急问大学分析化学：用NaOH溶液滴定H3PO4溶液至pH = 4.7时,溶液的简化质子条件为 ((H3PO4的pKa1~pKa3分别是2.12、7.20、12.36) (A) [H3PO4] = [H2PO4-] (B) [H2PO4-] = [HPO42-] (C) [H3PO4] = [HPO42-] (D) [H3PO4] = 2[PO43-] 选

Question

急问大学分析化学：用NaOH溶液滴定H3PO4溶液至pH = 4.7时,溶液的简化质子条件为 (
(H3PO4的pKa1~pKa3分别是2.12、7.20、12.36)
(A) [H3PO4] = [H2PO4-] (B) [H2PO4-] = [HPO42-]
(C) [H3PO4] = [HPO42-] (D) [H3PO4] = 2[PO43-]
选哪个?为什么?

Answer 1

感觉应该选C
PH=4.7时 溶液中H3PO4主要存在形式是H2PO4-
以H2PO4-和H2O为零水准 写质子守恒式：
[H3PO4]+[H+]=[HPO42-]+[OH-]+2[PO43-]
忽略[H+]和[OH-]和[PO43-](该PH下此形式几乎不存在）的话
答案就是C了