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1、1/13+第六章 酸碱滴定法 习题参考答案。写出下列各酸的共轭碱 H：H；H2CO4:42OHC;2442HPOPOH：;233COHCO：;C6OH：C6H5O；356NHHC：C6H5H；S:S2；362O)Fe(H：252)(OHO)Fe(H；-：COOCHNHRCOOHCHNHR2222。写出下列各碱的共轭酸 HO：H3O+；33HNONO：;424SOHHSO：;：HS-；C6H-:H5H；222)(OHO)Cu(H：)(OHO)Cu(H32；（CH2）6N4：(C2）6+；-：COOCHNHRCOONHCHR222；2-:HP 3。通过E 和B 写出下列溶液的BE（1）（N4)3
2、MBE：c2NHNH34 cCOHCOCOH23332 CBE:OH2COHCOHNH2334 PBE:NHOHHCOH2HCO3323（2）H4C3 B：cNHNH34 cCOHCOCOH23332 BE:OH2COHCOHNH2334:NHOHCOHCOH32332 4。写出下列物质的 ME、BE 和 PBE（设定质子参考水准直接写出)（1)KH MBE：cK cPHPPH22-CBE:OHP2HPKH2-质子参考水准 -HP H3O+H2O OH PBE：OHPPHH22-HHP H+H P2-2/13(2)NaNH4HP4 MB：cNa cNHNH34 cPOHPOPOHPOH3424
3、4243 CB：PO3HPO2POHOHHNHNa3424424-质子参考水准 4NH H3 24HPO 34PO H3O H2O PB：NHPOOHPOH2POHH3344342-（3)NH4H2PO4：cNHNH34 cPOHPOPOHPOH34244243 CE：PO3HPO2POHOHHNH3424424-质子参考水准 4NH NH3 42POH H3O+H2 H-PBE：NHPO2HPOOHPOHH3342443（)NH4CN MBE：cNHNH34 cCNHCN-CE：OHCNNHH4-质子参考水准 4NH H H3+H2O H BE：OHNHHCNH3(5)（NH4)2P4 M:
4、c2NHNH34 cPOHPOPOHPOH34244243 H+-H+H+2H+42POH H3PO4-+H+-H+H+HPO+H+H+34PO 24HPO HHCN+CN 3/13 C:PO3HPO2POHOHHNH3424424-质子参考水准 4NH 24HPO 34PO HO+H2O OH PE:NHPO2HPOOHPOHH3342443 8.下列酸碱溶液浓度均为。10 mo/L,能否采用等浓度的滴定剂直接准确进行滴定？若能滴定，应选择什么标准溶液和指示剂？（1)HF Ka 7210-根据判据 cspKa 0.。0=10-108,能采用等浓度的 NaOH标准溶液直接准确滴定；化学计量点为
5、 0.050 ol/L的a溶液。Kb Kw/K 01014/（7.2104）。4-11 Kb=0。050。411 7。01013 2 Kw，/Kb 。00(。4111）400 mol/L1037.8104.1050.0OH711bcK pO=。07 H=。92 选择酚酞指示剂。（)NaH3 81111asp101028.0106.5050.02Kc不能采用等浓度的 NOH 标准溶液直接准确滴定。8714awb104.2102.4/100.1/12KKK 888bsp101012.0104.2050.02Kc 能采用等浓度的 HCl 标准溶液直接准确滴定;化学计量点为 0040 mol/L的 H
6、2CO3溶液.21a47a40103.1102.4040.0KcK wa201KcK 400/1aKc L/mol103.1102.4040.0H47acK pH=9 选择甲基橙指示剂。（6）（C2)6N4 pKb=0。050。410-9=70-11 0-8,HA也能被滴定。若滴定至甲基橙由红色变为黄色为终点，这时溶液约为 4。,参与反应的 Hc即为Ac 31.0108.110108.1H54.45aaAcKK 15。解:（)根据题意得：1300.20221)(222HCl21HClCONaNaOH32VVVcVVcnn V2 .0 L(2)根据题意得:1200.2021HCl2HClNaOH
7、NaHCO3VVcVcnn V2=40。00 mL 或者：12NaOHNaHCO3nn N3和 NaO反应后 11323CONaNaHCOnn 1100.2000.20)(21HCl12HClCONaNaHCO323VVcVVcnn V 400 m（3）根据题意得：只含有 NaO 18.计算下列溶液的H值（1）2。010-7 mol/Hl 6/13 解：c106 o/L mol/L1041.22104)100.2(100.224H714277w2Kcc p 。2（）0。02 ol/L 24 解：mol/L1056.22100.1020.08)100.1020.0(100.1020.028)()
8、(H22222aaa222cKKcKc H=1.5（3）0.10 o/L Cl（NH的 K=1。810）解:K=wKb=1。0114/（.805）=561010 cKa 0.106100=。610-120 Kw，cKa 0。10/（5。61010)400 L/mol1048.7106.510.0H610acK =5.12(4）0.02 lL HCH 解：cK 0。251.8104 451020 Kw，c/K 0.05/（181-4)4 mol/L1004.22108.1025.04)108.1(108.124H34244a2aacKKK pH=269（5）014 mo CN 解：cKa=1。0
