ÎÞ»ú¼°·ÖÎö»¯Ñ§£¨µÚ¶þ°æ£©¸÷ÕÂÒªÇó¼°ÀýÌâ½²Îö5-9ÕÂ

Solution: When 99.9% of the Cuhas been precipitated,

c(S2?) = K?sp(CuS)/c(Cu2+)

= 6.3?10?36/(0.01?0.1%)

= 6.3?10?31

Qi = c(Cd2+)c(S2?) = 0.01?6.3?10?31 < K?sp(CdS) = 8.0?10?27

Therefor, 100% Cd2+ remains in the solution.

23£® Calculate the molar solubility of each of the following minerals from its K?sp

?10

(1) alabandite, MnS: K?sp=2.5?10

?8

(2) anglesite, PbSO4: K?sp=1.6?10

?11

(3) brucite, Mg(OH)2: K?sp=1.8?10 (4) fluorite, CaF2: K?sp=2.7?10?11

Solution: (1) s = (K?sp)1/2 = (2.5?10?10)1/2 = 1.6? 10?5(mol?L?1)

(2) s = ( K?sp)1/2 = (1.6?10?8)1/2 = 1.3?10?4(mol?L?1)

1/3?111/3?4?1

(3) s = ( K?sp/4) = (1.8?10/4) =1.7?10(mol?L)

(4) s =( K?sp/4)1/3 = (2.7?10?11/4)1/3 = 1.9?10?4(mol?L?1)

24£® Consider the titration of 25.00mL of 0.08230 mol?L?1 KI with 0.05110 mol?L?1 AgNO3.

Calculate pAg at the following volumes of added AgNO3: (1)39.00 mL ; (2) Vsp; (3) 44.30mL ;

Solution: (1) c(I?) = (0.08230?25.00?0.05110?39.00)mol/(25.00+39.00)L

?1

= 0.001009(mol?L) c(Ag+) = K?sp(AgI)/c(I?)

?17

= 8.3?10/0.001009 = 8.2?10?14

+

pAg = ?lg c(Ag) =13.09 +1/2

(2) c(Ag) ={K?sp(AgI)}

?171/2

= (8.3?10)= 9.1?10?9

pAg = 8.04 +

(3) c(Ag) = (0.05110?44.30?0.08230?25.00)/(25.00+44.30)

= 0.002976 pAg = 2.53

2+

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ÕÆÎÕÄÜ˹Ìع«Ê½¼°ÆäÓ¦Óã»ÀûÓõç³ØµÄµç¶¯ÊÆÅжϷ´Ó¦µÄ·½Ïò²¢¼ÆËã·´Ó¦µÄƽºâ³£Êý£»ÕÆÎÕÑõ»¯»¹Ô­·´Ó¦µÎ¶¨·¨£¨¸ßÃÌËá¼Ø·¨¡¢ÖظõËá¼Ø·¨¡¢µâÁ¿·¨£©µÄ»ù±¾Ô­Àí¼°ÆäÓ¦Óá£

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1£®ÇóÏÂÁÐÎïÖÊÖÐÔªËصÄÑõ»¯Öµ¡£ (1)CrO42?ÖеÄCr (2)MnO42?ÖеÄMn

(3)Na2O2 ÖеÄO (4)H2C2O4?2H2OÖеÄC

½â£º(1)Cr£º+6£»(2) Mn£º+6£» (3) O£º?1£» (4) C£º+3

2£®ÏÂÁз´Ó¦ÖУ¬ÄÄЩԪËصÄÑõ»¯Öµ·¢ÉúÁ˱仯£¿²¢±ê³öÑõ»¯ÖµµÄ±ä»¯Çé¿ö¡£

(1)Cl2 + H2O = HClO + HCl (2)Cl2 + H2O2 = 2HCl + O2

(3)Cu + 2H2SO4(Ũ) = CuSO4 + SO2 + 2H2O

(4)K2Cr2O7 + 6KI + 14HCl = 2CrCl3 + 3I2 + 7H2O + 8KCl ½â£º (1)Cl£º´Ó 0 ? +1 ºÍ ?1£»

