Salts product of neutralization reaction 1.00 M NaOH mol OH - L mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH -  H2OH2O+ Na + + Cl - NaCl (s) [Na + ]= mol.150 L+.075 L = M= [Cl - ] basestrong 150 mL MHCl = mol H mol= 0.15 L acid strong =

NaCl (s)HClNaOH conjugate baseconjugate acid Cl - Na + very weak Cl - + H 2 O  no reactionNa + + H 2 O  no reaction Salts product of neutralization reaction basestrong acid strong Li + K+K+ Rb + Ca 2+ Sr 2+ Ba 2+ negligible acidity Br - I-I- NO 3 - ClO 4 - negligible basicity

Salts product of neutralization reaction base acid strong weak 1.00 M NaOH mol OH - L mol1.00 mol L x L 150 mL MHFHF = mol H mol= 0.15 L= x = 0.075L H+H+ + F - + Na + + OH -  H2OH2O+ Na + + F - NaF (s) HFHF  K a = 1.5 x

Salts product of neutralization reaction base acid strong weakNaF (s) NaOH HFHF conjugate baseconjugate acidvery weakstrong Na + + H 2 O  no reaction F - + H 2 O  HF + OH - K a = 1.5 x K a xK b = Kw Kw 6.7 x hydrolysis “breaking water” Na + F-F- basic solution = [HF] [OH - ] [F - ]

Salts product of neutralization reaction base acid weak strong HClO 4 CH 3 NH 2 CH 3 NH 2 + H 2 O  CH 3 NH OH - + H + + ClO 4 - CH 3 NH 3 ClO 4 (s)  H2OH2O CH 3 NH 3 + ClO 4 - conjugate base weakconjugate acidstrong K b = 4.4 x K a = 2.3 x CH 3 NH H 2 O  CH 3 NH 2 + H 3 O + acidic solution hydrolysis

Salts product of neutralization reaction base acid weak NH3NH3 HNO 2 NH 4 NO 2 NH 4 + NO 2 -  NH 3 HNO 2 + H 2 O+ H 3 O + + H 2 O  + OH - K a = K b = 5.7 x x acidic solution

List the following species in order of C 5 H 5 NHCl K 2 CO 3 LiNO 3 NaHCO 3 cation anionpH C 2 H 5 NH + Cl - K+K+ CO 3 2- Li + NO 3 - Na + HCO 3 - w.b. s.b. s.a. w.a. acidic basic neutral basic I II III IV 0.20 M increasing pH

Calculate the pH of a C 2 H 5 NH 3 NO base acid C 2 H 5 NH H 2 O  C 2 H 5 NH 2 + H3O+H3O+ K b = 5.6 x K a x KbKb = K w = 1.0 x K a = 1.8 x x = [H 3 O + ][C 2 H 5 NH 2 ] [C 2 H 5 NH 3 + ] = x x = 1.34 x pH = M C 2 H 5 NH 3 NO 3 solution weakstrong