Salts product of neutralization reaction strong base strong acid 1.00 M NaOH 150 mL 0.500 M HCl mol OH- = mol H+ x L 1.00 mol = 0.075 mol = 0.15 L 0.500 mol L L x = 0.075L H+ + Cl- + Na+ + OH-  H2O + Na+ + Cl- [Na+] = 0.075 mol = 0.333 M = [Cl-] NaCl (s) .150 L + .075 L

Salts product of neutralization reaction strong base strong acid NaOH NaCl (s) HCl very weak conjugate acid very weak conjugate base Na+ Cl- Na ++ H2O  no reaction Cl- + H2O  no reaction Li+ Ca2+ Br- NO3- K+ Sr2+ I- ClO4- Rb+ Ba2+ negligible basicity negligible acidity

Salts product of neutralization reaction strong base weak acid 1.00 M NaOH 150 mL 0.500 M HF mol OH- = mol H+ x L 1.00 mol = 0.075 mol = 0.15 L 0.500 mol L L x = 0.075L HF  H+ + F- + Na+ + OH-  H2O + Na+ + F- Ka = 1.5 x 10-11 NaF (s)

Salts product of neutralization reaction strong base NaF (s) weak acid NaOH HF Ka = 1.5 x 10-11 very weak conjugate acid strong conjugate base Na+ F- Na ++ H2O  no reaction F- + H2O  HF + OH- [HF] [OH-] hydrolysis = Kb = 6.7 x 10-4 [F-] “breaking water” Ka x Kb = Kw basic solution

Salts product of neutralization reaction weak base strong acid CH3NH2 Kb = 4.4 x 10-4 HClO4 CH3NH2+ H2O  CH3NH3++ OH- + H+ + ClO4-  H2O CH3NH3ClO4 (s) CH3NH3+ Ka = 2.3 x 10-11 ClO4- strong conjugate acid weak conjugate base CH3NH3+ + H2O  CH3NH2 + H3O+ hydrolysis acidic solution

Salts product of neutralization reaction weak base weak acid HNO2 NH3 NH4NO2 NH4+ + H2O  NH3 + H3O+ Ka = 5.7 x 10-10  NO2- + H2O HNO2 + OH- Kb = 1.4 x 10-11 acidic solution

Calculate the pH of a 0.10 M C2H5NH3NO3 solution C2H5NH3 Kb = 5.6 x 10-4 NO3 + - weak base strong acid C2H5NH3+ + H2O  C2H5NH2 + H3O+ Ka x Kb = Kw = 1.0 x 10-14 Ka = 1.8 x 10-11 1.8 x 10-11 = [H3O+] [C2H5NH2] = x2 [C2H5NH3+] 0.1 x = 1.34 x 10-6 pH = 5.87