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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+