adding water to a stock solution. Moles of solute remain the same.

54. Example: What volume of 15.8mol/L HNO 3 is required to make 250 mL of a 6.0mol/L solution? V 1 = 95 mL of 15.8M HNO 3 is needed. C 1 = V 1 = C 2 = V 2 = 15.8 mol/L 250 ml0.250 L 6.0 mol/L ??? C 1 V 1 = C 2 V 2 V 1 = C 2 V 2 C 1 V 1 =(6.0 mol/L)(0.250L (15.8 mol/L) V 1 = 0.09493671 L ~ 0.095L or 95 ml

55. Preparation of Solutions: Materials Analytical Balance Pipette 1ml 5ml 10 ml 25 ml 50ml Filter Paper Mixing Beaker Distilled Water Graduated Cylinder Eye Dropper Pipette Bulb Volumetric Flask Scoopula

56. Preparing Solutions Prepare a 0.500L solution of 1.54 M (mol/L) NaCl. C= n= V= 1.54 mol/L 0.500 L ?????? C = n V n = CV n = (1.54 mol/L)(0.500L) n = 0.77 mol Can you measure 0.77 mols on a scale and mix it to make a solution? n = m M m = (n)(M)

57. Preparing Solutions Prepare a 0.500L solution of1.54 M (mol/L). m = (n)(M) M NaCl = 1 X 22.99 g/mol 1 X 35.45 g/mol + 58.44 g/mol m = (0.77 mol)(58.44 g/mol) m = 44.9988 g m = 45 g

58. Preparing Solutions 500 mL water 45.0 g NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 L of water 500 mL mark 500 mL volumetric flask Prepare a 0.500L solution of1.54 M (mol/L). A. B.

59. Preparing Solutions 250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!”

60. Solution Preparation Lab Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution. Make each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25M NaCl in 100.0 mL of water 3) 250.0 mL of 0.25M NaCl from 0.50M concentrate (1).

62. SATURATED SOLUTION no more solute can dissolve UNSATURATED SOLUTION more solute can dissolve SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

63. Solubility Solubility maximum grams of solute that will dissolve in 100 mL of solvent (units: g/100mL) varies with temp based on a saturated solution Solubility Curve Solubility Curve shows the dependence of solubility on temperature.

64. Solids are more soluble at... Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures & high pressures EX: nitrogen narcosis, the “bends,” soda, champagne, Oxygen levels in lakes

65. Net Energy Changes When a substance dissolves in a solvent and forms ions the original bonds of the substance are broken. Energy is required to break the bonds between Na (sodium) and Cl (chlorine). When the sodium and chlorine ions (Na+ and Cl-) form new bonds with water (aq), the news bonds release energy.

66. Net Energy Changes Net = (Energy of Bonds Broken) – (Energy of Bonds Formed) Net = Negative (-) Net = Positive (+) Exothermic Reaction Endothermic Reaction

67. Exo and Endo If the temperature of a solution decrease as a solute is dissolved the reaction is endothermic (entering the reaction) Energy is a reactant EQUATION: If the temperature of a solution rises as a solute is dissolved the reaction is exothermic (exiting the reaction) Energy is a product EQUATION: Reactant + Reactant + Energy  Products Reactant + Reactant  Products + Energy

68. Explaining Saturated Solutions Dynamic Equilibrium: the balance that exists when two opposing processes occur at the same time