Chapter 5 Chemical Reactions

Aqueous Solubility of Ionic Compounds John W. Moore Conrad L. Stanitski Peter C. Jurs Not all compounds dissolve in water. Solubility varies from compound to compound. http://academic.cengage.com/chemistry/moore Chapter 5 Chemical Reactions Stephen C. Foster Mississippi State University Soluble ionic compounds dissociate. Ions are solvated Molecular compounds usually stay associated. Soluble Ionic Compounds (Table 5.1) ALL ammonium and group 1A (Na +, K + and NH 4+ salts) nitrates (NO 3 ) acetates (CH 3 COO ) chlorates (ClO 3 ) perchlorates (ClO 4 ) MOST chlorides, bromides and iodides (not: AgX, Hg 2 X 2, and PbX 2 ; X = Cl, Br, I ). sulfates (SO 4 ) (not: CaSO 4, SrSO 4, BaSO 4, and PbSO 4 ) Insoluble Ionic Compounds (Table 5.1) ALL phosphates (PO 4 3 ) carbonates (CO 3 ) oxalates (C 2 O 4 ) oxides (O ) Except grp 1A, NH 4 + sulfides (S ) MgS, CaS & BaS are slightly soluble. hydroxides (OH ) Sr, Ba & Ca are slightly soluble Aqueous Solubility of Ionic Compounds Exchange Reactions: Precipitation (a) Nitrates (soluble) (b) Hydroxides (insoluble) Three types of exchange reactions: Form a precipitate an insoluble ionic compound AgNO 3 (aq) + KCl(aq) KNO 3 (aq) + AgCl(s) AgNO 3 Cu(NO 3 ) 2 Cu(OH) 2 AgOH Form a molecular compound Often water (c) Sulfides HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) CdS Sb 2 S 3 PbS (NH 4 ) 2 S soluble insoluble Form a molecular gas 2 HCl(aq) + Na 2 S(aq) 2 NaCl(aq) + H 2 S(g) 1

Precipitation Reactions When ionic solutions mix, a precipitate may form: AgNO 3 (aq) + NaCl(aq) KNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) No reaction A reaction occurs if a product is insoluble. The solubility rules help predict reactions. Precipitation Reactions Na 2 SO 4 (aq) and Ba(NO 3 ) 2 (aq) are mixed. Will they react? Products? Na + SO 4 Ba 2+ NO 3 NaNO 3 and BaSO 4. Insoluble? Yes. BaSO 4 A reaction occurs: Na 2 SO 4 (aq) + Ba(NO 3 ) 2 (aq) BaSO 4 (s) + 2 NaNO 3 (aq) Net Ionic Equations Soluble ionic compounds fully dissociate: AgNO 3 (aq) Ag + (aq) + NO 3 (aq) KCl(aq) K + (aq) + Cl (aq) On mixing: Ag + (aq) + NO 3 (aq) + K + (aq) + Cl (aq) AgCl(s) + K + (aq) + NO 3 (aq) K + and NO 3 appear on both sides NO 3 and K + are spectator ions. They are not directly involved in the reaction. Net ionic equation: Ag + (aq) + Cl (aq) AgCl(s) Net Ionic Equations General rules: 1. Write a balanced equation. 2. Solubility? 3. Dissociate soluble compounds. 4. Write the complete ionic equation. 5. Cancel ions appearing on both sides (spectators). 6. Check the charges are balanced. Net Ionic Equations Aqueous solutions of (NH 4 ) 2 S and Hg(NO 3 ) 2 react to give HgS and NH 4 NO 3. (a) Write the overall balanced equation (b) Name each compound (c) Write the net ionic equation (d) What type of reaction is this? (NH 4 ) 2 S(aq) + Hg(NO 3 ) 2 (aq) HgS + NH 4 NO 3 not balanced (NH 4 ) 2 S(aq) + Hg(NO 3 ) 2 (aq) HgS + 2 NH 4 NO 3 balanced Net Ionic Equations (NH 4 ) 2 S(aq) + Hg(NO 3 ) 2 (aq) HgS + 2 NH 4 NO 3 Product Solubility HgS insoluble (All sulfides except group 1A, 2A and NH 4+ salts) NH 4 NO 3 soluble (All ammonium salts; all nitrates) (b) name each compound ammonium sulfide mercury(ii) nitrate mercury(ii) sulfide ammonium nitrate 2

