文档介绍：无机及分析化学答案( 第二版) 第四章第四章酸碱平衡与酸碱滴定法 4-1 .将 mol?L?1HAc 溶液稀释到什么体积才能使解离度增加一倍。??L?300mLc? 解:设稀释到体积为 V ,稀释后 ?2??300?(2?)c?Ka?V?(1?2)1? 得: 1?由?5 ?1 因为 K?a =?10ca = mol?L caK?a > 20K?w ca/K?a>500 故由 1?2? =1?? 得V =[300?4/1 ]mL =1200mL 此时仍有 caK?a>20K?w ca/K?a>500 。 4-2 .奶油***后的分解产物之一为丁酸(C3H7COOH) ,有恶臭。今有 含 mol?L?1 丁酸的溶液, pH 为 , 求丁酸的 K?a 。解: pH= c(H+) =10? mol?L?1 ? =10? = ?10?2 ? ?22c?2??(?10)??10?5 ?21??10K?a= 1? ?1?4-3 . What is the pH ofa ?L solution of ammonium acetate at 25℃? pKa of acetic acid at 25℃ is , pK?a of the ammonium ion at 25℃ is , pKw is . ?10?10 解: c(H+)=Ka1Ka2?pH= ?logc(H+) = 4-4 . 已知下列各种弱酸的 K?a 值, 求它们的共轭碱的 K?b 值, 并比较各种碱的相对强弱。(1)HCN K?a = × 10?10 ; (2)HCOOH K?a = × 10?4 ; (3)C6H5COOH( 苯甲酸) K?a = × 10?5 ; (4) C6H5OH ( 苯酚) K?a = × 10?10 ; (5)HAsO2 K?a = × 10?10 ; (6) H2C2O4 K?a1=?10?2 ; K?a2=?10?5 ; 解: (1)HCN Ka= ?10?10 Kb=Kw/?10?10 =?10 ?5 (2)HCOOH Ka= ?10?4 Kb=Kw /?10?4 =?10?11 (3)C6H5COOH Ka= ?10?5 Kb=Kw /?10?5 = × 10?10 (4)C6H5OH Ka=?10?10 Kb=Kw /?10?10 =?10?5 (5)HAsO2 Ka=?10?10 Kb=Kw /?10?10 =?10?5 (6)H2C2O4 Ka1=?10?2 Kb2=Kw /?10?2 =?10?13 Ka2=?10?5 Kb1=Kw /?10?5 = × 10?10 ??? ? 碱性强弱: C6H5O 4-5 . 用质子理论判断下列物质哪些是酸?并写出它的共轭碱。哪些是碱?也写出它的共轭酸。其中哪些既是酸又是碱? ----- H2PO4 , CO32 , NH3 , NO3 , H2O , HSO4 , HS , HCl 解: 酸共轭碱碱共轭酸既是酸又是碱--- H 3PO4 H2PO4 H2PO4 H2PO4 HPO42? NH3 NH3 NH4+ NH3 NH2? +H2O H2O H3O H2O OH? --- H2SO4 HSO4 HSO4 HSO4 SO42? --- H2S HS HS HS S2? - HCl HNO3 NO3 Cl? CO32? HCO3? 4-6 .写出下列化合物水溶液的 PBE : (1) H3PO4 (2) Na2HPO4 (3) Na2S (4)NH4H2PO4 (5) Na2C2O4 (6) NH4Ac (7) HCl+HAc (8)NaOH+NH3 解: > AsO2? > CN? > C6H5COO?>C2O42? > HCOO? > HC2O4? (1) H3PO4 : c( H+) = c(H2PO4? )+ 2c( HPO42?) + 3c (PO43?) + c(OH?) (2) Na2HPO4 : c(H+) + c(H2PO4? )+ 2c(H3PO4) = c(PO43?) + c(OH ?) (3) Na2S : c(OH?) = c(H+) + c(HS?) + 2c(H2S ) (4)NH4H2PO4 :