無機化學(xué)1、2347章chapter4-溶液與電離平衡

溶液與解離多相體系StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 (導(dǎo)電性StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 難揮發(fā)非電解質(zhì)稀溶液的依數(shù)StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

iI1/2b i

溶液與解離StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離子間相互作用的濃度。以a(activity)表示。ab/bγ活度系數(shù)。稀溶液中，γ1；，γ接近1StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離①離子強度越大，離子間相互作用越顯著，②離子強度越小，活度系數(shù)趨于1③離子電荷越大，相互作用越強，活度系數(shù)1lg0.509Z I：活度系數(shù)；Z1、Z2分別表示離子電荷；I為溶液離子強度StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離Arrhenius )??1887離子全部都是H+ NH4Cl顯酸性的事 如液氨中：NH+NH? StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 質(zhì)子(H+)+堿[Al(H2O)6]3+ [Al(H2O)5(OH)]2++酸酸

質(zhì) 質(zhì)StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離越弱，反之亦然。StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離44HSO–NH+44

4SO2–4 質(zhì)

NH+44StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離質(zhì)子理論無法解釋如SnCl4、AlCl3StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離H++:OH–=H2OAlCl3+ =AlCl4–Cu2++4:NH3=[Cu(NH3)4]2+金屬陽離子及缺電子的分子都是酸與金屬離子結(jié)合的陰離子或中性分子都有些分子和離子的中心原子可擴大其配位層以接的電子對，從而顯 酸性。如SiF4+2F–= 酸性。如CO2+OH–=[HCO3]– StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離軟硬酸堿理論—— 化學(xué)家★半徑較小和(或)正電荷較高的陽離子、離子團(tuán)在外電場中易變形而被稱之為軟酸(Softacids)(Softbases)優(yōu)點：StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 酸酸

, 堿NH3,CO2–,NO堿 SO2–,PO3–,ClO

NO2 N

H–,R–, CO,I–, C6H6, 門當(dāng)戶

Cl– Br– I– R2Se 軟酸與之形成的化合物穩(wěn)定性StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離除A之外的其他軟酸金屬離子的毒性都很強。例如， 中毒可能是由于2(軟酸)與蛋白質(zhì)分子中的S2(軟堿)結(jié)合，從而改變了蛋白質(zhì)分子的結(jié)構(gòu)。由于S是比更軟的原子從而對g顯示出更強的結(jié)合力。根據(jù)這一原理，藥物化學(xué)家有可能設(shè)計出含S藥物，從 患者的白質(zhì)S原子上除去2離子。盡管S–是個毒性很大的軟堿，但一定條件下卻能化害為利。StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離以反應(yīng)(C2H5)3Si－IAgBr→(C2H5)3Si－Br為例，反應(yīng)能夠向右進(jìn)行，是因為兩個產(chǎn)物都滿足了硬–硬和軟–軟結(jié)合的規(guī)高價的金屬化合物[AgF4]2–、[FeO4]2–和OsO4中的、Fe(+6)和Os(+8)就是用F–和O2–這樣的硬堿使之穩(wěn)定StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離AlF3AlCl3AlBr3 AgF>AgCl>AgBr>堿的硬度：堿的硬度：F–Cl–Br–>酸的硬度：Al3酸的硬度：Ba2+Fe2+溶解度：BaSFeSStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離 SolutionO2–和都是硬堿，而S2–硬酸Li+,Na+,Ti2+/4+,Al3+,Cr3+等主要以與硬堿O2–結(jié)合狀態(tài)存在。軟酸Ag+,Zn2+,Cd2+,Pb2+等主要以與軟堿S2–結(jié)合狀StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離第三節(jié)H2O+

H3O++OH–H++OH–[H3O+]=[OH–]=1.0×10–7moldm[HO

[OH ( )( )1.010 Kw

pHpOHStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離H2O(l)=H+(aq)+OH– 79.89kJ lg KK w1.0 HH m55.90kJ KKwStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離 ([H]/

[In]/c)/([HIn]/c當(dāng)c(HInc(In–)時，溶液呈紅色，是酸性當(dāng)c(HIn)c(In–)[H] [HIn]/ [In]/在[HIn]/[In–10或[HIn]/[In–]0.1時，指示劑[H+在0.1~10之間。StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 /kJa a

27.15kJ KaT KaT

RTlnStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離 /kJ

NH3H2O(aq)=NH4(aq)+OH 27.12RTln 1.76 電離度(離解度α％)：平衡時弱電解質(zhì)αStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen 初始濃 平衡濃 c?

