كتاب التطورات الرتيبة

‫א א‪ ª‬א – א א‬ ‫‪2007‬‬ ‫د رة‬ ‫‪:‬‬ ‫‪(F) :‬‬ ‫‪. 1 µF = 10-6 F :‬‬ ‫) ‪( µF‬‬ ‫‪. 1 nF = 10-9 F :‬‬ ‫) ‪( nF‬‬ ‫‪. 1 pF = 10-12 F :‬‬ ‫) ‪( pF‬‬ ‫)‪: « Q = f (UAB‬‬ ‫»‬ ‫‪:‬‬ ‫‪d‬‬ ‫‪C = 0,5×10-6 F = 0,5 µF‬‬ ‫–‬ ‫(‪d‬‬ ‫)‪: S‬‬ ‫(‬ ‫)‪: d‬‬ ‫‪ε = ε0 . εr‬‬ ‫–‪7‬‬ ‫‪.S‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪:‬‬ ‫◉‬ ‫◉‬ ‫ـــــ‪S‬ـــ ‪C = ε‬‬ ‫‪d‬‬ ‫‪S‬‬ ‫«‬ ‫)‪» : (ε‬‬ ‫‪:‬‬ ‫‪ε0 :‬‬ ‫‪:7 :‬‬ ‫‪ε0‬‬ ‫=‬ ‫ــــــــــ‪1‬ـــــــــ‬ ‫=‬ ‫‪8,85×10-12‬‬ ‫‪F.m-1‬‬ ‫‪36π × 109‬‬ ‫‪.‬‬ ‫‪εr‬‬ ‫‪C‬‬ ‫=‬ ‫ـــ‪S‬ـ‪.‬ـــ‪ε‬ـ‬ ‫=‬ ‫ــ‪S‬ـــ‪r‬ــ‪ε‬ـ‪.‬ــ‪0‬ــ‪ε‬‬ ‫=‬ ‫ــــــــــ‪S‬ـ‪.‬ــ‪r‬ــ‪ε‬ــــــــ‬ ‫=‬ ‫×‪8,85×10-12‬‬ ‫ـــ‪S‬ـ‪.‬ـ‪r‬ــ‪ε‬ـ‬ ‫‪:‬‬ ‫‪:‬‬ ‫‪d‬‬ ‫‪d‬‬ ‫‪36π × 109. d‬‬ ‫‪d‬‬ ‫‪:‬‬ ‫– ‪εr‬‬ ‫‪80 24 7 4 – 6 2,2 1 εr‬‬ ‫‪U1 U2 U3‬‬ ‫‪B‬‬ ‫‪:‬‬ ‫‪.2.1.1‬‬ ‫‪C1 C2 C3‬‬ ‫‪U‬‬ ‫‪:‬‬ ‫(‬‫‪A +Q -Q +Q -Q +Q -Q‬‬ ‫‪U G1 D1 G2 D2 G3 D3‬‬ ‫‪8‬‬ ‫–‬ ‫‪BA‬‬ ‫⊖⊕ ◉◉‬ ‫‪. C3 C2 C1‬‬ ‫‪Céq‬‬ ‫) (‪.‬‬‫‪A‬‬ ‫‪+Q -Q‬‬ ‫‪B‬‬ ‫‪( ) D1G2‬‬ ‫‪U‬‬ ‫‪.U‬‬ ‫‪– Q C1 D1‬‬‫⊖⊕ ◉◉‬ ‫‪.+Q‬‬ ‫‪C2 G2‬‬ ‫‪:8 –Q‬‬ ‫‪.+Q‬‬ ‫← ‪+Q‬‬ ‫‪.Q‬‬ ‫←‪: :‬‬ ‫( ‪. U = UAB = U1 + U2 + U3 :‬‬ ‫‪. Q = C3U3 Q = C2U2 Q = C1U1‬‬ ‫←)‬ ‫‪⑴ ..........‬‬ ‫‪) :‬ـــ‪1‬ــ ‪+‬ـــ‪1‬ــ ‪+‬ـــ‪1‬ــ ( ‪= Q‬ــــ‪Q‬ـ ‪+‬ــــ‪Q‬ـ ‪+‬ــــ‪Q‬ـ = ‪ ⇐ U‬ـــ‪1‬ــ ‪+‬ـــ‪1‬ــ ‪+‬ـــ‪1‬ــ = ــــ‪U‬ـ‬ ‫‪Q‬‬ ‫‪C1‬‬ ‫‪C2‬‬ ‫‪C3‬‬ ‫‪C1‬‬ ‫‪ =2‬ـ‪C‬ـــ‪U‬ـ‬ ‫‪ 3‬ـ‪C‬ــ‪1‬ــ‬ ‫‪C1‬‬ ‫‪2‬ـــ‪C‬ـ‪Q‬ـ‬ ‫‪C3‬‬ ‫⑵‬ ‫‪..........‬‬ ‫= ‪Céq‬‬ ‫‪U‬‬ ‫‪Q Céq‬‬ ‫⇐‬ ‫‪:‬‬ ‫←‬ ‫‪51‬‬

