题目内容

常温下,下列各溶液的叙述中正确的是
A.pH=7的醋酸钠和醋酸混合液中:c(Na)=c(CH3COO
B.0.1mol/L的醋酸的pH=a,0.01mol/L的醋酸的pH=b,则a+1>b
C.0.1mol/L的醋酸钠溶液20mL与0.1mol/L盐酸10mL混合后溶液显酸性
c (CH3COO)>c (Cl)>c (H)>c (CH3COOH)
D.已知酸性HF>CH3COOH,pH相等的NaF与CH3COOK溶液中,
[c(Na+)-c(F)]<[c(K+)-c(CH3COO)]
AB
A中醋酸钠考虑水解平衡、醋酸考虑电离平衡,则溶液中存在的离子有:Na、H+、CH3COO、OH-,由电荷守恒,得:c(Na)+ c(H)=c(CH3COO)+c(OH-
pH=7,则c(H)=c(OH-),故c(Na)=c(CH3COO),A正确;
B、“0.1mol/L的醋酸的pH=a”,则c(H)=10-amol.L-1,醋酸浓度由0.1mol/L变到0.01mol/L(相当于稀释了10倍),若醋酸的电离平衡不移动,则c(H)=10-a+1mol.L-1,但溶液越稀,电离程度越大,所以实际上c(H)>10-a+1mol.L-1,即PH(b =)<a+1,故B正确;
C、CH3COONa+HCl= CH3COOH+NaCl,反应后变为等浓度的CH3COONa、CH3COOH、NaCl的混合物,
醋酸钠:CH3COONa=CH3COO-+Na+,CH3COO-+ H2O CH3COOH+ OH-
醋酸: CH3COOH CH3COO-+ H,NaCl: NaCl= Na+ Cl
利用守恒思想:C(CH3COO)+C(CH3COOH)=( 0.1mol/L×0.02L)/0.03L="0.2/3" mol/L
“显酸性”说明醋酸的电离程度大于醋酸根的水解程度,则c (CH3COO)>c (CH3COOH),
即:c (CH3COO)>0.1/3 mol.L-1c (CH3COOH)<0.1/3 mol.L-1,而c (Cl) ="0.1/3" mol.L-1
c (CH3COO)>c (Cl) >c (CH3COOH) >c (H)(醋酸的电离程度很小),故C错误;
D、NaF溶液:NaF=Na++F-,F-+ H2OHF+ OH-
电荷守恒:c(Na)+ c(H)=c(F)+c(OH-),
c(Na+)-c(F)= c(OH-)- c(H
CH3COOK溶液:CH3COOK=CH3COO-+K+,CH3COO-+ H2O CH3COOH+ OH-
电荷守恒:c(K)+ c(H)=c(CH3COO)+c(OH-
c(K+)-c(CH3COO)= c(OH-)- c(H
两溶液“pH相等”即c(H)相等、在相同温度下,则c(OH-)也相等,
故[c(Na+)-c(F)]=[c(K+)-c(CH3COO)],D错误;
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网