Net Ionic Equations (c) Write a net ionic equation (NH 4 ) 2 S(aq) + Hg(NO 3 ) 2 (aq) HgS(s) + 2 NH 4 NO 3 (aq) 2 NH 4+ (aq) + S (aq) + Hg 2+ (aq) + 2 NO 3 (aq) HgS(s) + 2 NH 4+ (aq) + 2 NO 3 (aq) (d) Type of reaction? Exchange reaction Precipitation reaction S (aq) + Hg 2+ (aq) HgS(s) Acids Increase the concentration of H + ions in water. Protons (H + ) always combine with water to form H 3 O + (hydronium ion). Sour tasting. Change the color of pigments (indicators) Litmus, phenolphthalein Strong acids dissociate (>99%) in water (strong electrolytes). Weak acids partially dissociate (weak electrolytes). Acids Hydrochloric acid (HCl) 100% dissociated in H 2 O Very few HCl molecules exist in solution. HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) HCl is a strong acid. Acetic acid (CH 3 COOH) is 5% dissociated in H 2 O. 95% intact. CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq) Acetic acid is a weak acid. Bases Increase the concentration of OH (hydroxide ion) in water. Bases: Counteract an acid (neutralize an acid). Change an indicator s color (phenolphthalein ). Have a bitter taste. Feel slippery. Bases can be strong or weak. H NaOH(s) 2 O Na + (aq) + OH (aq) strong NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH (aq) weak Common Acids and Bases Strong Acid Strong Base HCl Hydrochloric acid LiOH Lithium hydroxide HBr Hydrobromic acid NaOH Sodium hydroxide HI Hydroiodic acid KOH Potassium hydroxide HNO 3 Nitric acid Ca(OH) 2 Calcium hydroxide H 2 SO 4 Sulfuric acid Ba(OH) 2 Barium hydroxide HClO 4 Perchloric acid Sr(OH) 2 Strontium hydroxide Weak Acid Weak Base HF Hydrofluoric acid NH 3 Ammonia H 3 PO 4 Phosphoric acid CH 3 NH 2 Methylamine CH 3 COOH Acetic acid H 2 CO 3 Carbonic acid HCN Hydrocyanic acid HCOOH Formic acid C 6 H 5 COOH Benzoic acid Neutralization Reactions acid + base salt + water. salt = ionic compound made from an acid anion and base cation. HX(aq) + MOH(aq) HBr(aq) + KOH(aq) H 3 PO 4 (aq) + 3 NaOH(aq) MX(aq) + H 2 O(l) Neutralizations are exchange reactions. KBr(aq) + H 2 O(l) Na 3 PO 4 (aq) + 3 H 2 O(l) 3

Net Ionic Equations for AcidBase Reactions Strong Acid + Strong Base Overall: HX + MOH MX + H 2 O Full ionic: H + (aq) + X (aq) + M + (aq) + OH (aq) M + (aq) + X (aq) + H 2 O(l) net ionic: H + (aq) + OH (aq) H 2 O(l) Net Ionic Equations for AcidBase Reactions Weak Acid + Strong Base Similar: HA + MOH MA + H 2 O but the weakacid remains undissociated: Full ionic HA(aq) + M + (aq) + OH (aq) M + (aq) + A (aq) + H 2 O(l) Net ionic: HA(aq) + OH (aq) A (aq) + H 2 O(l) GasForming Exchange Reactions Metal carbonate/acid exchange Overall: CaCO 3 (s) + 2 HCl(aq) CaCl 2 (aq) + H 2 CO 3 (aq) Tums H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) Carbonic acid (unstable) often written: CaCO 3 (s) + 2 HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) Net ionic: CaCO 3 (s) + 2 H + (aq) Ca 2+ (aq) + H 2 O(l) + CO 2 (g) GasForming Exchange Reactions Metal hydrogen carbonate/acid exchange Overall: NaHCO 3 (aq) + HCl(aq) NaCl(aq) + H 2 O(l) + CO 2 (g) AlkaSeltzer Net ionic HCO 3 (aq) + H + (aq) H 2 O(l) + CO 2 (g) GasForming Exchange Reactions Metal sulfite/acid exchange Similar. Overall: Na 2 SO 3 (aq) + 2 HCl(aq) 2 NaCl(aq) + H 2 SO 3 (aq) H 2 SO 3 (aq) H 2 O(l) + SO 2 (g) Sulfurous acid (unstable) or Na 2 SO 3 (aq) + 2 HCl(aq) 2 NaCl(aq) + H 2 O(l) +SO 2 (g) GasForming Exchange Reactions Metal sulfide/acid exchange Overall: Na 2 S(aq) + 2 HCl(aq) 2 NaCl(aq) + H 2 S(g) net ionic: S (aq) + 2 H + (aq) H 2 S(g) net ionic: SO 3 (aq) + 2 H + (aq) H 2 O(l) + SO 2 (g) 4

