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Download Chapter 4 Reactions in Aqueous Solution 4.1 Aqueous Solutions
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4.1 Aqueous Solutions Chapter 4 Reactions in Aqueous Solution Electrolytes • Electrolyte – substance that gives an aqueous solution that conducts electricity • Nonelectrolyte – substance that does not produce conducting solution when dissolved in water • Mobile ions conduct electricity – electrolytes break apart (dissociate or ionize) into ions when dissolved in water • Solution – homogeneous mixture of 2 or more substances • Solute – the substance present in a smaller amount (usually solid in Chap. 4) • Solvent – the substance present in the larger amount (usually liquid in Chap. 4) • Aqueous solution – solvent is water Electrical Conduction by Electrolyte Solution e- e- M+ X- Cations (positive ions) migrate to cathode (negative electrode) Anions (n egative ions) migrate to anode (positive electrode) Electrochemistry at electrodes (Chap. 19) Strong Electrolytes • Dissociate completely into ions in H2O • Salts like NaCl, KNO3, MgSO4 KNO3(s) H 2O K+ (aq) + NO3 - (aq) •Strong acids like HCl, HNO3 HCl(g) H2O H + (aq) + Cl-(aq ) Weak Electrolytes • Ionize partially into ions in H2O • Weak acids (HF) HF(aq) H2 O H + (aq) + F-(aq) •Weak bases (NH3) H2O NH 3(aq) NH 4+ (aq) + OH - (aq) •Double arrow indicates that the reaction is reversible, i.e., runs both ways • Ionic compounds are strong electrolytes • When KCl dissolves: KCl(s) d K+(aq) + Cl-(aq) • Ions are surrounded by polar molecules – in general, solvation – in water, hydration + end of dipole points to anion - end of dipole points to cation H2O NH3(aq) NH 4+(aq) + OH-(aq) • Reversible reaction - can occur in both directions • Reactants form products as soon as reaction begins • Once products are formed, they in turn react to re-form reactants • Chemical Equilibrium - when reactants form products as fast as products form reactants, no further net change in concentrations Reversible Ionization of a Weak Base Principal Types of Reactions • Precipitation Reactions (4.2) • Acid - Base Reactions (4.3) • Oxidation-Reduction Reactions (4.4) •Both NH3 and NH4 + exist in rapid equilibrium = 1% is NH 4+ (Chap. 15) 4.2 Precipitation Reactions • Reactions that result in the formation of precipitate, insoluble solid that separates from the solution • Often involve ionic compounds AgNO3(aq) + 2NaBr(aq) d AgBr(s) + NaNO 3(aq) Precipitate Solubility • Maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature • Table 4.2 classifies substances in one of 3 categories –insoluble –slightly soluble –soluble Simple solubility rules Soluble salts Exceptions Alkal i meta l (Li + toC s+), ammonium ( NH4+) Using the solubility rules, classify as soluble or insoluble. • Ag2SO4 N itrate ( NO 3-), bicar bon ate ( HCO 3-), sulf ate ( SO 42- ), etc. Sulfates of Ag +, C a2+, Sr2+, Ba 2+, Pb 2+ H alide (Cl- , Br- , I -) Halides of Ag +, Hg 22+, Pb 2+ insoluble • Li2S soluble • Pb(NO3)2 Insoluble salts Exceptions C arbonat e (CO32- ), Alkal i metal (Li + to Cs+), phosphat e (PO 43- ), sulfide ammoni um ( NH4+) (S 2-), etc. H ydroxide (OH- ) soluble • AgCl insoluble Alkal i metal (Li + to Cs+), Ba 2+ Ionic equations • “Molecular” equation: write full formula of each species AgNO3(aq) + NaBr(aq) d AgBr(s) + NaNO3(aq) •Ionic equation: write dissolved species as free ions Net ionic equations • Spectator ions are not involved in the overall reaction: Na+(aq) and NO3-(aq) Ag+(aq) + NO3-(aq) + Na+(aq) + Br-(aq) d AgBr(s) + Na+(aq) + NO3-(aq) •Net ionic equation: write only species that take part in reaction Ag+(aq) + NO3-(aq) + Na+(aq) + Br-(aq) d AgBr(s) + Na+(aq) + NO3-(aq) Ag+(aq) + Br-(aq) d AgBr(s) Predicting and writing precipitation reactions PbCl2 + Na2SO4 1 Write “molecular” equation 2 Dissociate electrolytes d ionic equation 3 