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1¡¢ÈܶȻý£»2¡¢Í¬Àë×ÓЧӦ£»3¡¢ÑÎЧӦ£»4¡¢·Ö²½³Áµí£»5¡¢³ÁµíµÄת»¯£»6¡¢ÒøÁ¿·¨£»7¡¢Îü¸½Ö¸Ê¾¼Á

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1£®Ð´³öÏÂÁÐÄÑÈܵç½âÖʵÄÈܶȻý³£Êý±í´ïʽ£º AgBr£¬Ag2S£¬Ca3(PO4)2£¬MgNH4AsO4 ½â£º K?sp(AgBr) = c(Ag+)?c(Br?)

K?sp(Ag2S) = c2(Ag+)?c(S2?)

K?sp[Ca3(PO4)2] = c3(Ca2+)?c2(PO43?)

2++3?

K?sp(MgNH4AsO4) = c(Mg)?c(NH4)?c(AsO4)

2£®ÇóCaC2O4ÔÚ´¿Ë®Öм°ÔÚ0.010mol?L?1µÄ(NH4)2C2O4ÈÜÒºÖеÄÈܽâ¶È¡£ ½â£ºÔÚ´¿Ë®ÖÐ

s = {K?sp(CaC2O4)}1/2 = (4?10?9)1/2 = 6?10?5 mol?L?1

?1?1

ÔÚ0.010mol?LµÄ(NH4)2C2O4ÈÜÒºÖУ¬ÉèÈܽâ¶ÈΪx mol?L,

2+2?

Ôò c(Ca) = x£» c(C2O4) = x +0.010 ¡Ö 0.010

x?0.010 = 4?10?9

ËùÒÔ s = x = 4?10?7 mol?L?1

3£® ¼Ù¶¨Mg(OH)2ÔÚ±¥ºÍÈÜÒºÖÐÍêÈ«µçÀ룬¼ÆË㣺

(1) Mg(OH)2ÔÚË®ÖеÄÈܽâ¶È£»

?

(2) Mg(OH)2±¥ºÍÈÜÒºÖÐOHµÄŨ¶È; (3) Mg(OH)2±¥ºÍÈÜÒºÖÐMg2+µÄŨ¶È£»

?1

(4) Mg(OH)2ÔÚ0.010mol?LNaOHÈÜÒºÖеÄÈܽâ¶È£» (5) Mg(OH)2ÔÚ0.010mol?L?1MgCl2ÈÜÒºÖеÄÈܽâ¶È¡£ ½â£º (1)ÉèMg(OH)2ÔÚË®ÖеÄÈܽâ¶ÈΪx

K?sp(Mg(OH)2) = c(Mg2+)?c2(OH?) = x? (2x)2 = 4x3

ËùÒÔ x ={ K?sp(Mg(OH)2)/4}1/3

= (1.8?10?11/4)1/3 = 1.7?10?4(mol?L?1)

(2) c(OH?) = 2?1.7?10?4 mol?L?1 = 3.4?10?4 mol?L?1 (3) c(Mg2+) = 1.7?10?4 mol?L?1

(4)ÉèMg(OH)2ÔÚ0.010mol?L?1NaOHÈÜÒºÖеÄÈܽâ¶ÈΪx

c(Mg2+) = x c(OH?) = 2x +0.010 ¡Ö 0.010

x?(0.010)2 = 1.8?10?11

ËùÒÔ s = x = 1.8?10?7 mol?L?1

(5)ÉèMg(OH)2ÔÚ0.010mol?L?1MgCl2ÈÜÒºÖеÄÈܽâ¶ÈΪx

c(Mg2+) = x +0.010¡Ö 0.010, c(OH?) = 2x

2?11

0.010 ?(2x)= 1.8?10 ?5?1

ËùÒÔ s = x = 2.1?10 mol?L

?1

4£®ÒÑÖªAgClµÄÈܽâ¶ÈÊÇ0.00018g?(100gH2O)(20¡æ)£¬ÇóÆäÈܶȻý¡£ ½â£º c(AgCl) ¡Ö b(AgCl) = 0.00018?10/143.32 = 1.3?10?5 (mol?L?1)