溶液與解離 K(c/c

/{c(1)/c當(dāng)α5%1α (c/c) /c/cStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離[注c/Ka380[H+5%·c時，可用上式進(jìn)行 Kb /(c/cαKiαc而變，而Ki不隨c而變；StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 平 c0-

x

[Ac/c][H+/c][HAc/c

(x/c0c/cx/0當(dāng)c0/ c0xxx/c[H+]/cKc/ pHlg([H]/c),[H]/c?,[OH]0StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen0

溶液與解離平 00c0–([H]/c0 (c[H])/0

(先不考慮c這一項 [H+]2K[H+]Kc [H[H+]a(K)24Ka2 StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 溶液與解離moldm (2)1.0×10?5moldm解：(1)c0/

a x2/0.1xx2/a[H]1.3103(moldm3

c0/

380x2/(1.0105x)1.8

x[H]7.16StateKeyLaboratoryStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

(moldm3溶液與解離例對于一元弱堿，其pH值計算如下c0–=0x+0xc0–+=HA+0x0x

[B+/c][OH/c][BOH/c

(x/cc0/cx/當(dāng)c0/ c0xKb(c0/cx/cKb(c0/c當(dāng)c0/

x/ b(K)24Kc/b StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離Ka、Kb

HA+H2O=A–+

H3O++ HA+

2H2O=H3O++ [HO][A [HA][OH 3=[HO][OH3=

[A StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen例：已知0.10moldm?3HB的pH3，求0.10moldm?3NaB的pH值。解：首先寫出HB和NaB的電離

溶液與解離 HB+H2O B–+

H3O++ HB+ moldmpHlg[HO

(10pH1 [HO] Kc

aa

StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離根據(jù) Kb可得②的 Kw/ Kc(BKc(B 22

lg[OH2pH214pH2= StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 HB+H2O

H3O++

B–+

HB+

pH2=7+lg(c1c2)1/2+StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 配成500ml水溶液，計算溶液的pH值。(已知HCN的Ka=4.9×10?10)解 的濃度=(2.45/49)÷0.5=0.1(moldmCN?+=+c0xx c KW/ 2.0c0x=[OH?]=1.4×10?3moldmpOH=2.85,pH=

xStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離H2S=H++ Hc)([HS]/c)/([HS]/c)1.1 =H++ Hc ]/c)/([HS]/c)1.02 H/c2([S2]/c)/([HS]/c 1.1StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離溶液中同時存在H2S、HS?、H+、S2?([H+]/c2S2?]/c)/([H2S]/c1.110?21moldmmoldm液)的[H+]、[HS?]、[S2?]及pOH。StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離000.1-xxc/ 0.1x1ax2/0.1 x[H][HS]1.05104(moldm3a1

[H ]/[HS 2y 1.02

[H][OH]K,[OH]9.5wpOH

(moldm3StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離moldmmoldm解

0x[HS[HS]0.1/0.33.67108(moldm3

y[S2]y [HS]/[H]1.010143.67108/1.221021(moldm3StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離多元弱酸中，若Ka1Ka2Ka3，通常Ka1/Ka2105,求[H+]時，可做一元弱酸處理。StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離 Ka1>>Ka2[H+和[HS?表示溶液中總的[H+和[HS?Ka1Ka2，所以第一步解離出來的[H+大大超過第二步解離出來的[H+]；并且由于同離子，第一步解離出來的HHS解離H+。因此溶液中的[H+]可按第一步解離來算，并且[H+]≈StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離0.1moldm + a Kw/a

1.01014/1.01014HS?+ =

/

1.01014/1.11079.09 Kb1Kb2，溶液[OH?StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 +

(0.1x)2/x 0.1xx[S2]0.01(moldm3[OH]0.10.019102(moldm3)[HS(0.10.01)/0.1100%StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離磷酸溶液中的[H3PO4H2PO?],[HPO2?],[PO3?], [Al(H2O)6]3++H2O=[Al(H2O)5(OH)]2++H3[Al(OH)(H2O)5]2++H2O=[Al(H2O)4(OH)]++H3[Al(OH)2(H2O)4]2++H2O=[Al(H2O)4(OH)]++H3

