3829787 توازن مجموعة كيميائية الدرس مع تمارين مصححة وفق...

8/17/2019 توازن مجموعة كيميائية الدرس مع تمارين مصححة وفق البرنامج التعليمي الجزائري 3829787

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78 I : : 04 óÝ» Ším óï÷bïáïØI - PH - PH - 1 - : o(H - O) = 1,013 A

o(H - H) = 1,63 A 104 , 5

) .(- ) O H ( .- :

- : ⎛ ⎞⎜ ⎟⎝ ⎠v molC =l - : ⎛ ⎞⎜ ⎟

⎝ ⎠m gC = l - : ( )mC C= mol/l = M

- : . - : . -(M) = (ne ) d +d +


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HH104, 5°-dd + d +HH104, 5°-do104, 5

79 - : - ne × - : ) ( C .V=C V¢ ¢) ( ) (eV =V+V¢3 - :

- : C 25

-6SMw = 5, 5×10 ) Auto protolyse ( . - : H + OH

- 3H O+ .+HH O -HH O -HH O H

+⎡ ⎤⎢ ⎥⎢ ⎥- -⎣ ⎦HO-⎡ ⎤


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⎢ ⎥⎢ ⎥⎣ ⎦+

: 3H O+ OH -

: 2+

3 21H O +OH 2H O⎯⎯®¬⎯⎯ .4

PH : PH sorvensen

⎧ ⎡ ⎤⎪ ⎣ ⎦⎨⎡ ⎤⎪⎣ ⎦⎩+3+ PH3PH= log H O :

H O =10 :

PH 3H O +⎡ ⎤⎣ ⎦ .

: 2

⎡ ⎤£ £⎣ ⎦12

+mol moll l 310 H O 105

PH : PH : PH . PH .


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) .( 806 : :+

2 32H O H O +OH⎯⎯®¬⎯⎯

⎡ ⎤ ⎡ ⎤×⎣ ⎦ ⎣ ⎦⎡ ⎤⎣ ⎦

+3C22OH H OK =H O

:⎡ ⎤ ⎡ ⎤⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦2

+e C 2 3 = × H O = OH × H O ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

+

e 3 = OH × H O : 25 C

:

+

14e 3

= OH × H O =10⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦7

: (

( )3H O+: 1+

2 322H O H O +OH


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⎯⎯®¬⎯⎯ ( )3H O Cl+ -+ 3H O

+ (2) ( )OH - :⎡ ⎤⎣ ⎦

7OH 10 mol/l ( ( )OH -: ( )Na OH+ -+

(2) ( )OH - :

⎧⎡ ⎤⎪⎣ ⎦⎨⎡ ⎤⎪

⎣ ⎦⎩

7moll+

7moll 3OH >10H O


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( : ⎯⎯®2 +

3 2H O +OH 2H O

⎯⎯®+

2 312H O H O +OH (1)

3H O+ OH - C 25

14e

>10 .

81II

1 : fX maxX ) (

f maxX X < fZ 1 ffmaxXZ 1X= < . .2

: :

12a A + b B dC + l D⎯⎯®¬⎯⎯ ) 1 ( ) 2 ( 1V


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2V 2V 1V 1 2V V = .

) ( : : 1 mol 1 mol ) ( :13 2 5 3 2 5 22CH COOH C H OH CH COOC H H O⎯⎯®- + - +¬⎯⎯ :

0 0 1 1t 0 =23231313eqt :

) 82 1 2V V = .

) 1 ( ) 2 ( ) .(

3

: IUPAC : a A + b B d C + l D⎯⎯®- - - - - - ¬⎯⎯ a A + b B = dC + l D - - - - - - 4 ) Quotient (


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( )f : rf ) ([ ] [ ][ ] [ ]d lr

a bC DfA B×=× : rf ) Système ermé .(5 C

K : ) A, B, C, D ( : a A + b B dC + l D⎯⎯®¬⎯⎯ : [ ] [ ][ ] [ ]l d

C a b

D CKA B×=× :1

r Cf K < .1 mo

0 ,610 ,34

t ( h)n( mo

)

832 r Cf K > .3


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X f CK

[ ] [ ][ ] [ ]d l

C

a b C Df KA B×= =× :1 CK K .2

CK .3 4CK 10 > .4 1 mo

2H O 1 = ⎡ ⎤⎣ ⎦. :

C : 3 2 2 3CH COOH H O H O + CH COO -⎯⎯®+ ¬⎯⎯ :3 3C3

H O . CH COO

CH COOH+ -⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=⎡ ⎤⎣ ⎦III


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1

:

: : 1 ) 1890 ) ( Arrhenius :( ) ( H + . 2 ) 1923 ) ( Lewis :( ) ( . 3 ) 1923 ) ( Brouste et Lory :( ) . AH A H- +

⎯⎯® +¬⎯⎯ : :1

: ( )OH - . 842

: .3

:

B H B H+ ++ ⎯⎯®2 : : : 1 : 3H O+ ) ( :2 3

AH H O A H O- ++ ⎯⎯® + : ( )3 2 3 3HNO H O H O NO+ -+ ⎯⎯® + ( )22 4 2 3 4

H SO 2H O 2H O SO+ -+ ⎯⎯® + : 1 a = % 100 : a = 1 a 0 ³ ³ 1 a = ) ( 0 a = 2 :


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3H O+ 1 a 0 > > .

3

:

AH H O A H O

2 3- ++ + ® AH H O A H O2 3- ++ + ®

2

3H O , A , OH ,HO- -+⎡ ⎤ ⎡ ⎤⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤⎣ ⎦[ ]2

3H O , A , OH ,H O , AH- -+⎡ ⎤ ⎡ ⎤⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤⎣ ⎦ HCl H Cl

+ -⎯⎯® +3 4NH H NH+ ++ ⎯⎯®22 4 4H SO 2H SO+ -


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⎡ ⎤= ⎣ ⎦e3

( ) OHH O -+ ⎡ ⎤

=⎣ ⎦⎡ ⎤⎣ ⎦3H O A+ -⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

[ ]AH C A-

⎡ ⎤= - ⎣ ⎦

3H Oa 1C+⎡ ⎤⎣ ⎦

= =3H Oa 1C+⎡ ⎤⎣ ⎦= < ) :( : 1 : ) ( :2

B H O B H OH+ -+ ⎯⎯® +B

25 PH =14 l og (C ) C OH a=100% a=1- ⎫


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⎡ ⎤= ⎪⎣ ⎦⎬ ⎪⎭

OH OH+ -⎯⎯® +

( ) ( )22Ca OH Ca 2 OH+ -⎯⎯® + ( )Na OH Na OH+ -⎯⎯® +

2

: 2B H O B H OH+ -⎯⎯®+ +¬⎯⎯ B 0 a 1 < < 3 :

B H O BH2OH+ -+ + ®B H O BH2OH+ -+ +

23HO , BH , OH ,HO+ -+⎡ ⎤ ⎡ ⎤⎡ ⎤⎣ ⎦


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⎣ ⎦ ⎣ ⎦⎡ ⎤⎣ ⎦[ ][ ]23H O , B , OH ,H O , BH-+

⎡ ⎤⎡ ⎤⎣ ⎦ ⎣ ⎦⎡ ⎤⎣ ⎦ 86

PH

PH 14 logC C OH-

= + Ü =⎡ ⎤⎣ ⎦

PH 14 logC C OH-< + Ü ⎡ ⎤>⎣ ⎦

PH

310 H O- +⎡ ⎤= ⎣ ⎦) (PH 1410 OH+ - -⎡ ⎤= ⎣ ⎦

OH BH- -⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

PH310 H O- +


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⎡ ⎤= ⎣ ⎦) (PH 1410 OH+ - -⎡ ⎤= ⎣ ⎦OH BH

- -⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

[ ]B C BH +⎡ ⎤= -⎣ ⎦

OHa 1C-⎡ ⎤⎣ ⎦= =1 a 0 > >3 / : AH A- :

