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

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

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)

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!

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

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

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

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

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

(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

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

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!