(2)Cl£º´Ó 0 ? ?1£»O£º´Ó ?1 ? 0£» (3)Cu£º´Ó 0 ? +2£»S£º´Ó +6 ? +4£» (4)Cr£º ´Ó+6 ? +3£» I£º´Ó ?1 ? 0¡£

3£®ÓÃÀë×Óµç×Ó·¨ÅäƽÏÂÁÐÔÚ¼îÐÔ½éÖÊÖеķ´Ó¦Ê½¡£

(1)Br2 + OH? ¡ú BrO3? + Br? (2)Zn + ClO? ¡ú Zn(OH)42? + Cl? (3)MnO4? + SO32? ¡ú MnO42? + SO42? (4)H2O2 + Cr(OH)4? ¡ú CrO42? + H2O ½â£º(1) Br2+12OH?=2BrO3?+6H2O+10e?

( 2e? + Br2 = 2Br?) ¡Á 5

?

6Br2 + 12OH= 2BrO3? + 6H2O +10 Br?

?2? ?

(2) Zn + 4OH = Zn(OH)4+ 2e

H2O + ClO? + 2e? = 2OH? + Cl?

?? 2? ?

Zn + H2O + 2OH + ClO= Zn(OH)4+ Cl(3) (MnO4? + e? = MnO42?) ¡Á 2

? 2?

2OH+ SO3= H2O + SO42? + 2e?

2MnO4? + 2OH? + SO32? = 2MnO42? + H2O + SO42? (4) (H2O2 + 2e? = 2OH?) ¡Á3

? ?

(4OH+ Cr(OH)4= CrO42? + 4H2O + 3e?) ¡Á 2

? ? 2?

3H2O2 + 2OH+ 2Cr(OH)4= 2CrO4+ 8H2O

4£®ÓÃÀë×Óµç×Ó·¨ÅäƽÏÂÁÐÔÚËáÐÔ½éÖÊÖеķ´Ó¦Ê½¡£

(1) S2O82? + Mn2+ ¡ú MnO4? + SO42?

(2) PbO2+ HCl ¡ú PbCl2 + Cl2 + H2O (3) Cr2O72? + Fe2+ ¡ú Cr3+ + Fe3+

?

(4) I2+ H2S ¡ú I+ S

½â£º (1) (S2O82? + 2e? = 2SO42?) ¡Á 5

2+ ? + ?

(4H2O + Mn= MnO4+ 8H+ 5e) ¡Á2

2? 2+ ?

5S2O8+ 8H2O + 2Mn= 2MnO4+ 16H+ + 10SO42?

(2) PbO2 + 4H+ + 2e?= Pb2+ + 2H2O

??2Cl=Cl2 +2e

PbO2 + 4HCl = PbCl2 + Cl2 + 2H2O (3) Cr2O72? +14H+ + 6e? = 2Cr3+ + 7H2O

(Fe2+ = Fe3++e?) ¡Á 6

2? +

Cr2O7+ 14H+ 6Fe2+ = 2Cr3+ + 7H2O + 6 Fe3+

? ?

(4) I2 + 2e= 2I

+ ?

H2S = S + 2H+ 2e ? + I2+ H2S = 2I+ S + 2H

5£®Diagram galvanic cells that have the following net reactions.

(1) Fe + Cu2+ = Fe2+ + Cu (2) Ni + Pb2+ = Ni2+ + Pb (3) Cu + 2Ag+ = Cu2+ + 2Ag (4) Sn + 2H+ = Sn2+ + H2 ½â£º(1) (?) Fe | Fe2+¡¬Cu2+ | Cu(+)

2+2+

(2) (?)Ni | Ni¡¬Pb | Pb(+)

2++

(3) (?)Cu| Cu¡¬Ag |Ag(+) (4) (?)Sn| Sn2+¡¬H+ |H2| Pd(+) 6£®ÏÂÁÐÎïÖÊÔÚÒ»¶¨Ìõ¼þ϶¼¿ÉÒÔ×÷ΪÑõ»¯¼Á£ºKMnO4, K2Cr2O7, CuCl2, FeCl3, H2O2, I2, Br2, F2,