9、10-4 7201=7.10-14 400 mol/L1086.210102.7100.1H714104waKcK pH =。5（6）1010-4 ml/L NaCN 解：Kb=Kw=1.01014（7。21010)=1415 cKb=1.0104 1。10=。40-9 2 K，c/Kb 1。014（.47/13 0)2 Kw，c/Kb=0。10/(1。410-）400 mol/L1018.1104.110.0OH59bcK pOH=493 p=。07(8）0。1 mol/NH4CN 解：)a(NH4cK=0。10。1010=.101120 Kw，c 20K(HC）=207.210-1 L/m
10、ol1035.6106.5102.7H101010)a(NHa(HCN)4KK H=920（9）。010 ol/KH 解：2acK=0。0103.9-6=。10 2 Kw，c=0.00 1a20K=201.103=0.2 mol/L1024.6101.1010.0109.3101.1010.0H5363aaa121KcKcK pH=4。20(0）0.10 moL Na2 解:21awb/KKK 1。01014/（1.2-)=8。3 12awb/KKK 1.0101/（.710-)1.80-21bb4091.03.810.0KcK wb201KcK 400/1bKc mol/L099.023.8
11、10.043.83.824OH2b2bb111cKKK OH=。00 p=13。00（11).10 mol/COOHCHNH23(氨基乙酸盐)8/13 解:1aK=03 2aK=2.101 21a23a401012.2105.410.0KcK wa201KcK 400/1aKc mol/L1091.12105.410.04)105.4(105.424H23233a2aa111cKKK pH=。72 19.计算 0。010 mol/L 3P溶液的(1）24HPO，（2）34PO的浓度.解：1aK=7。1 2aK=6.3108 3aK=.4103 21a33a401072.8106.7010.0K
12、cK wa201KcK 400/1aKc mol/L107.52106.7010.04)106.7(106.724H33233a2aa111cKKK由于 HPO4的第二、第三级解离及水的解离均可忽略，POHH42 mol/L103.6HPOHHPO8a42a2422KK mol/L109.41071.5103.6104.4HHHPOPO183813aa24a34233KKK 20.（1）20 mg NaC2O4溶解并稀释至500 m，计算pH40 时该溶液中各种型体的浓度。解：mol/L1073.3105000.134/10250333c 001.0104.6109.5109.510)10()
13、10(HHH52242424aaa22OCH211422KKK mol/L107.3001.01073.3OCH63OCH422422c 61.0104.6109.5109.510)10(109.510HHH52242424aaa2aOHC211142KKKK-mol/L103.261.01073.3OHC33OHC4242-c 39.0104.6109.5109.510)10(104.6109.5HH52242452aaa2aaOC2112142KKKKK-2 9/13 mol/L104.139.01073.3OC33OC4242-2-2c（2）计算 pH.00 时，。10 mol/L H2
14、S 溶液中各种型体的浓度。解：1102.1107.5107.510)10()10(HHH158812121aaa22SH2112KKK mol/L10.0110.0SHSH22c 7158812181aaa2aSH107.5102.1107.5107.510)10(107.510HHH2111KKKK-mol/L107.5107.510.0HS87HSc-211588121158aaa2aaS108.6102.1107.5107.510)10(102.1107.5HH21121KKKKK-2 mol/L108.6108.610.0S2221S2c-2 21.解：原来mol143.02.140/
15、0.20462N)(CHn,mol0480.0100.4123HCln 反应后mol0950.00480.0143.0462N)(CHn，mol0480.0HClN)(CH462n(CH2）N4（CH2）6N4+缓冲溶液，a=K/Kb=14(1.19)=716，p 51 45.5100.0/0480.0100.0/0950.0lg15.5lgppHHN)(CHN)(CHa462462ccK 22 解：Ka=/Kb=1.04(1810-5)=。61 根据43NHNHalgppHccK 和 mol/L 1.043NHNHcc 得：44NHNH0.1lg26.910.00cc mol/L 0.154
16、NHc mol/L 0.853NHc mL 56.7/15101.00.85-3NH3V，g 0.849.531.00.15ClNH4m 答：略 24.解：(1)未加 HC前 74.40.10.1lg74.4lgppHHAcAcaccK-加 HCl后 69.405.074.4101/)0.10.61000.1(101/)0.10.61000.1(lg74.4lgppHHAcAcaccK-溶液的H减小了 0。05 个单位。10/13(2）根据题意得:HAcAclg74.45.00cc-101/)0.60.1100(101/)0.60.1100(lg74.45.10HAcAccc-解联立方程得：m
