C(HAc)=0.1mol/LHAc和C(NaOH)=0.1mol/LNaOH等体积混合溶液的PH是多少?
人气:384 ℃ 时间:2019-08-20 06:12:19
解答
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-xx x所以 x^2 /(0.05-x) = 5.56×10^-10x = 5.27 ×10^-6pOH=...
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