Presentation on theme: "Chapter 4 Solutions and Chemical Reactions I.Water A.Importance 1.Life (as we know it) depends on water 2.Human civilization requires water for many purposes."— Presentation transcript:
Chapter 4 Solutions and Chemical Reactions I.Water A.Importance 1.Life (as we know it) depends on water 2.Human civilization requires water for many purposes 3.Many important chemical reactions occur in Aqueous Solutions, where other compounds are dissolved in water B.The nature of water 1.Bent shape and unequal sharing of electrons makes water polar 2.This aids water in dissolving ionic compounds (cations and anions) 3.Water hydrates the ions by interacting with its oppositely charged ends
4.The ionic substance breaks up into independent cations and anions 5.Nonionic compounds can also dissolve in water if they are polar 6.Nonpolar substances generally don’t dissolve in water: grease, oils, skin II.Electrolytes A.Solutions 1.A solution is a homogeneous mixture the same throughout 2.We can vary the composition by adding more or less of the components 3.Solvent = usually a liquid; the most abundant component of a solution 4.Solute = the lesser abundant component(s) of a solution Ethanol NaCl(s) -----> Na + (aq) + Cl - (aq)
B.Solutions and Electrical Conductance 1.A substance allowing current to flow through it is electrically conductive 2.Pure water does not conduct electricity 3.Different solutes dissolved in water help it to be conductive a.Strong electrolyte = completely ionized; strongly conductive solution b.Weak electrolyte = partially ionized; somewhat conductive solution c.Nonelectrolyte = not ionized; nonconductive solution Arrhenius ( ) found that the more ions present, the better the conductivity
C.Strong Electrolytes 1.Completely ionized when dissolved in water 2.Many salts (ionic compounds) are strong electrolytes 3.Strong Acids are strong electrolytes a.Acid = substance that produces H + when dissolved in water b.Strong Acids completely ionize in solution i.Hydrochloric Acid HCl(g) > H + (aq) + Cl - (aq) ii.Nitric Acid HNO 3 (g) > H + (aq) + NO 3 - (aq) iii.Sulfuric Acid H 2 SO 4 (l) > H + (aq) + HSO 4 - (aq) 4.Strong Bases are strong electrolytes a.Base = substance that produces OH - when dissolved in water b.Strong bases completely ionize in solution NaOH(s) > Na + (aq) + OH - (aq) KOH(s) > K + (aq) + OH - (aq)
D.Weak Electrolytes 1.Only partially ionized when dissolved in water 2.Weak Acids are weak electrolytes a.Weak acid only produces a few H + ions b.Acetic acid is a weak acid c.HC 2 H 3 O 2 (aq) > H + (aq) + - C 2 H 3 O 2 (aq) d.Only 1 molecule in a 100 dissociates 3.Weak Bases are weak electrolytes a.Weak base produces only a few OH - ions b.Ammonia is a weak base c.NH 3 (aq) + H 2 O(l) -----> NH 4 + (aq) + OH - (aq) d.Only 1 molecule in 100 reacts E.Nonelectrolytes 1.Does not ionize when dissolve in water 2.Sugar is a nonelectrolyte 3.C 12 H 22 O 11 (s) > C 12 H 22 O 11 (aq)
III.Solution Concentration A.The Stoichiometry of Chemical Reactions 1.We must know what the reactants and products are 2.We must know the amounts of the reactants and products 3.How do we describe the amounts in a solution? B.Molarity 1.Unit for the concentration of a solute in a solution 2.M = moles solute/liters of solution M NaCl = 1 mole of NaCl dissolved in 1 L of solution a.Any volume having the same concentration is also 1.0 M NaCl b.500 ml (0.500 L) of 1.0 M NaCl would contain 0.5 mol NaCl 4.Example: Calculate M of 11.5 g NaOH in 1.5 L of total solution.
5.Example: M = ? for 1.56 g HCl in a total of 26.8 ml of solution? 6.Molarity descriptions of a solution reflect composition before dissolution a.1.0 M NaCl actually contains no NaCl b.1.0 M NaCl is 1.0 M in Na + and 1.0 M in Cl - c.1.0 M CaCl 2 is 1.0 M in Ca 2+ and 2.0 M in Cl - d.CaCl 2 (s) > Ca 2+ (aq) + 2Cl - (aq) 7.Example: Give the concentration of each ion a.0.5 M Co(NO 3 ) 2 = 0.5M in Co 2+ and 1.0 M in NO 3 - b.Co(NO 3 ) 2 (s) > Co 2+ (aq) + 2NO 3 - (aq) c.1M Fe(ClO 4 ) 3 = 1M Fe 3+ and 3M ClO Example: ??? moles of Cl - in 1.75L of M ZnCl 2
9. Example: What volume of 0.14 M NaCl contains 1.0mg NaCl? 10. Standard Solution = concentration is accurately known a.Accurate masses come from an analytical balance (0.4563g) b.Accurate volumes are obtained using a Volumetric Flask c.Example: How much K 2 Cr 2 O 7 needed for 1.00 L of M?
C.Dilution 1.Chemicals are often purchased or prepared as concentrated stock solutions 2.Dilution = adding water to stock solution to make a less concentrated one 3.M 1 V 1 = M 2 V 2 is a useful equation to calculate dilutions 4.Example: ??? volume of 16 M H 2 SO 4 is needed for 1.5 L 0.10M H 2 SO 4
IV.Precipitation Reactions A.Definitions 1.When two solutions are mixed and a solid forms 2.Precipitate = solid that forms from a precipitation reaction 3.K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq) = 2K + (aq) + CrO 4 2- (aq) + Ba 2+ (aq) +2NO 3 - (aq) a.K 2 CrO 4 and Ba(NO 3 ) 2 are both soluble (all dissolve in water) b.A yellow precipitate forms when these solutions are mixed c.K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq) -----> BaCrO 4 (s) + 2KNO 3 (aq) 4.AgNO 3 (aq) + KCl(aq) > AgCl(s) + KNO 3 (aq) + = Precipitate Spectator Ions
B.Solubility Rules Example: predict what will happen when you mix: a.KNO 3 (aq) + BaCl 2 (aq) > b.Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq) > c.3KOH(aq) + Fe(NO 3 ) 2 (aq) >
C.Describing Reactions in Solution 1.Molecular Equation shows what compounds the ions came from a.Does not give clear picture of what happens in solution b.K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq) -----> BaCrO 4 (s) + 2KNO 3 (aq) 2.Complete Ionic Equation represents the form of the ions in solution a.All strong electrolytes are represented as their ions 2K + (aq) + CrO 4 2- (aq) + Ba 2+ (aq) +2NO 3 - (aq) ----> BaCrO 4 (s) + 2K + (aq) + 2NO 3 - (aq) 3.Net Ionic Equation shows only the ions participating in the reaction a.The K + and NO 3 - ions occur on both sides of the complete ionic eqn. b.These spectator ions can be cancelled out of each side (algebra) c.Ba 2+ (aq) + CrO 4 2- (aq) > BaCrO 4 (s) 4.Example a.3KOH(aq) + Fe(NO 3 ) 3 (aq) -----> Fe(OH) 3 (s) + 3KNO 3 (aq) b. 3K + (aq) + 3OH - (aq) + Fe 3+ (aq) + 3NO 3 - (aq) ----> Fe(OH) 3 (s) + 3K + (aq) + 3NO 3 - (aq) c. Fe 3+ (aq) + 3OH - (aq) > Fe(OH) 3 (s)