Lecture 6. Classes of Chemical Reactions
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1 Lecture 6 Classes of Chemical Reactions Lecture 6 Outline 6.1 The Role of Water as a Solvent 6.2 Precipitation Reactions 6.3 Acid-Base Reactions 1
2 Electron distribution in molecules of H 2 and H 2 O The electrical conductivity of ionic solutions 2
3 The dissolution of an ionic compound A substance that conducts a current when dissolved in water is an electrolyte. Ionic compounds are strong electrolytes Sample Problem: Determining Moles of Ions in Aqueous Ionic Solutions PROBLEM: How many moles of each ion are in the following solutions? (a) 5.0 mol of ammonium sulfate dissolved in water (b) 78.5g of cesium bromide dissolved in water (c) 7.42x10 22 formula units of copper(ii) nitrate dissolved in water (d) 35mL of 0.84M zinc chloride n = m/m c = n/v SOLUTION: (a) (NH 4 ) 2 SO 4 (s) 2NH 4 + (aq) + SO 4 2- (aq) 5.0mol (NH 4 ) 2 SO 4 10.mol NH mol SO 4 2-3
4 Sample Problem continued Determining Moles of Ions in Aqueous Ionic Solutions CsBr: (b) CsBr(s) Cs + (aq) + Br - (aq) 78.5g/ 212.8g/mol = 0.369mol CsBr = 0.369mol Br = 0.369mol Cs + (c) Cu(NO 3 ) 2 (s) Cu 2+ (aq) + 2NO 3 - (aq) 7.42x10 22 formula 6.022x10 23 formula units = 0.123mol Cu 2+ = 0.123mol Cu(NO 3 ) 2 = 0.246mol NO 3 (d) ZnCl 2 (aq) Zn 2+ (aq) + 2Cl - (aq) n = c V ZnCl L x 0.84mol/L = 2.9x110-2 mol ZnCl 2 = 2.9x110-2 mol Zn 2+ = 5.8x110-2 mol Cl - Precipitation Reactions 4
5 Precipitation Reactions Precipitate insoluble solid that separates from solution molecular equation ionic equation Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb NO Na + + 2I - PbI 2 (s) + 2Na + + 2NO - 3 net ionic equation Pb I - PbI 2 (s) Na + and NO 3- are spectator ions Sample Problem PROBLEM: Determining the Molarity of H + Ions in Aqueous Solutions of Acids Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H + ion. What is the molarity of H + (aq) in 1.4M nitric acid? PLAN: Use the formula to find the molarity of H +. SOLUTION: Nitrate is NO 3 -. HNO 3 (l) H + (aq) + NO 3 - (aq) 1.4M HNO 3 (aq) should have 1.4M H + (aq). 5
6 Solubility Rules For Ionic Compounds in Water Soluble Ionic Compounds 1. All common compounds of group 1A(1) ions (Li +, Na +, K +, etc.) and ammonium ion (NH 4+ ) are soluble. 2. All common nitrates (NO 3- ), acetates (CH 3 COO ) and most perchlorates (ClO 4- ) are soluble. 3. All common chlorides (Cl - ), bromides (Br - ) and iodides (I - ) are soluble, except those of Ag +, Pb 2+, Cu +, and Hg Insoluble Ionic Compounds 1. All common metal hydroxides are insoluble, except those of group 1A(1) and the larger members of group 2A(2)(beginning with Ca 2+ ). 2. All common carbonates (CO 2-3 ) and phosphates (PO 3-4 ) are insoluble, except those of group 1A(1) and NH All common sulfides are insoluble except those of group 1A(1), group 2A(2) and NH 4+. Solubility Rules for Common Ionic Compounds In water at 25 0 C Soluble Compounds Compounds containing alkali metal ions and NH 4 + NO 3-, HCO 3-, ClO 3 - Cl -, Br -, I - SO 4 2- Insoluble Compounds CO 3 2-, PO 4 3-, CrO 4 2-, S 2- OH - Exceptions Halides of Ag +, Hg 2 2+, Pb 2+ Sulfates of Ag +, Ca 2+, Sr 2+, Ba 2+, Hg 2+, Pb 2+ Exceptions Compounds containing alkali metal ions and NH + 4 Compounds containing alkali metal ions and Ba 2+ 6
7 A precipitation reaction and its equation AgNO 3 Na 2 CrO 4 AgCrO 4 Formation of Silver Chloride 7
8 Formation of Silver Chloride, Silver Bromide, Silver Iodide Ag(NO 3 ) 3 + NaCl AgCl + Na + (aq) + NO 3 (aq) white Dissolves again in diluted NH 3 Ag(NO 3 ) 3 + NaBr AgBr + Na + (aq) + NO 3 (aq) slightly yellow Dissolves again in concentrated NH 3 Ag(NO 3 ) 3 + NaI AgI + Na + (aq) + NO 3 (aq) yellow Does not dissolve in NH 3 Writing Net Ionic Equations 1. Write the balanced molecular equation. 2. Write the ionic equation showing the strong electrolytes 3. Determine precipitate from solubility rules 4. Cancel the spectator ions on both sides of the ionic equation Write the net ionic equation for the reaction of silver nitrate with sodium chloride. AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) Ag + + NO 3- + Na + + Cl - AgCl (s) + Na + + NO 3 - Ag + + Cl - AgCl (s) 8
