1. Acid- Base Reactions Chapter 4 part IV

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

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

6. 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-

7. Weak electrolytes An example of a weak base is NH 3. An example of a weak base is NH 3. So what is the net ionic equation for hydrochloric acid and ammonia? So what is the net ionic equation for hydrochloric acid and ammonia? HCl + NH 3  NH 4 + + Cl - HCl + NH 3  NH 4 + + Cl - H + + Cl - + NH 3  NH 4 + + Cl - H + + Cl - + NH 3  NH 4 + + Cl - H + + NH 3  NH 4 + H + + NH 3  NH 4 +

9. Stoichiometry of an acid/base reaction: The steps. 1. L ist species present in the combined solution, before any reaction occurs. 2. D ecide what reaction will occur. 3. W rite out net ionic equation. 4. C alculate moles of reactant in solution. Use volume of the original solution and its molarity. 5. D etermine the Limiting reagent.

10. Stoichiometry of an acid/base reaction: The steps. 6. Calculate the moles of required reactant or product formed. 7. Convert to grams or volume as required. 8. In other words, after you answer the question, reread the question and make sure.

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

12. Answer List the species: H+ Cl- Na+ OH- List the species: H+ Cl- Na+ OH- What are the possible products? What are the possible products? NaCl (s) And H 2 O (l) NaCl (s) And H 2 O (l) NaCl is soluble therefore it is not a possible product in solution. NaCl is soluble therefore it is not a possible product in solution. Write the balanced equation. Write the balanced equation. H+(aq) + OH-(aq)  H 2 O(l) H+(aq) + OH-(aq)  H 2 O(l)

13. Answer Calculate moles of reactants: Calculate moles of reactants: OH- = 25 mL NaOH x (1L/1000mL)x (0.350 mol OH-/L NaOH) = OH- = 25 mL NaOH x (1L/1000mL)x (0.350 mol OH-/L NaOH) = 8.75 x10 -3 mol OH- 8.75 x10 -3 mol OH- No limiting reagent, finding volume of HCl 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. 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 Convert to volume: V x M=mol

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