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1第五章酸碱平衡课后习题参考答案1解:共轭碱:CN-;H2AsO4-;NO2-;F-;H2PO4-;IO3-;H4IO6-;[Al(OH)2(H2O)4]2+;[Zn(OH)(H2O)5]+。2解:共轭酸:HCOOH;PH4+;HClO;HS-;HCO3-;H2SO3;HP2O73-;HC2O4-;C2H4(NH2)(NH3+);CH3(NH3+)。3解:酸:H3AsO3;H3PO3;碱:SO32-;Cr2O72-;NH2-NH2;BrO-;两性:H2C2O4-;HCO3-;H2PO4-;HS-。4解:(1)查表得:50℃时的Kθw=5.31×10-14;∴7141030.21031.5][WKHpH=6.64同理:100℃时的Kθw=5.43×10-13;∴7131037.71043.5][WKHpH=6.13(2)[H+]=0.20pH=0.699(3)[OH-]=8.0×10-3pH=11.90(4)[H+]=0.05×0.1÷1=5×10-3pH=2.30(5)333102.15.0100.14.01021.0][HpH=2.92(6)[H+]=0.1/2=0.05pH=1.30(7)5461005.521010][OHpH=14-4.30=9.70(8)[OH-]=(0.1-0.01)/2=0.045pH=12.658解:(1)酸HClO2---共轭碱ClO2-;碱NO2----共轭酸HNO2;7.16100.6100.1)()(]][][[]][][[]][[]][[422222222222HNOKHClOKHNOHClOHClOHNONOHClOClOHNOKaa(2)HPO42-(酸)—PO43-(碱);HCO3-(碱)—H2CO3(酸)671332143332434321007.1102.4105.4)()(]][[]][[COHKPOHKHCOHPOPOCOHKaa(3)NH4+(酸)—NH3(碱);CO32-(碱)—HCO3-(酸)82.11108.1107.4100.1)()(]][[]][[51114322423433COHKNHKCONHNHHCOKaa(4)HAc(酸)—Ac-(碱);OH-(碱)—H2O(酸)9145108.110108.1)(]][[][WaKHAcKOHHAcAcK(5)HAc(酸)—Ac-(碱);NH3(碱)—NH4+(酸)414554341024.310108.1108.1)()(]][[]][[NHKHAcKHAcNHNHAcKaa(6)H2PO4-(酸)—HPO42-(碱);PO43-(碱)—HPO42-(酸)25138433432344224241038.1105.4102.6)()(]][[]][[POHKPOHKPOPOHHPOHPOKaa15解:由缓冲溶液公式得:∴C酸=0.28(mol/L)所需体积为250×0.28÷6=11.7(ml)18解:(1)刚好中和为NH4Cl溶液。其浓度为0.05mol/L∴pH=5.28(2)混合后为NH3-NH4Cl缓冲液,C碱=0.05(mol/L);C盐=0.05(mol/L)∴pH=9.25(3)混合后为NH3-NH4Cl缓冲液,C碱=0.05(mol/L);C盐=0.05(mol/L)∴pH=9.25(4)混合后组成为HAc-NaAc缓冲液,C酸=0.05(mol/L);C盐=0.05(mol/L)pH=pKa=4.75(5)混合后组成为HAc-NaAc缓冲液,C酸=0.05(mol/L);C盐=0.05(mol/L)pH=pKa=4.75(6)混合后成为NH3·H2O溶液,C碱=0.05(mol/L);45105.905.0108.1[碱CKOHbpH=10.98(7)混合后组成为H3PO4-H2PO4-缓冲液,C酸=(0.15-0.075)/0.55=0.136(mol/L);C碱=0.136(mol/L)对于平衡:H3PO4==H2PO42-+H+平衡浓度0.136-x0.136+xx代入表达式解一元二次方程得:x=6.11×10-3(mol/L)pH=2.21(8)混合后组成为H2PO4--HPO42-缓冲液,C酸=0.0625(mol/L);C碱=0.125(mol/L)(9)混合后组成为HPO42--PO43-缓冲液,C酸=0.0714(mol/L);C碱=0.143(mol/L)对于平