?5?5?10

K?sp(AgCl) =1.3?10 ?1.3?10 = 1.7?10

5£®ÒÑÖªZn(OH)2µÄÈܶȻýΪ1.2?10?17 (25¡æ)£¬ÇóÈܽâ¶È¡£

½â£ºÉèÈܽâ¶ÈΪx

3

K?sp{ Zn(OH)2} = 4x

ËùÒÔ s = x = (1.2?10?17/4)1/3 =1.4?10?6(mol?L?1)

6. 10mL 0.10mol?L?1MgCl2ºÍl 0mL0.010mol?L?1°±Ë®»ìºÏʱ£¬ÊÇ·ñÓÐMg(OH)2³Áµí²úÉú? ½â£º K?sp(Mg(OH)2)=1.8?10?11 c(Mg2+) = 0.050 mol?L?1 c(NH3) = 0.0050 mol?L?1

c(OH?) = (cb?K?b)1/2 = (0.0050?1.74?10?5)1/2 = 2.9?10?4(mol?L?1) Qi = c(Mg2+)?c2(OH?) = 0.050 ?(2.9?10?4 )2 =4.4?10?9 > K?sp=1.8?10?11 ËùÒÔ£¬Éú³É³Áµí¡£

?1?1

7£®ÔÚ20mL0.50mol?LMgCl2ÈÜÒºÖмÓÈëµÈÌå»ýµÄ0.10mol.LµÄNH3.H2OÈÜÒº£¬ÎÊÓÐÎÞ

Mg(OH)2Éú³É?ΪÁ˲»Ê¹Mg(OH)2³ÁµíÎö³ö, ÖÁÉÙÓ¦¼ÓÈë¶àÉÙ¿ËNH4Cl¹ÌÌå(Éè¼ÓÈëNH4Cl¹ÌÌåºó£¬ÈÜÒºµÄÌå»ý²»±ä)¡£

½â£º c(Mg2+) =0.50mol?L?1/2 = 0.25 mol?L?1 c(NH3) = 0.10mol.L?1/2 = 0.050 mol?L?1

?1/2?51/2?4?1

c(OH) = (cb?K?b) = (0.050?1.74?10) = 9.3?10(mol?L)

Qi = c(Mg2+)?c2(OH?) = 0.25?(9.3?10?4 )2 =2.2?10?7 > K?sp=1.8?10?11

ËùÒÔ£¬Éú³É³Áµí¡£

Ϊ²»Éú³É³Áµí£¬Ôò

c(OH?)¡Ü[K?sp/ c(Mg2+)]1/2 = [1.8?10?11/0.25]1/2 = 8.5?10?6(mol?L?1)

¸ù¾Ý K?b = c(NH4+?)?c(OH?)/ c(NH3?H2O) Ôò c(NH4+)= K?b? c(NH3?H2O)/ c(OH?)

?5?6

=1.74?10?0.050/8.49?10 = 0.10(mol?L?1)

?1?3?1

m(NH4Cl) = 0.10 mol?L?40?10L?53.5g? mol

= 0.21g 2+

8£®¹¤Òµ·ÏË®µÄÅŷűê×¼¹æ¶¨Cd½µµ½0.10mg?L?1ÒÔϼ´¿ÉÅÅ·Å¡£ÈôÓüÓÏûʯ»ÒÖкͳÁ

µí·¨³ýCd2+£¬°´ÀíÂÛ¼ÆË㣬·ÏË®ÈÜÒºÖеÄpHÖµÖÁÉÙӦΪ¶àÉÙ?

?15

½â£º K?sp(Cd(OH)2)= 5.25?10

c(Cd2+) = 0.10?10?3g?L?1/112.4g? mol?1 = 8.9?10?7 mol?L?1

?2+1/2

c(OH) = [K?sp/c(Cd)]

= (5.25?10?15/8.9?10?7)1/2

?5?1

=7.7?10 (mol?L)

