溶液PH值计算0.1 mol/l NaH2PO4H3PO4 pKa1=2.12 pKa2=7.2 pKa3=12.36
问题描述:
溶液PH值计算
0.1 mol/l NaH2PO4
H3PO4 pKa1=2.12 pKa2=7.2 pKa3=12.36
答
质子条件式:[H+]+[H3PO4]=[HPO4-2]+2[PO4-3]+[OH]则:[H+]+[H+][H2PO4-1]/Ka1+Kw/[H+]=[HPO4-2]+2Ka3[HPO4-2]/[H+]+Kw/[H+]=[HPO4-2](1+2Ka3/[H+])+Kw/[H+]因为2Ka3/[H+]《1,所以[H+]+[H+][H2PO4-1]/Ka1=[HPO4-2]...