OxidationReduction Reactions Oxidation Originally: add oxygen. 2 Cu(s) + O 2 (g) 2 CuO(s) 2 CO(g) + O 2 (g) 2 CO 2 (g) Cu and CO are oxidized. O 2 is the oxidizing agent for both. Reduction Originally: reduce ore to metal; reverse of oxidation. CuO(s) + H 2 (g) Cu(s) + H 2 O SnO 2 (s) + 2 C(s) Sn(s) + 2 CO(g) CuO and SnO 2 are reduced. H 2 and C are the reducing agents (note: H 2 is oxidized, so CuO is an oxidizing agent) OxidationReduction Reactions In all cases: If something is oxidized, something must be reduced. Oxidationreduction = redox. Redox reactions move e. Here: 2 Ag + (aq) + Cu(s) 2 Ag(s) + Cu 2+ (aq) Cu changes to Cu 2+. Cu loses 2 e ; each Ag + gains one e Ag + is reduced (ore turned to metal) Gain of e = reduction +2 e 2 e (so, loss of e = oxidation) Redox Reactions and Electron Transfer Loss of electrons is oxidation Gain of electrons is reduction Redox Reactions and Electron Transfer e Oxidation is loss Reduction is gain Leo says ger Oil rig M M loses electron(s) M is oxidized M is a reducing agent X X gains electron(s) X is reduced X is an oxidizing agent Common Oxidizing and Reducing Agents Oxidizing Agent Reaction Product O 2 (oxygen) O (oxide ion) H 2 O 2 (hydrogen peroxide) H 2 O(l) F 2, Cl 2, Br 2, I 2 (halogen) F, Cl, Br, I (halide ion) HNO 3 (nitric acid) nitrogen oxides (NO, NO 2..) Cr 2 O 7 (dichromate ion) Cr 3+ (chromium(iii) ion) MnO 4 (permanganate ion) Mn 2+ (manganese(ii) ion) Reducing Agent Reaction Product H 2 (hydrogen) H + or H 2 O C CO and CO 2 M (metal: Na, K, Fe..) M n+ (Na +, K +, Fe 3+..) Oxidation number Compares the charge of an uncombined atom with its actual or relative charge in a compound. General rules: Pure element = 0. Monatomic ion = charge of ion. Some elements have the same oxidation number in almost all their compounds The sum of the oxidation numbers of all atoms in any species = the charge of the species. 5

Element Ox# Exceptions F 1 None O 2 Metal peroxides (MO 2 ) = 1 Halogen oxides (OX n ) (varies) Cl, Br, I 1 Interhalogens (strongest = 1) H +1 Metal hydrides (e.g. NaH) = 1 Find the oxidation number for all elements in SO 3 Sulfur? O = 2 ( not a peroxide, not bonded to halogen) Sum of the oxidation numbers = charge = 2 (ox. no. for S) + 3(ox. no. for O) = 2 (ox. no. for S) + 3(2) = 2 (ox. no. for S) 6 = 2 Sulfur = +4 Compound Known Unknown SO 2 SO 4 NH 4 + NO 2 NO 3 OF 2 ClF 5 KMnO 4 H 2 O 2 O = 2 O = 2 H = +1 O = 2 O = 2 F = 1 F = 1 K = +1 O = 2 H = +1 S = +4 S = +6 N = 3 N = +3 N = +5 O = +2 Cl = +5 Mn = +7 O = 1 Ox. numbers always change during redox reactions Increase ox. number = Oxidation Decrease ox. number = Reduction If an element reacts to form a compound, it is a redox reaction. Oxidation and reduction must both occur. S 8 (s) + 8 O 2 (g) 8 SO 2 (g) S : 0 +4 (oxidized: added O; lost e ; increased ox. no.) O : 0 2 (reduced: lost O; gained e ; decreased ox. no.) A more complex example: 0 +1 +5 2 Cu(s) + 4 H + (aq) + 2 NO 3 (aq) Cu 2+ (aq) + 2 NO 2 (g) + 2 H 2 O(l) +2 +4 2 +1 2 Cu is oxidized (ox. no. ; loss of e ). H is unchanged. O is unchanged. N is reduced (ox. no. ; gain of e ). Exchange reactions are not redox no change in oxidation state occurs. e.g. +1 +1 1 +1 +1 1 AgNO 3 (aq) + KCl(aq) KNO 3 (aq) + AgCl (s) NO 3 : N is in a +5, and O is in a 2 ox. state. 6

