用0.l mol/L NaOH溶液滴定100mL0.l mol/L的HCl溶液时,如果滴定误差在±0.1%以内,

4.3--9.7

正常情况下,滴定到终点时,需要NaOH溶液100mL,误差为0.1%,即多滴或少滴了0.1mL.
(1)少滴0.1mL,溶液呈酸性.
c(H+)=0.1 x 0.1/(100+99.9) = 0.00005(mol/L)
pH=5-lg5 = 4.3
(2)多滴0.1mL,溶液呈碱性
c(OH-)=0.1 x 0.1/(100+100.1) = 0.00005(mol/L)
则c(H+)=Kw/0.00005
pH= 10-lg2 = 9.7