C(HAc)=0.1mol/LHAc和C(NaOH)=0.1mol/LNaOH等体积混合溶液的PH是多少?
问题描述:
C(HAc)=0.1mol/LHAc和C(NaOH)=0.1mol/LNaOH等体积混合溶液的PH是多少?
答
混合后,溶液为NaAc溶液,浓度是0.05 mol/LAc- + H2O = HAc + OH- K= Kw/Ka = 1×10^-14 ÷ 1.8 ×10^-5 = 5.56 ×10^-100.05-x x x所以 x^2 /(0.05-x) = 5.56 ×10^-10x = 5.27 ×10^-6pOH=...