答：
【1】0.1000mol/L NaOH滴定20.00mL0.1000mol/LHAc,终点时pH值计算：
[OH-] = (0.05x5.6x10^-10)^1/2 = 5.3X10^-6 mol/L pOH= 5.28 pH = 8.72
【2】终点时pH值比计量点时高0.5个单位,计算Et
pH = 8.72 + 0.5 = 9.22
【3】当滴定20.02 ml 时,pH = 9.70 ,其Et 为+0.1%
所以终点时pH值比计量点时高0.5个单位,计算Et = +0.05%,
符合滴定分析的要求.