PbO2¡£ÊÔ¸ù¾Ý±ê×¼µç¼«µçÊƵÄÊý¾Ý£¬°ÑËüÃÇ°´Ñõ»¯ÄÜÁ¦µÄ´óС˳ÐòÅÅÁУ¬²¢Ð´³öËüÃÇÔÚËáÐÔ½éÖÊÖеĻ¹Ô­²úÎï¡£

½â£ºÑõ»¯ÄÜÁ¦ÓÉ´óµ½Ð¡ÅÅÁÐÈçÏ£º

F2 > H2O2 > KMnO4 > PbO2 > K2Cr2O7 > Br2 > FeCl3 > I2 > CuCl2

ÔÚËáÐÔ½éÖÊÖеĻ¹Ô­²úÎïÒÀ´ÎΪ£º

F?, H2O, Mn2+, Pb2+, Cr3+, Br?, Fe2+, I?, Cu

7£®Calculate the potential of a cell made with a standard bromine electrode as the anode and a

standard chlorine electrode as the cathode.

Solution: E?(Cl2/Cl?)=1.358V, E?(Br2(l)/Br?) =1.065V

the potential of a cell: E? = E?(+) ? E?(?)=1.358V?1.065V=0.293V

8. Calculate the potential of a cell based on the following reactions at standard conditions.

(1) 2H2S +H2SO3 ¡ú 3S +3H2O (2) 2Br?+2Fe3+¡úBr2 +2Fe2+ (3)Zn +Fe2+¡úFe+Zn2+

(4)2MnO4?+5H2O2+6HCl¡ú2MnCl2 +2KCl+8H2O+5O2

Solution: (1)0.308V; (2)?0.316V; (3)0.323V; (4)0.828V.

?+2+

9£®ÒÑÖª MnO4+8H +5e = Mn+4H2O E?=1.51V

Fe3++e =Fe2+ E?=0.771V

(1)ÅжÏÏÂÁз´Ó¦µÄ·½Ïò

MnO4? + 5Fe2+ + 8H+ ¡ú Mn2+ + 4H2O +5Fe3+

(2)½«ÕâÁ½¸ö°ëµç³Ø×é³ÉÔ­µç³Ø£¬Óõç³Ø·ûºÅ±íʾ¸ÃÔ­µç³ØµÄ×é³É£¬±êÃ÷µç³ØµÄÕý¡¢¸º¼«£¬²¢¼ÆËãÆä±ê×¼µç¶¯ÊÆ¡£

(3)µ±ÇâÀë×ÓŨ¶ÈΪ10mol?L?1£¬ÆäËü¸÷Àë×ÓŨ¶È¾ùΪ1mol?L?1ʱ£¬¼ÆËã¸Ãµç³ØµÄµç¶¯ÊÆ¡£ ½â£º(1) MnO4?+5Fe2++8H+¡ú Mn2++4H2O+5Fe3+

E?+ > E?? ·´Ó¦ÕýÏò½øÐÐ (2) (?)Pt| Fe3+(c1), Fe2+(c2)¡¬MnO4?(c3), Mn2+(c4) | Pt£¨+£©

E? =1.51? 0.771= 0.739V

c(Ñõ»¯ÐÍ)0.0592Vlg nc(»¹Ô­ÐÍ)? E? (3) E = E+? E? = E?++?

80.0592Vlg105= 1.51V + ? 0.771V

= 0.83V

10£®ÒÑÖª Hg2Cl2(s)+2e?=2Hg(l)+2Cl? E? =0.28V Hg22++2e?=2Hg(l) E? =0.80V

ÇóK?sp(Hg2Cl2)¡£ (Ìáʾ£º Hg2Cl2(s) Hg22++2Cl?)