12 AH A H- +⎯⎯® +¬⎯⎯ :3 3CH COOH CH COO-

4 3NH NH+ 4

a ) / ( : :2 3


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AH H O A H O- +⎯⎯®+ +¬⎯⎯ / :

AH A-

a

:[ ]3aA H O

AH+⎡ ⎤ ⎡ ⎤×

⎣ ⎦⎣ ⎦=[ ]3a C 2A H O

. H OAH+⎡ ⎤ ⎡ ⎤×

⎣ ⎦⎣ ⎦= = ⎡ ⎤⎣ ⎦ :

a

.

a

.

a

. 875

aP

: aP :a a


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P

=

log

P

aa

10-=6

: a

a

P

. :2 31 + H O 1 H O +⎯⎯® +¬⎯⎯ [ ]3a1

1 H O

1 +⎡ ⎤ ⎡ ⎤×⎣ ⎦ ⎣ ⎦=2 32 + H O 2 H O +

⎯⎯® +¬⎯⎯ [ ]3a22 H O

2 +

⎡ ⎤⎡ ⎤×⎣ ⎦ ⎣ ⎦= a a2 1P P < a a


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1 2

< 7

PH aP

: :2 3AH H O A H O- +⎯⎯®+ +¬⎯⎯

PH

aP

:

a[ A ]PH P

log[ AH ]= + : 1 aPH P

> . 2 a

PH P

< . 3

aPH P

= . : 0,4 m 200 ml ( )HCOOH 0,2 mol/l PH 4 , 3 = . aP = 3,8 ( )HCOOH HCOO

-. : :( ) ( )2Na HCOO H O HCOOH Na OH+ - + -+ + ⎯⎯® + + 2 3HCOOH H O HCOO H O- +

⎯⎯®+ +¬⎯⎯ :+ PH 4,3 53[ H O ] = 10 = 10 = 10 :+

53H O = 5×10 mol/l


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⎡ ⎤⎣ ⎦1410e53

10OH 2 10 mol/l5 10H O

-- --+⎡ ⎤ = = = ×⎣ ⎦ ⎡ ⎤ ×⎣ ⎦ 3Na H O HCOO . OH+ + - -

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 88 Na HCOO+ -⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

+3

OH H O⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ 1VHCOO NaV 200- +⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ +

:[ ]aHCOOlog P

PHHCOOH-⎡ ⎤⎣ ⎦


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+ = [ ]

HCOO = 3,16HCOOH⎡ ⎤⎣ ⎦ :[ ] [ ]0

HCOOH HCOO HCOOH-⎡ ⎤+ =⎣ ⎦ :1 V 0, 4 200 0 , 21 HCOO3 ,16 V 200- × + × ⎛ ⎞ ⎡ ⎤+ × =

⎜ ⎟ ⎣ ⎦ +⎝ ⎠ :4 ,16 V 4 V 0 ,4 200 0 , 23 , 16 200 V V 200× × + ×× =+ + :400,526 V = 0,4V + 40 V =

0,126⇒ : V = 317 ml8 -

K a :

K a : :20

C aK

a 1×=- : 1 a :2

KK a C a


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C= × ⇒ = : AH AH :2 3AH H O A H O- +⎯⎯®+ +¬⎯⎯

3H O , AH , A+ - PH :

a[ A ]Log + P

=PH[ AH] :1 ( :[ ] [ ]

aAPH P log 0 AH AAH--⎡ ⎤⎣ ⎦ ⎡ ⎤= Ü = = ⎣ ⎦2 ( :[ ]

[ ]aAPH P

log 0 A AHAH--⎡ ⎤⎣ ⎦ ⎡ ⎤> Ü > >⎣ ⎦3 ( :

[ ] [ ]aAPH P log 0 A AHAH--⎡ ⎤⎣ ⎦


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⎡ ⎤< Ü < <⎣ ⎦

89 :

aP PH < A- .

aP

PH > AH .

aP

PH = AH A-.

: : . :n nI H I

- :1n 2 n 32 I H H O I H O 2 1- +⎯ ⎯®+ +¬⎯⎯

3H O+ OH - nI H .


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: aP

PH =

nI H nI - n nI I H-⎡ ⎤ = ⎡ ⎤⎣ ⎦⎣ ⎦

) .(AH [ ]AH A-⎡ ⎤⎣ ⎦ aP

1 + aP 1 -

AP [ ]AH A-⎡ ⎤⎣ ⎦ [ ]AH A-⎡ ⎤=

⎣ ⎦ [ ]AH A-⎡ ⎤= ⎣ ⎦ A


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- AH n nI H I -⎡ ⎤⎡ ⎤⎣ ⎦ ⎣ ⎦

aP 1 + aP

1 -AP

n nI I H-⎡ ⎤ ⎡ ⎤⎣ ⎦

⎣ ⎦ PH nI - nI H ) (

90 I

:1

: : . 2

: 3H O+ OH -

. :3OH H On n - + = B BOHC Vn


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1000- = 3A AH OC Vn1000+ = A A B B

C V C V =

. :1 (

1V 1C 2V .2 (

1S

moll 1C 0 , 8 = . 1S 1 l moll0 ,02 molL0,5

3200cm 1S . : n V C . n C .V = ) ( ( ) ( )1 1 2 2 C V C V = :1 12

2C VCV=2

1 2 2C =0,8 mol/l , V = 1 l , C = 0,02 mol/l :

2 3


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21 21C 0,02V = V = 1× = 2,5 × 10 l = 25 cmC 0,8 :31V 25 cm =

3mol molL L 1 1 2V 200 cm , C 0 ,8 , C 0 , 5 = = = 312 12C 0 ,8V V 200 320cmC 0 , 5= = × =

91 V :32 1V = V

V = 320

200 = 120cm :3V=120 cm3 : PH 7 :3H O OH+ -

⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ 14310 H O OH- + -⎡ ⎤ ⎡ ⎤= ×⎣ ⎦ ⎣ ⎦4

:

1M 2M ) ( 1M


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2M 2M 2M1M eq

E

( )eq eq eqE V , PH 7 = . : 1V 1C

22C =10 mol/l . PH : eqE .

. :

( )3eqE 20cm , PH 7 = :( )2 3 H O HCl H O Cl+ -+ ⎯⎯® + ( ) ( )

Na ,OH Na OH+ - + -⎯⎯® +0 2 4 6 8 10 121210 8 6 4 2


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⎣ ⎦ 147molL e 3710OH 10

H O . OH10-

- - + --⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = Ü =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Na+⎡ ⎤⎣ ⎦:31 2V V V 20 20 40cm = + = + =

:2 22 2C VNa V V C NaV+ +⎡ ⎤ ⎡ ⎤× = × ⇒ =⎣ ⎦ ⎣ ⎦23mol

L20 10Na 5 1040-+ -×⎡ ⎤ = = ×⎣ ⎦ :3mol

LNa 5 10+ -⎡ ⎤ = ×⎣ ⎦ Cl -⎡ ⎤⎣ ⎦


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M3M2MPHVeqV 93

:A A B B C V C V =3

: .2AH OH A H O- -+ ⎯⎯® + :[ ]A AH-

⎡ ⎤ =⎣ ⎦ 0//a[ A ]PH P [ Log ][ AH ]-= +

aP PH = eqBVV 2 2 2= =4 :

III

:1

: . .2

:


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3H O+ OH - A A B BC V C V =3

:

[ ]B BH +⎡ ⎤= ⎣ ⎦ AP

PH = . eqPH

1M2MPHVeqVeqV /2eqEA

PH = P

944 : : ( )6 , 2 4 ,4 - .

1 :

+H3O

5


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2 × 10 mol/l .

PH C 25 :⎡ ⎤ ⇒⎣ ⎦+

53PH=

log H O PH =

log 2×10 : PH= 4, 7

2 :

PH 8 ,6 =

C 25 .