衡:PO43-+H2O==HPO42-+OH-平衡浓度0.143-x0.0714-xx代入表达式x2+0.0934x–3.15×10-3=0x=0.0263即:[OH-]=0.0263pH=12.4219解:配合物形成体配位体配位原子配位数配离子电荷形成体电荷命名[CrCl2(H2O)4]ClCr3+H2O,Cl-O,Cl6+1+3氯化二氯·四水合铬(Ⅲ)[Ni(en)3]Cl2Ni2+enN6+2+2二氯化三(乙二胺)合镍(Ⅱ)K2[Co(NCS)4]Co2+NCS-N4-2+2四异硫氰酸合钴(Ⅱ)酸钾Na3[AlF6]Al3+F-F6-3+3六氟合铝(Ⅲ)酸钾[PtCl2(NH3)2]Pt2+Cl-,NH3Cl,N40+2二氯·二氨合铂(Ⅱ)2501125lg745.40.5酸C65141027.505.0108.110][盐CKHh75.4lg盐碱CCpKpOHb75.4lg盐碱CCpKpOHb51.7125.00625.0lg102.6lglg82盐酸CCpKpHa213141022.2105.4100.1143.0)0714.0(xxx31107.6136.0)136.0(aKxxx3[Co(NH3)4(H2O)2]2(SO4)3Co3+NH3,H2ON,O6+3+3硫酸四氨·二水合钴(Ⅲ)[Fe(EDTA)]-Fe3+EDTAN,O6-1+3乙二胺四乙酸合铁(Ⅲ)离子[Co(C2O4)3]3-Co3+C2O42-O6-3+3三草酸合钴(Ⅲ)配离子Cr(CO)6CrCOC600六羰基合铬(0)HgI42-Hg2+I-I4-2+2四碘合汞(Ⅱ)配离子K2[Mn(CN)5]Mn3+CN-C5-2+3五氰合锰(Ⅲ)酸钾[FeBrCl(en)2]ClFe3+Br-,Cl-,enBr,Cl,N6+1+3氯化溴·氯·二(乙二胺)合铁(Ⅲ)21解:(1)(2)(3)22解:查表知:Kθf(HgI42-)=5.66×1029,因Hg2+与I-反应的平衡常数非常大,则完全反应形成HgI42-。I-离子过量,则生成的HgI42-的浓度为0.0100(mol/L).设平衡时自由的[Hg2+]=xHg2++4I-==HgI42-平衡浓度x0.783-0.0400+4x0.0100-x代入稳定常数表达式:x=5.80×10-32∴c(Hg2+)=5.80×10-32mol/L;c(HgI42-)=0.010mol/L;c(I-)=0.743mol/L23解:对于反应:Cr3++H2Y2-====CrY-+2H+初始浓度0.00100.05平衡浓度0.0010-x0.05-xx1.0×10-661172322321007.4109.5109.6100.1]][[]][[22YHCrYKKYHCrHCrYK平因平衡常数非常大,可认为反应向右进行较彻底,则x≈0.0010,[H2Y2-]=0.049(mol/L)设达平衡时[Cr3+]=y∴y=5.0×10-21(mol/L)25解:反应:Ni(NH3)62++3en===Ni(en)32++6NH39818)()(326363231034.21097.8101.2]][)([]][)([26323NHNienNiKKenNHNiNHenNiK因平衡常数非常大,可认为反应向右进行完全,生成的Ni(en)32+为0.10,剩下的C(en)=2.3-0.10×3=2.0(mol/L),C(NH3)=1.0+0.1×6=1.6(mol/L).设平衡时Ni(NH3)62+的浓度为x,则x=8.96×10-11(mol/L)422424]/)(][/)([/)()(cIccHgccHgIcHgIKf429)4743.0(0100.01066.5xxx67132332322232232332321074.11067.1109.2))(())((]][)([]][)([NHAgKOSAgKOSNHAgNHOSAgK稳稳平3220523342361056.2106.1101.4)(())((OCFeKCNFeKK平235242431016.1100.11016.1)(())((NCSCoKNHCoKK平)/(783.05000.00.1660.65)(0LmolIc6261007.4049.0)100.1(0010.0y9361034.226.11.0x
本文标题:第五章酸碱平衡课后习题参考答案
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