ËùÒÔ pH = 14+lg7.7?10?5 = 9.89

9£®Ò»ÈÜÒºÖк¬ÓÐFe3+ºÍFe2+Àë×Ó £¬ËüÃǵÄŨ¶È¶¼ÊÇ0.05mol?L?1¡£Èç¹ûÒªÇóFe(OH)3³ÁµíÍê

È«¶øFe2+Àë×Ó²»Éú³ÉFe(OH)2³Áµí£¬ÎÊÈÜÒºµÄpHÓ¦¿ØÖÆΪºÎÖµ? ½â£º Fe(OH)3³ÁµíÍêȫʱµÄpH

c(OH?) = [K?sp(Fe(OH)3)/c(Fe3+)]1/3

= (4?10?38/1.0?10?6)1/3 = 3?10?11

pH = 3.5

²»Éú³ÉFe(OH)2³ÁµíʱµÄpH

c(OH?) = [K?sp(Fe(OH)2)/c(Fe2+)]1/2

?161/2

= (8.0?10/0.05) = 1?10?7 pH = 7.1

ËùÒÔÓ¦¿ØÖÆ 3.5 < pH < 7.1

?12+

10£®ÔÚ0.1mol?LFeCl2ÈÜÒºÖÐͨÈëH2S±¥ºÍ£¬ÓûʹFe²»Éú³ÉFeS³Áµí£¬ÈÜÒºµÄpH×î¸ßΪ

¶àÉÙ?

½â£ºÎª²»Éú³É³Áµí£¬Ôò

c(S2?)¡ÜK?sp(FeS)/ c(Fe2+)

= 6.3?10?18/0.1 = 6?10?17 mol?L?1

c(H+) =[c(H2S) ?K?a1 K?a2/c(S2?)]1/2

=(0.1?1.1?10?7?1.3?10?13/ 6.3?10?17)1/2

?3

=5?10 pH = 2.3

11£®º£Ë®Öм¸ÖÖÑôÀë×ÓŨ¶ÈÈçÏ£º

Àë×Ó Na+ Mg2+ Ca2+ Al3+ Fe2+

Ũ¶È/mol?L?1 0.46 0.050 0.01 4?10?7 2?10?7

?

(1) OHŨ¶È¶à´óʱ£¬Mg(OH)2¿ªÊ¼³Áµí£¿ (2) ÔÚ¸ÃŨ¶Èʱ,»á²»»áÓÐÆäËûÀë×Ó³Áµí£¿

?2+

(3) Èç¹û¼ÓÈë×ãÁ¿µÄOHÒÔ³Áµí50%Mg£¬ÆäËûÀë×Ó³ÁµíµÄ°Ù·ÖÊý½«ÊǶàÉÙ£¿ (4) ÔÚ(3)µÄÌõ¼þÏ£¬´Ó1Lº£Ë®ÖÐÄܵõ½¶àÉÙ³Áµí£¿ ½â£º(1) Mg(OH)2¿ªÊ¼³Áµíʱ

c(OH?) = [K?sp{Mg(OH)2}/c(Mg2+)]1/2

?111/2

=(1.8?10/0.050)

?5?1

=1.9?10(mol?L)

2+2??52?6

(2) Qi=c(Ca)c(OH) = 0.01?(1.9?10) < K?sp{Ca(OH)2}= 5.5?10 £¬ÎÞ³Áµí£»

Qi=c(Al3+)c3(OH?) = 4?10?7?(1.9?10?5)3> K?sp{Al(OH)3} = 1.3?10?33£¬ÓгÁµí£» Qi=c(Fe2+)c2(OH?) = 2?10?7?(1.9?10?5)2< K?sp{Fe(OH)2} = 8.0?10?16£¬ÎÞ³Áµí£»

ͬÀí£¬ÎÞNa(OH) ³Áµí¡£

(3) 50%Mg2+³Áµíʱ£¬c(Mg2+) = 0.025 mol?L?1 ?2+1/2

c(OH) = [K?sp(Mg(OH)2/c(Mg)]

=(1.8?10?11/0.025)1/2

?5

=2.7?10

c(Al3+) = [K?sp{Al(OH)3}/c3(OH?)]

