7 沉淀-溶解平衡习题解答(p180-182)1. 解答：(1)解：AgI(2)解：Mg(OH)22. 解答：(1) CaF 2 ⇌ Ca 2+ + 2F - s+0.0010 2sK sp =(s+0.0010)(2s)240.0010s 2(2) Ag 2CrO 4 ⇌ 2Ag + + CrO 42-2s+0.010 sKsp =(2s+0.010)2s 0.0102s3. 解答： M 2X = 2M + + X 2- X 2-有酸效应：4. 解答：(1) CaF 2 ⇌ Ca 2+ + 2F -(2) BaSO 4 ⇌ Ba 2+ + SO 42-(3) CuS ⇌ Cu 2+ + S 2-)L mol (104.1)L mol ()5.077.234104.1(1161262----⋅⨯=⋅⨯⨯==s K sp )L m ol (102.1)L m ol ()32.581105.8(44)2(11113332----⋅⨯=⋅⨯⨯⨯==⋅=s s s K sp 15111L mol 102.8L mol 0010.04107.20.00104----⋅⨯=⋅⨯⨯=⨯=θspK s 182122L mol 100.2010.0100.2010.0---⋅⨯=⨯==θspK s 192122X(H)100.1Ka Ka ][H Ka ][H 1⨯=++=++α)L (mol 100.14100.1100.44)2(110319493X(H)sp X(H)sp 'sp 2---⋅⨯=⨯⨯⨯=⋅=⋅==⋅ααθθθK s K K s s )L (mol 102.14)10(107.24)2(10108.6101][113322.11132)(2)(22.142)(-----+⋅⨯=⨯⨯=⋅=⋅=⋅=⨯+=+=H F sp H F sp a H F K s K s s K H αααθθ)L (mol 104.110101.110102.10.21][1142.210)()(22.22)(2424224----+⋅⨯=⨯⨯=⋅=⋅==⨯+=+=---H SO sp H SO sp a H SOK s K s K H αααθθ5. 解答：(1) AgBr ⇌ Ag + + Br -328.0Ag(NH )13233-11[NH ]+[NH ]=107.110mol Ls αββ-=+==⨯•(2) BiI 3 ⇌ Bi 3+ + 3I --3-4-513.8Bi(I)3453sp Bi(I)11[I ]+[I ]+[I ]=10s(3s+0.10):s 0.017mol L K θαβββα-=+=⨯=•用逼近法求得(3) BaSO 4 ⇌ Ba 2+ + SO 42-由于K BaY 较大且BaSO 4的K sp 较大，所以Ba 2+消耗的EDTA 不能忽略 c Y =[Y]-[BaY]=0.010-s6. 既考虑配位效应，又考虑酸效应1510][,5914,9---⋅==-=∴=L mol OH pOH pH -++⋅OH NH O H NH 4235234θb101.75O]H [NH ][OH ][NH --+⨯=⋅⋅=K52323θb101.75O]H [NH ][OH O]}H [NH {2.8--⨯=⋅⋅⋅-=K1θb23L 1.02mol ][OH ][OH 2.8O]H [NH ---⋅=+⋅=⋅⇒K )L (mol 102.21010610][][1189.1936)()(29.192)(222122---++⋅⨯=⨯⨯=⋅=⋅==++=---H S sp H S sp a a a H SK s K s K K H K H αααθθ1475256.556.310)(256.556.33.286.7)(3.2)(L mol 1013.60104104)1010(101.1101010010.0101][110010.0-----⋅⨯==⨯-⨯+-⨯⨯=⋅=-=-⨯+=+=-=s s s s K s s sY K sc Y Ba sp BaY Y Ba H Y Yαααθ313829139θa2θa22θa2(H)S7.427.4 3.423231)(NH Ag 107.880.977692.31101.3109.5)10(1101.31011][H ][H 1101.02101.02101][NH ][NH 123⨯=++=⨯⋅⨯⨯+⨯⨯+=⋅++==⋅+⋅+=⨯+⨯+=-----++-+K K K αββα(H)S 2)(NH Ag 2θsp (H)S 2)(NH Ag 22(H)S 22)(NH Ag 222θsp 232323S}{Ag ][S ][Ag ][S }]{[Ag ')'(2]'[S ]'[Ag '-+-+-+⋅⋅=⋅⋅⋅=⋅⋅⋅=⋅=⋅=-+-+-+ααK ααααs s K113327.4493(H)S2)(NH Ag 2θsp 106.34107.8)(101024S}{Ag '23--⨯=⨯⋅⋅⨯=⋅⋅=-+ααK s7. 解答： CaCl 2 + 2NaOH = Ca(OH)2 + 2NaCl 1.11/111=0.1 0.12平衡时：[Ca 2+]=0.1-0.12/2=0.04 molL -1设Ca(OH)2的溶解度为s ，则：(s+0.04)(2s)2=K sp =5.510-6用逼近法求得：s=5.510-3,[OH -]=2s=0.012mol/L, pH=12.04, [Ca 2+]=0.046 mol L -18. 解答： BaSO 4 = Ba 2+ + SO 42- s s+0.01 s(s+0.01)=Ksp =1.110-10 s=1.110-8mol L -1 BaSO 4沉淀的损失=1.110-8200233.4=5.110-4mg9. 解答： BaSO 4的溶解度为：10. 解答：Ca 3(PO 4)2 = 3Ca 2+ + 2PO 43- 3s 2s11. 解答： AgCl = Ag + + Cl -Ag +与NH 3会形成配合物，影响AgCl 沉淀的溶解平衡或者：AgCl + 2NH 3 = Ag(NH 3)2+ + Cl -12. 解： [Ba 2+]=0.010/1.0=0.010molL -1 [F -]=0.020/1.0=0.020molL -1 Ba 2+ + 2F - = BaF 2 J=[Ba 2+][F -]2=0.0100.0202=4.010-6Ksp有沉淀生成s (2s)2= Ksp =1.010-6[Ba 2+]=6.310-3mol L -1 [F -]=2s=0.013 molL -113. 解答：)L (mol 106.166.067.0101.11510242---⋅⨯=⨯⨯=⋅=-+SOBa apK s γγθ29.3]H []H []H [1]H []H []H [11233232333221(H)PO-34=+++=+++=++++++Ka Ka Ka Ka Ka Ka βββα165229523(H)P O2(H)P O223L mol 101.142729.3100.223)2()3(-34-34---⋅⨯=⨯⨯⨯=⨯⋅=⋅=ααθθsp sp K s K s s 140.710)Ag(NH 40.7240.7232)Ag(NH L mol 069.010108.1'101101]NH [133--⋅=⨯⨯=⋅==⨯+=+=αβαθsp K s 12240.71022323L mol 069.0110108.1][NH ]][Cl )[Ag(NH ---+⋅==⨯⨯⋅==s s K K sp βθθ133L mol 103.64--⋅⨯==θspK sJ=[Mg 2+][OH -]2= 1.010-3(1.810-6)= 3.210-15Ksp所以：无Mg(OH)2沉淀生成14. 解： MnS(s) + 2HAc(aq) = Mn 2+(aq) + 2Ac -(aq) + H 2S(aq) x 0.20 0.40 0.20所以，溶解0.20molMnS 需HAc 的浓度为：0.032+0.40=0.43molL -115. 解：必须同时考虑同离子效应和酸效应。