9 Sample Problem PROBLEM: Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) sodium sulfate(aq) + strontium nitrate(aq) (b) ammonium perchlorate(aq) + sodium bromide(aq) SOLUTION: (a) Na 2 SO 4 (aq) + Sr(NO 3 ) 2 (aq) 2NaNO 3 (aq) + SrSO 4 (s) 2Na + (aq) +SO 4 2- (aq)+ Sr 2+ (aq)+2no 3 - (aq) 2Na + (aq) +2NO 3 - (aq)+ SrSO 4 (s) SO 4 2- (aq)+ Sr 2+ (aq) SrSO 4 (s) (b) NH 4 ClO 4 (aq) + NaBr (aq) NH 4 Br (aq) + NaClO 4 (aq) All reactants and products are soluble so no reaction occurs. Reactions in Water.. Solvation in Water Hydration: Formation of ions that are surrounded by water Dissociation I: Dissociation of ionic compounds (hydration energy) Dissociation II: Heteropolar dissociation caused by the reaction with water (protolysis). H δ+ Cl δ Hydronium Ion Electrolytes: Substances that show electrical conductivity after dicossiation 9
10 Acid and Base Reactions Equilibrium H 2 O NH 3 HCl NaOH 10
11 Arrhenius (1883) Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water Acid-Base Theory of Brönsted (1923) Acid: donate protons (H + ) Base: accept protons (H + ) Corresponding acid and base Acid 1 Base 1 Acid 2 Base 2 HCl + H 2 O H 3 O + + Cl Acid Strength H 2 SO 4 + H 2 O H 3 O + + HSO 4 HSO 4 + H 2 O H 3 O + + SO 4 2 NH H 2 O H 3 O + + NH 3 Base Strength HCO 3 + H 2 O H 3 O + + OH 11
12 Neutralization Reaction acid + base salt + water HCl (aq) + NaOH (aq) H + + Cl - + Na + + OH - H + + OH - NaCl (aq) + H 2 O Na + + Cl - + H 2 O H 2 O Monoprotic acids HCl H + + Cl - HNO 3 H + + NO 3 - (H 2 O neglected for clarity) Strong electrolyte, strong acid Strong electrolyte, strong acid CH 3 COOH H + + CH 3 COO - Weak electrolyte, weak acid Diprotic acids (H 2 O neglected for clarity) H 2 SO 4 H + + HSO - 4 Strong electrolyte, strong acid HSO 4 - H + + SO 4 2- Weak electrolyte, weak acid Triprotic acids H 3 PO 4 H + + H 2 PO 4 - H 2 PO - 4 H + + HPO 2-4 HPO 2-4 H + + PO 3-4 (H 2 O neglected for clarity) Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid 12
13 Selected Acids and Bases Acids Strong hydrochloric acid, HCl hydrobromic acid, HBr hydroiodic acid, HI nitric acid, HNO 3 sulfuric acid, H 2 SO 4 Bases Strong sodium hydroxide, NaOH potassium hydroxide, KOH calcium hydroxide, Ca(OH) 2 strontium hydroxide, Sr(OH) 2 barium hydroxide, Ba(OH) 2 perchloric acid, HClO 4 Weak hydrofluoric acid, HF Weak ammonia, NH 3 phosphoric acid, H 3 PO 4 acetic acid, CH 3 COOH (or HC 2 H 3 O 2 ) ph -value ph = log c H + poh = log c OH - ph + poh = 14 Acid c(h 3 O + ) > c(oh ) Neutral c(oh ) = c(h 3 O + ) Basic c(oh ) > c(h 3 O + ) 13
14 Strength of an acid Strength of a base pks value: Sample Problem Writing Ionic Equations for Acid-Base Reactions PROBLEM: Write balanced molecular, total ionic, and net ionic equations for each of the following acid-base reactions and identify the spectator ions. (a) strontium hydroxide(aq) + perchloric acid(aq) (b) barium hydroxide(aq) + sulfuric acid(aq) PLAN: SOLUTION: reactants are strong acids and bases and therefore completely ionized in water products are water spectator ions (a) Sr(OH) 2 (aq)+2hclo 4 (aq) Sr 2+ (aq) + 2OH - (aq)+ 2H + (aq)+ 2ClO - 4 (aq) 2H 2 O(l)+Sr 2+ (aq)+2clo - 4 (aq) 2OH - (aq)+ 2H + (aq) (b) Ba(OH) 2 (aq) + H 2 SO 4 (aq) 2H 2 O(l) 2H 2 O(l)+Sr(ClO 4 ) 2 (aq) 2H 2 O(l) + BaSO 4 (s) Ba 2+ (aq) + 2OH - (aq)+ 2H + (aq)+ SO 4 2- (aq) 2H 2 O(l)+BaSO 4 (s) 2OH - (aq)+ 2H + (aq) 2H 2 O(l) 14
15 An acid-base titration Solution with known concentration Solution with an unknown concentration HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O neutral point addition of base An acid-base titration ph indicator 15
16 Sample Problem PROBLEM: Finding the Concentration of Acid from an Acid-Base Titration You perform an acid-base titration to standardize an HCl solution by placing 50.00mL of HCl in a flask with a few drops of indicator solution. You put M NaOH into the buret, and the initial reading is 0.55mL. At the end point, the buret reading is 33.87mL. What is the concentration of the HCl solution? SOLUTION: NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O(l) At the neutral point: 1 mol (NaOH) = 1 mol (HCl) mol NaOH: c = n/v Molar ratio is 1:1 n = c V L X M = 5.078x10-3 mol c (HCl): c = n/v HCl: 5.078x10-3 mol 0.050L = M As previously indicated: for these tiration exercises you often apply the two simple formula: c = n/v n = m/m. see practical course 16
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