OH - : :3PH log H O 8 ,6+⎡ ⎤= - =⎣ ⎦

:8,6 93H O 10 2 , 5 10+ - -⎡ ⎤ = = ×⎣ ⎦ :ee 33

H O OH OHH O+ - -+⎡ ⎤ ⎡ ⎤ ⎡ ⎤= × ⇒ =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤⎣ ⎦ 1469

10OH 4 10 mol / l2 , 5 10-- --⎡ ⎤ = = ×⎣ ⎦ ×


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:

6OH =4×10 mol/l⎡ ⎤⎣ ⎦ 3 : m ( )

OH PH 12, 6 m [ ]

39 , O=16, H=1 =

:

:PH 12,63H O 10 10+ - -⎡ ⎤ = =⎣ ⎦

14

1,4

2e

12,6+

3

10OH = = =10 =4×10 mol/l10H O⎡ ⎤⎣ ⎦ ⎡ ⎤⎣ ⎦ :3H O

OH+ + -

⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1M3M2MPHVeq

VeqV /2eqPH( ) ∆ 95 2


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OH 4 10 mol/l+ - -⎡ ⎤ ⎡ ⎤ = ×⎣ ⎦ ⎣ ⎦ :

OHM 56 , m = C M = × :2

m 56 4 10 2, 24 g-= × × = : m = 2,24 g 4 : PH 12 ,7 = .

( )2Ba OH : :+

PH +

12,73 3H O = 10 H O =10 mol/l⎡ ⎤ ⎡ ⎤

⇒⎣ ⎦ ⎣ ⎦ :

14 1,3 2

12,710OH = = 10 = 5×10 mol/l10⎡ ⎤⎣ ⎦

2

OH = 5×10 mol/l⎡ ⎤⎣ ⎦ :( )+22Ba OH Ba 2OH -⎯⎯® + 2 1

25 10 mol/l × X :

2 25×10X= =2,5×10 mol/l2 :( )


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22Ba OH = 2,5×10 mol/l⎡ ⎤⎣ ⎦ 5 : 1 (

20 . HCl moll0,2 3

100 cm

3H O+ PH 2 (

moll0,1 .

( )

3H O+ PH . : NaOH HCl ) .(1

5

2 31V 5 10 l = 10 5 cm20= × × = = :

[ ]HCl =0 , 2 mol/l n[ ]

5

5n = HCl × V = 0,2 × 5 × 10 = 10 mol 3100 cm 1l ) 10 ( : [ ]

4H3O 10 mol/l PH= 4 + = ⇒

962 - -5V 5 10¢ = × ( )OH-


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5 5n 5 10 0 , 1 0 , 5 10 mo

- -= × × = × ( ) ( )5 5 3H O 10 mo

+0,5 10 mo

OH+ -

×

3H O+ :5 5 53H O 10 0 , 5 10 0, 5 10 mo + - - -= - × = × 3

100 cm 1

+ 5 53H O = 0,5×10 ×10 = 5×10 mo

/

⎡ ⎤⎣ ⎦ : PH 4 , 3 =

: ) ( 5

( mol ) 10 0 , 5 2 × × ( )

OH- PH = 7

:

50 ,5 10 mol × ( )OH- 3100 cm ) 10 (

5OH = 5×10 mol/l⎡ ⎤⎣ ⎦ 14103510


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H O 2 105 10-+ --⎡ ⎤ = = ×⎣ ⎦ ×

10

PH = log[ 2×10 ]=9,7

: PH 9,7 = 6 : 35 cm 2 4H SO 3100 cm 310 cm 0,1 320 cm

. S 32 , O 16 , H 1 = = = . . . : 1 1 2 2( ) C V C V = () 1 122C V 20 0 , 1

C 0 , 2V 10×= = = ( ) :c c D DC V C V = 97 D Dcc

C V 100 0 , 2C 4V 5×= = = ( ) 1 : ( )2 41


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1 mol H SO mol2⎯⎯® () ( )2 44 mol H SO X ⎯⎯® : X 2 mol = m X M 2 98 196 g = × = × = 2 : = 4 × 49 = 196 g × = m 7 :

320 cm . 322 cm .

HCl 1 .

.

. Cl 35 ,5 , H 1 = :

:( ) ( )1 1 2 2C V HCl C V NaOH =

2 211C V 22 1C 1,1V 20×= = = ( )

: V 10 V

D D c c( ) C V C V = ( )

c10V 1,1C 11V×= = ( ) :C C = 11 mol/l HCl : m C M 11 36 , 5 401, 5 g = × = × = HCl 36 , 5 g 22,4 ⎯⎯®

401, 5 g V ⎯⎯®22, 4 401 ,5

V 246.436 , 5×= = : V = 246,4 l 8 :I

1 mol 1 mol 0 ,33 mol (1/3) . . 98


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.

rf . CK .II 1 mo 0 ,152 mo

. 1 mo

. .

: ( :13 2 5 3 2 5 22CH COOH + C H OH CH COOC H + H O⎯⎯®¬⎯⎯ ( :H2O CH

3COOC2H5 C2H5OH CH3COOH0 0 1 1

2/3 2/3 1/3 1/3 rf :2 2 5 20 0r3 2 50 0CH COOC H H Of

CH COOH C H OH× ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=× ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ :2 2 5 20 0(2/3)


42/114

CH COOC H H OV= =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ 3 2 50 0(1/3)CH COOH C H OHV

= = ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ( )( )2r2( 2/3) / v4f = = =41(1/3) / v

:rf 4 =CK : C rK f = :CK = 4II 1 mo

:2H O

3 2 5CH COOC H2 5C H OH3CH COOH0 0 1 1 2/3 2/3 1/3 1/3 2/3 2/3 4/3 1/3 1mo

:[ ] [ ]

[ ] [ ]r2/3 2/3f = 14/3 1/3= C rK f >

99


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13 2 5 3 2 5 22 CH COOH C H OH CH COOC H H O⎯⎯®⎯⎯®+ +¬⎯⎯ :

2H O3 2 5CH COOC H2 5C H OH3CH COOH2/3 2/3 4/3 1/3 1mol(2/3) + x (2/3) + x (4/3)

x (1/3)

x 0 ,152 mol :1

x = 0,1523 : x = 0,178 mol :2H O3 2 5CH COOC H2 5C H OH3CH COOH0,844 0,844 1,155 0,152

:( )( ) ( )2'C0 , 844 / V

40 ,152 /V 1,155 / V=

'

C C

= .


44/114

100

. :1 ( ( )A 54 , 54 % .

. 2 ( 0,2 mol 0,2 mol ( )B 10, 56 g . ( . ( ( )B .

( . ( ( )A ( C .3 ( 2 mol ( )B 1 mol . ( . ( . H = 1 g/mol , C = 12 g/mol , O = 16g/mol

:1 ( ( )A : n 2n 2C H O :54 , 54 12nn 4100 14n 32= Ü =+ :

4 8 2C H O2 (

( :3 7 3 7 2HCOOH C H OH HCOOC H H O⎯⎯®+ - +¬⎯⎯ ( :


45/114


46/114

3 7HCOOC H3 7C H OH - HCOOH0 0 2 1 t 0 =x x 2

x 1

x :( ) ( )2

Cx

= = 2,251

x 2

x :21,25 x

6,75 x + 4,5=0 et 0 x 1 £ £ II :2x = 0,78 mol ) ( 1x = 4,62 mol ) (

:2H O3 7HCOOC H3 7C H OH - HCOOH0,78 0,78 1,22 0,22 14,04 68,64 73,2 10,12 . .1 ( 0 ,1 mol

( )A 8,96 2CO . ( )A 4 .Û 102

.2 (

(A) (B) (C)

n 2n 2C H O (

(C) ( (B) .3 ( 14 ,8 g (A) 12 g (B) 13, 92 g (C) . ( ( C . ( 1 mol (A) 2 mol (B) .


