812滴定分析習題庫

1、16章后習題解答TOP習題1. 下列數據，各包括幾位有效數字？(1) 2.0321 0.0215(3) pKHin = 6.30-5 0.01 %(5) 1.0 10(解 5; 3; 2; 1; 22. 分析天平的稱量誤差為).0001 g，用減差法稱取樣品時，兩次稱量的最大誤差可達).0002 g，欲使稱量的相對誤差小于0.1%，求至少應稱取樣品質量(g)?ERE =X100%T0皿2100TT = 0.2 g所稱樣品至少要大于0.2 g。3. 兩人分析同一試樣中的含銅量，其結果質量分數(3)如下:甲 0.36100.36120.3603乙 0.36410.36420.3643巳知此試樣中含

2、銅量的真實值為0.3606。試問何人所得結果的準確度高？何人的結果精密度高?為什么？解 甲的 X =(0.3610 + 0.3612 + 0.3603 ) /3 = 0.3608乙的 X =(0.3641 + 0.3642 + 0.3643 ) /3 = 0.3642甲的準確度高，因平均測量值接近于真實值；乙的精密度高，因它的重現性好。4. 在用鄰苯二甲酸氫鉀標定NaOH溶液的濃度時，若在實驗過程中發(fā)生下列過失的情況，試說明每種情況下NaOH溶液所測得的濃度是偏大還是偏小。(1) 滴定管中NaOH溶液的初讀數應為 1.00 mL,誤記為0.10 mL。(2) 稱量鄰苯二甲酸氫鉀的質量應為0.3

3、518 g，誤記為0.3578 g。解可利用下式計算NaOH溶液的濃度：c(NaOH)=m(KHC 8H4O4)V(NaOH) M (KHC 8H4O4)(1) 初讀數1.00 mL誤計為0.10 mL,使V(NaOH)偏大，由上式可知測定的NaOH溶液濃度偏低；(2) 由于將稱量的鄰苯二甲酸氫鉀的質量0.3518 g誤計為0.3578 g，使m(KHC 8日4。4)偏大，由上式可知測定的NaOH溶液的濃度偏高。5. 某一弱堿型指示劑的KIn- = 1.3 X0-5，此指示劑的變色范圍是多少？解pK|n- = -lg Kin- = -lg(1.3 W5) = 4.89pKHin = 14 4.

4、89 = 9.11故變色范圍為pH 8.1110.116. 稱取分析純 Na2CO3 1.3350 g，配成一級標準物質溶液250.0 mL ,用來標定近似濃度為0.1_1mol L HCl溶液，測得一級標準物質溶液25.00 mL恰好與HCI溶液24.50 mL反應完全。求此 HCI溶液的準確濃度。解一級標準物質溶液 25.00 mL中Na2CO3的質量m(Na2CO3)=0.1335 g1.335 0g 25.00 mL250.0 mL根據Na2CO3標定HCI溶液的計量關系Na2CO3 + 2HCI = 2NaCI + CO 2 + H2Oc(HCI)V(HCI) =2 n( Na2CO

5、3)=2 m(Na2C3)M (Na2CO3)-0.1028moI LJ2 m(Na2CO3)2 0.1335gC(HCI )13M(Na2CO3)漢V(HCI) 106.0g mol 漢 24.5010 L因此HCI溶液的準確濃度為 0.1028 moI L-1。7. 標定濃度為0.1 moI L-1的NaOH溶液，欲消耗25 mL左右的NaOH溶液，應稱取一級標準物質草酸(H2C2O4 2H2O)的質量約為多少？能否將稱量的相對誤差控制在).05%范圍之內？若改用鄰苯二甲酸氫鉀(KHC8H4O4)，結果又如何？2n(H2C2O4 2H2O) =c(NaOH)v(NaOH)2m(H2C2O4

6、 2H2O)ML2C2O4 2H2O)= c(NaOH)V(NaOH)m(H2C2Q g CZOHNZOH) M (H266 2H2。)2_0.1mol Lx25d0( L 匯126g mol- 2=0.15g若稱量草酸采取差減法進行，則分析天平引起的絕對誤差為0.0001 g X20.0002RE100% =0.133%，故不能控制在 0.05%范圍內。0.15若改用鄰苯二甲酸氫鉀，M(KHC 8H4O4)= 204 g mol-1n(KHC 8H4O4) =c(NaOH)V(NaOH)m(KHC8H4O4) =c(NaOH)V(NaOH) M (KHC 8H4O4) =0.1 mol L