αS 2-（H ）=1+β1[H +]+β2[H +]2β1=1/k a2=121069.7⨯; β2=1/k a1 k a2=191010.8⨯ 所以αS 2-（H ）=131010.8⨯θsp θsp K 'K =αS 2-（H ）=41086.4-⨯所以s ’= 'K θsp /[ S 2-]=31086.4-⨯1-⋅L mol16. 解：氯化钡过量所以[Cl -]=13302.01.0202.01050204.01050---⋅=⨯⨯⨯-⨯⨯⨯L mol )L (mol 05.015.00.150.5]NH [)L (mol 5.0100.151322495.0]NH [)L (mol 100.10.50.100.100015.0]Mg [13134132---+--+⋅=⨯=⋅=⨯⨯⨯=⋅⨯=+⨯=)L (mol 108.15.005.0108.1][NH ][NH ][OH 16543---+-⋅⨯=⨯⨯=⋅=θb K 1138251022)()(2)(2222L mol 032.0103.1105.9)1075.1(105.220.040.020.0)([HAc]S][H ]][Ac [Mn 2221------+⋅=⨯⨯⨯⨯⨯⨯=⨯⨯⋅⋅==x x K K K K K S H aS H a HAc a sp θθθθ同理，[Ba 2+]=101.0-⋅L mol[BaSO4]K θsp = ][Ba ][24-SO 得][24-SO =8101.1-⨯1-⋅L mol 同理，[Ag +]=9109.9-⨯1-⋅L mol17. 解：（1）由pV=nRT ，得n （CO2）=mol 31033.6-⨯n （CO2）=n （SrCO3），所以s=133104.444.11033.6---⋅⨯=⨯L mol Lmol（2）[SrCO3]K θsp = ][2+Sr ][23-CO = 3104.4-⨯ ⨯ 3104.4-⨯= 1.9×10-5（3）CO2不能完全溢出回收，所以测定值小于实际值。

18. 解：解法一，当Mg(OH)2完全溶解时12121043.1502005.020][---+⋅⨯=+⋅⋅=L mol mLmL L mol mL Mg由于11222108.1][OH 1043.1][Mg(OH)K ---⨯=⋅⨯=θsp ∴当Mg(OH)2恰好完全溶解时，溶液中[OH-]浓度为：521122θsp 103.55101.43101.8101.43][Mg(OH)][OH -----⨯=⨯⨯=⨯=K 当加入与NaOH 等当量的NH4Cl 时，完全生成NH3)(357.00.7050.00.50][13-⋅=⨯=L mol NH353θb 102.50101.750.357)(NH ][OH ---⨯=⨯⨯=⨯=K c∴必须再加入NH4Cl 形成缓冲溶液，再降低溶液pH)L 0.176(mol 103.550.357101.75)(NH )(NH 0.357lg)10lg(1.8)10lg(3.55)(NH )(NH lg)(NH p pOH 1554455433θb ---++--+⋅=⨯⨯⨯=-⨯-=⨯--=c c c c K ∴加入的总的NH4Clg 00.249.53)357.0176.0(100070=⨯+⨯解法二， 0.357101.43 50.020.00.5050.0 50.020.00.05020.0O H 2NH Mg 2NH (s)Mg(OH)223242=⨯=+⨯+⨯⋅+=+-++x)L 0.176(mol (0.357)101.43)10(1.75101.8)](NH [][Mg(OH)O]H [NH ][NH ][OH ][OH ][Mg ][NH ][OH O]H [NH ][OH ][Mg ][NH O]H [NH ][Mg 1222251123θb 2θsp 2232422224222322242232----+--++--+++⋅=⨯⨯=⨯⨯==⋅⋅⋅=⋅⋅⋅⋅=⋅⋅=x xK K K∴加入的总的NH4Clg 00.249.53)357.0176.0(100070=⨯+⨯19. 解答：J=[Zn 2+][OH -]=2.610-16K sp =1.210-171543110.610.5)Zn(NH '2210.523231)Zn(NH L mol 108.1][NH ][NH ]OH [L mol 101010.0]Zn [][Zn 10][NH ]NH [1)33--+---++⋅⨯=⋅=⋅====+++=b K αββαL计算说明有Zn(OH)2沉淀生成。