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. 0V =22,4 mol , O=16 g/mol , C=12 g/mol , H=1 g/mol :1 (

( :n 2n 1C H OH+ - :

n 2n+1 2 2 23C H

OH + nO nCO + (n+1) H O2 ⎯⎯® 2CO :0V 8 , 86n 0 ,4 molV 22, 4= = =

n 2n 1 2C H OH nCO 1mol nmol 0 , 1mol 0 ,4 mol+ - ⎯⎯®⎯⎯®⎯⎯® n 4 = ( )A :4 9 4 10

C H OH C H O - º :

1 ) (3 2 2 2CH CH CH CH OH - - - - 2 2 ) ( 3 23CH C H CH OHCH- - -|

2 ) ( 3 2 3CH C H CH CH OH- - -|

2 2 ) (

3 3


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3OHCH C H CHCH- -|| 1032 (

( )

C

:n 2n 2C H O

M 29

116 g/mol = × = M 14 n 32 116 n 6 = + = ⇒ = ( )C :6 12 2C H O ( ( )B4 9 n 2n 2 6 12 2 2

(B) (A) (C)C H OH + C H O C H O + H O⎯⎯®¬⎯⎯ n 2 = 2 4 2C H O 3CH COOH ) ( 14 , 8(

) n 0 , 2 mol74= = 12( ) n 0 , 2 mol60= = 13 , 92() n 0 , 12 mol

116= =2H O6 12 2C H O4 9C H OH3CH COOH


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0 0 0,2 0,2 0,12 0,12 0,08 0,08

:0, 120 ,6 60%0,08= = = n ()R

n ()=

:5 11 2C3 4 9C H COOH . H O

2 , 25CH COOH . C H OH⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= =

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ H2O C5H11COOH C4H9

OH CH3COOH0 0 1 2 x x 1 x 2 x :( ) ( )2Cx

= = 2,25

2

x 1

x : x = 0,78 mol / H2O C5H11COOH C4


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H9

OH CH3COOH

0 0 1 2

0,78 0,78 0,22 1,22

104. : 30 g 30 g ( )A x mol .1

) ( .2

C

4 = . ( C

x y y ( )A . ( 1x = mol3 y ( )A . ( . ( 3 ( .

:1 ( :3 3 2CH COOH + R

OH CH COOR + H O⎯⎯®¬⎯⎯2 ( x y :30= = 0,5 mol60) (

nH20 CH3COOR R

OH CH3COOH0 0 y 1 / 2 = 0,5


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x x y

x 0,5

x :( ) ( )2Cx

=(1/2)

x y

x ( y : C

4 =

( )( ) ( )21/3= 4 y = 0,5 mol(1/2)

(1/3) y

(1/3) ⇒ : m m 30n = y = 0,5 = M = = =60 g/molM y 0,5

⇒ :(1/3) 2R = = = 67%(1/2) 3 3 2 2CH CH CH OH - - - 1 105 : 3 2 2 3

CH COO CH CH CH - - - - ) (3 ( : ) .(

) .( . . :1 ( 2,30 g 3 g 2 2,64 g ( . ( . ( CK .

2 ( 0 ,3 mo

0 ,5 mo

2 1 mo

.

. : +C HO HO


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C H3 CH CH3OHC HO CHOCH3CH

3+ O H2 : ( : : 2 , 3n 0 ,05 mo

46= = :

3n 0 ,05 mo

60= = : 2 ,64n 0 ,03 mo

88= = 0 0

05 , 0 05 , 003 , 0 03 , 0 02 , 0 02 , 0 106 nR 0 ,6 60%n= = =

( CK :C[ ] [ ]K 2 , 25[ ] [ ]= =


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2 ( ( ) ( . 1 0 0,5 0,31 + x x 0,5

x 0,3

x :( )( ) ( )C

x x + 1K = = 2,250,3 x 0,5 x :21 , 25 X 2 ,8 X 0 , 3375 00 X 0 , 3⎧ - + =⎪⎨ £ £⎪

⎩ :1x = 0,13 mol ) ( 2x = 2,1 mol ) ( 1,13 0,13 0,37 0,17 . : ( )A m 3 g =

1 litre 3V = 40 cm ( )A molLC 0 ,1 = . PH 1V .

E D

.

. . .

D a

.


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246851020 154, 78,7DE

2, 9 107 :

E .

D ( )aP

PH =

:A Bn n = ) (

3eqV 20cm = A A B BC V C V = :B BmolL AAC V 0, 1 20C 0 ,05V 40

×= = = :molL AC 0 ,05 = : m mM nn M= Û = gmol

3 gM 60 M0,05 mol= Ü =

:n 2n 2C H O : n 2 60 32 14n = Ü = + 3CH COOH .


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molLC 0 ,05 = molL 3H O C 0, 05+⎡ ⎤ = =

⎣ ⎦ PHmolL 3H O 10 0 ,05 PH 1 ,3+ -⎡ ⎤ = = ⇒ =⎣ ⎦

PH 2 , 9 = 2,9mol

L 3H O 10+ -⎡ ⎤ =⎣ ⎦ 2,9 1,3mol molL L 3H O 10 10+ - -

⎡ ⎤ = <⎣ ⎦ . : .

.3 2 3 3CH COOH H O CH COO H O- +⎯⎯®+ +¬⎯⎯

( )D . ( )D PH 4 ,7 = 5 4,7molL 32 10 10 H O


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- - +⎡ ⎤× = = ⎣ ⎦ 5molL 3H O 2 10+ -⎡ ⎤

= ×⎣ ⎦14

10emolL53

10OH 5 102 10H O

---+⎡ ⎤ = = = ×⎣ ⎦ ⎡ ⎤ ×⎣ ⎦ 108 3 3

H O OH CH COO+ - -⎡ ⎤ ⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3 3 CH COO H O OH- + -⎡ ⎤ ⎡ ⎤ ⎡ ⎤= -⎣ ⎦ ⎣ ⎦ ⎣ ⎦

5molL 3CH COO 2 10- -⎡ ⎤ = ×⎣ ⎦


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: PH = P A5molL 3 3CH COO CH COOH 2 10- -⎡ ⎤ = = × ⎡ ⎤⎣ ⎦⎣ ⎦

P

4,7aa aP

4 ,7

10 10- -= ⇒ = = :5

K 2 10-= ×

. :310cm 2mo

AC 10-= 340cm N OH

3mo

BC 5 10-= ×

PH : :3N , C

, OH , H O

+ - - +

: :34OH N 5 10 40n n 2 . 10 mo 1000


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- +--× ×= = =324C

H O10 10n n 10 mo

1000- +--×= = = :3 2H O OH 2 H O+ -+ ⎯⎯® 3

H O 1 mol 1mol OH+ -⎡ ⎤ ⎡ ⎤⎯⎯®⎣ ⎦ ⎣ ⎦) (2 410 mol x x 10 mol- -⎯⎯® Û = OH - :

4 4 42 .10 10 10 mol- - -- = :3 2A BV=V +V = 10 + 40 = 50 cm = 5 × 10 L : 10943

moll22 .10Na 4 105 10-+ --⎡ ⎤


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= = ×⎣ ⎦ × 43moll210Cl 2 10

5 10-- --⎡ ⎤ = = ×⎣ ⎦ ×43moll2

10OH 2 105 10-- --⎡ ⎤ = = ×⎣ ⎦ ×1412mol

l e 3 3310

H O OH H O 5 102 10-+ - + --⎡ ⎤ ⎡ ⎤ ⎡ ⎤= × ⇒ = = ×⎣ ⎦ ⎣ ⎦ ⎣ ⎦ × PH :

123PH log H O log 5 10 11, 3+ -⎡ ⎤= - = - × =⎣ ⎦ . :1 ( ( )


60/114


61/114

⎣ ⎦ 3moll 3H O 2 ,5 10+ -⎡ ⎤ = ×⎣ ⎦ OH

-⎡ ⎤⎣ ⎦ 110 1412emoll33

10OH 4 102 ,5 10H O-- --+⎡ ⎤ = = = ×⎣ ⎦ ⎡ ⎤ ×⎣ ⎦

6 5C H COO -⎡ ⎤⎣ ⎦ :6 5 3C H COO OH H O- - +⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ =⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3OH H O- +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ OH -⎡ ⎤⎣ ⎦


