25℃,0.4mol/lch3cooh溶液和0.2mol/lNaOH各100mL混合后pH=5

题目

25℃,0.4mol/lch3cooh溶液和0.2mol/lNaOH各100mL混合后pH=5
c(ch3coo-)+c(ch3cooh)= c(ch3coo-)-c(ch3cooh)=

答案

C(CH3COOH)=0.4mol/L*0.1L=0.04mol
根据物料守恒得:C(CH3COO^-)+C(CH3COOH)=0.04mol/0.2L=0.2mol/L
C(NaOH)=0.2mol/L*0.1L=0.02mol
CH3COOH+NaOH=CH3COONa+H2O
1 1 1
X 0.02mol Y
X=0.02mol
Y=0.02mol
剩余醋酸的物质的量=0.04mol-0.02mol=0.02mol
混合后溶液中含n(H^)=10^-5mol/L*0.2L=2*10^-6mol
CH3COOH﹤=﹥CH3COO^- + H^+
1 1 1
2*10^-6mol 2*10^-6mol 2*10^-6mol
混合后溶液中含n(CH3COOH)=0.02mol-2*10^-6mol
混合后溶液中含C(CH3COOH)=(0.02mol-2*10^-6mol )/0.2L=(0.1-10^-5)mol/L
混合后溶液中含n(CH3COO^-)=0.02mol+2*10^-6mol
混合后溶液中含C(CH3COO^-)=(0.02mol+2*10^-6mol)/0.2L=(0.1+10^-5)mol/L
C(CH3COO^-)-C(CH3COOH)= (0.1+10^-5)mol/L - (0.1-10^-5)mol/L=2*10^-5mol/L