1、第六章絡合滴定法4、解： 當 ph=9.0 時，oh-=1.0 10-5 mol?l-1 , nh3=1.0 102 mol ?l-10m(nh3) = 1+ 1nh3 + 2nh32 + 3nh33 + 4nh34=1 + 102.0 x 102.0 + 105.0 x 10-2.0 x2 + 107.0 x 10-2.0 x3 + 1010.0 x 10-2.0 x= 1+1+10+10+ 102.0 = 122om(oh)= 1+ 1oh- + 2oh-2 + 3oh-3 + 4oh-4=1 + 104.0 x 10-5.0 + 108.0 x 10-5.0 x2 + 1014.0 x

2、10-5.0 x3 + 1015.0 x 10-5.0 x=1 + 10-1.0+ 10-2.0+ 10-1.0 + 10-5.0=1.21由此可知，4nh34項數值最大，所以m(nh 3)42+為主要存在形式0m = om(nh3) + om(oh) -1=12210 2.0m(nh 3)42+=cm 布(nh 3)4+ = (0.10 x百)mol ?l = 0.082mol ?l-1當 ph=13.0, oh-=1.0 10-1.0 mol?l-1 ,nh3=1.0 10-2 mol ?l-1cm(nh3) = 1+ 1nh3 + 2nh32 + 3nh33 + 4nh34= 1220m

3、(oh) = 1+ 1oh- + 2oh-2 + 3oh-3 + 4oh -4=1 +104.0x 10-1.0 + 108.0 x10-1.0 x2+ 1014.0 x 10-1.0x3 + 1015.0x 10-1.0x4=1 +103.0+ 106.0 + 1011.0 +1011.0 = 2.0 x 1011由此可知，3oh-3和4oh-4這兩項數值最大，所以m(oh)3-和m(oh)42-為主要存在形式11cm = om(nh3) + om(oh) -1=2.0 x 101011.01m(oh) 3-=cm m(oh) 3 = (0.10 x2.0 x101.0) mol ?l= 0

4、.050mol ?l-110 11.0m(oh)42=cm moh) 2- = (0.10 x2.0x101.0) mol ?l = 0.050mol ?l-15、解：用en代表乙二胺，由題意知：en= 0.010mol ?l-1cag+=ag +ag(en) +ag(en) 2+=0.10 mol l-1§ag(en) +ag(en) +)=4 7104t 10-2：= 00909 1 + 1 en + %en21 + 104.710-2.0 + 107.710-2.0 x2= 107.7 x10-2.0 _= 0 9091 + 1en + feen21 + 104.7 x 10-

5、2.0 + 107.7 x 10-2.0 x2cen = en + ag(en)+ + 2ag(en) + )=(0.010+0.0909 x0.10+2 x0.909 x0.10) mol ?l-1 =0.20 mol ?l-16、解：(1)由題意可知，cd2+除了與y的反應作為主反應之外，還存在與灑石酸的絡合效應,y有酸效應和由zn2+產生的共存離子效應，而 zn2+又存在與灑石酸的絡合效應。ph=6.0 時，lgoy(h)=4.65,加入等體積的 edta 之后有tart=0.10 mol ?l-1則有cd(tart) = 1+ 1tart=1+102.8x0.10=64.1=101.8

6、1y(zn) = 1 +kzpiy zn2+ = 1 +kzpiyzn = 1 +kzpiyaftr+ 12y(zn)zny l zny azn(tart)zny 1+01 tart +02tart 2=1 + 1016.5 x2 4 0.010 8 32= 10881+10 2.4 x 0.10+10 8.32 x 0.102oy = ocy(h) + ocy(zn) - 1 = 104.65 + 108.18 - 1 = 108.18同理，可得:y(cd)lgkcdy = lgkcdy - lg cd - lg y = 16.46 - 1.81 - 8.18 = 6.47 zn(tart)

7、 = 1+ 1tart+ 2 tart2=106.32八 xccd2+0.010=1 + kcdycd2+ = 1 + 2 t)= 1 + 1016.46 x正同=1012.65配=ay(h) + 碉cd) - 1 = 104.65 + 1012.65 - 1 = 1012.65lgkzny = lgkzny - lg zn - lg y = 16.50 - 6.32 - 12.65 = -2.478、解：查附錄表8,碘離子可與鎘離子絡合。zn + yzny由題意可知，y有酸效應和cd2+產生的共存離子效應，且i- = 1mol ?l-1ccd2+ay(cd)= 1 + kcdy cd2+ =

8、 1 + kcdyocd(i)k ccd±cdy 1 + mi-+ mi-2+ fei-3+ mi-4=108.701016.46 x0.00500101 + 102.10 + 103.43 + 104.49 + 105.41當 ph=5.0 時，lg oy(h)= 6.45y = oy(h) + oy(cd) - 1 = 108.70_ lgkzny = lgkzny - lg y = 16.50 - 8.70 = 7.80pznsp = 2 (lgk zny + pczn) = 2 x (7.80 + 2.30) = 5.05以二甲酚橙作指示劑時，pznep=4.80,（附錄表1

