Question

SOS.1.将50mL0.1mol/L的BaCl2溶液中加入0.1mol/L的AgNO3溶液10mL,反应后c(Ag+)

Answer 1

n(BaCl2)=0.05*0.1=0.005mol

n(AgNO3)=0.01*0.1=0.001mol

BaCl2 + 2AgNO3=2AgCl↓+Ba(NO3)2

0.0005```0.001`````````0.0005

所以剩余n(Ag+)=0mol c(Cl-)=0.005-0.001=0.004mol

n(Ba2+)=0.005mol n(NO3-)=0.001mol

溶液总体积0.05+0.01=0.06

c(Ag+)=0

c(Cl-)=0.004/0.06=0.0667mol/L

c(Ba2+)=0.005/0.06=0.0833mol/L

c(NO3-)=0.001/0.06=0.01667mol/L