4.1 Aqueous Solutions. Chapter 4. Reactions in Aqueous Solution. Electrolytes. Strong Electrolytes. Weak Electrolytes

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1 Chapter 4 Reactions in Aqueous Solution 4.1 Aqueous Solutions 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 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 Electrical Conduction by Electrolyte Solution Cations (positive ions) migrate to cathode (negative electrode) e - e - M + X - Anions (negative ions) migrate to anode (positive electrode) Electrochemistry at electrodes (Chap. 19) Strong Electrolytes Dissociate completely into ions in H 2 O Salts like NaCl, KNO 3, MgSO 4 KNO 3 (s) HCl(g) H 2 O H 2 O K + (aq) + NO 3 - (aq) Strong acids like HCl, HNO 3 H + (aq) + Cl - (aq) Weak Electrolytes Ionize partially into ions in H 2 O Weak acids (HF) H 2 O HF(aq) H + (aq) + F - (aq) Weak bases (NH 3 ) NH 3 (aq) H 2 O NH 4 + (aq) + OH - (aq) Double arrow indicates that the reaction is reversible, i.e., runs both ways

2 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 NH 3 (aq) H 2 O 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 NH 3 and NH 4 + 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 AgNO 3 (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

3 Simple solubility rules Soluble salts Alkal i meta l (Li + to C s + ), ammonium ( NH4 + ) Nitrate ( NO 3- ), bicar bon ate ( HCO 3- ), sulfate ( SO 4 2- ), etc. Exceptions Sulfates of Ag +, Ca 2+, Sr 2+, Ba 2+, Pb 2+ Halide (Cl -, Br -, I - ) Halides of Ag +, Hg 2 2+, Pb 2+ Insoluble salts Carbonat e (CO 3 2- ), phosphat e (PO 4 3- ), sulfide (S 2- ), etc. Exceptions Alkal i metal (Li + to Cs + ), ammoni um ( NH 4+ ) Hydroxide (OH - ) Alkal i metal (Li + to Cs + ), Ba 2+ Using the solubility rules, classify as soluble or insoluble. Ag 2 SO 4 insoluble Li 2 S soluble Pb(NO 3 ) 2 soluble AgCl insoluble Ionic equations Molecular equation: write full formula of each species AgNO 3 (aq) + NaBr(aq) d AgBr(s) + NaNO 3 (aq) Ionic equation: write dissolved species as free ions Ag + (aq) + NO 3- (aq) + Na + (aq) + Br - (aq) d AgBr(s) + Na + (aq) + NO 3- (aq) Net ionic equations Spectator ions are not involved in the overall reaction: Na + (aq) and NO 3- (aq) Ag + (aq) + NO 3- (aq) + Na + (aq) + Br - (aq) d AgBr(s) + Na + (aq) + NO 3- (aq) Net ionic equation: write only species that take part in reaction Ag + (aq) + Br - (aq) d AgBr(s) Predicting and writing precipitation reactions 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 PbCl 2 + Na 2 SO 4 1. Write molecular equation, using solubility rules to predict precipitate(s) PbCl 2 (aq) + Na 2 SO 4 (aq)d PbSO 4 (s) + 2NaCl (aq) 2. Dissociate electrolytes d ionic equation Pb 2+ (aq) + 2Cl - (aq) + 2Na + (aq) + SO (aq)d PbSO 4 (s) + 2Na + (aq) + 2Cl - (aq) 3. Cancel spectator ions d net ionic equation Pb 2+ (aq) + SO 4 2- (aq)dpbso 4 (s)

4 More examples (on your own) K 2 SO 4 + AgNO 3 (NH 4 ) 2 S + FeSO Acid-Base Reactions Acid-base reactions are very important in chemistry Long history many models for acids and bases details in Chaps. 15 and 16 Empirical criteria Acid sour taste corrodes metals, often producing H 2 (g) H 2 SO 4 (aq) + Fe(s) d FeSO 4 (aq) + H 2 (g) turns litmus (plant dye) red gives carbon dioxide (g) w/ carbonates 2HCl(aq) + CaCO 3 (s) d CaCl 2 (aq) + H 2 O(l) + CO 2 (g) electrolyte Empirical criteria Base (alkali) bitter taste (NaHCO 3 = baking soda) feels slippery (soap) turns litmus (plant dye) blue electrolyte Arrhenius concept (1887) Acid = Proton (H + ) donor in H 2 O HCl(aq) d H + (aq) + Cl - (aq) H 2 SO 4 (aq) d H+(aq) + HSO 4- (aq) Base = Hydroxide (OH - ) donor in H 2 O NaOH(aq) d Na + (aq) + OH - (aq) MgO(s) + H 2 O(l) d Mg 2+ (aq) + 2 OH - (aq) Neutralization H + (aq) + OH - (aq)dh 2 O(l) Acid + Based Water Brønsted-Lowry concept (1923) Brønsted Acid = Proton donor HCl(g) + HF(l) qeh 2 F + (HF) + Cl - (HF) Brønsted Base = Proton acceptor NH 3 (aq) + H 2 O(l) qenh 4+ (aq) + OH - (aq) Neutralization NH 3 + HClqeNH 4+ + Cl - Base 1 + Acid 2 qeacid 1 + Base 2 Not necessarily in water