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ ‫ــ‬1‫ــ = ـــ‬1‫ ـــ‬+ ‫ــ‬1‫ ـــ‬+ ‫ــ‬1‫ـــ‬ Céq C1 C2 C3 :⑵ ⑴ . : . Q1 C1 :(Qt U C2 ◉B Céq U A◉ Q2 C3 A Qt ◉B 9 – : Q3 :9 ◉ U Q3 = C3U Q2 = C2U Q1 = C1U : Qt = Q1 + Q2 + Q3 = C1U + C2U + C3U = U( C1 + C2 + C3) ‫ـ‬Q‫ـــ‬t‫ـ‬ = C1 + C2 + C3 .......... ⑴ : U ‫ـ‬Q‫ـــ‬t‫ـ‬ Céq = U .......... ⑵ : Céq : ⑵⑴ Céq = C1 + C2 + C3 .: uC R : .3.1.1CV : – : 10 : – : U0 = 30 V ( ) C = 100 µF : . R = 300 kΩ . 10 – w = 10 kΩ/V (150 ) 30 V uC30 V . i = ‫ـ‬u‫ــ‬C‫ــ‬ (U0 = 30 V) R ←: t=0 . : 150 s : uC (V) . ←30 – 11 i(t) = u‫ـــ‬C‫ــ(ــ‬t‫ ـ)ـ‬: – R20 – uC (t) : ‫ا ن‬ t (s) 0 15 30 60 90 120 150 uC (V) 30 18,2 11 4 1,5 0,55 ε i = uC/R (µA) 100 61 37 13,5 5 1,8 ε'10 – t (s) uC (t) : 110 15 30 60 90 52

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ f1 – (Q0) +++ +++ A 12 –K (f3 f2 f1) (R) :U0 C B R U0 K --- --- f2 + Q0 = CU0 : - Q0 = - CU0 : A f3 ◉ K ◉ .« B » Q0 ← K : 12 I0 = ‫ـ‬U‫ـــ‬0‫ــ‬ : U0 R . . {f1 R f2 K f3} A B ( ) : 13 q f1 –K . . (q) i . ++ ++ (q ) uC A (uC ) i = ‫ـ‬u‫ــ‬C‫ــ‬ uC C R - - e-- B f2 R - ‫و‬ ‫ا‬ ‫ت‬ f3 ◉ ◉ . 11 i ~ uC ~ q : K K : 13 q = 0 ⇒ uC = 0 ⇒ i = 0 : BA i : B (+) A : ⊕ A R Ri i uC q . .( ) uC B . 14 – – ⊖ ∀t , i > 0 : – ∀t , uC > 0 : A B : 14 ∀t , q > 0 : C.uC ( ) : : ) uC – Ri = 0 ⇐ |dq| dt . |dq| : |dq| () dqi = - ‫ــ‬d‫ــ‬q‫ ــــ‬: : q i = ‫ـ|ـ‬d‫ـــ‬q‫ ـ|ــ‬: ( dt uC = ‫ـ‬q‫ ـــ‬: dt ‫ـ‬q‫ـــ‬ ‫ــ‬d‫ــ‬q‫ــــ‬ C C dt + R = 0 » ‫ــ‬d‫ــ‬q‫ ــــ‬q : .« dt q = Q0 e – t /RC : . (s) « » τ = RC : : – ‫ـ‬q‫ـــ‬ ‫ــ‬Q‫ــــ‬0‫ــ‬e‫ــ‬C‫ــ–ــ‬t‫ــ‬/‫ـ‬R‫ــ‬C‫=ـــ‬U0 uC = C = e – t /RC : i = ‫ـ‬u‫ــ = ـــ‬U‫ــــ‬0‫ــ‬e‫ــ–ــــ‬t‫ــ‬/‫ـ‬R‫ــ‬C‫ = ـــ‬I0 e – t /RC : RR 53

