已知三元弱酸H3A的pKa1 = 2.16,pKa2 =7.12,pKa3 =12.32,则最大时溶液的pH为
问题描述:
已知三元弱酸H3A的pKa1 = 2.16,pKa2 =7.12,pKa3 =12.32,则最大时溶液的pH为
答
问那个最大时?
答
= H+ + A2- Ka2=[H+]*[A2-]/[HA-]=10-5 Ka1*Ka2=[H+]2*[A2-]/H2A=10-7 [H2A]=[A2-] [H+]2=10-7 [H+]=10-3.5 pH=
答
[H+]=(C*Ka1)1/2次方 pH=-lg[H+]