1.等当点pH计算
pKb=3.7==>pKa=10.3==>Ka=5.0x10^-11
设到达等当点时,[H+]=Xmol/L.有
Ka=X^2/0.0500(假设:酸碱等体积=25.00mL,总体积=50.00mL)
X=[H+]=1.58x10^-6
pH=5.8
2.过量0.10mLHCl
[H+]=0.1000x0.100)/50.1.00=0.0001996mol/L
pH=3.7
3.比等当点少0.10mLHCl(缓冲溶液)
[HA]=24.90x0.1000/(24.90+25.00)=0.04990mol/L
[A]=0.10x0.1000/49.90=2.004x10^-4
pH=pKa+log(2.004x10^-4/4.990x10^-2)=10.3-2.4=7.9
所以,等当点pH=5.8;pH突跃范围=3.7~7.9