?33?53

= [1.3?10/(2.7?10)]

?203+

=6.6?10 AlÒѳÁµíÍêÈ«¡£

(4) m = m{ Mg(OH)2} + m{ Al(OH)3}

= [0.025?1?58.32 + 4?10?7?1?78]g= 1.5g

?12+

12£®ÎªÁË·ÀÖ¹ÈÈ´øÓã³ØÖÐË®ÔåµÄÉú³¤, ÐèʹˮÖб£³Ö0.75mg?LµÄCu £¬Îª±ÜÃâÔÚÿ´Î»»

³ØˮʱÈÜҺŨ¶ÈµÄ¸Ä±ä£¬¿É°ÑÒ»¿éÊʵ±µÄÍ­ÑηÅÔڳصף¬ËüµÄ±¥ºÍÈÜÒºÌṩÁËÊʵ±µÄ2+2+CuŨ¶È¡£¼ÙÈçʹÓõÄÊÇÕôÁóË®£¬ÄÄÒ»ÖÖÑÎÌṩµÄ±¥ºÍÈÜÒº×î½Ó½üËùÒªÇóµÄCuŨ¶È£¿

CuSO4£¬ CuS£¬ Cu(OH)2£¬ CuCO3£¬ Cu(NO3)2

?10?

½â£º Ksp(CuCO3) = 1.4?10

c(Cu2+) =(1.4?10?10)1/2 = 1.18?10?5 mol?L?1

?(Cu2+) = 64?103mg?mol?1?1.18?10?5 mol?L?1 = 0.76mg?L?1 CuCO3 µÄ±¥ºÍÈÜÒº×î½Ó½üËùÒªÇóµÄCu2+Ũ¶È¡£

13£®Ïּƻ®ÔÔÖÖijÖÖ³¤ÇàÊ÷,µ«ÕâÖÖ³¤ÇàÊ÷²»ÊÊÒ˺¬¹ýÁ¿ÈܽâÐÔFe3+µÄÍÁÈÀ,ÏÂÁÐÄÇÖÖÍÁÈÀÌí

¼Ó¼ÁÄܺܺõĽµµÍÍÁÈÀµØÏÂË®ÖÐFe3+µÄŨ¶È?

Ca(OH)2(aq)£¬ KNO3 (s)£¬ FeCl3(s)£¬ NH4NO3(s)

½â£ºCa(OH)2 ¡£

14£®·Ö±ð¼ÆËãÏÂÁи÷·´Ó¦µÄƽºâ³£Êý,²¢ÌÖÂÛ·´Ó¦µÄ·½Ïò¡£

(1) PbS + 2HAc = Pb2+ + H2S + 2Ac?

(2) Mg(OH)2 + 2NH4+ = Mg2+ + 2NH3.H2O (3) Cu2+ + H2S = CuS + 2H+

½â£º(1) K? = c(Pb2+)c(H2S)c2(Ac?)/c2(HAc)

2

= c(Pb2+)c(H2S)c2(Ac?)c2(H+)c(S2?)/c2(HAc)c2(H+)c(S2?) = K?sp(PbS) K?a(HAc)/ K?a1(H2S)K? a2(H2S)

?28?52?7?13

=1.3?10 ?(1.74?10 )/(1.1?10?1.3?10) = 2.8?10?18 ·´Ó¦ÄæÏò½øÐÐ 2+2.2+

(2) K? = c(Mg)c(NH3H2O )/c(NH4)

?2+2?

= c(Mg2+)c2(NH3.H2O )c2(OH)/c(NH4)c(OH) = K?sp{Mg(OH)2}/ K?b 2(NH3.H2O) =1.8?10?11 /(1.74?10?5 ) = 5.9?10?2 ·´Ó¦ÄæÏò½øÐÐ

(3) K? = c(H )/c(Cu) c(H2S )

= c2(H+ ) c(S2?)/c(Cu2+) c(H2S )c(S2?) = K?a1(H2S)K?a2(H2S)/K?sp(CuS) = 1.1?10?7?1.3?10?13/6.3?10?36 = 2.3?1015 ·´Ó¦ÕýÏò½øÐÐ

15£®¼ÆËãÏÂÁвⶨÖеĻ»ËãÒòÊý(Ö»Áгöʽ×Ó) ²â¶¨Îï ³ÆÁ¿Îï

(1) P2O5 (NH4)3PO4?12MoO3 (2) MgSO4?7H2O Mg2P2O7 (3) FeO Fe2O3

(4) A12O3 Al(C9H6ON)3

½â£º(1) M(P2O5)/2M[(NH4)3PO4?12MoO3]