7、1 25 10 J L 204 g mol J =0.51gRE二空002 100% =0.039%，故可以控制在0.05%范圍內。 0.518. 化學計量點和滴定終點有何不同？在各類酸堿滴定中，計量點、滴定終點和中性點之間的關系如何？解化學計量點是化學反應的等量點，而滴定終點則是指示劑的變色點。在各類酸堿滴定中，所選指示劑的變色點與該反應的化學計量點應盡可能接近。中性點的pH值永遠等于7而化學計量點和變色點的 pH值則根據滴定反應類型的不同和所選指示劑的不同而不同。9用NaOH溶液滴定某一元弱酸時，已知加入40.00 mLNaOH溶液時達到化學計量點，而加入NaOH 標準溶液16.00 mL

8、時，溶液的pH值為6.20。求此弱酸的離解常數。解設一元弱酸為HA ,pH二 pKa lgn(NaA) .,16.00 mL c(NaOH)n(HA) _p a g (40.0016.00)mL 漢 c(NaOH )6.20 二 pKa lg16.00 mL24.00 mLpKa = 6.3810.下列各酸（假設濃度均為 0.1000 mol L，體積均為20.00 mL），哪些能用0.1000 molNaOH溶液直接滴定？哪些不能？如能直接滴定，應選用什么指示劑，各有幾個滴定突躍？(1)蟻酸(HCOOH) pKa = 3.755.61 琥珀酸（H2C4H4O4） pKai = 4.16 ；

9、pKa2(3) 順丁烯二酸(H2C4H2O4)pKai =1.83；pKa2=6.07(4) 鄰苯二甲酸(H2C8H4O4)pKa1 =2.89;pKa2=5.51(5) 硼酸(H3BO3) pKa1 = 9.27解 (1 )能，因為pKa = 3.75, Ka = 1.7 10-4,cKa 10,由于滴定突躍范圍為6.759.70,因此可選用酚酞作指示劑。(2) 能，因為CKa1 1-0 , CKa2 1&但K/Ka2 10 , CKa2 1-8，且K/Ka2 10：故有2個滴定突躍，可進行分步滴定。 由于第一化學計量點時的 pH = 3.95,第二化學計量點時的 pH = 9.30,因此可

10、分別選用甲基橙和百里 酚酞作為兩步滴定的指示劑。(4) 能，因為cKa1 105, cKa2 1(8,但Ka1/Ka2 104，故只有一個滴定突躍，可作為一元弱酸進行滴定。由于化學計量點時的pH 9.0左右，故可選用酚酞或百里酚酞作指示劑。(5) 不能，因為cKa1 1-8、cKa2 1& cKa3 10但Ka1/Ka2 104、 Ka2/Ka3 o 20.00mLc(EDTA) 0.05000mol LCaCO3 0.2100 g，用HCI溶液溶解后，配成250.0 mL溶液，吸取25.00 mL，在pHT為指示劑用EDTA標準溶液滴定，用去標準溶液20.15 mL ,求EDTA標準溶n(C

11、aCO3)= n(EDTA) = c(EDTA) V(EDTA)c(EDT A)=n( CaCO3)V (EDT A)m(CaCO3)M(CaCO3) V(EDT A)0.2100g250 mL-1-3100g mol 1 x20.15x10 L1=0.01042mol L-120.取100.00 mL水樣，在pH值為10左右，以鉻黑 T為指示劑，用 0.010 48 mol L- -EDTA標準溶 液滴定至終點，共消耗 14.20 mL ;另取100.00 mL該水樣，調節(jié)pH值至13,使Mg2+形成Mg(OH) 2沉 淀，以紫脲酸胺為指示劑， 用上述EDTA標準溶液滴定，消耗EDTA標準溶

12、液10.54 mL，計算水中Ca2+、Mg2+的濃度各為多少。c(Ca )-c(EDT A)V2(EDT A)V(水樣)-10.01048 mol L 10.54 mL二 0.001 105 mol L100.00 mLc(Mg2)c(EDT A)V,EDTA)V(水樣)-0.000 383 6mol LJ0.01048 mol L-1(14.20 -10.54) mL100.00 mL21.取醫(yī)院放射科含銀廢液10.00 mL，經HNO3酸化后，以 NH4Fe(SO4)2為指示劑，用 0.043 82mol L-1NH4SCN標準溶液滴定，消耗 23.48 mL ,求廢液中 Ag +的含量(

13、g解Ag+(aq) + SCN-(aq) = AgSCN( s)+ -n(Ag ) = n(SCN ) = c(NH 4SCN) V(NH 4SCN)pg+u n(Ag+)M(Ag+) _c(NH4SCN)V(NH4SCN)M(Ag*)g - v(試樣)-V(試樣)A-10.04382mol L x 23.48 m107.87g mol . . _. j=11.09g L10.00mLExercises1. If you had to do the calculation of (29.837-29.24)/32.065, what would be the correct result of