½â£º Hg2Cl2(s) +2e=2Hg(l)+2Cl E?= 0.28V Hg22++2e?=2Hg(l) E? = 0.80V

½«ÉÏÊöÁ½µç¼«·´Ó¦×é³ÉÔ­µç³Ø£º

(?)Pt|Hg(l)|Hg2Cl2(s)|Cl?¡¬Hg22+|Hg(l)|Pt(+)

µç³Ø·´Ó¦Îª£º Hg22++2Cl?= Hg2Cl2(s) E? = 0.80V ?0.28V = 0.52V

?2?0.52Vn?E? 0.0592V= 17.57 K? =1017.56 ·´Ó¦µÄƽºâ³£Êý lgK= 0.0592V?

??

K?sp(Hg2Cl2) = 1/K?= 1/10 = 2.8?10

11£®ÒÑÖªÏÂÁеç³Ø Zn| Zn2+(x mol?L?1)¡¬Ag+(0.10 mol?L?1)|AgµÄµç¶¯ÊÆE =1.51V£¬ÇóZn2+Àë

×ÓµÄŨ¶È¡£

½â£º E = E+? E? = 1.51V

[E(Ag/Ag) + 0.0592Vlgc(Ag)] ?[E(Zn/Zn) +

0.05922?

+

+

?

2+

17.56?18

0.05922Vlgc(Zn2+)] =1.51V

[0.799V+0.0592Vlg0.10] ? [?0.763V +Vlgc(Zn2+)] =1.51V

c(Zn2+) = 0.57(mol?L?1)

12£®ÎªÁ˲ⶨµÄÈܶȻý£¬Éè¼ÆÁËÏÂÁÐÔ­µç³Ø

(?)Pb| PbSO4| SO42?(1.0 mol?L?1)¡¬Sn2+(1.0 mol?L?1)|Sn(+)

ÔÚ25?Cʱ²âµÃµç³Øµç¶¯ÊÆE?= 0.22V£¬ÇóPbSO4ÈܶȻý³£ÊýK?sp¡£ ½â£º²é±í E?( Sn2+/Sn) = ?0.136V E?(Pb2+/ Pb) = ?0.126V

E?= E?+ ? E?? 0.22V = ?0.136 ? E?? E?? = E?( PbSO4/ Pb)= ?0.356V

2?0.0592lgc(Pb)2+2+??2E(PbSO4/ Pb)= E(Pb/ Pb) = E(Pb/ Pb) +

0.0592?0.356V = ?0.126V +20.0592?0.356V = ?0.126V +2lgc(Pb)

lg K?sp(PbSO4)/c(SO4)

2?

2+

lgK?sp(PbSO4) = ?7.77 K?sp = 1.7?10?8

13£®¼ÆËã298KʱÏÂÁеç³ØµÄµç¶¯ÊƼ°µç³Ø·´Ó¦µÄƽºâ³£Êý¡£

(1) (?) Pb | Pb2+(0.1 mol?L?1)¡¬Cu2+(0.5 mol?L?1)|Cu(+)

(2) (?) Sn | Sn2+(0.05 mol?L?1)¡¬H+(1.0 mol?L?1)|H2(105Pa) |Pt(+) (3) (?) Pt | H2(105Pa)|H+(1mol?L?1)¡¬Sn4+(0.5 mol?L?1), Sn2+(0.1 mol?L?1 )|Pt(+) (4) (?) Pt | H2(105Pa)|H+(0.01 mol?L?1)¡¬H+(1.0 mol?L?1)|H2(105Pa) | Pt(+)

0.0592½â£º(1) E+ = 0.337V +2Vlg0.5 =0.33V

0.0592E? = ?0.126V +2Vlg0.1 = ?0.16V

E = E+ ? E? = 0.33V ?(?0.16V) = 0.49V

2?(0.337??0.126)nE ??0.0592V0.0592lgK== 15.64 K? = 4.38¡Á1015

(2) E+ = E?(H+/H2) = 0

E? = ?0.136V +

E = E+ ? E? = 0.17V

0.05922Vlg0.05 = ?0.17V

2?(?0?0.136)nE ?4.594 ?0.0592V0.0592lgK== K?=3.9¡Á10

(3) E+ = 0.151V +

0.05922E? = E?(H+/H2) =0

E = E+ ? E? = 0V ? 0.17V = 0.17V

0.5Vlg0.1= 0.17V