Question

求算pH的问题 100mL,0.01mol/L H3PO4 与 100ml,0.02mol/L NaOH 溶液混合,求此时pH

Answer 1

按酸与碱的物质的量比,发生下列反应：
--------------------H3PO4 + 2NaOH = Na2HPO4
物质的量 ：100*0.01 100*0.02 1
= 1 = 2
此时生成0.005 mol / L的Na2HPO4,根据多元弱酸的酸式盐的相关公式,最简式有：
c(H+) = [ Ka2 *Ka3 ]^1/2
即：pH = 1/2 (pKa2 + pKa3)
查出H3PO4的二级和三级电离平衡常数知：Ka2＝6.23×10^－8,Ka3＝4.50×10^－13
c(H+) = [ Ka2 *Ka3 ]^1/2
= [6.23×10^－8 * 4.50×10^－13]^1/2
= 1.674 *10^-10
pH = 9.8