9、4）則有?pzn = 4.80 - 5.05 = -0.25 10。25 i。25et =x 100% = -0.22%v0.00500x 107.8012、解:a、ph=5.0 時，lg oy(h)= 6.45r ,ka八 1ac- = c 汩=0.31 x-_- = 0.20 mol ?l-1 ka + h opb(ac)= 1 + mac- + 也ac- 2 = 102.43igkpby = igkpby - lg pb - lg y = 18.04 - 6.45- 2.43 = 9.16_1,spppbsp = 2(igkpby+ pcpb) = 6.08' . _ _ _ _

10、 _ _ _ pbep = igk'pbin = 7.0- lg pb = 7.0- 2.43 = 4.57?ppb = 4.57- 6.08 = -1.51et =10?ppb10-?ppb，記kpby10-1.51 . 101.51x100% = x100% = -2.7%v0.0010x 109.16lgkpby = lgkpby - lg y =118.04 - 6.45 = 11.59ppbsp = 2(lgkpby +sp一pcpb) = 7.30pbep = lgk'pbin = 7.010?p pb -10-?p pbet =:，喘 kpby?ppb = 7.0

11、- 7.3 = -0.310-0.3 - 100.3x100% = x100% = -0.008%v0.0010x 1011.5913、解：由題意可知,cu(nh3)= 1+ 1nh3+ 2nh32 + 3nh33 + 4nh34= 109.36ph=10 時, ocu(oh) = 1.70lg oy(h)= 0.45%u = «cu(oh) + «cu(nh3) - 1 = 1.70 + 109.36 - 1 = 109.36igkcuy = igkcuy - lg cu - lg 級h) = 18.80 - 9.36 - 0.45 = 8.99一, _ . ，一一 .p

12、cuep = pcuep - 1g cu(nh3) = 13.8 - 9.36 = 4.44pcusp = 2(1gkcuy + pccu) = 2 x (8.99+ 2.00) = 5.50?pcu = 4.44- 5.50 = -1.0610?pcu10-?pcuet = 一，盤kcuy10-1.06 - 101.06x100% = x100% = -0.36%v0.010x 108.9914、解：a.在只含la3+溶液中，當?pla = 0.2 , et <0.3%,則有lg(k 、lay csp) >5.0igk'lay >7.0 1g y(h)= 15.5-

13、7.0=8.5ph=4.0最低酸度為la3+水解產生沉淀時的ph值，則r ,3 ksp 3/10-18.81561oh- = v-p = ( v 001 ) mol ?l-1 = 10-5.6 mol ?l-1ph = 8.4因此，滴定la3+的酸度范圍為ph=4.08.4o在兩種金屬離子la3+和mg2+共存混合溶液中，則有7.100y(h) = oy(mg) = 10查表可知，ph =4.7,滴定酸度控制在ph4.7 8.4b.若指示劑不與mg2+顯色，lg(kmgycmg) = 7.1.在la3+獨立存在溶液中，ph4.0 8.4, ph=4.0時,lg y(h) = 8.4, 0y(h

14、)oy(mg)。 mg2+不干擾測定，滴定酸度范圍為ph=4.08.4oc、當 0y(h) = 0.1 y(mg)時,則有 0y(h)= 106.1 ,查表得 ph = 5.27.1w = 0y(h) + ay(mg) - 1 ocy(mg) = 10igklay = igklay - lg y= 15.5- 7.1 = 8.41 ,、1/、plasp =2(1gk- igcla)= 2 x(8.4+2.0)= 5.2當 ph = 5.2 時，plaep = 4.7,則有?pla = 4.7 - 5.2 = -0.5etx100% =10?pla - io?plavcep klay10。5 -

15、 100.5v 102.0 x 108.4x100% = -0.2%17、解：滴定后 cfe3+=fe3+fey+fef+fef2+fef3fe(f)= 1+ 1f-+ 2f- x 1.000 x5000 x1000+ 3f-22、解：由題意可知，ph=1.0時絡合滴定的是bi3+,則有=1+104.28+107.30+109.06fe(edta) = 1+k'y查表，lgkcuy = 18.80, lgkfey=25.1,ph=4.5 時，lgoy(h)= 7.44,pcuep=8.3則有lgkcuy = 18.80 - 7.44 = 11.36， _ lgkfey = 25.1 -

16、 7.44 = 17.66對于cu2+與edta的絡合，終點時cuy= 0.010 mol.l -1,則有)mol ?l-1 = 10-5.06 mol ?l-1、,_ cuy _0.010yep = cuep kcuy = (10-8.3 x1011.36因此有fe(edta) =1+k'y=1012.60fe= fe(f) + fe(edta)-1fe(edta)=1012.60fef3fey109.06w1x100% = 0.029%19、解：ph=0.8時絡合滴定的是zr(iv),則有wzro2 =1.000 x10-2 x 10.00 x 123.221.000 x1000 x50.00200.0x100% = 4.93%ph=1.5時絡合滴定的是fe3+,則有wfe2o3 =x100% = 6.39%1.000 x10-2 x 20.00 x 159.69wbi0.02479