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3moll 6 5 3C H COO H O 2 , 5 10- + -⎡ ⎤ ⎡ ⎤= = ×⎣ ⎦ ⎣ ⎦ :6 5 6 5

C H COOH C H COO C-⎡ ⎤+ = ⎡ ⎤⎣ ⎦ ⎣ ⎦ :6 5 6 53 3mollC H COOH C C H COO 0 ,1 2 , 5 10 97 , 5 10

-- -⎡ ⎤= - ⎡ ⎤⎣ ⎦ ⎣ ⎦= - × = × :36 52C H COO2 , 5 10

a 2 , 5 10C 0 ,1- --⎡ ⎤ ×⎣ ⎦= = = × 2a 2 ,5 10 1-

= × <

K3 36 5 35

36 5


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C H COO H O2, 5 10 2, 5 10K = 6 ,4 10C H COOH 97 , 5 10- + - ---⎡ ⎤ ⎡ ⎤ × × ×⎣ ⎦ ⎣ ⎦

= = ×⎡ ⎤ ×⎣ ⎦5a a a

6 ,4 10 P

log

4, 2-= × Û = - =3 ( :6 5 6 5 2 3C H COO , C H COOH , H O , H O , OH , Na- + - +

4,2 5moll 3H O 10 6 , 3 10+ - -⎡ ⎤ = = ×⎣ ⎦ 149,8 10e

moll4, 23

10OH 10 1, 6 1010H O-- - --+⎡ ⎤

= = = = ×⎣ ⎦ ⎡ ⎤⎣ ⎦ Na+⎡ ⎤⎣ ⎦: ( )


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a B B BNa V V C V+⎡ ⎤ + =⎣ ⎦ B BA BC V

Na V V+⎡ ⎤ = ¬⎯⎯⎣ ⎦ +moll0 , 1 5Na 0 ,03315

+ ×⎡ ⎤ = =⎣ ⎦

111 :3 6 5Na H O OH C H COO+ + - -⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

+ = +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ moll 6 5Na C H COO 0 ,033+ -⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦moll 6 5C H COO 0,033

-⎡ ⎤ =⎣ ⎦ 6 5C H COOH ⎡ ⎤⎣ ⎦ 6 5 6 5


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C H COOH C H COO C-⎡ ⎤+ = ⎡ ⎤⎣ ⎦ ⎣ ⎦ A AmollA B

C V 10C 0, 1 0 ,066V V 15= = × =+ 6 5 6 5C H COOH C C H COO 0 ,066 0,033-⎡ ⎤= - = - ⎡ ⎤⎣ ⎦ ⎣ ⎦

moll 6 5C H COOH 0 ,033 = ⎡ ⎤⎣ ⎦ 6 5 6 5C H COOH C H COO -⎡ ⎤= ⎡ ⎤⎣ ⎦ ⎣ ⎦

6 5a6 5C H COOPH P

logC H COOH0-⎡ ⎤⎣ ⎦= +⎡ ⎤⎣ ⎦

log 1 0 =

aPH P 4 , 2 = = . : n 2n+3C H N % 24


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210- / ( )moll PH ) 25 C

(

PH 11, 4 = .1 2

.3

.4

aP

n 2n 3 n 2n 3C H NH C H N++ +. H = 1 / C = 12 / N = 14 /

12 11,410 3, 98 10- -× = :

: ( ) ( )n 2n 3M C H N 14n 17 g mol+ = + 112

( )% 100 14n 17% 24 14 n¬⎯⎯⎯ +¬⎯⎯⎯ :( )100 14 24 14 n 17 × = + 3 n 3 9C H N

3 2 13 2 2 2CH CH CH NH - - -23 2 13 3NHCH CH CH - -


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2 5 3C H N CHH- - 3 33CH N CHCH

- -

:3 9 2 3 9C H N H O C H NH OH+ -⎯⎯®+ +¬⎯⎯

a

P

3 9 3 9C H NH C H N+ 3 9 3 9a a3 9 3 9C H N C H NPH log P log P PHC H NH C H NH+ +

⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦- = Û + =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ 11,4 PH = 3 9C H NH +⎡ ⎤⎣ ⎦ :

12 11,4mol moll l 310 3 , 98 10 H O- - +⎡ ⎤× = = ⎣ ⎦ :e


68/114

e 33

OH H O OHH O- + -+⎡ ⎤ ⎡ ⎤ ⎡ ⎤= ⇒ =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤

⎣ ⎦ :143moll11,410OH 2, 5 1010-- --

⎡ ⎤ = = ×⎣ ⎦ :3 9 3C H NH H O OH+ + -⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ :3 12 3

gmol 3 9 2 ,5 10 10 3 , 88 10 2, 5 C H NH- - - +⎡ ⎤× = × - × = ⎣ ⎦3gmol 3 9C H NH 2 , 5 10+ -

⎡ ⎤ = ×⎣ ⎦ ) ( ) ( 2 ) ( 1


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) ( 113 : ) (3 9 3 9C H N C H NH C+⎡ ⎤+ = ⎡ ⎤⎣ ⎦ ⎣ ⎦

3 9 3 92 3 3mollC H N C C H NH10 2 ,5 10 7 , 5 10+- - -⎡ ⎤= - ⎡ ⎤⎣ ⎦ ⎣ ⎦

= - × = × aP

:3a37 ,5 10P 11 ,4 log 11,4 log 3 10 , 92 , 5 10-

-×= - = - =× . : . 2mollC 10-=( )

A : ( )B : .( )C : ( )D : . . PH :


70/114

1 ( ) ( ) ( ) ( )D C B A .2

) 1 ( .3

.4

a

( )HCOOH HCOO-.5

a

( )3 3CH COOH CH COO -

510 1 ,6-× : . 3 2,910 1, 26 10- -× = :1 ( : 1 2 3 4PH

7 2 12 2,9 1142moll 3PH 2 10 C H O- +⎡ ⎤= Ü = = ⎣ ⎦2 (

( )C :2emoll 3

OH C 10 H OOH- - +


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-⎡ ⎤ ⎡ ⎤= = ⇒ =⎣ ⎦ ⎣ ⎦ ⎡ ⎤⎣ ⎦1412moll 32

10H O 10 PH 1210-+ --⎡ ⎤ = = ⇒ =⎣ ⎦3 ( ( )B 7 PH = Ü .4 (

( )D 3C H O +⎡ ⎤> ⎣ ⎦ 2 PH 7 < < 1 2 3 4PH7 2 12 2,9 B A

CD2

:( ) ( ) ( )3 2Na ,OH H O ,Cl Na ,Cl H O+ - + - + -+ ⎯⎯⎯® +3

2 3HCOOH H O HCOO H O- +⎯⎯®

+ +¬⎯⎯ a :2 ,930,92HCOO H O10a 10 1


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C C 10- + ---⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = = = < . 2 3

HCl H O H O Cl+ -+ ⎯⎯® +232H O10a 1C 10+ --

⎡ ⎤⎣ ⎦= = = .4

a

( )HCOOH HCOO- :[ ]3a

H O HCOO

HCOOH+ -⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= HCOO -⎡ ⎤⎣ ⎦: 3 2, 9

mol moll l 31, 26 10 10 H O- - +⎡ ⎤× = = ⎣ ⎦ ( )14


73/114

11,1 12emoll2,93

10OH 10 7 , 94 1010H O-

- --+⎡ ⎤ = = = = ×⎣ ⎦ 1153 3OH HCOO H O HCOO H O OH- - + - + -⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ = ⇒ = -

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3 12 3mollHCOO 1, 26 10 7 , 94 10 1, 26 10( )- - - -⎡ ⎤ = × - × = ×⎣ ⎦ :

