2.直接在反应体系中加HAc和NaAc, 按缓冲容量算得c=0.19molL n(HAc)=0.19×50×0.365=3.5mmol, n(NaAc=6.0mmol m(NaAc)=6.0X82=0.49g 3. pHo=pKa +lg nb nb =1.74 174n-1 pH,=pa+lg +1 解得:n(HAc)=33mmol, n(naAc)=5.7 mmol m(NaAc)=5.7X82=0.419 99 2. 直接在反应体系中加HAc 和NaAc, 按缓冲容量算得c=0.19mol·L-1 n(HAc)=0.19×50×0.365=3.5mmol, n(NaAc)=6.0mmol m(NaAc)=6.0×82=0.49g 3. 1 a a a 1.74 1 pH p lg 1 n K n − = + + b 0 a a pH p lg 1.74 b a n K n n n = + = 解得: n(HAc)=3.3mmol, n(NaAc)=5.7mmol m(NaAc)=5.7×82=0.47g