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使用教材:许力等编著, 《 工程化学 》 ,兰州大学出版社 授课对象:非化学类各专业学生 主讲教师:董文魁、许力、李静萍等 PowerPoint PPT Presentation


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工程化学 —— 第三章 溶液与离子平衡. 使用教材:许力等编著, 《 工程化学 》 ,兰州大学出版社 授课对象:非化学类各专业学生 主讲教师:董文魁、许力、李静萍等. 学习目的和要求 : 掌握溶液的通性,包括气压下降、沸点上升、凝固点下降和渗透压四个方面的内容。 掌握酸碱的质子理论、同离子效应、弱电解质的解离平衡和配离子的解离平衡,能熟练计算各类溶液的 PH 值,能计算含配离子的溶液中各种物质的浓度。 掌握有关溶度积和溶解度的基本计算 、 溶度积规则及其应用。 了解有关分散系统方面的基本知识。. 本章节重点、难点 各类溶液的 PH 值的计算.

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使用教材:许力等编著, 《 工程化学 》 ,兰州大学出版社 授课对象:非化学类各专业学生 主讲教师:董文魁、许力、李静萍等

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:PH


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PH


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13


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3-1

WB=mBm

WBm+mBBWB150.0g7.70g

WNaCl=7.7050.0100=15.4


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1LmolL-1,SImolm-3

cB=nBV

VnBBcBB


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xB=nBn(A)+nB

nBnAxB 1


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1kgmmolkg-1

mB=n(B)mA

mBBnBBmAAmA


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3-12515.4%=1.109gcm-3

100g15.40g

mH2O=100.0g-15.40g=84.60g=84.60103kg

nNaCl=15.40g/58.44gmol-1=0.264mol

nH2O=84.6g/18.02gmol-1=4.964mol

V=m=100.0g1.109gcm-3=90.17cm3=90.17103L


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xNaCl=nNaClnNaCl+nH2O=0.264mol0.264mol+4.694mol=0.0532

mNaCl=nNaClmH2O=0.264mol84.6103kg=3.12molkg1

cNaCl=nNaClV=0.264mol90.17103L=2.92molL1

0.05323.12molkg12.92molL1


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  • 3-2

  • BA

  • (. Ostwald .


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Tb / Tf / 20/(gcm-3)

100.00 0.00 0.9982

0.5molkg-1 100.27 -0.93 1.0687

0.5molkg-1 100.24 -0.94 1.0012

1

1 : 202339Pa

100101.325kPa


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(2)

()


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3Raoults law

ppAxA

xAnA /(nA+nB) xBnB/(nA+nB)

ppA-ppAxB

B


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1 20 2.33 kPa. 17.1gC12H22O113.00g[CO(NH2)2]100g. .


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2

(1)

(2)

Tb.


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TbKb.mb

TfKf.mb

2 80.2 2.67 g(C10H8) 100g80.731 .

m(C10H8)=128.0g.mol-1

mb(2.67/128)/(100/1000)mol.kg-1

=0.2086 mol.kg-1

Kb Tb/mb

0.531K/0.2086 mol.kg-1


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()C2H4(OH)2


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  • 1

  • ,


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(mol.dm-3)

(Pac(mol.m-3)T(K)n(mol)Vm3)


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cRT1000

3 Tf= 0.56 37 .

=1000 (0.56/1.86) 8.314(273+37

=7.756105Pa

(2)

2000kPa


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273.15K373.15K


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(m)

i

iTf'/Tf


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1. c(C6H6O6)=0.10mol.dm-3

2. c(CaCl2)=0.05 mol.dm-3

3. c(Na3PO4)=0.033 mol.dm-3

4. c(KNO3)=0.10 mol.dm-3


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1. C(C6H6O6)=0.10mol.dm-3

2. c(CaCl2)=0.15 mol.dm-3

3. c(Na3PO4)=0.13 mol.dm-3

4. c(KNO3)=0.20 mol.dm-3

4231

1324


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PA=pAxB

TbKbmB

KbR(Tb.A)2MA/VAPHm.A

TfKfmB

KfR(Tf.A)2MA/subHm.A

V= nBRT

p157: 6


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  • 3-3

  • 1.