(2) 2M (MgSO4?7H2O)/M(Mg2P2O7) (3) 2M(FeO)/M( Fe2O3)

(4) M(A12O3)/2M{Al(C9H6ON)3}

16£®²â¶¨FeSO4?7H2OÖÐÌúµÄº¬Á¿Ê±£¬°Ñ³ÁµíFe(OH)3×ÆÉÕ³ÉFe2O3×÷Ϊ³ÆÁ¿ÐÎʽ¡£Îª

ÁËʹµÃ×ÆÉÕºóµÄFe2O3ÖÊÁ¿Ô¼Îª0.2g£¬ÎÊÓ¦¸Ã³ÆÈ¡ÑùÆ·¶àÉÙ¿Ë? ½â£º m(FeSO4?7H2O) = m(Fe2O3)?2M(FeSO4?7H2O)/M(Fe2O3)

= 0.2g?2?278.01/159.69 = 0.7g

17£®³Æȡij¿ÉÈÜÐÔÑÎ0.3232g£¬ÓÃÁòËá±µÖØÁ¿·¨²â¶¨ÆäÖк¬ÁòÁ¿£¬µÃBaSO4³Áµí0.2982g¡£

¼ÆËãÊÔÑùÖÐSO3µÄÖÊÁ¿·ÖÊý¡£ ½â£º w = m(SO3)/ms

= [0.2982g?(80.06/233.39)]/0.3232g = 0.3165

?1

18£®³ÆÈ¡ÂÈ»¯ÎïÊÔÑù0.1350g£¬¼ÓÈë0.1120mol?LµÄÏõËáÒøÈÜÒº30.00mL£¬È»ºóÓÃ0.1230

?1

mo1?LµÄÁòÇèËáï§ÈÜÒºµÎ¶¨¹ýÁ¿µÄÏõËáÒø£¬ÓÃÈ¥10.00mL¡£¼ÆËãÊÔÑùÖÐC1?µÄÖÊÁ¿·ÖÊý¡£

½â£º w = (0.1120?30.00?0.1230?10.00) ?10?3?35.45/0.1350

= 0.5593

19£®³ÆÈ¡º¬Òø·ÏÒº2.075g, ¼ÓÈëÊÊÁ¿ÏõËᣬÒÔÌú立¯ÎªÖ¸Ê¾¼Á£¬ÏûºÄÁË0.04600mol?L?1

µÄÁòÇèËáï§ÈÜÒº25.50mL¡£¼ÆËã´Ë·ÏÒºÖÐAgµÄÖÊÁ¿·ÖÊý¡£ ½â£º w = 0.04600?25.50?10?3?107.9/2.075

= 0.06100

20£®Calculate the molar solubility of Sn(OH)2 .

(1)In pure water (pH=7);

+

(2) In a buffer solution of containing equal concentrations of NH3 and NH4.Solution: (1) s = c(Sn2+)= K?sp{Sn(OH)2}/c2(OH?)

?28?72

= 1.4?10/(10) = 1.4 ?10?14

(2) c(OH?) = K?b(NH3)?cb/ca

=1.74?10?5

s = K?sp{Sn(OH)2}/c2(OH?) = 1.4?10?28/(1.74?10?5)2 = 4.62 ?10?19

21£® Even though Ca(OH)2 is an inexpensive base, its limited solubility restricts its use. What is

the pH of a saturated solution of Ca(OH)2?

2+2?32+

Solution: K?sp{ Ca(OH)2 }= c(Ca)c(OH) = 4 c(Ca)

c(OH?)= 2c(Ca2+) = 2[K?sp{ Ca(OH)2 }/4]1/3

?61/3

=2(5.5?10/4) = 0.022(mol?L?1)

pH = 14+lg0.022 = 12.34 ?1

22£® A solution is 0.010 mol?L in both Cu2+ and Cd2+. What percentage of Cd2+ remains in the

solution when 99.9% of the Cu2+ has been precipitated as CuS by adding sulfide?

2+2+

ÁªÏµ¿Í·þ£º779662525#qq.com(#Ì滻Ϊ@)