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Neutralization Reactions with Strong Acids and Strong Bases

Neutralization Reactions with Strong Acids and Strong Bases. Naming Acids. Binary acid (i.e. H & a non-metal) the prefix hydro is used the root of the anion is used the suffix -ic is used the word acid is used as the second word of the name Example HCl = hydro chlor ic acid

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Neutralization Reactions with Strong Acids and Strong Bases

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  1. Neutralization Reactions with Strong Acids andStrong Bases

  2. Naming Acids • Binary acid (i.e. H & a non-metal) • the prefix hydro is used • the root of the anion is used • the suffix -ic is used • the word acid is used as the second word of the name • Example • HCl = hydrochloricacid • HBr = hydrobromic acid • HI = hydroiodic acid • HF = hydrofluoric acid

  3. Naming Acids • Polyatomic acids • oxyacids: (acids with oxygen in the polyatomic anion) • change suffix of –ate with -icOR • change suffix -ite to -ous • These acids have the general formula HaXbOc where X = an element other than Hydrogen or Oxygen • Examples • HNO3 (nitrate) (-ate  -ic) • HSO3 (sulfite) (-ite  -ous) • H2SO4 (sulfate) (-ate  -ic) Nitric acid Sulfurous acid Sulfuric acid

  4. Acid Assumptions • Strong Acid: • an acid that completely dissociates into ions. • (100 molecules of HCl → 100 H+ ions) • The six strong acids to be memorized • HCl • HBr • HI • H2SO4 • HNO3 • HClO4 (Perchloric acid)

  5. Base Assumptions • Strong Bases • a base that completely dissociates into ions. • (100 formula units of NaOH → 100 OH- ions) • When combined with the OH_ (hydroxide) ion, elements found in group 1 (IA) and 2 (IIA) form strong bases • Examples • KOH • CsOH • Ba(OH)2 • Ca(OH)2

  6. Identifying Strong Acids & Bases • H+ (Hydrogen ion) indicates strong acid • pH scale with a value of 2 or less • OH- (Hydroxide ion) indicates strong base • pH scale with a value of 12 or more

  7. Assignment • Read p.271-273 Questions: P.273 #’s 29-33

  8. Neutralization • Current Assumption • When strong acid and strong base are combined, all H+ and OH- ions join to form HOH (H2O)

  9. Balancing Neutralization Reactions • Neutralization reaction is a double displacement • Example • For NaOH and HCl • Predict the products of the reaction • balance the equation • NaOH(aq) + HCl(aq) NaCl + HOH • Use solubility rules to confirm whether each product will be aqueous, solid or liquid • NaOH(aq) + HCl(aq) NaCl(aq) + HOH(l) • Write the total ionic equation, showing all ions that are in solution

  10. Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l) • Cancel the spectator ions and write the net ionic equation • OH-(aq) + H+(aq) → H2O(l)

  11. Example • Write the molecular, ionic, net ionic equations for • Sulfuric acid & potassium hydroxide H2SO4(aq) + KOH(aq) → K2SO4 + HOH 2 H+(aq)+ SO42-(aq)+ 2 K+(aq) + 2 OH-(aq) → 2 K+(aq) + SO42-(aq) +2 HOH(l) 2 H+(aq) + 2 OH-(aq) → + 2 HOH(l) Simplify H+(aq) + OH-(aq) → + HOH(l) 2 2 (aq) (l)

  12. Neutralization Reaction: Calculating Volume or Concentration • Recall: • Concentration is calculated as moles per litre • mol/L • M • [NaOH] refers to the concentration of sodium hydroxide • Equation • [ ] =mol/L

  13. Example 1 • In the reaction of 35.0 mL of liquid drain cleaner containing NaOH, 50.08 mL of 0.409 mol/L HCl must be added to neutralize the base. What is the concentration of the base in the cleaner? • Write a balanced equation and the chart NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) mol 0.0205mol 0.0205mol Mm m V 0.0350L 0.05008L [ ] 0.586M 0.409M

  14. Example 2 • Calculate the volume of 0.256 mol/L Ba(OH)2 that must be added to neutralize 46.0 mL of 0.407 mol/L HClO4. Ba(OH)2(aq) + 2 HClO4(aq) → BaCl2(aq) + 2H2O(l) mol 0.00935mol 0.0187mol Mm m V 0.0365L 0.0460L [ ] 0.256M 0.407M

  15. Assignment • P. 614 • #’s 30-31 • P. 616 • #’s 32-33, 36-37 • Aqueous Reactions Worksheet #1 • #’s 5-7 • Aqueous Reactions Worksheet #2

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