2. Neutralization Reactions
acid + base  salt + water
HX(aq) + MOH(aq)  MX(aq) + H2O(l)
DR rxn: liquid water is product

3. hydrochloric acid + sodium hydroxide
 sodium chloride + water
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Is it balanced??

4. complete chemical equation: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O
complete ionic equation:
H+1(aq)+Cl-1(aq)+Na+1(aq)+OH-1(aq)  H2O(l)+Na+1(aq)+ Cl-1(aq)
net ionic equation:
H+1(aq) + OH-1(aq)  H2O(l)

5. nitric acid + calcium hydroxide
2HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2H2O(l)
2H+1(aq) + 2NO3-1(aq) + Ca+2(aq) + 2OH-1(aq) 
Ca+2(aq) + 2NO3-1(aq) + 2H2O(l)
2H+1(aq) + 2OH-1(aq)  2H2O
REDUCE!

6. for any neutralization reaction between an acid & base:
net ionic equation is always
H+1(aq) + OH-1(aq)  H2O(l)

7. pH changes during neutralization
Start with an acid
Add a base
At neutralization
Start with a base
Add an acid
pH < 7
pH 
pH = 7
pH > 7
pH 
pH = 7

8. H+1(aq) + OH-1(aq)  H2O(l)
1-to-1 relationship between H+1and OH-1
At neutralization point:
# moles H+1 = # moles OH-1

9. Molarity of H+1 Molarity of OH-1 Molarity of H+1 = moles H+1
liters of soln
Molarity of OH-1
Molarity of OH-1 = moles OH-1
liters of soln

10. At neutralization point
Moles H+1 = Moles OH-1
(MH+1) (VH+1) = (MOH-1) (VOH-1)

11. (MA)( VA) = (MB)( VB) MA = molarity of H+1 VA = volume of acid
MB = molarity of OH-1
VB = volume of base
true for:
monoprotic acid with monohydroxy base
diprotic acid with dihydroxy base
triprotic acid with trihydroxy base

12. the # H’s in the acid DO NOT EQUAL the # of OH’s in the base?
So what to do if:
the # H’s in the acid
DO NOT EQUAL
the # of OH’s in the base?
must multiply acid/base sides by # H’s/# OH’s

13. When # H’s ≠ # OH’s (MA)(VA) = (MB)(VB) (#H’s) (#OH’s)
this is an extra step you will need to do:
neither Titration formula on Table T nor word problem will tell you to do this – MUST MEMORIZE!