Chapter 4. Reactions In Aqueous Solution

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Chapter 4 Reactions In Aqueous Solution I) General Properties of Aqueous Solutions Homogeneous mixture on a molecular level - prop. same throughout - separable by physical means - variable composition Has at least 2 components: Solvent : Solute(s) : component present in greatest amount - same phase as soln. other component(s) 1

A) Dissolution Process of dissolving Solute is solvated by solvent Solvent surrounds solute Solvation 1) Aqueous H 2 O as solvent 2

3

B) Nonelectrolytes Substances which dissolve in molecular form - Present as molecules in solid, liquid & gas states and in soln. - do NOT form ions in soln. Soln. does NOT conduct electricity Pure H 2 O 4

C) Electrolytes Substance whose aqueous soln. conducts electricity - due to ions in the soln. 1) Dissociation Ionic solutes dissolve as ions NaCl (s) H 2 O Na + (aq) + Cl! (aq) Dissociation All soluble ionic substances dissociate completely - strong electrolytes 5

2) Ionization Some polar covalent molecules dissolve w. formation of ions. 1) Strong Acids HCl (g) H 2 O HCl (aq) HCl (aq) H 2 O H + (aq) + Cl! (aq) Ionization HCl - strong electrolyte (completely ionizes) 6

2) Weak Electrolytes Weak conductors - Partial ionization a) Weak Acids & Bases Chemical Equilibrium - some molecular form remains 7

II) Precipitation Reactions Precipitate / insoluble solid formed 2 KI(aq) + Pb(NO 3 ) 2 (aq) 2 KNO 3 (aq) + PbI 2 (s) A) Solubility Rules Which ionic cmpds. are soluble or insoluble in H 2 O 1) Solubility amt. of solid that dissolves in a given amt. of H 2 O at a given temp. NOTE: ALL ionic cmpds. of group IA & NH 4 + are soluble 8

EMPIRICAL RULES FOR THE SOLUBILITY OF IONIC SOLIDS IN H 2 O COMPOUNDS CONTAINING SOLUBILITY IMPORTANT EXCEPTIONS alkali metal (grp 1A) ammonium (NH 4+ ) Soluble None nitrates (NO 3 - ), acetates (C 2 H 3 O 2 - ), chlorates (ClO 3 - ), perchlorates (ClO 4 - ), permanganates (MnO 4 - ) Soluble None chlorides (Cl - ), bromides (Br - ), iodides (I - ) Soluble Cmpds of Ag +, Hg 2 2+, Pb 2+, Hg 2+ iodide and Hg 2+ bromide sulfates (SO 4 2- ) Soluble Cmpds of Sr 2+, Ba 2+, Hg 2 2+, Pb 2+ hydroxides (OH - ), oxides (O 2- ), sulfides (S 2- ) Insoluble Cmpds of alkali metals (grp 1A), NH 4+, Ca 2+, Sr 2+, Ba 2+ sulfites (SO 3 2- ), carbonates (CO 3 2- ), phosphates (PO 4 3- ), chromates (CrO 4 2- ) Insoluble Cmpds of alkali metals (grp 1A), NH 4 + 9

B) Exchange (Metathesis) Reactions Also called, Double-Replacement Rx s Cations & Anions exchange partners AX + BY AY + BX or AX + BY BX + AY Must be a driving force for a rx. to occur (remove ions) - formation of precipitate - formation of gas - formation of weak or nonelectrolyte 10

1) Precipitation Reactions 2 KI(aq) + Pb(NO 3 ) 2 (aq) 2 KNO 3 (aq) + PbI 2 (s) a) Molecular Eqn: written as if all substances were molecules - tells what soln s are combined - must use for quantitative calc. b) Ionic Eqn: shows each substance as it exist in soln. 2 K + (aq) + 2 I! (aq) + Pb 2+ (aq) + 2 NO 3! (aq) 2 K + (aq) + 2 NO 3! (aq) + PbI2 (s) 11

