Properties of Solutions

1. Properties of Solutions

2. Classification of Matter Solutions are homogeneous mixtures

3. A. 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 • molality - used by chemists

4. Solute A solute is the dissolved substance in a solution. Salt in salt water Sugar in soda drinks Carbon dioxide in soda drinks Solvent A solvent is the dissolving medium in a solution. Water in salt water Water in soda

5. MOLARITY

6. MOLARITY A unit on concentration that is the ratio between moles of DISSOLVED substance and liters of solution

7. Molarity (M) =moles of solute (mol)volume of solution (L)

8. There are many ways to represent molarity: • Molarity • M • Molar • mol / L

9. MAKE SURE YOUR UNITS ARE CORRECT!!!ALWAYS mol/L

10. MOLARITY A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute (n) per liter of solution M = n / V = mol / L Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in 500.0 mL of water. First calculate the moles of solute: 1.5 g NaCl (1 mole NaCl) = 0.0257 moles of NaCl 58.45 g NaCl Next convert mL to L: 0.500 L of solution Last, plug the appropriate values into the correct variables in the equation: M = n / V = 0.0257 moles / 0.500 L = 0.051 mol/L

11. MOLARITY M = n / V = mol/ L How many grams of LiOH is needed to prepare 250.0 mL of a 1.25 M solution? First calculate the moles of solute needed: M = n / V , now rearrange to solve for n: n = MV n = (1.25 mol / L) (0.2500 L) = 0.3125 moles of solute needed Next calculate the molar mass of LiOH: 23.95 g/mol Last, use deminsional analysis to solve for mass: 0.3125 moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH

12. Problem • 4:67 pg 153 a,b

13. C. Dilution • Preparation of a desired solution by adding water to a concentrate. • Moles of solute remain the same.

14. C. Dilution • What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M)V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

15. MOLARITY & Dilution Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar). M1 = 0.05 mol/L M2 = ? V1 = 25.0 mL V2 = 50.0 + 25.0 = 75.0 mL M1V1 = M2V2 M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI V2 75.0 mL

16. Problem • 4:75 pg 153 a,b

17. Chapter 4Aqueous Reactions and Solution Stoichiometry

18. Electrolytes • Substances that dissociate into ions when dissolved in water. • Anonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

19. Electrolytes and Nonelectrolytes Soluble ionic compounds and strong acids tend to be electrolytes. • There are only seven strong acids: • Hydrochloric (HCl) • Hydrobromic (HBr) • Hydroiodic (HI) • Nitric (HNO3) • Sulfuric (H2SO4) • Chloric (HClO3) • Perchloric (HClO4

20. Electrolytes and Nonelectrolytes Molecular compounds tend to be nonelectrolytes, except for acids and bases.

21. Electrolytes • A strong electrolyte dissociates completely when dissolved in water. • A weak electrolyte only dissociates partially when dissolved in water.

22. Strong Electrolytes Are… • Strong acids • Strong bases • Soluble ionic salts

23. Electrolytes? • Pure water • Tap water • Sugar solution • Sodium chloride solution • Hydrochloric acid solution • Lactic acid solution • Ethyl alcohol solution • Pure sodium chloride

24. Answers to Electrolytes NONELECTROLYTES: ELECTROLYTES: • Pure water • Sugar solution • Ethanol solution • Pure NaCl • Tap water (weak) • NaCl solution • HCl solution • Lactate solution (weak)

25. Problem • 4:15 pg 150

26. Precipitation Reactions When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

27. Double replacement Metathesis comes from a Greek word that means “to transpose” It appears the ions in the reactant compounds exchange, or transpose, ions AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq) Metathesis (Exchange) Reactions

28. Need to know which ionic bonds are soluble and which form precipitates Need to use solubility chart Net Ionic Equation

29. Writing Net Ionic Equations • Write a balanced molecular equation. • Dissociate all strong electrolytes. • Cross out anything that remains unchanged from the left side to the right side of the equation. • Write the net ionic equation with the species that remain.

30. Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions. Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) AgCl(s) + K+(aq) + NO3-(aq) Net Ionic Equation AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq) Not soluble

31. Net Ionic Equation • Ag+(aq) + Cl-(aq) AgCl(s)

32. Problem • 4.19pg 151 • 4.39 b and C

33. Acids There are only seven strong acids: • Hydrochloric (HCl) • Hydrobromic (HBr) • Hydroiodic (HI) • Nitric (HNO3) • Sulfuric (H2SO4) • Chloric (HClO3) • Perchloric (HClO4)

34. Bases The strong bases are the soluble salts of hydroxide ion (OH-): • Alkali metals • Calcium • Strontium • Barium

35. When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l) H+ (aq)+ Cl- (aq)+ Na+ (aq) + OH-(aq) Na+ (aq)+ Cl- (aq)+ H2O (l) H+ (aq)+ Cl- (aq)+ Na+ (aq) + OH- (aq) Na+ (aq) + Cl- (aq) + H2O (l) Neutralization Reactions

36. When a strong acid reacts with a strong base, the net ionic equation is… Neutralization Reactions H+ (aq) + OH- (aq) H2O (l)

37. Neutralization Reactions Observe the reaction between Milk of Magnesia, Mg(OH)2, and HCl.

38. Problem • 4.81g 153