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Acid- Base Reactions

Acid- Base Reactions. Chapter 4 part IV. Characteristics of acids. Sour Red litmus test Low pH <7 Arrhenius: Produce H+ in water. Brømsted Lowry: produce protons. Lewis: Electron acceptor. Characteristics of bases. Slippery & bitter Blue litmus test High pH >7

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Acid- Base Reactions

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  1. Acid- Base Reactions Chapter 4 part IV

  2. Characteristics of acids • Sour • Red litmus test • Low pH <7 • Arrhenius: Produce H+ in water. • Brømsted Lowry: produce protons. • Lewis: Electron acceptor.

  3. Characteristics of bases • Slippery & bitter • Blue litmus test • High pH >7 • Arrhenius: forms OH- in water. • Brømsted Lowry: accept protons. • Lewis: Electron donor.

  4. Net Ionic Equation • Of all strong acids plus strong bases: • H+ + OH- H2O • But what about weak acids and bases? • In a strong acid or base, it dissociates completely in water (strong electrolyte) • Weak acids and bases do not dissociate completely in water

  5. Weak electrolytes • In fact about 90% of acetic acid remains intact in aqueous solution. • Therefore the net ionic equation of a weak acid or base includes the intact weak acid or base. • Example: HC2H3O2 +Na++ OH- H2O+ C2H3O2- +Na+ Net Ionic: HC2H3O2 + OH- H2O + C2H3O2-

  6. Weak electrolytes • An example of a weak base is NH3. • So what is the net ionic equation for hydrochloric acid and ammonia? • HCl + NH3 NH4+ + Cl- • H+ + Cl- + NH3 NH4+ + Cl- • H+ + NH3 NH4+

  7. Stoichiometry of an acid/base reaction: The steps. • List species present in the combined solution, before any reaction occurs. • Decide what reaction will occur. • Write out net ionic equation. • Calculate moles of reactant in solution. Use volume of the original solution and its molarity. • Determine the Limiting reagent.

  8. Stoichiometry of an acid/base reaction: The steps. • Calculate the moles of required reactant or product formed. • Convert to grams or volume as required. • In other words, after you answer the question, reread the question and make sure.

  9. Example: • What volume of 0.1 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH?

  10. Answer • List the species: H+ Cl- Na+ OH- • What are the possible products? • NaCl (s) And H2O (l) • NaCl is soluble therefore it is not a possible product in solution. • Write the balanced equation. • H+(aq) + OH-(aq)  H2O(l)

  11. Answer • Calculate moles of reactants: • OH- = 25 mL NaOH x (1L/1000mL)x (0.350 mol OH-/L NaOH) = • 8.75 x10-3 mol OH- • No limiting reagent, finding volume of HCl • Moles of reactant needed: molar ration is 1:1 therefore need 8.75 x 10-3 mols acid. • Convert to volume: V x M=mol

  12. Therefore • V x 0.100 mol H+ /L = 8.75 x 10-3 mol H+ • V= 8.75 X 10-2 L

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