5 Dissociation of HCl in water Acid 1 Base 2 Acid 2 Base 1 H 3 O + = Hydronium (oxonium) ion H + (aq) + H 2 O(l) H+ associates with several H 2 O molecules e.g. H 7 O 3 + or [H(H 2 O) n ] + (n = 3) Often abbreviated H + (aq) H 3 O + (aq) H 3 O + = Hydronium ion NH 3 (aq) + H 2 O(l) qenh 4+ (aq) + OH-(aq) Base Weak base in water Acid Very little NH 3 (=1%) is ionized typical weak base in water H 2 O acts as a Brønsted acid in this reaction Some substances can act as either acid or base depending on reaction Polyprotic acids Diprotic supply 2 H + in 2 steps H 2 S(aq) qe H+(aq) + HS-(aq) SH - (aq)qe H + (aq) + S 2- (aq) Triprotic supply 3 H + in 3 steps H 3 PO 4 (aq)qe H+(aq) + H 2 PO 4- (aq) H 2 PO 4- (aq)qe H+(aq) + HPO (aq) HPO 4 2- (aq)qe H + (aq) + PO (aq) Each step is successively weaker 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 + basedsalt + water HBr(aq) + KOH(aq)dKBr(aq) + H 2 O(l) net: H + (aq) + OH - (aq) d H 2 O(l) Examples of Neutralization Reactions HF(aq) + NaOH(aq) d NaF(aq) + H 2 O(l) 2HNO 3 (aq) + Ba(OH) 2 (aq) d Ba(NO 3 ) 2 (aq) + 2H 2 O(l) H 2 SO 4 (aq) + 2LiOH(aq) d Li 2 SO 4 (aq) + 2H 2 O(l)

6 4.4 Oxidation-Reduction Reactions Redox reactions electron transfer reactions 2K(s) + Cl 2 (g) d 2KCl(s) K loses an electron to become K + 2K d 2K + + 2e - Cl 2 gains 2 electrons to become 2Cl - Cl 2 + 2e - d 2Cl - Each step is a half-reaction explicitly shows electron transfer Redox mnemonic LEO Loss of Electrons = Oxidation GER Gain of Electrons = Reduction LEO GER Redox Definitions 2K(s) + Cl 2 (g) d 2KCl(s) K is oxidized to K + oxidation half-reaction 2K d 2K + + 2e - K is the reducing agent (reductant) Cl 2 is reduced to 2Cl - reduction half-reaction Cl 2 + 2e - d 2Cl - Cl 2 is the oxidizing agent (oxidant) The Activity Series for Metals Displacement Reactions M + n H 2 O M(OH) n + n/2 H 2 M + n H + M n+ + n/2 H 2 Examples: Ca + 2H 2 O Ca(OH) 2 + H 2 Pb + 2H 2 O Pb(OH) 2 + H 2 Short Activity Series Li most reactive K Ba Na Zn H Cu Hg Au least reactive Oxidation Number Reactions of molecular, not just ionic, compounds are redox reactions P 4 (s) + 5O 2 (g) d P 4 O 10 (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

7 Ox# Rules 1. In free elements each atom has Ox # = 0 2. For monatomic ions, Ox # = charge 3. Ox # of oxygen is usually Ox # of hydrogen is usually +1 except metal hydrides, Ox # (H) = Ox # of fluorine is always Σ Ox # = charge on molecule or ion 7. Ox # can be fractional H 3 PO 4 SO 2 SF 6 CO Re 2 Cl 2-8 SH 3 + Calculate Ox#s Types of Redox Reactions Combination Reaction substances combine to form 1 product Fe(s) + O 2 (g) d Fe 2 O 3 (s) Decomposition Reaction breakdown of compound into components NH 4 NO 2 (s) d N 2 (g) + H 2 O(l) Displacement Reactions Hydrogen displacement Mg(s) + H 2 O(l) d Mg(OH) 2 (aq) + H 2 (g) Metal displacement more active metal displaces less active metal (Chap. 19) Al(s) + Fe 2 O 3 (s) d Fe(l) + Al 2 O 3 (s) Halogen displacement reactivity F 2 > Cl 2 > Br 2 > I 2 Cl 2 (g) + CaBr 2 (s) d Br 2 (l) + CaCl 2 (s) Disproportionation Reaction Element in one oxidation state is simultaneously oxidized and reduced 2CuCl(s) d Cu(s) + CuCl 2 (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

8 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 Conc. NaCl= 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 BaCl 2 solution. [Ba 2+ ] =? 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 Convert molarity to mass of solute needed a Transfer solute to volumetric flask b Add enough solvent to dissolve solute c Dilute to mark 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 Dilution Start with a concentrated stock solution Add more solvent to produced solution of lower concentration M = mol/ L mol contained = M V

9 Dilution Formula How many ml H 2 O are required to dilute 205 ml of 1.15 M HCl solution to 0.81 M? Add solvent M i V i = M f V f V f = M i V i / M f V f = (1.15 M)(0.205 L)/(0.81 M) = L Same # mol solute in each beaker mol = M V in each beaker V H2O = V total - V in V H2O = L L = L = 86 ml M i V i = M f V f 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) a. Acid-Base Titrations (Section 4.7) b. Redox Titrations (Section 4.8) How many grams of NaCl are required to precipitate all the Ag + ions from 125 ml of M AgNO 3 solution? NaCl (aq) + AgNO 3 (aq) d AgCl (s) + NaNO 3 (aq) [work on blackboard] g 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 M H 2 SO 4 are required to neutralize 20.0 ml of a KOH solution. Calculate the molarity of the KOH solution. H 2 SO KOH dk 2 SO H 2 O M KOH = ( L H 2 SO 4 )(0.383 mol H 2 SO 4 /L H 2 SO 4 )(2 mol KOH/ 1 mol H 2 SO 4 )/ ( ml KOH) = M KOH

10 4.8 Redox titrations Read on your own Won t be on exam