1) soluble ionic substances written as ions(aq) Na + (aq) + Cl! (aq) 2) insoluble ionic substances written as formula(s) XY(s) AgBr(s) 3) molecular subst. that are strong electrolytes ions(aq) H + (aq) + Cl! (aq) 4) molecular weak electrolytes or nonelectrolytes molecular formula (phase) CH 3 CO 2 H(aq), H 2 O(R) 12

b) Net Ionic Eqn: Note : K + & NO 3! do not undergo change & appear on both sides of arrow Spectator Ions Eliminate spectator Ions Y Net Ionic Eqn. Pb 2+ (aq) + 2 I! (aq) PbI 2 (s) 2 Advantages 1) focuses attention on change that is occurring 2) more general: Any subst. that produces Pb 2+ + I! in soln. will produce a ppt. of PbI 2 (s) 13

2) Ex: Mix KI (aq) & NaNO 3 (aq) soln s. KI (aq) + NaNO 3 (aq) NaI (aq) + KNO 3 (aq) Ionic Eqn: K + (aq) + I! (aq) + Na +! (aq) + NO 3 (aq) Na + (aq) + I! (aq) + K + (aq) + NO! 3 (aq) Eliminate ALL spectator ions - nothing left No driving force ˆ If we know solubilities of cmpds then we can predict when a ppt. will occur. 14

III) Acids, Bases & Neutralization Rxns. A) Acids Ionize in water to form H + (aq) - Proton donors HCl, H 2 SO 4, HC 2 H 3 O 2 (CH 3 CO 2 H) 1) 7 Strong Acids Ionize Completely HCl, HBr, HI, HNO 3, HClO 3, HClO 4, H 2 SO 4 (1 st H + only) HCl (aq) H + (aq) + Cl! (aq) 15

2) Weak Acids Partial ionization - weak electrolyte HC 2 H 3 O º 2 H +! + C 2 H 3 O 2 Chemical Equilibrium Acid that s NOT a strong acid! HF, HNO 2, HClO 2, HSO 4 16

B) Bases Produce OH! (aq) in water - Proton acceptors NaOH, Ca(OH) 2, NH 3 1) Strong Bases soluble ionic cmpds. which dissolve to give OH! NaOH (s) Na + (aq) + OH! (aq) Dissociate completely Hydroxides & Oxides of Groups IA & IIA (except Be, Mg) 17

2) Weak Bases Partially ionized a) Ammonia, NH 3 b) Amines 18

C) Neutralization Rx s Formation of Weak or Nonelectrolyte 1) Strong Acid & Strong Base HCl (aq) + NaOH(aq) NaCl(aq) + HOH(R) acid base salt water Driving Force formation of H 2 O Net Ionic Eq. H + (aq) + OH! (aq) H 2 O (R) True for rx. between any strong acid & strong base 19

2) Weak Acid & Strong Base HF (aq) + KOH (aq) v KF (aq) + H 2 O (R) Net : HF (aq) + OH! (aq) v F! (aq) + H 2 O (R) Generally, Rx. of weak monoprotic acid w. strong soluble base HA (aq) + OH! (aq) v A! (aq) + H 2 O (R) 20

3) Weak Base & Strong Acid HCl (aq) + NH 3 (aq) v NH + 4 (aq) + Cl! (aq) Net : H + (aq) + NH 3 (aq) v NH + 4 (aq) 4) Insoluble Hydroxides & Acids Cu(OH) 2 (s) + 2 HCl (aq) v CuCl 2 (aq) + 2 H 2 O (R) Net: Cu(OH) 2 (s) + 2 H + (aq) v Cu 2+ (aq) + 2 H 2 O (R) 21

5) Formation of Weak Electrolyte NaC 2 H 3 O 2 (aq) + HCl(aq) v HC 2 H 3 O 2 (aq) + NaCl(aq) Net: C 2 H 3 O 2! (aq) + H + (aq) º HC2 H 3 O 2 (aq) formation of HC 2 H 3 O 2 removes ions from soln. 22

D) Neutralization Rx s w. Gas Formation Gaseous product which escapes from the soln. 1) Carbonates & Bicarbonates Na 2 CO 3 (aq) + 2 HCl(aq) v 2 NaCl(aq) + H 2 CO 3 (aq) Carbonic acid decomposes: H 2 CO 3 (aq) H 2 O(R) + CO 2 (g) Na 2 CO 3 (aq) + 2 HCl (aq) Net: 2 NaCl (aq) + H 2 O (R) + CO 2 (g) CO 3 2- (aq) + 2 H + (aq) H 2 O(R) + CO 2 (g) 23

