文档介绍：分析化学习题酸碱滴定法解(1) NH 4H 2 PO 4 MBE c = [NH 3 ]+[NH 4 +] c = [H 3 PO 4 ]+[H 2 PO 4 - ]+[HPO 4 2- ]+[PO 4 3-] CEB [H + ]+[NH 4 + ]= [H 2 PO 4 - ]+ 2[HPO 4 2- ] + 3[PO 4 3- ]+[OH -] PBE [H + ]= [NH 3 ]+[ HPO 4 2- ]+2[PO 4 3- ] +[OH -] -[H 3 PO 4] 1. 写出 NH 4H 2 PO 4和 NH 的 MBE 、 CEB 及 PBE (浓度为 c mol/L )。(2) NH MBE c = [NH 3 ] + [NH 4 + ] c = [CN - ] + [HCN] CEB [H + ] + [NH 4 + ] = [OH - ] + [CN -] PBE [H + ] = [NH 3 ] + [OH - ] - [HCN] 解: (1) 溶液, NH 是弱酸弱碱组成的盐?-10 = 10 ?-10 a(HCN) K = 10 ? ?+)4 -14 -5 w b a(NH K = K /K = 10 / 10 2. 计算 、 KHP 和 Na 2S溶液的 pH 值。+4 a(HCN) w a(NH ) + a(HCN) K (cK +K ) [H ] = c +K ? ?????-10 -10 -14 + -10 10 ( 10 + 10 ) [H ] = + 10 ?-10 = 10 (mol/L) pH = +4 W a(HCN) a NH cK > 20K c >> K () 由于;?+4 + -10 a(HCN) a(NH ) [H ] = K K = 10 mol/L 计算结果相同(2)KHP 是两性物质,查表知: ? ?-3 -6 1 2 Ka = 10 , Ka = 10 2 w 1 cKa > 20K , c < 20Ka , + 1 2 1 cKa Ka [H ] = c +Ka ? ????-3 -6 -3 10 10 = + 10 pH = ?-5 = 10 mol/L (3) Na 2S溶液 Na 2 S → 2Na + + S 2- S 2-是二元碱,查表知其解离常数分别为??-14 w -15 b12K 10 K = = = Ka 10 ???-14 -7 w -8 b21K 10 K = = = 10 Ka 10 由于 K b1?K b2,故可按一元碱处理 1 1 1 2 b b b - -K + K +4cK [OH ] = 2 ? ? 2 -+ +4 =2 pOH = pH = c/K b1<400 , cK b1>20K w = (mol/L) 解: 2 1 1 2 + 2 H S + 2 + a a a [H ] δ= [H ] +[H ]K +K K ? ? ???-8 2 2 -8 -15 10 + 10 () = (0. 1 00 1 ) + = 1 - 1 1 2 +a + 2 + HS a a a [H ]K δ= [H ] +[H ]K +K K 3. 计算 pH= 时, H 2S溶液中各型体的浓度。?? ? ?????- -8 -2 -1