Use solubility rules to predict precipitate 4 Cancel spectator ions d net ionic equation 1. Write “molecular” equation, using solubility rules to predict precipitate(s) PbCl2(aq) + Na2SO 4(aq) d PbSO4(s) + 2NaCl (aq) 2. Dissociate electrolytes d ionic equation Pb2 +(aq) + 2Cl-(aq) + 2Na +(aq) + SO42 -(aq) d PbSO4(s) + 2Na +(aq) + 2Cl-(aq) 3. Cancel spectator ions d net ionic equation Pb2 +(aq) + SO4 2-(aq) d PbSO4(s) More examples (on your own) • K2SO4 + AgNO 3 4.3 Acid-Base Reactions • Acid-base reactions are very important in chemistry • Long history • (NH4)2S + FeSO 4 – many models for acids and bases – details in Chaps. 15 and 16 Empirical criteria Empirical criteria • Acid – sour taste – corrodes metals, often producing H2(g) H2SO4(aq) + Fe(s) d FeSO4(aq) + H2(g) – turns litmus (plant dye) red – gives carbon dioxide (g) w/ carbonates 2HCl(aq) + CaCO 3(s) H2O(l) + CO2(g) • Base (alkali) – bitter taste (NaHCO3 = baking soda) – feels slippery (soap) – turns litmus (plant dye) blue – electrolyte d CaCl2(aq) + – electrolyte Arrhenius concept (1887) • Acid = Proton (H+) donor in H2O HCl(aq) d H+(aq) + Cl-(aq) H2SO4(aq) d H+(aq) + HSO4-(aq) • Base = Hydroxide (OH-) donor in H2O NaOH(aq) d Na +(aq) + OH-(aq) MgO(s) + H2O(l) d Mg2+(aq) + 2 OH -(aq) • Neutralization H+(aq) + OH-(aq) d H2O(l) Acid + Base d Water Brønsted-Lowry concept (1923) • Brønsted Acid = Proton donor HCl(g) + HF(l) qeH2F+ (HF) + Cl-(HF) • Brønsted Base = Proton acceptor NH3(aq) + H2O(l) qeNH4+(aq) + OH-(aq) • Neutralization NH3 + HCl qeNH4+ + ClBase1 + Acid2 qeAcid1 + Base2 • Not necessarily in water Dissociation of HCl in water Acid1 Base 2 Acid2 Base 1 H3O+ = Hydronium (oxonium) ion + H (aq) + H2O(l) + H 3O (aq) • H+ associates with several H2O molecules • e.g. H7 O3+ or [H(H2O) n]+ (n = 3) • Often abbreviated H+(aq) H3O+ = Hydronium ion Weak base in water NH3(aq) + H2O(l) qeNH4+(aq) + OH-(aq) Base Acid • Very little NH3 (=1%) is ionized – typical weak base in water • H2O acts as a Brønsted acid in this reaction • Some substances can act as either acid or base depending on reaction Acid - Base Neutralization Reaction • Neutralization reaction - reaction between acid and base to produce a salt and water • Salt - ionic compound w/ cation besides H+ • acid + base d salt + water HBr(aq) + KOH(aq) d KBr(aq) + H2O(l) net: H+(aq) + OH-(aq) d H2O(l) Polyprotic acids •Diprotic – supply 2 H+ in 2 steps H2S(aq) qe H+(aq) + HS-(aq) SH-(aq) qe H+(aq) + S2-(aq) •Triprotic – supply 3 H+ in 3 steps H3 PO4(aq) qe H+(aq) + H2 PO 4-(aq) H2 PO4-(aq) qe H+(aq) + HPO42 -(aq) HPO42-(aq) qe H+(aq) + PO43 -(aq) •Each step is successively weaker Examples of Neutralization Reactions • HF(aq) + NaOH(aq) d NaF(aq) + H2 O(l) • 2HNO3(aq) + Ba(OH) 2(aq) d Ba(NO3) 2(aq) + 2H2O(l) • H2SO4(aq) + 2LiOH(aq) d Li2SO4(aq) + 2H2O(l) 4.4 Oxidation-Reduction Reactions Redox mnemonic • “Redox reactions” – electron transfer reactions • 2K(s) + Cl2(g) d 2KCl(s) • K loses an electron to become K + 2K d 2K+ + 2e • Cl2 gains 2 electrons to become 2ClCl2 + 2e- d 2Cl • Each step is a half-reaction • LEO Loss of Electrons = Oxidation GER • GER Gain of Electrons = Reduction – explicitly shows electron transfer Redox Definitions • 2K(s) + Cl2(g) d 2KCl(s) • K is oxidized to K + – oxidation half-reaction 2K d 2K+ + 2e – K is the reducing agent (reductant) - • Cl2 is reduced to 2Cl – reduction half-reaction – Cl2 + 2e- d 2Cl – Cl2 is the oxidizing agent (oxidant) Short Activity Series • • • • • • • • • Li most reactive K Ba Na Zn H Cu Hg Au least reactive LEO The Activity Series for Metals Displacement Reactions M + n H2O M + n H+ M(OH)n + n/2 H2 Mn+ + n/2 H2 Examples: Ca + 2H2O Ca(OH)2 + H2 Pb + 2H2O Pb(OH) 2 + H2 Oxidation Number • Reactions of molecular, not just ionic, compounds are redox reactions P4(s) + 5O2(g) d P4 O10 (s) – There are no ionic charges shown, but it is a redox reaction • Oxidation number (state) – charge an atom would have if e- were transferred completely P(+5) and O(-2) – remember: Σ Ox# = charge on ion (molecule) • Chap. 9 explains direction of transfer Ox# Rules Calculate Ox#s 1. In free elements each atom has Ox # = 0 2. For monatomic ions, Ox # = charge 3. Ox # of oxygen is usually -2 4. Ox # of hydrogen is usually +1 except metal hydrides, Ox # (H) = -1 5. Ox # of fluorine is always -1 6. Σ Ox # = charge on molecule or ion 7. Ox # can be fractional Types of Redox Reactions • Combination Reaction – substances combine to form 1 product – Fe(s) + O2(g) d Fe2O3(s) • Decomposition Reaction – breakdown of compound into components – NH4 NO2(s) d N2(g) + H2O(l) H3PO4 SO2 SF6 CO Re2Cl82SH3+ Displacement Reactions • Hydrogen displacement Mg(s) + H2 O(l) d Mg(OH) 2(aq) + H2(g) • Metal displacement – more active metal displaces less active metal (Chap. 19) Al(s) + Fe 2O3(s) d Fe(l) + Al2O3(s) • Halogen displacement – reactivity F2 > Cl2 > Br 2 > I2 Cl2(g) + CaBr 2(s) d Br 2(l) + CaCl2(s) Disproportionation Reaction • Element in one oxidation state is simultaneously oxidized and reduced 2CuCl(s) d Cu(s) + CuCl2(s) • Cu(+1) goes to Cu(0) and Cu(+2) 4.5 Concentration of Solutions • Amount of solute present in a given quantity of solvent or solution • Molarity (M) – the number of moles of solute in 1 liter of solution M = molarity = mol solute L solution Conc. NaCl = 0.34 M Calculating molarity What is the molarity of a solution made by dissolving 5.0 g of NaCl in enough water to make 250.0mL of solution? Conc. = 0.34 M • What is the conc. of Na + in solution? [Na+] = 0.34 M • What is the conc. of Cl- in solution? [Cl-] = 0.34 M Consider a 0.50 M BaCl2 solution. • [Ba2+] = ? 0.50 M • [Cl-] = ? 1.0 M • Molarity (mol/L) can be used to convert – mol solute to L solution – L solution to mol solute How many moles of HCl are in 500 mL of 0.30 M HCl? 0.50 L x (0.30 mol/ L) = 0.15 mol • a b c Prepare 250 mL of 2.0 M KOH solution • Need (2.0 mol KOH/ L)(0.25 L) = 0.50 mol KOH • (0.50 mol KOH)(56 g /mol) = 28 g KOH • Dissolve 28 g KOH in enough water to make 250 mL of solution Convert molarity to mass of solute needed Transfer solute to volumetric flask Add enough solvent to dissolve solute Dilute to mark Dilution • Start with a concentrated stock solution • Add more solvent to produced solution of lower concentration • M = mol/ L • mol contained = M•V Dilution Formula Add solvent How many mL H2O are required to dilute 205 mL of 1.15 M HCl solution to 0.81 M? MiVi = MfVf Vf = MiVi / Mf Vf = (1.15 M)(0.205 L)/(0.81 M) = 0.291 L • Same # mol solute in each beaker • mol = M•V in each beaker VH2O = Vtotal - Vin VH2O = 0.291 L - 0.205 L = 0.086 L = 86 mL • Mi Vi = M fVf i = initial, f = final Quantitative Analysis • • Determination of the amount or concentration of a substance in a sample Two Quantitative Analysis Techniques 1. Gravimetric Analysis – based on the measurement of mass (Section 4.6) 2. Titration - solution of known concentration is added to solution of unknown concentration until chemical reaction is complete (endpoint) How many grams of NaCl are required to precipitate all the Ag+ ions from 125 mL of 0.0081 M AgNO3 solution? NaCl (aq) + AgNO 3 (aq) d AgCl (s) + NaNO3 (aq) [work on blackboard] 0.059 g a. Acid-Base Titrations (Section 4.7) b. Redox Titrations (Section 4.8) 4.7 Acid-Base Titration • Equivalence point – point at which acid is completely neutralized by added base (or vice versa) • Equivalence point (or end point) is signaled by an indicator – substance that is different color in acid and base 22.5 mL of 0.383 M H2SO4 are required to neutralize 20.0 mL of a KOH solution. Calculate the molarity of the KOH solution. H2SO4 + 2 KOH d K2SO4 + 2 H2O M KOH = (0.0225 L H2SO4)(0.383 mol H2SO4/L H2SO4)(2 mol KOH/ 1 mol H2SO4)/ (0.0200 mL KOH) = 0.862 M KOH 4.8 Redox titrations • Read on your own • Won’t be on exam