Insoluble Carbonates also work: Net: CaCO 3 (s) + 2 H + (aq) Ca 2+ (aq) + H 2 O(R) + CO 2 (g) Bicarbonates NaHCO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (R) + CO 2 (g) 24

2) Sulfites & Bisulfites Na 2 SO 3 (aq) + 2 HCl(aq) v 2 NaCl(aq) + H 2 SO 3 (aq) Sulfurous acid decomposes: H 2 SO 3 (aq) H 2 O(R) + SO 2 (g) Na 2 SO 3 (aq) + 2 HCl (aq) Net: 2 NaCl (aq) + H 2 O (R) + SO 2 (g) SO 3 2- (aq) + 2 H + (aq) H 2 O(R) + SO 2 (g) Bisulfites NaHSO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (R) + SO 2 (g) 25

3) Sulfides K 2 S(aq) + 2 HCl(aq) 2 KCl(aq) + H 2 S(g) Net : S 2 - (aq) + 2 H + (aq) H 2 S(g) 26

IV) Oxidation - Reduction Rx s (Redox) Involves loss of e! by one element & gain of e! by another element Oxidation: lose e! (inc. in oxidation #) Reduction: gain e! (dec. in oxidation #) Oxidizing agent: Reducing agent: substance that is reduced substance that is oxidized 27

Zn + Br 2 Zn Br 2 Zn Zn 2+ + 2 e! Zn lost e! Y oxidized Br 2 + 2 e! 2 Br! Br gained e! Y reduced 28

A) Oxidation Numbers Charge that an atom would have if e! in cmpds. are assigned in a certain way. 1) Elemental Form Cu, H 2, O 2, S 8 ox. # = 0 (zero) 2) Monatomic Ion ox. # = charge Na +, Zn 2+, Al 3+, O 2!, Br! Group IA + 1 Group IIA + 2 Group IIIA + 3 (Always) (Always) (usually) 29

3) Hydrogen H + 1 (usually) 4) Oxygen exceptions hydrides, H! 1 NaH, CaH 2 O! 2 (usually) exceptions peroxides, O 2 2! O! 1 H 2 O 2 Na 2 O 2 superoxides, O 2 1! O! 1/2 KO 2 30

5) Fluorine F! 1 Always 6) Halogens: Cl, Br, I & 1 except when combined w. a more E.N. element CBr 4 : Br! 1 can be : +1, +3, +5, +7 ClO 4! : +7 31

7) Sum of ox. no. s of atoms in neutral cmpds. = 0 (zero) 8) Sum of ox. no. s of atoms in a polyatomic ion = charge 9) Ox. no. can t be: more positive than the group # or more negative than (group #! 8) 32

B) Examples 1) Ex 1: What is ox. # of N in NH 3? # x N + 3(+1) = 0 # x N =!3! 2) Ex 2: What is ox. # of N in NO 3? 3) Ex 3: What is ox. # of N in NO 2!? 33

4) Ex 4: What is ox. # of Xe in XeOF 4? 5) Ex 5: What is ox. # of Cr in K 2 Cr 2 O 7? 34

C) Redox Reactions Oxidation: inc. in ox. # (lose e! ) Reduction: dec. in ox. # (gain e! ) 1) Combination 2 Ca + O 2 2 CaO 2) Decomposition 2 NH 3 N 2 + 3 H 2 3) Combustion CH 4 + 2 O 2 CO 2 + 2 H 2 O 35

4) Displacement Rxs. Single! Replacement Rxs one element replaces another element in a cmpd. Active metals displace less active metals or hydrogen the most active metals lose their e! most easily! most easily oxidized (strongest reducing agent) 36

a) Metals 2 K(s) + 2 H 2 O(R) v 2 KOH(aq) + H 2 (g) metal water hydroxide hydrogen Zn(s) + 2 HCl(aq) v ZnCl 2 (aq) + H 2 (g) metal acid salt hydrogen Fe(s) + CuSO 4 (aq) v Cu(s) + FeSO 4 (aq) metal salt metal salt 3CuCl 2 (aq) + 2Al(s) 2AlCl 3 (aq) + 3Cu(s) +2 0 +3 0 37

b) Nonmetals Halogens order of F > Cl > Br > I activity 2 KI(aq) + Cl 2 (g) 2 KCl(aq) + I 2 (s) 2 NaCl(aq) + F 2 (g) 2 NaF(aq) + Cl 2 (g) NaBr(aq) + I 2 (s) c) Metal Salt & Hydrogen or Carbon CuO(s) + H 2 (g) v Cu(s) + H 2 O(R) NiO(s) + C(s) v Ni(s) + CO(g) 38

