Đáp án:
CMHNO3=0,123(M)
CMHCl dư=0,316(M)
C%HCl dư=1,11%
C%HNO3=0,744%
Giải thích các bước giải:
AgNO3+HCl->AgCl+HNO3
VddHCl=150,15/1,05=143(ml)
=>nHCl=0,143x0,5=0,0715(mol)
nAgCl=0,02(mol)
=>HCl dư
CMHNO3=0,02/(0,02+0,143)=0,123(M)
CMHCl dư=(0,0715-0,02)/0,163=0,316(M)
mdd AgNO3=20x1,1=22(g)
mdd spu=22+150,15-0,02x(108+35,5)=169,28(g)
C%HCl dư=0,0515x36,5/169,28x100=1,11%
C%HNO3=0,02x63/169,28x100=0,744%