答案a - 广西民族大学

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1、滴定分析概论答案A110101(C)0102(B)0103(C)1101040.018g1[K+]=────×─────=4.60×10-3mol/L0.1L39.1g/mol0.365g1[Cl-]=────×──────=1.03×10-1mol/L0.1L35.5g/mol(C)0105(A)01061.08×103g1[Na+]=───────×1.02g/mL×─────×1000mL/L106g23g/mol=4.79×10-2mol/L270g1[SO42-]=───×1.02g/mL×─────×1000mL/L106

2、g96g/mol=2.87×10-3mol/L(D)110107(D)0108(C)0109(C)0110(B)0111(A)0112(C)0113(A)0114(D)0115(D)0116(A)0117(C)0118(C)0119(D)1101201；0.101222.0×10-301231000124500125突跃小,终点不明显,终点误差大；造成浪费0126Na2CO3Na2B4O7·10H2O；H2C2O4·2H2O邻苯二甲酸氢钾01270.2~0.3;2~3;称草酸约0.3g,定容于100mL容量瓶中,移取25.00mL3份

3、进行标定。0128偏高；偏高0129无影响；因Na2CO3和Na2O与HCl反应的化学计量比均是1:2。0130偏高；偏低；硼砂；HCl与两者均按化学计量比1:2反应,但硼砂摩尔质量大,称量误差小。0131109.5%硼砂部分失去结晶水0132Na2S2O3、KMnO4;K2Cr2O7、草酸钠(或草酸)。0133H2C2O4·2H2O、Na2C2O4、K2Cr2O7、Zn(或邻苯二甲酸氢钾)(或草酸)0134溶液NaOHEDTAAgNO3Na2S2O3基准物H2C2O4·2H2OZnNaClK2Cr2O70135Na2S2O3;AgN

4、O3;Na2S2O3;AgNO301361.B2.C3.A4.D0137锌片,二甲酚橙或铬黑T；Na2C2O4,自身指示剂K2Cr2O7淀粉；As2O3淀粉；NaClK2CrO401381:2、2:1、1:1、2:11101390.0266701401.D2.C3.B4.A01410.04000014267.04%32.96%w(Fe2O3)=(1-79.85%)×(159.7×2)/(32.00×3)×100%=67.04%w(Fe)=(1-67.04%)×100%=32.96%01430.5014450.30%01450.2014

5、6c(HCl)=0.03814×1000/381.37=0.1000(mol/L)(50.00×0.1660-0.1000×27.14)×180.16试样纯度=─────────────────────×100%=91.7%2×0.5490×1000滴定终点时,邻羟基苯甲酸钠的酸碱性都极弱,溶液pH可由NaAc确定[OH-]=pH=14-5.3=8.7所以可用酚酞指示剂0147P→PO43-→MgNH4PO4→H3PO4→H2PO4-1P2HCl(0.1012×25.00-0.1008×24.50)×30.97w(P)=──────

6、──────────────×100%=0.47%20001481.c(NaOH)=0.1143×44.62/50.00=0.1020(mol/L)m(CO2)=(0.1026-0.1020)×44.00=0.026(g)0.1020-0.10202.Er=────────×100%=0.6%0.10200149溶液中：n(H)=25×0.40×3=30(mmol)n(PO4)=30×0.50+25×0.40=25(mmol)所以混合后生成HPO42-20mmol,H2PO4-5mmol20/41.V(HCl)=────=50(mL)

7、0.105/42.V(NaOH)=────=12.5(mL)0.100150(0.1000×50.00-0.2100×11.03)×180.2w(乙酰水杨酸)=───────────────────×100%=88.1%0.2745×10000151c(H2SO4)=(0.2940g/100.09g/mol)×(1/10.00mL)×1000mL/L=0.2937mol/Lc(NaOH)×(11.20mL/1000mL/L)=2c(H2SO4)×(10.00mL/1000mL/L)c(NaOH)=2×0.2937mol/L/1.12

8、=0.5245mol/L[c(NaOH)V(NaOH)-2c(H2SO4)V(H2SO4)]M(KHC4H4O6)w(KHC4CH4O6)=───────────────────────×100%=95.5%m(试样)×