14、 sig nifica ntfigure?Answer The significant figure would be 0.019-1 -12. Two monoprotic acid, both 0.1000 mol L in concentration, are titrated with 0.1000 mol L NaOH. ThepH at the equivalence point for HX is 8.8 and that for HY is 7.9 .(a)Which is the weaker acid?(b) Which in dicators in followi ng

15、could be used to titrate each of these acids?Methyl orange pH ran ger for3.1 4.4color cha ngeMethyl red4.4 6.2Bromthymol blue6.0 7.5Pheno Iphthale in8.0-9.6Solution (a)HX is weaker. The higher the pH at the equivale nee point, the stron ger the conjugate base(X ) and the weaker the conjugate acid(HX

16、).(b) PhenoIphthalein, which changes color in the pH 89.6 range, is perfect for HX and probably appropriate for HY.3. A sample of 0.1276 g of an unknown mono protic acid, HX, was dissolved in 25.00 mL of water and-1titrated with 0.0633 mol L NaOH solutio n. The volume of base required to reach the e

17、quivale nee point was 18.40 mL.(a) What is the molar mass of the acid?(b) After 10.00 mL of base had been added in the titration, the pH was determined to be 5.87. What is theKa for the unknown acid?SolutionHA + NaOH = NaA + H 2O(a)n( HA) = n( NaOH) = c(NaOH) V(NaOH)m(HA)=c(NaOH)V(NaOH)M (HA)0.1276g

18、_430.0633 mol L 18.40 10 L M (HA)M(HA) =109.5g molNaAV(NaA)(b) pH = pKa lgpKa lgHAV(HA)10.00 mLpKa =5.87-lg5.87 - 0.076 = 5.79(18.40 -10.00) mLKa = 1.6 1 氐61 c 2-4. How many grams of H 2C2O4 2H2O are need to prepare 250.0 mL of a 0.1000 mol L C2O4 standard-1 +solution and a 0.1000 mol L H3O standard

19、 solution?Solutionx126g mol=250.0 10： L 0.1000 mol LJx = 3.150 g_12-3.150 g of H 2C2O4 2H2O are needed to prepare 250.0 mL of a 0.1000 mol L C2O4 standard solution.Ka15.6 10_KaZ 1.5 10-102+ +Almost all the H 3O in the soluti on comes from the first io nizatio n react ion. The concen tratio n in H 3O

20、 can be determined by considering only K a1. +H2C2O4 = HC2O4+ H3Oy-0.1 0.250.1 0.250.1 )0.2512HC2O4 H 3O (0.1000 mol Lx0.2500 L)/K a1 =1 = 5.6 漢 10H2C2O4(y0.1000 mol 匸一漢0.2500 L)y =0.036 mol-1 m(H2C2O4 2H2O) = y M(H2C2O4 2H2O) = 0.036 mol 120g mol = 4.5 g-1 +4.5 g of H 2C2O4 2H2O are needed to prepa

21、re 250.0 mL of a 0.1000 mol L H3O standard solution.5. A student titrated 25.00 mL of NaOH solution with standard sulfuric acid. It took 13.40 mL of 0.05550-1mol L H2SO4 to neutralize the sodium hydroxide in the solution. What was the molarity of the NaOH solution?The equation for the reaction is 2N

22、aOH + H 2SO4 = Na2SO4 + 2H 2OSolution2NaOH + H2SO4 = Na2SO4 + 2H 2O=n( NaOH) = n (H 2SO4)2Ic(NaOH) V(NaOH) = c(H2SO4)V(H2SO4)2c(NaOH)=2 c(H2SO4)V(H2SO4)V(NaOH)2 0.05550mol LJ 13.40 mL25.00mL_1=0.05950mol L-16. A freshly prepared soluti on of sodium hydroxide was sta ndardized with 0.1024 mol L H2SO4

23、 .(a)If 19.46 mL of the base was n eutralized by 21.28 mL of the acid, what was the molarity of the base?(b)How many grams of NaOH were in each liter of this soluti on?Solution (a)1-c(NaOH )V(NaOH) = c(H 2SO4N (H2SO4)2c(NaOH)二0.1024 mol L21.28 mL19.46 mL2 =0.2240 mol L，(b)m(NaOH) =c(NaOH)V(NaOH)M (NaOH)= 0.2240mol L，1L 40g molJ = 8.958g7. A 0.5720 g sample of a mixture con tai ning Na 2CO3, NaHCO