0.1mol naoh 滴定100ml 0.1mol/l盐霜,如果滴定误差在+-0.1%以内,求反应完毕后ph范围 9.

恰好完全反应时,V(NaOH)=V(HCl)=100ml
若NaOH过量,显碱性,C(OH-)=0.1mol/L*100ml*0.1%/200ml=10^-4*1/2
C(H+)=2*10^-10
PH=-logC(H+)=-log(2*10^-10)=10-log2=9.3
若NaOH少量,显酸性,C(H+)=0.1mol/L*100ml*0.1%/200ml=10^-4*1/2
PH=-logC(H+)=-log(10^-4*1/2)=4+log2=4.7