[ ] [ ]HCOOH HCOO C HCOOH C HCOO- -⎡ ⎤ ⎡ ⎤+ = ⇒ = -⎣ ⎦ ⎣ ⎦[ ]2 3 3mollHCOOH 10 1, 26 10 8 ,75 10- - -= - × = ×

a

:( )( )234a3


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1 , 26 10

1,78 108 ,75 10---×= = ××5

( ) ( )

a 3 a CH COOH HCOOH < . . . :

1 .1

2

PH ( )25 C

3, 1 a

P

/ 4 , 9 : [ ][ ]. .3

V mollC 0 ,1 = PH

4 , 9 . 3 23 2[ CH CH COO ][ CH CH COOH ]- V . :2 1,8 4 3,110 1, 58 10 10 7 , 9 10- - - -

× = × = g g gmol mol molH 1 , O 16 , Na 23 = = = : : 1162 5 2 2 5 3


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C H COOH H O C H COO H O- +⎯⎯®+ +¬⎯⎯ 2 5C H COO - 2

[ ][ ] : [ ]a Log P

PH ⎡ ⎤⎣ ⎦

Ü + =[ ]a Log PH P ⎡ ⎤⎣ ⎦ = - :[ ]PH P

3,1 4,9 1.8 2

a 10 10 10 1, 58 10 - - - -⎡ ⎤⎣ ⎦ = = = = × :2 5 3 2 2 5C H COO , OH , H O , H O , C H COOH

- - + :PHmoll 3H O 10+ -⎡ ⎤ =


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⎣ ⎦ 3,1 4moll 3H O 10 7 , 94 10+ - -⎡ ⎤ = = ×⎣ ⎦

ee 33

H O OH OHH O+ - -+⎡ ⎤ ⎡ ⎤ ⎡ ⎤= × Û =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤⎣ ⎦

1410,9 11moll3,110OH 10 1, 26 1010-- - --⎡ ⎤

= = = ×⎣ ⎦ 2 5C H COO -⎡ ⎤⎣ ⎦: : 3 2 5H O OH C H COO+ - -

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3 2 5H O OH C H COO+ - -⎡ ⎤ ⎡ ⎤ ⎡ ⎤- =


77/114

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ OH -⎡ ⎤⎣ ⎦ 3H O +⎡ ⎤⎣ ⎦

4moll 2 5C H COO 7 , 94 10- -⎡ ⎤ = ×⎣ ⎦ 2 5C H COOH ⎡ ⎤⎣ ⎦

: [ ]2 10 1, 58 - ⎡ ⎤⎣ ⎦× = 2 522 5

C H COO1, 58 10C H COOH--⎡ ⎤⎣ ⎦ = ×⎡ ⎤⎣ ⎦ 117

42 52moll 2 52 2C H COO7 , 94 10C H COOH = 5 101 ,58 10 1, 58 10


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- --- -⎡ ⎤ ×⎣ ⎦= = × ⎡ ⎤⎣ ⎦ × ×3

aP PH Û = :2 52 5 2 52 5C H COO1 C H COOH C H COOC H COOH--⎡ ⎤⎣ ⎦ ⎡ ⎤

= Û = ⎡ ⎤⎣ ⎦ ⎣ ⎦⎡ ⎤⎣ ⎦ :2 5 32 4 2mollC C H COO OH H O5 10 7 , 94 10 5 10

- - +- - -⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + =⎣ ⎦ ⎣ ⎦ ⎣ ⎦= × + × ×

: 31 0 ,05V 500 cm0 ,1

×Û = =( ) ( )C V( ) ( )C V = 31


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eq2V V 250 cm = = . ± 98 BAC ± . / ( )1 1A H A- ( )

2 2A H A- ( )1S ( )2S ( )3

S PH 25 C

:1PH = 3,9 2PH = 3,0 3PH = 3,0 10 10 PH :1PH = 4,4

2PH = 3,5 3PH = 4,01

( )3S .2

310cm ( )

NaOH 31V 1 ,0cm = ( )1S 32V 16 ,0 cm =


80/114

( )2S 33V 1,0 cm = ( )3S . . .

a1

a2

. 118 0,110 1, 26 = :

* 3C H O +⎡ ⎤= ⎣ ⎦ : ( )3S : C + 3

3[H O ] = 10 mol/l ( )3S : C ¢ 4moll 3H O 10+ -

¢⎡ ⎤ =⎣ ⎦ :PH3PH'3H O


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1010 1010H O+ --+⎡ ⎤⎣ ⎦ = Û =

¢⎡ ⎤⎣ ⎦ C10C =¢ PH'=PH+1 ( )3S

+

3moll 3C= H O =10⎡ ⎤⎣ ⎦2 ( : : n n = :( ) ( )

C V( ) ( )C V = 32moll 10 10C 101-

-×= = ( )1S ( )2S


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:( ) ( )C V( ) ( )C V = ( )123

moll S C1 10C 10V 10--× ×

= = = V :( )222moll S16 10

C 1,6 1010--×= = × a1 a2

:

[ ]3aA H O

AH- +⎡ ⎤ ⎡ ⎤×⎣ ⎦ ⎣ ⎦


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= A-⎡ ⎤⎣ ⎦ [ ]AH : 1A H :

3.9moll 310 H O 3 , 9 PH- +⎡ ⎤= Ü =⎣ ⎦ 4moll 3H O 1, 26 10

+ -⎡ ⎤ = ×⎣ ⎦ 1410,1 11emoll3,93 10

OH 10 7 , 94 1010H O-- - --+⎡ ⎤ = = = = ×⎣ ⎦ ⎡ ⎤⎣ ⎦

119 OH -⎡ ⎤⎣ ⎦ 3H O +⎡ ⎤


84/114

⎣ ⎦ +

+

4moll 3 3H O = OH + A A = H O =1,26×10⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤Û⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3

4

3

moll 1 1 1A + A H = C A H = 10

1,26×10 = 8,74×10 Û ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ( )244a411 , 26 10

= = 0 ,18 108 ,74 10---××× :3 5moll 2 a2

A H 15 10 ,

6 ,7 10- -= × = × ⎡ ⎤⎣ ⎦ 3moll 2A 10- -⎡ ⎤ =⎣ ⎦

a a1 2

< 2A H 1A H .


85/114

. : HCOOH ( )30 , 3 10,0 cm

( )molL0 ,02 0, 15

.1

.

2

PH : 3, 1 4 ,4 8 , 2 10, 0 6 ,0 7 ,6 .3

( )320 ,0 0, 1 cm

. .4

molLC 0 ,1 = PH 8 ,4 ) 25

.( 9 8,410 4 10- -× = . a

/ . :1 ( :2 3

HCOOH H O HCOO H O- +⎯⎯⎯®+ +¬⎯⎯⎯ 2H ONaOH Na OH+ -⎯⎯⎯® + 120

:( ) ( )2 HCOOH Na ,OH Na , HCOO H O+ - + -⎯⎯®+ +¬⎯⎯2 (


86/114

( )10 ,0 8 , 2 - .3 ( : n n = :( ) ( )C V

(

) (

)C V = : (C ) (V )20,0 0,15C = = = 0,3 mol/lV 10× :

C CC C∆ ∆= VV∆+

VV∆+ molL 0,02 0 ,1 0 , 3C 0, 3 0 ,050 ,15 20 10 , 3⎛ ⎞

∆ = + + =⎜ ⎟⎝ ⎠ ( )molL C 0 , 3 0 ,05 = ±4 ( - : :


87/114

2 3H O HCOOH HCOO N

OH H O- + - +

PH 8,4 9mo

mo

L L 3H O 10 10 3 , 98 10+ - - -⎡ ⎤

= = = ×⎣ ⎦9molL 3H O 3, 98 10+ -⎡ ⎤ = ×⎣ ⎦ 145 ,6 6

emolL8,43

10OH 10 2 , 5 1010H O-- - --+

⎡ ⎤ = = = = ×⎣ ⎦ ⎡ ⎤⎣ ⎦Na+⎡ ⎤= ⎣ ⎦ molL

Na 0 ,1+⎡ ⎤ =⎣ ⎦ :3H O Na OH HCOO+ + - -⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤


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+ = +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ :39 6molL HCOO H O Na OH 0 ,1 3 , 98 10

2,5 10 0,1- + + -

- -⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + -⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦= + × ×

121 [ ] C HCOO HCOOH-⎡ ⎤

= +⎣ ⎦ :[ ]molL HCOOH C HCOO 0 , 3 0 ,1 0 , 2-⎡ ⎤= - = - =⎣ ⎦ :[ ]

[ ]934aHCOO H O3 , 98 10 0 ,1

1 ,6 10HCOOH 0 , 2+ --⎡ ⎤

- × × ×⎣ ⎦= = = ×4a

1, 6 10-= × . .


