Neutralization Reactions acidbaseacid + base  salt + water HX(aq)MOH(aq)HX(aq) + MOH(aq)  MX(aq) + H 2 O(l) this is a DR rxn

hydrochloric acid + sodium hydroxide HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) Is it balanced??  sodium chloride + water

HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O H +1 (aq) + Cl -1 (aq) + Na +1 (aq) + OH -1 (aq)  H 2 O(l) + Na +1 (aq) + Cl -1 (aq) Na +1 and Cl -1 are spectator ions: H +1 (aq) + OH -1 (aq)  H 2 O(l)

nitric acid + potassium hydroxide HNO 3 (aq) + KOH(aq)  KNO 3 (aq) + H 2 O(l) K +1 (aq) + NO 3 -1 (aq) + H 2 O(l) H +1 (aq) + NO 3 -1 (aq) + K +1 (aq) + OH -1 (aq)  H +1 (aq) + OH -1 (aq)  H 2 O

sulfuric acid + calcium hydroxide H 2 SO 4 (aq) + Ca(OH) 2 (aq)  CaSO 4 (aq) + H 2 O(l) H 2 SO 4 + Ca(OH) 2 → CaSO H 2 O

for any neutralization reaction between an acid & base: net ionic equation net ionic equation is always H +1 (aq) + OH -1 (aq)  H 2 O(l) H +1 (aq) + OH -1 (aq)  H 2 O(l)

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

H +1 (aq) + OH -1 (aq)  H 2 O(l) 1-to-1 relationship between H +1 and OH -1 neutralization point: At neutralization point: # moles H +1 = # moles OH -1

Molarity of H +1 Molarity of H +1 = moles H +1 liters of soln liters of soln

Molarity of OH -1 Molarity of OH -1 = moles OH -1 liters of soln liters of soln

At neutralization Moles H +1 = Moles OH -1 (M H+1 ) (V H+1 ) = (M OH-1 ) (V OH-1 )

(M A )( V A ) = (M B )( V B ) M A = molarity of H +1 V A = volume of acid M B = molarity of OH -1 V B = volume of base True for: monoprotic acid with monohydroxy base monoprotic acid with monohydroxy base diprotic acid with dihydroxy base diprotic acid with dihydroxy base triprotic acid with trihydroxy base triprotic acid with trihydroxy base

Molarity of H +1 In HCl: molarity H +1 = molarity HCl 2In H 2 SO 4: 2 molarity H +1 = 2(molarity H 2 SO 4 ) 3In H 3 PO 4 : 3 molarity H +1 = 3(molarity H 3 PO 4 )

Molarity of OH -1 In KOH: molarity OH -1 = molarity KOH 2In Ca(OH) 2 : 2 molarity OH -1 = 2(molarity Ca(OH) 2 )

When # H’s ≠ # OH’s (M A )(V A ) = (M B )(V B ) (#H ’ s) (#OH ’ s)