  • A Arrhenius

1

H+

OH-


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H+ + OH- = H2O

2

Arrhenius

NH3 HClNH4Cl


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B (J.N.Brnsted)

1

.

.

.


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2

.

NH4Cl NH4+ Cl.

HB = H+ + B- +


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2. pH

1.

H2O(l)+H2O(l)H3O+(aq)+OH-(aq)

Kw=c(H3O+)cc(OH-)c

Kc=1molL-1

K=c(H3O+)c(OH-)

K=c(H+)c(OH-)


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KWWaterK25K=1.00910-14


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(2)

KaKb


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1Kb


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K

pH

HAc

HAc(aq) + H2O(l)= H3O+(aq) + Ac-(aq)

HAc(aq) = H+(aq) + Ac-(aq)


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c


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1- 1

Ka c Ka c


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.

3 25 0.200molL-10.934%c(OH), pH.


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KW.

4 0.100mol.dm-3NH4ClPH,NH4+Ka5.6510-10


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7.510-6.mol.dm-3

p H -lg(7.510-6)=5.12

5 1.0L0.10mol.L-1HAc0.10molNaAc(s)pHKa1.810-5

1.0L 0.10mol.L-1HAcpH

c(H+)


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p H(1)=2.87

0.10molNaAc(s)H+x

HAc + H2O = H3O+ + Ac-

0.10-x x 0.10+x

x0.10+x0.100.10-x0.10

x1.810-5

p H(2)=4.75

p H=4.75-2.87=1.87


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pH

H+pH


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6 0.10mol.dm-3NaAcpH.

Ac-1 + H2O = HAc + OH-1


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H+c(H+)pH .


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7 0.100mol.dm-3H2SpH


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3 pH

1 the buffered solution

pH.

2

HAc-NaAc


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HAcHAcH

HH

HAcAc -NH4+-NH3 , H2PO4HPO42-


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10-4

3 pH(calculate the pH of buffered solutions)

NH4+-NH3 , H2PO4HPO42-


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:(1)HAc


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(2) 1.00mo1.dm-3 HCl

H+Ac-HAc


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4

PH

SiO2

Si02 6HF H2[SiF6] 2H20

HFHHpHHFNH4F


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H2PO4-1-HPO4-2pH

pH H2CO3NaHCO3NaH2PO4Na2HPO4pH

pH H2CO3-NaHCO3pH7.4PH0.5


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1

pKa1


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p155 1315


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1.

()()


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2

[Ag(NH3)2+]


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9


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3

Cu(NH3)4H2SO4,

H(aq)NH3NH4(aq)[Cu(NH3)42+]


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p158159 8 9 11 14 17


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.

(solubility product constant and solubility)

(ion product & rule of solubility product)

-

mobile of precipitation dissolution equilibrium

1. Ksp

()100(g/100gH2O).


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-S(mol/dm-3) .


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Ksp


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1Ksp

- RTlnKsprGm

rGm

rGm=BfGm(B)

rGm=rHm-TrSm

rHm=BfHm(B)

rSm=BS(B)


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2


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2

Ag2CrO4


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(C)


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Ksps;

2.

1 (rule of solubilit product)

AnBmsnAm+ (aq) + mBn- (aq)

Q


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()


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  • 2

().

C.

3 25Ag2CrO40.010 mol L-1 K2CrO4


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4

HCl BaCl2Na2CO3

HCl

Ba+2CO3-2

BaCO3


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KSP

5 25 C(SO42-)6.010-4 molL-1. 40.0L0.010 molL-1 BaCl210.0LBaSO4


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B.

CaSO4Na2CO3CaSO4CaCO3

KSP(CaSO4)=7.110-5; KSP(CaCO3)=4.9610-9


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C.

a)

CaCO3(s)+2H+(aq)Ca2+(aq)+CO2(g)+H2O(l)


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b)

()

AgBr

c)


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6 0.20L 0.50mol L-1 MgCl20.10 mol L-1Mg (OH)2Mg (OH)2NH4Cl(s) (NH4Cl(s) )


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p160 181921


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