作业-酸碱滴定法

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,*,4,写出下列酸碱组分的,MBE,、,CEB,和,PBE,（设定质子参考水准直接写出），浓度为,c,（,molL,-1,）。,（,1,）,KHP,（,2,）,NaNH,4,HPO,4,（,3,）,NH,4,H,2,PO,4,（,4,）,NH,4,CN (5)(NH,4,),2,HPO,4,（,1,）,KHP,MBE,：,K,+,=c,H,2,P+HP,-,+P,2-,=c,CBE,：,K,+,+H,+,=2P,2-,+OH,-,+HP,-,PBE,：,H,+,+H,2,P=P,2-,+OH,-,（,2,）,NaNH,4,HPO,4,MBE,：,Na,+,=c,NH,4,+,+NH,3,=c,H,2,PO,4,-,+H,3,PO,4,+HPO,4,2-,+PO,4,3-,=c,CBE,：,Na,+,+NH,4,+,+H,+,=OH,-,+2HPO,4,2-,+,3PO,4,3-,+H,2,PO,4,-,PBE,：,H,+,+H,2,PO,4,-,+2H,3,PO,4,=OH,-,+NH,3,+PO,4,3-,（,3,）,NH,4,H,2,PO,4,MBE,：,NH,4,+,+NH,3,=c,H,3,PO,4,+H,2,PO,4,-,+HPO,4,2-,+PO,4,3-,=c,CBE,：,NH,4,+,+H,+,=H,2,PO,4,+2HPO,4,2-,+3PO,4,3-,+OH,-,PBE,：,H,+,+H,3,PO,4,=OH,-,+NH,3,+HPO,4,2-,+2PO,4,3-,（,4,）,NH,4,CN,MBE,：,NH,4,+,+,NH,3,=c,CN,-,+HCN=c,CBE,：,NH,4,+,+H,+,=OH,-,+CN,-,PBE,：,HCN+H,+,=NH,3,+OH,-,(5)(NH,4,),2,HPO,4,MBE,：,NH,4,+,+,NH,3,=2 c,H,3,PO,4,+H,2,PO,4,-,+HPO,4,2-,+PO,4,3-,=c,CBE,：,NH,4,+,+H,+,=H,2,PO,4,-,+2HPO,4,2-,+3PO,4,3-,+OH,-,PBE,：,H,+,+2H,3,PO,4,+H,2,PO,4,-,=OH,-,+NH,3,+PO,4,3-,18,计算下列各溶液的,pH,：,（,1,）,2.010,-7,molL,-1,HCl,（,3,）,0.10 molL,-1,NH,4,Cl,（,4,）,0.025 molL,-1,HCOOH,（,6,）,1.010,-4,molL,-1,NaCN,解：（,1,）,2.010,-7,molL,-1,HCl,选择公式：,精确式,pH=-lgH,+,=6.62,条件的判断：,（,3,）,0.10 molL,-1,NH,4,Cl,（,4,）,0.025 molL,-1,HCOOH,（,6,）,1.010,-4,molL,-1,NaCN,K,b,(NH,3,)=1.8,10,-5,条件的判断：,pH=-lgH,+,=5.13,K,a,=1.8,10,-4,判断：,CK,a,=0.025,1.8,10,-4,20K,w,C/K,a,=0.025/(1.8,10,-4,)=138.9 V,1,试样中含有,NaHCO,3,Na,2,CO,3,根据,2HCl+CaO=CaCl,2,+H,2,O T,HCl/CaO,C,HCl,滴定,NaHCO,3,消耗,HCl,的量为,:31.40mL-213.30mL=4.80mL,M,：,CaO,、,Na,2,CO,3,、,NaHCO,3,56.08,、,105.99,、,84.01,有效数字：四位,37,某试样中仅含,NaOH,和,Na,2,CO,3,。称取,0.3720 g,试样用水溶解后，以酚酞为指示剂，消耗,0.1500molL,-1,HCl,溶液,40.00,mL,，问还需多少毫升,HCl,溶液达到甲基橙的变色点？,解：,到酚酞变色点时：,n,NaOH,+n,Na2CO3,=0.040000.1500 mol,根据题意：,n,NaOH,40.00,+n,Na2CO,3,105.99=,0.37,2,0g,解方程组：,n,NaOH,=0.004000mol n,Na2CO,3,=0.002000mol,错误：,0.3750,？,继续滴定至甲基橙变色点，即,Na,2,CO,3,滴定产物,NaHCO,3,消耗的,HCl,的量,