39

f) Ex : If iron is placed in a soln. of silver nitrate will silver plate out? 40

V) Concentration of Solutions amount of solute dissolved in a given quantity of solvent or solution A) Molarity (M) Conversion factor between moles solute & L soln. 41

1) Ex 1: How many moles of CaCl 2 are present in 200.0 ml of 2.0 M CaCl 2? 42

2) Ex 2: A rx. req. 1.94 g Ca 2+ ions. What vol. of 2.00 M CaCl 2 is req. to provide this amt. of Ca 2+? 43

B) Dilution Add more solvent & get a more dilute soln. moles solute before dilution = moles solute after dilution M 1 V 1 = M 2 V 2 44

1) Ex 1: A 100.0 ml sample of 0.200 M sucrose was diluted to 500.0 ml. Calc. the new molarity. 45

2) Ex 2: How much water must be added to 100.0 ml of 0.300 M NaCl to prepare a 0.0500 M NaCl soln? 46

VI) Solution Stoichiometry When working w. solutions How much may refer to either volume required or concentration. A) Ions and Ionic Compounds Na 2 SO 4 (aq) v 2 Na + (aq) + SO 4 2! (aq) 1.0 M 2.0 M 1.0 M Use stoichiometry of balanced eqn. to determine molarity of ions in soln. 47

1) Ex : What is the molarity of OH! in a 0.20 M Ba(OH) 2 soln.? a) Use dimensional analysis mol OH! 0.20 mol Ba(OH) 2 2 mol OH!? ------------- = ------------------------ x ------------------- L soln L soln 1 mol Ba(OH) 2 = 0.40 M OH! b) Use ICE table in Molarity 1 Ba(OH) 2 v 1 Ba 2+ + 2 OH! Initial 0.20 0 0 Change - 0.20 + 0.20 + 0.40 Equilibrium 0 0.20 0.40 48

B) Reaction Stoichiometry Stoichiometry problems involving solns 1) Ex : What volume (ml) of 0.215 M HCl is req. to completely react w. 10.00 ml of 0.400 M Ba(OH) 2? Ba(OH) 2 + 2 HCl v BaCl 2 + 2 H 2 O 0.400 mol Ba(OH) 2 2 mol HCl 1 L soln? L = 0.01000 L x ------------------------ x ------------------ x -------------------- 1 L soln 1 mol Ba(OH) 2 0.215 mol HCl = 0.0372 L = 37.2 ml Could also calc. moles of Ba(OH) 2 first by using M*V and then do a mole-to-mole stoich. problem to find moles of HCl and then use M*V to solve for the volume of HCl. 49

C) Titrations (Volumetric Analysis) neutralization, precipitation, redox 1) Titration Method for determining conc. of a substance, A, by adding a measured volume of a soln. of known conc., S, (standard soln.). Rxn. between A and S must go to completion. 50

a) Equivalence Point Point in titration at which stoichiometrically equivalent amts of reactants have reacted. b) Endpoint Point at which an indicator changes color Want endpoint and eq. pt. to be close to each other 1) Indicator Substance which changes color as close as possible to the eq. pt. 51

2) Acid-Base Titration At eq. pt.: mol Acid = mol Base M a (V a = M b (V b In this eqn. the molarities are the molarities of H + and OH!. The volumes can be left in ml. 52

a) Ex : It takes 60.00 ml of 0.1000 M NaOH to titrate 20.00 ml of HCl to the eq. pt. What is the conc. of the original HCl solution? HCl (aq) + NaOH(aq) v NaCl(aq) + HOH(R) 0.1000 mol NaOH 1 mol HCl? mol HCl = 0.06000 L NaOH x ------------------------ x ----------------- 1 L soln 1 mol NaOH = 0.006000 mol HCl 0.006000 mol HCl M (HCl) = ------------------------ = 0.3000 M HCl 0.02000 L soln Could use M a *V a = M b *V b M b *V b (0.1000 M) (60.00 ml) M a = ----------- = ------------------------------ = 0.3000 M V a 20.00 ml 53