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100g NaAc·3H2O的物质的量=100g/(136g/mol)=0.735mol
即n(Ac-)=0.735mol
13ML6.0MOL/LHAC的物质的量=0.013L×6.0mol/L=0.078mol
即n(HAc)=0.078mol
HAc Ac- +H+
Ka(HAc)=[Ac-]×[H+]/[HAc]
=约=c(Ac-)×[H+]/c(HAc)
=n(Ac-)×[H+]/n(HAc)
=0.735×[H+]/0.078
[H+]=1.75×10^-5 ×0.078/0.735=1.86×10^-6
pH=-lg[H+]=5.73
附加答案中c( )代表起始浓度,而[ ]代表平衡浓度
此溶液中加入0.1mol HCl
则有:
n(Ac-)=0.635mol
n(HAc)=0.178mol
同样是缓冲溶液
HAc Ac- +H+
Ka(HAc)=[Ac-]×[H+]/[HAc]
=约=c(Ac-)×[H+]/c(HAc)
=n(Ac-)×[H+]/n(HAc)
=0.635×[H+]/0.178
[H+]=1.75×10^-5 ×0.178/0.635=4.9×10^-6
pH=-lg[H+]=5.30
1年前
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