17、ol/L 0.76Acc mol/L 0.42HAcc 此缓冲溶液中 HAc、A的分析浓度为 0。2 mol/L、07 o/。2 计算下列标准缓冲溶液的 pH(考虑离子强度影响）(1)034 ml/饱和酒石酸氢钾溶液 解：034.0)1(034.01034.021)(21222HAHA2KKZcZcI 29.0)034.030.0034.01034.0()2(50.0lg2A2 51.02A mol/L1077.251.0/103.4101.9/454AaaH221KKa pH=56（）0。10 mol/L 硼砂溶液 33322274BO2HBO2HOH5OB 020.0)1(020.0102
18、0.021)(21222BOHBOH2NaNa3232ZcZcI 059.0)020.030.0020.01020.0()1(50.0lg2BOH32 87.032BOH 18.9020.01020.087.0lg108.5lglgppH10BOHBOHa3232aaK 27。解:（1）苯甲酸的浓度为 c=0.1002。702.=0。80 o/L（）计量点时苯甲酸钠的浓度为 0.082025。0/（25.00+0.7）.04 o/L =Kwa=14/（6。215）=1.61010 cKb=0.0401。100=.21012 0 Kw /Kb=.0530（1。61）40 mol/L107.210
19、6.104530.0OH610bcK O=557 H=.4（3）选择酚酞指示剂。8.解：（1）计量点为。050 mol/NCl溶液。Ka Kw 1.01014（1。815）=。610 11/13 cK=0.05056101=11120 Kw c/0。050/(560-10)40 L/mol103.5106.5050.0H610aspcK Hp=。28(2).相对误差时 26.600.326.998.39/00.2010.098.39/)98.1900.20(10.0lg26.9lgppH43NHNHaccK 0。1%相对误差时 H=0。10(20.0220.00）/40.0=10-5 ol/L
20、 H=40(3）选择甲基红指示剂。29.解：根据反应式,HCl 和a2B47等体积反应。计量点时 H3BO3的浓度为 c=40。050V/2=0。1 ol/L,H3BO为一元弱酸，K 5。1010 cKa=10。8110 51-1 Kw，c/K .10/(5.81010)00 L/mol106.7108.510.0H610aspcK pHs=.2 选择酚酞指示剂。30。解：()根据题意得：50.6aa50.1aa1050.6p10 1.50p2211KKKK即；即（2）5aa1021KK 能分步滴定.850.6aps850.1aps101010.031101010.0212211KcKc，可以
21、用100 ol/L的 NaH标准溶液分别滴定至第一计量点和第二计量点。（3)第一化学计量点为 0.50 olL 的 NaB 溶液。2acK=0。05010。50 20 Kw c .050 1a20K 01。5 .63 mol/L108.710050.01010050.0H550.150.650.1aaa121KcKcK pH=41 第二化学计量点为 0。3 olL 的 Na2B 溶液。21awb/KKK=1。104/10-。50=0-7.50 12awb/KKK 1。01014/1.50=1012.5 12/13 21b50.7b4010033.0KcK wb201KcK 400/1bKc m
22、ol/L102.310033.0OH550.7b1cK O .49 p=91 (4）分别选择甲基橙、酚酞指示剂.31。计算下述情况时的终点误差 解:（)%006.0%1002/10.01010%100HOH5.55.8epHCl,epeptcE(）酚酞p ep 8。5 15.0108.111010OHOH55.55.5bepepep,NH3K%15%100)15.02/10.01010(%100)OHH(5.55.8,epNH,epNHepept33cE 甲基橙p ep=。0 651010bepepep,NH106.5108.1100.1100.1OHOH3K%2.0%100)106.52/1
23、0.0100.1100.1(%100)OHH(6104,epNH,epNHepept33cE 解：13CaOCaOHCl3HClLmol4993.0208.5601400.010210MTc 根据题意得:V1=1330 m，2=31.40 1。=18。1，V2 V 混合碱为 Na2CO3和 NO3%99.63%100100.10.1061030.1324993.021%100102213CONa31CONa3232mMVcw%30.18%100100.101.8410)30.1310.18(4993.0%10010)(3NaHCO312NaHCO33mMVVcw杂质=9918。30%=1.1
24、39 解:根据题意，2=3。2 m,V=48.36-33。2=14。64 mL，2 V 混合溶液的组成为 H3P4和 NaHP 14.64mmol1.00014.6443POHn 19.08mmol1.00014.64)72.33(42PONaHn 46.解：COHHCOlgppH323a1 K 13/13 COH103.2lg10.67.20322-mol/L108.1COH332 8.12108.1/103.2CO/HHCO32323 49.解:mol/L05002.02 mol/L05002.01010015.1326610.04424NH3SO)NH(cc O6H3HHN)(CH6HCHO4NH24624 滴定反应:(H2）N4H +O-=（2)6N4 +H2O H OH =HO 计量点产物为:(C)64 4462NH4N)1(CH mol/L01000.050/00.2005002.0241462N)(CHc 0.01000。1106.110-8 20 w，c/Kb 0.00/（106)400 mol/L107.3104.101000.0OH69bcK pOH=5。43 pH 87 选择酚酞指示剂。