如题所述
pH = pKa -lg(n酸/n碱)
pKa=-lg(1.8*10^-5)=4.74 > 4.44
HAc-NaAc缓冲系,
设加入NaAc为x mol
4.44=4.74-lg[0.01/(x-0.01)]
x=0.015mol,
即1.231g