89/114

4,6 g .1

.2

.

.

.

1 .

3

PH

2 ,4

25 C.

.

a

/ aP

.

2,4

3g g gmol mol mol

C=12 , H=1 , O=16 ,Log 6=0,8 , 10 =4×10 :1 ) .(2

( ) ( )3H O HCOO+ -. :+ 2 3HCOOH + H O H O + HCOO

⎯⎯⎯®¬⎯⎯⎯ ( )HCOO-. 122

mn

M= nCV= :gmol HCOOH4 ,6M 46 n 0 ,1 mol


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46= ⇒ = = :molL HCOOH0 ,1 molC 0, 11 L= =3 - - :+ -

2 3HCOOH+ H O H O + HCOO⎯⎯⎯®¬⎯⎯⎯ 2 32H O OH H O- +⎯⎯⎯® +¬⎯⎯⎯ :3

HCOOH, HCOO , OH , H O- - +

2 ,4 PH = 2,4molL 3H O 10+ -⎡ ⎤ =⎣ ⎦ :

+

33[H O ]=4 10 mol/l × OH -⎡ ⎤⎣ ⎦ : e 3 H O OH+ -⎡ ⎤ ⎡ ⎤= ×

⎣ ⎦ ⎣ ⎦ :1414 2 ,4e2,43

10OH 1010


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H O-- - +-+⎡ ⎤ = = =⎣ ⎦ ⎡ ⎤⎣ ⎦11,6 11

molLOH 10 0 , 25 10- - -⎡ ⎤ = = ×⎣ ⎦ :

11molL[OH ]=0,25×10 HCOO

-⎡ ⎤⎣ ⎦ : 3H O OH HCOO+ - -⎡ ⎤ ⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3

OH H O- +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

:3molL 34 10 H O HCOO- + -⎡ ⎤ ⎡ ⎤× = =

⎣ ⎦ ⎣ ⎦ :

3molL[HCOO ]=4×10 :[ ] [ ]


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HCOOH HCOO C HCOOH-⎡ ⎤+ = =⎣ ⎦ :[ ]3

3molLHCOOH C HCOO 0 ,1 4 10 =96 10

- -⎡ ⎤= - = - × ×⎣ ⎦ a

:[ ] ( )2333

a34 10 HCOO H O

0 ,166 10HCOOH 96 10- - +--⎡ ⎤ ⎡ ⎤ ×⎣ ⎦ ⎣ ⎦= = = ××

aP

:4a aP Log Log 1 ,66 10-⎡ ⎤= - = - ×⎣ ⎦ :aP

3,78 =

123 . :( 25 C

:

0.6Log 24 = 1,4 , 0,25 = 10 2


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molLC 10-= PH 10, 6 .I

1

( )3NH .2

3

.

II

1

( )a

( )4 3NH NH+ . 2

( )aP

.

3

:a1

( )3 3 3 2CH NH CH NH+- - 1110 1, 9-×

a2

( )4 3NH NH+ 1010 5 ,6-×

( )3NH ( )3 2CH NH - . :1

: 2


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molLC 10 , PH 10, 6-= 10,63H O 10+ -⎡ ⎤ =

⎣ ⎦

143,4

2emolL10,63

10molOH 10 < 10

l10H O-- --+⎡ ⎤ = = =⎣ ⎦ ⎡ ⎤⎣ ⎦

3 2 4NH H O NH OH+ -⎯⎯®+ +¬⎯⎯2

:3 2 4 3NH , H O, NH , OH , H O+ - + 3

:

10 ,6 10molL 3PH 10 ,6 H O 10 0 , 25 10+ - -⎡ ⎤= ⇒ = = ×⎣ ⎦ OH -


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⎡ ⎤= ⎣ ⎦ :ee 33

H O OH OHH O+ - -

+⎡ ⎤ ⎡ ⎤ ⎡ ⎤= × Û =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤⎣ ⎦143molL1010OH 0,4 10

0 , 25 10-- --⎡ ⎤ = = ×⎣ ⎦ × 1244 NH +

⎡ ⎤= ⎣ ⎦: :3 4OH H O NH- + +⎡ ⎤ ⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ :

34 3

molNH OH H O OH = 0,4 10l+ - + -⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= - ×⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3


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NH = ⎡ ⎤⎣ ⎦ :+ + 2 3 4molL 3 4 3 4 C= NH + NH NH = C

NH =10

0,4×10 = 96×10⎡ ⎤ ⎡ ⎤⇒ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎣ ⎦ ⎣ ⎦II 1 :[ ]3aH O

+⎡ ⎤ ⎡ ⎤⎣ ⎦⎣ ⎦

= :3 3a+4NH H O

NH+⎡ ⎤⎡ ⎤⎣ ⎦

⎣ ⎦=⎡ ⎤⎣ ⎦2

aP

( )4 3NH NH+ :a a

P

Log

= -3 33a+ +4 4NH H ONHP Log PH LogNH NH


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+⎡ ⎤⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦= - = -⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦4a

496 10P

10 ,6 Log 9, 24 10--×= - =×3

3NH 3 2

CH NH - a a2 1 > ( )3 2CH NH - ( )3NH . . : 1 litre PH

( )2 ,4 .1

.2

( )a

HCOOH HCOO-.3 ( )m a

P

HCOOH HCOO- 3, 8 . 3 2,4 2 1,410 4 10 , 10 4 10- - - -× = × :


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1 ( ( )HCOOH . 1252 3HCOOH H O HCOO H O- +⎯⎯⎯®+ +¬⎯⎯⎯

2 (

( )a

( )HCOOH HCOO- [ ]

3aHCOO H O

HCOOH+⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=3 ( m ( )1g :a aP Log = -[ ]a 3

P

Log H O Log HCOO Log HCOOH+ -⎡ ⎤ ⎡ ⎤= - - +⎣ ⎦ ⎣ ⎦

[ ]a HCOOP

PH LogHCOOH

-⎡ ⎤⎣ ⎦= -[ ] HCOOH :1 OH -


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⎡ ⎤= ⎣ ⎦ 1412emolL33

10OH = 2, 5 104 10H O-- --+⎡ ⎤ = = ×⎣ ⎦ ⎡ ⎤ ×⎣ ⎦

3H O +⎡ ⎤⎣ ⎦ :2,4molL 3H O HCOO 10+ - -⎡ ⎤ ⎡ ⎤

=⎣ ⎦ ⎣ ⎦

:[ ]a2,4Log HCOOH PH P

Log HCOO2 ,4 3 , 8 Log 10 1--⎡ ⎤- = - -

⎣ ⎦= - - =

[ ]Log HCOOH 1 = - [ ]1mol


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L HCOOH 10-= :[ ] [ ] C HCOOH HCOOH HCOO-⎡ ⎤

= = + ⎣ ⎦1 3

3molLC 10 4 10 = 104 10 mol/l- -= + × ×

( )m ( )1g :

gmol HCOOHM 46 = :1 mol 46 g0 ,104 mol m⎯⎯⎯®⎯⎯⎯® : m 4,784 g =. : 25 C

.