Displacement Reactions + + A XZ AZ X Redox: Fe(s) + CuSO 4 (aq) FeSO 4 (aq) + Cu(s) 0 +2 +2 0 Cu(s) + 2 AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2 Ag(s) 0 +1 +2 0 Not all metals can displace another from its salts: Cu(s) + ZnSO 4 (aq) no reaction An activity series was developed Activity Series of Metals Displace H 2 from H 2 O (l), steam or acid Displace H 2 from steam or acid Displace H 2 from acid No reaction with H 2 O, steam or acid Li K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H Sb Cu Hg Ag Pd Pt Au Ease of oxidation increases Displacement Reactions Powerful reducing agents at the top. Higher elements displace lower ones: Zn(s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu(s) Mg(s) + 2 HCl(aq) MgCl 2 (aq) + H 2 (g) Metals at the bottom are unreactive. Coinage metals Their ions are powerful oxidizing agents. Li K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H Sb Cu Hg Ag Pd Pt Au Displacement Reactions Displacement Reactions Potassium + water 2 K(s) + 2 H 2 O(l) 2 KOH(aq) + H 2 (g) Solution Concentration Relative amounts of solute and solvent. solute substance dissolved. solvent substance doing the dissolving. There are several concentration units. Most important to chemists: Molarity 7

Molarity Molarity = V of solution not solvent. moles solute liters of solution = mol L Molarity Calculate the molarity of sodium sulfate in a solution that contains 36.0 g of Na 2 SO 4 in 750.0 ml of solution. n Na2SO4 = 36.0 g = 0.2534 mol 142.0 g/mol Shorthand: [NaOH] =1.00 M Brackets [ ] represent molarity of [Na 2 SO 4 ] = 0.2534 mol 0.7500 L Unit change! (ml to L) [Na 2 SO 4 ] = 0.338 mol/l = 0.338 M Molarity 6.37 g of Al(NO 3 ) 3 are dissolved to make a 250. ml aqueous solution. Calculate (a) [Al(NO 3 ) 3 ] (b) [Al 3+ ] and [NO 3 ]. (a) Al(NO 3 ) 3 FM = 26.98 + 3(14.00) + 9(16.00) = 213.0 g mol 6.37 g n Al(NO3)3 = = 2.991 x 10 2 mol 213.0 g/mol Molarity 6.37 g of Al(NO 3 ) 3 in a 250. ml aqueous solution. Calculate (a) the molarity of the Al(NO 3 ) 3, (b) the molar concentration of Al 3+ and NO 3 ions in solution. (b) Molarity of Al 3+, NO 3? Al(NO 3 ) 3 (aq) Al 3+ (aq) + 3 NO 3 (aq) 1 Al(NO 3 ) 3 1 Al 3+ 1 Al(NO 3 ) 3 3 NO 3 [Al 3+ ] = 0.120 M Al(NO 3 ) 1 Al 3+ 3 = 0.120 M Al 3+ 1 Al(NO 3 ) 3 [Al(NO 3 ) 3 ] = 2.991 x 10 2 mol 0.250 L = 0.120 M [NO 3 ] = 0.120 M Al(NO 3 ) 3 NO 3 3 = 0.360 M NO 3 1 Al(NO 3 ) 3 Solution Preparation Solutions are prepared either by: 1. Diluting a more concentrated solution. or 2. Dissolving a measured amount of solute and diluting to a fixed volume. Solution Preparation by Dilution M conc V conc = M dil V dil Example Commercial concentrated sulfuric acid is 17.8 M. If 75.0 ml of this acid is diluted to 1.00 L, what is the final concentration of the acid? M conc = 17.8 M M dil =? M dil = M conc V conc = V dil V conc = 75.0 ml V dil = 1000. ml 17.8 M x 75.0 ml 1000. ml = 1.34 M 8

