(1) n(HCl)=a/22.4 mol V(溶液)=m(溶液)/溶液密度=(a/22.4+1000×1) / b
C(HCl)=a/22.4 / [(36.5×a/22.4+1000×1) / b]= ab/(36.5a+22400) mol/L
(2) A C1(Na+)=0.2mol/L C2(Na+)=0.1mol/L
B C1(Na+)=0.1mol/L C2(Na+)=0.1mol/L
C C1(Na+)=1 mol/L C2(Na+)=0.1mol/L
D C1(H+)=2mol/L C2(H+)=1mol/L
选B