存在酸式電離及堿式

,

–既是質(zhì)，又是質(zhì)子H2PO4?溶液呈 6.2108, 1.31012, a[H](

StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離Ka(Kb相差很大，只考慮電離常數(shù)大的弱酸(堿，相差.1mldmNH3HHAc、H+的濃 解 NH++Ac?= + 平衡時：c0- c0- [NH][HAc][H K

/(

4[NH+][Ac [H][OH 4

3.21K

c0/ x2/(0.1StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 c0xc0 x/0.1 x5.66104(moldm33

HO][HAc]5.662

(moldm3 平衡時5.66×10-

0.1-5.66×10- 1.8x(0.15.66104)/5.661041.8x1.0107(moldm3)[HStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離NH++ NH+ NH++HO=NH+H +H2O=HAc+

Ka(NH+4K4b(Ac

和H3O++OH–= 1/ ha(NH+4b(AcK1wKb(NH3K1Kw=Kb(NH)3StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 b(NH例：已知 4.931010, 1.77 b(NH3 )=0.010moldm?3時 →根據(jù)Kh值求 StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen求NH4CN的解離常

溶液與解離 NH++ NH+ Kw

1.0

4h(NH+,CN4

(4.931010)(1.77105b(NH3 由解離常數(shù)求酸堿的濃度(簡單起見，設(shè) 1.0NH + + 1.0不能 0.010–x0.010– 1.0x5.0103(moldm3) (0.010由酸堿濃度和電離常數(shù)求pHNH +HO= +H 0.010–5.0× 5.0×

Ka(NH+45.01034

/

40.0105.04

a(NH+

b(NH3

1.77x=5.65×10?10,pH=StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離

HAc=H++StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離平StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離平0.1moldm和0.1moldm 1moldm?3HCl或)0.1moldmNaAcNa

cAc?+x

x[H]

cca apHa

lgc

4.75lg0.1StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離滴加 1moldm 0.1+新平衡0.1+1×10?4–

0.1–0.1–1×10?4+ax[H] a

0.111 pH StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離滴加 1moldm 新平

ax[H]a

0.111040.11

pH StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離加入酸，如HCl，H+與Ac?結(jié)合，生成cAc?/cHAc變化不大，溶液的pH值變化不加入堿，如NaOH，HAc與OH?與結(jié)合，生成，cAc?/cHAc變化不大，溶液pHStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen緩沖溶液[H+

溶液與解離平 平 酸(x/c)(x鹽/c)K / 酸

ax/ a

pHpH lgcpOHpOHpKblgcStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離平StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen

溶液與解離平 酸/pH 例：欲配制pH5.0例：欲配制pH9.0

應(yīng)選擇NH3-NH4Cl體系 4.74,pHStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離平moldmNH3Hmoldmmoldm解：(1)pOH / lg1.8105lg(1/ pHStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,XiamenNH3H

溶液與解離平+各為0.05—NHH—3H

x0.05pOH lg / 4.74lg 4.73 0.05 pHpH=StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離平StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離節(jié)沉AnBm(s)=nAm+(aq)+mBn?(aq) RTlnQRTln

Q{[Am]/c}n{[Bn]/c}m(s

QKsp時，沉淀從溶液中析出(溶液過飽和)QKsp時，沉淀與飽和溶液的平衡QKspStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離平當(dāng)Q (1)往盛有1.0dm3純水中加入0.1cm3濃度為0.01moldm?3的CaCl2和Na2CO3： ]= ]==10?6mol

×Q[Ca2][CO2]1012 8.7 因此無CaCO3StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離 moldm[Ca2+]=[CO2?]=10?4mol3Q[Ca2][CO2]1083

sp(CaCO3StateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen1.010?3moldm?3的K2CrO4溶液中滴加4的[Ag+]=?CrO2?沉淀完全時，[Ag+4解

=2Ag++ [Ag]2[CrO2 [Ag]{ 4[CrO24

3.5

5moldm

moldm

1.2 [Ag]4[CrO24

}2 1.0

1.1

3moldmStateKeyLaboratoryofPhysicalChemistryofSolidSurfaces,Xiamen溶液與解離平例：向0.1moldm?3的ZnCl2溶液中通H2S氣體至溶液的[H+]=?。解 Zn2++

[Zn2][S2

[Zn2

1.2

(