Solution Preparation from Pure Solute Prepare a 0.5000 M solution of potassium permanganate in a 250.0 ml volumetric flask. Mass of KMnO 4 required n KMnO4 = [KMnO 4 ] x V = 0.5000 M x 0.2500 L (M mol/l) = 0.1250 mol KMnO 4 Mass of KMnO 4 = 0.1250 mol x 158.03 g/mol = 19.75 g Solution Preparation from Pure Solute Weigh exactly 19.75 g of pure KMnO 4 Transfer it to a volumetric flask. Rinse all the solid from the weighing dish into the flask. Fill the flask ⅓ full. Swirl to dissolve the solid. Fill the flask to the mark on the neck. Shake to thoroughly mix. Grams of A Liters of A solution Use molar mass of A Moles of A Use solution molarity of A Use mole ratio n A = [ A ] x V Moles of B Use molar mass of B Use solution molarity of B [product] = n product / (total volume). Grams of B Liters of B solution What volume, in ml, of 0.0875 M H 2 SO 4 is required to neutralize 25.0 ml of 0.234 M NaOH? H 2 SO 4 (aq) + 2 NaOH(aq) Na 2 SO 4 (aq) + 2 H 2 O(l) n NaOH = 0.0250 L 0.234 mol L 2 NaOH 1 H 2 SO 4 = 5.850 x 10 3 mol n H2SO4 = 5.850 x 10 3 mol NaOH 1 H 2SO 4 2 NaOH = 2.925 x 10 3 mol H 2 SO 4 (aq) + 2 NaOH(aq) Na 2 SO 4 (aq) + 2 H 2 O(l) 0.002925 mol 0.00585 mol V? 25.0 ml V acid needed: mol H V = 2 SO 4 = [H 2 SO 4 ] 2.925 x 10 3 mol 1 L 0.0875 mol = 0.0334 L = 33.4 ml A 4.554 g mixture of oxalic acid, H 2 C 2 O 4 and NaCl was neutralized by 29.58 ml of 0.550M NaOH. What was the weight % of oxalic acid in the mixture? H 2 C 2 O 4 (aq) + 2 NaOH(aq) Na 2 C 2 O 4 (aq) + 2 H 2 O(l) n NaOH = 0.02958 L x 0.550 mol/l = 0.01627 mol 1 H 2 C 2 O 4 2 NaOH Oxalic acid reacted = 0.01627 mol NaOH 1 H 2 C 2 O 4 2 NaOH = 8.135 x10 3 mol 9

A 4.554 g H 2 C 2 O 4 / NaCl mixture Wt % of oxalic acid in the mixture? Mass of acid consumed, m acid = 8.135 x10 3 mol x (90.04 g/mol acid) = 0.7324 g m acid Weight % = sample mass x 100% 25.0 ml of 0.234 M FeCl 3 and 50.0 ml of 0.453 M NaOH are mixed. Which reactant is limiting? How many moles of Fe(OH) 3 will form? FeCl 3 (aq) + 3 NaOH(aq) 3 NaCl (aq) + Fe(OH) 3 (s) n FeCl3 = 0.0250 L x 0.234 mol/l = 0.005850 mol 0.7324 g Weight % = x 100% 4.554 g = 16.08% n NaOH = 0.0500 L x 0.453 mol/l = 0.02265 mol FeCl 3 (aq) + 3 NaOH(aq) 3 NaCl (aq) + Fe(OH) 3 (s) 0.005850 mol 0.01925 mol n Fe(OH)3? 1 Fe(OH) 0.00585 mol FeCl 3 3 = 0.00585 mol Fe(OH) 3 1 FeCl 3 1 Fe(OH) 0.02265 mol NaOH 3 = 0.00755 mol Fe(OH) 3 3 NaOH Aqueous Solution Titrations Titration = volumebased method used to determine an unknown concentration. A standard solution (known concentration) is added to a solution of unknown concentration. Monitor the volume added. Add until equivalence is reached stoichiometrically equal moles of reactants added. An indicator monitors the end point. FeCl 3 is limiting; 0.00585 mol Fe(OH) 3 produced. Often used to determine acid or base concentrations. Aqueous Solution Titrations Titrant: Base of known concentration Buret = volumetric glassware used for titrations. Slowly add standard solution. End point: indicator changes color. Determine V titrant added. Unknown acid + phenolphthalein (colorless in acid) turns pink in base 10