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

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 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+

9. 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.

10. 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.

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

12. 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)

13. 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

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