126

( )A HCOOH 13, 8 g 3 litre PH 2 ,4 = .1

.2

( )A .3

a

( )HCOOH HCOO-. : Log4 0 ,6 , 16 O , 12 C , 1 H = = = = :1

:2 3HCOOH H O HCOO H O- +


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⎯⎯⎯®+ +¬⎯⎯⎯2 ( )A : :gmoln m 13 ,8 gC

V M .V 46 3= = =× :molLC 0, 1 =3

a

( )HCOOH HCOO-

:2 3H O , HCOOH, HCOO , OH , H O- - + :

3a[ HCOO ] [H O ]

[HCOOH]

+= PH 3molL 3H O 10 4 10+ - -⎡ ⎤ = = ×⎣ ⎦ :3 3

H O OH HCOO H O HCOO+ - - + -⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + ⇒⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3mol


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LHCOO 4 10- -⎡ ⎤ = ×⎣ ⎦ :[ ] C HCOOH HCOO-

⎡ ⎤= + ⎣ ⎦ :[ ]3 3molLHCOOH C HCOO 0 ,1 4 10 96 10- - -⎡ ⎤= - = - × = ×⎣ ⎦

:[ ] ( )23 34a34 10 HCOO H O 1,67 10HCOOH 96 10- +

--⎡ ⎤ ⎡ ⎤ ×⎣ ⎦ ⎣ ⎦= = = ×× 127 . : 2 ,3 g 100% .1

.

2

3

( )A 3250cm . . . ) ( 1 a = PH ( )


103/114

A .

1 a > PH ( )A : Log 2 0, 3 = :1 :2 5 2 5 2C H OH O CH COOH H O + ⎯⎯® +2

:( ) 46 g 60 g2 ,3 g m⎯⎯®⎯⎯®A2 ,3 60m 3 g46×= =

3

( )A :molLm 3C= = =0,2M .V 60×0,25 :3 2 3 3CH COOH H O CH COO H O- +⎯⎯⎯®

+ +¬⎯⎯⎯ 1 a = :( )3 3H O C H O 0 , 2 PH Log 0 , 2+ +⎡ ⎤ ⎡ ⎤= ⇒ = ⇒ = -⎣ ⎦ ⎣ ⎦2PH Log 1 Log 2 PH 0 ,710

= - = - ⇒ = - 1 a > PH . . 3NH molL0 ,1 .


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3BV 20 cm = 1M2M3MeqE

(∆ )9 ,35 , 319 , 23V(cm )PH 128 AV

: ) .( . . . aP a

. :

:( )3eqE 19 , 2 cm , PH 5 ,3 = :A A B BC V C V = :A ABBC V

CV= 3B30 ,1 19 , 2 10C20 10-


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-× ×=× 2molL BC 9 ,6 10-= ×

:( )3 3 4 2 NH H O NH H O+ ++ ⎯⎯® + PH 5 , 3 = 5 ,3 PHmolL 3 3H O 10 10 H O+ - - +

⎡ ⎤ ⎡ ⎤= Ü =⎣ ⎦ ⎣ ⎦ 148 ,7emolL5 ,33

10OH 10

10H O-- --+⎡ ⎤ = = =⎣ ⎦ ⎡ ⎤⎣ ⎦ 3

OH H O- +⎡ ⎤ ⎡ ⎤<⎣ ⎦ ⎣ ⎦ . aP D ( )


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3eqE 9 ,6 cm , PH 9 , 3 = aP

PH = P

9 ,3 10aa

10 10 5 10

- - -= = = × . . ( )310 cm ( )1A H

129

( )NaOH ( )molL0 , 1 ( )2A H ) .( PH ( )1

n ( )1A H .1 : ( )2n ( )2A H .

( )aP

( )1 1A H A- ( )


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2 2A H A- .2

3H O +⎡ ⎤⎣ ⎦

Na+⎡ ⎤⎣ ⎦ ( )35 cm ( )2A H . :

0,8

10 = 6,3 :1 ( )2A H ( )NaOH :32 n2n V 10 cm , PH 8, 4⎛ ⎞

= =⎜ ⎟⎝ ⎠. - ( )

PK ) / ( e1V V

2 = PH ( )

PK ( )

PH PK =e


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1V V2⎛ ⎞=⎜ ⎟⎝ ⎠ : ( )1A H .

3 3eeVV 8 cm 4cm2= ⇒ = 6,2=PH( )34cm : ( )

2A H .3 3eeVV 10 cm 5 cm2= ⇒ =2A H3V(cm )

PH22246810121n2n

1A H8,42 46 81012 1304,2=PH


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( )35 cm 16,2=PK

24,2=PK-

PK ( )

2A H ( )1A H .2 - - 3H O +⎡ ⎤⎣ ⎦ ( )

2A H : : 4,2=PH PH310 H O- +⎡ ⎤= ⎣ ⎦ :4 ,23

10 H O- +⎡ ⎤= ⎣ ⎦5molL 3H O 6 ,3 10+ -⎡ ⎤ = ×⎣ ⎦

Na+ : 3V 10 5 15 cm = + = :3NaOH3n =0,1 mol 1000 cm


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n 5 cm⎯⎯®¢ ⎯⎯®45 0 , 1n 5 10 mol1000 -×¢

= = × Na+ Na+ 4molL3n 5 10Na 0 ,033V 15 10

-+-¢ ×⎡ ⎤ = = =⎣ ⎦ × . :3500cm 3 g

n 2 n 1C H COOH+ - ( )25 C

320 cm

molL0 , 2 PH .1

: PH . .2 a


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P

.3

. : PH : PH ( )V 0 = PH218,5

103V(cm )2,7 131 PH 2 ,7 = . : A A B BC V C V = :( )3

n 10 cm 8 , 5 , :molL A10 0 , 2C 0 ,120×= = : 2 ,7molL 3

H O 10 2,7=PH+ -⎡ ⎤ = Û⎣ ⎦ 1moll 310 H O- +⎡ ⎤>

⎣ ⎦ . 2 : n 2 n 1C H COOH+ - ⎡ ⎤⎣ ⎦ . :


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2 ,7molL 32,7=PH H O 10+ -⎡ ⎤Û =⎣ ⎦3molL 3

H O 1, 99 10+ -⎡ ⎤ = ×⎣ ⎦ :14e33

10OH1, 99 10

H O---+⎡ ⎤ = =⎣ ⎦ ⎡ ⎤ ×⎣ ⎦ 11mol

LOH 0 ,5 10- -⎡ ⎤ = ×⎣ ⎦ :3 n 2 n 1H O C H COO OH+ - -+⎡ ⎤ ⎡ ⎤ ⎡ ⎤= +

⎣ ⎦ ⎣ ⎦ ⎣ ⎦n 2 n 1 3 3 C H COO H O OH H O- + - ++⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= -⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦


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3molL n 2 n 1C H COO 1 , 99 10- -+⎡ ⎤ = ×⎣ ⎦ :n 2n 1 n 2 n 1

C C H COOH C H COO-+ +⎡ ⎤= + ⇒ ⎡ ⎤⎣ ⎦ ⎣ ⎦n 2 n 1 n 2n 13 3molLC H COOH C C H COO0 ,1 1 , 99 10 98 10

-+ +- -⎡ ⎤= - ⎡ ⎤⎣ ⎦ ⎣ ⎦= - × = ×3molL n 2n 1C H COOH 98 10-

+ = × ⎡ ⎤⎣ ⎦ 3molL98 10-× .

aP

:

aPH P

= 4 ,8 PH = aP

4 ,8 = . : 132 :nC n C V 0 ,5 0, 1 0,05 mol


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V= Û = × = × = :m mn MM n= ⇒ = :gmol3M 60

0 ,05= = : 14n 46 60 n 1 + = Û = :3CH COOH .

3829787 توازن مجموعة كيميائية الدرس مع تمارين مصححة وفق...

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Transcript of 3829787 توازن مجموعة كيميائية الدرس مع تمارين مصححة وفق...