Ac- + H2O ~~> OH + AcH Kb = [OH-][AcH]/[Ac-] Kb=10^-14/Ka = 5.556x10^-10 [OH-]______[AcH]________[Ac-] initial_____0_______0_________1M change___+x______+x_________-x final______x_______x__________0.052-x Kb = x2/(0.052-x) = 5.556x10-10 Solve for x: x=5-log(0.535) = [OH-] pOH = -log[OH] = 5.27 pH=14-pOH=8.73 Solution Ac- + H2O ~~> OH + AcH Kb = [OH-][AcH]/[Ac-] Kb=10^-14/Ka = 5.556x10^-10 [OH-]______[AcH]________[Ac-] initial_____0_______0_________1M change___+x______+x_________-x final______x_______x__________0.052-x Kb = x2/(0.052-x) = 5.556x10-10 Solve for x: x=5-log(0.535) = [OH-] pOH = -log[OH] = 5.27 pH=14-pOH=8.73.