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬( I0 U0 Q0 ) :. t ↦ i t ↦ uC t ↦ q : :– . U0 = 10 V R = 2 MΩ C = 4 µF : : (1 : Q0 = CU0 = 40 µC : τ = RC = 8 s : t (s) 0 4 8 16 24 32 40 I0 = U‫ـــ‬0‫ـ‬ = 5 µA t/τ 0 0,5 1 2 3 4 5 R e-t/τ 1 0,607 0,135 q = 40 e – t /8 : (µC)q (µC) 40 24,28 0,368 5,4 0,050 10 6,07 14,78 1,35 2 0,018 0,007 uC = 10 e – t /8 : (V)uC (V) 5 3,04 3,68 0,68i (µA) 1,84 0,5 0,72 0,028 i = 5 e – t /8 : (µA) .e 0,25 0,18 0,007 : (2 0,09 0,004 : (3 ← e –t/τ = e -1 = ‫ـ‬1‫ـــ‬ q (µC) e t=τ:40 – ‫[ ؛‬t ↦ ∞] ⇒ [e –t/τ ↦ 0] : e –t/τ :30 – Q0 1 % 5τ : ‫ز ا ا‬ .20 – . i uC 5 τ tT10 – :0 8 16 t (s) τ = 1×10-12 = 10-12 s ⇐ C = 1 pF R = 1 Ω : 24 tT = 5 τ = 5×10-12 s : .← q (t) : 15 τ = 107×10-3 = 104 s ⇐ C = 1000 µF R = 10 MΩ : (13 heures ) tT = 5 τ = 5×104 s : .← : . « la durée relative : »x x = ‫ــ‬t‫ ــ‬: τ ‫ــ‬q‫= ــــ‬ ‫ـ‬u‫ـــ‬C‫= ــ‬ ‫ـــ‬i‫ـــ‬ :y y= Qx0 U0 I0 = e – x : . « variables réduites : y . » «U0 Q0 : » y=e–x : . τ I0 i (t) uC (t) q (t)i (µA) q (µC) uC (V) y . 16 _ Ox Oy τ I0 U0 Q0 5 – 40– 10 – 1– tT = 5 τ : : ‫ـ‬τt‫ــ‬ : ‫ـ‬Q‫ـــ‬0‫⇐ ــ‬ 10 . tT = 2,3 τ ⇐ = ln 10 ⇐ e –t/τ = 10 ⇐ q = 10% . tT = 4,6 τ ⇐ ‫ـ‬t‫ــ‬ = ln 100 ⇐ e –t/τ = 100 ⇐ q = ‫ــ‬Q‫ـــ‬0‫ـــ‬ ⇐ 1% τ 100 ⇐ q = ‫ـــ‬Q‫ـــ‬0‫ــــ‬ . tT = 6,9 τ 1000 ⇐ 0,1%00 00 y = e -x : 16 0 1 2 3x 8 16 24 t (s) 54

‫ א – א א‬ª‫א א‬ 2007⊕ R V Ri ‫د رة‬ : : :–U0 C uC «» R = 300 Ω C = 100 µF ⊖ U0 . U0 = 30 V 17 – . uR = Ri = U0 – uC : uC : 17 . t=0 U0 = 30 V : . .← : . uC (V) . (150 ) 150 s . U0 .←30 – uC ( t ) :20 –10 – t (s) 0 15 30 60 90 120 150 uR (V) 30 18,2 11 4 1,5 0,55 ε 0 30 i (µA) 100 61 37 13,5 5 1,8 ε' uC (t) uC (V) 0 11,8 19 26 28,5 29,5 30 : i(t) t (s) uC ( t ) . 11 – 90 . 18 – q (t) = C.uC (t) : 60 : 18 . uC ( t ) : (R) ( ) (G) . 19 – f1 K (f4 f3 f2 f1) K :← P●G R . uC U0 N● A f2 I0 = ‫ـ‬U‫ـــ‬0‫ــ‬ : R f4 B C A ( ): ← f3 ( ◉K ◉ B⊕ : 19 ( ) :⊖ )B A B {f3 K f4 G f1 R f2} f1 20 . A : P● i (+ q) AU0 G R Ri . : (q) . (– q) B N● (q ) uC A ++ +f+2 . e- ‫و ت‬ ‫ا‬C Ri = U0 – uC : i(t) B- - - - uC . (i) ( uC ~ q ) : ← : ← f4 ◉ K ◉ f3 . K : 20 55

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ vA = vP ⇒ uC = U0 vB = vN uC = U0 ⇒ i = 0 : QT = C.U0 : : i i uC q . . 21 – ← ⊕ R Ri U0 – Ri – uC = 0 : ( )U0 ⊖ ) : uC = ‫ـ‬q‫ـــ‬ ← A dq > 0 : dt C ( C uC B i = ‫ـ‬d‫ــ‬q‫ ـــ‬: ( ) dt : 21 U0 – R ‫ـ‬d‫ــ‬q‫ـ – ـــ‬q‫ =ـــ‬0 : dt C : q = QT( 1 – e – t /RC ) « » τ = RC : x↦y=(1–e–x) : t↦q: ∴ . (s) :– uC = U0( 1 – e – t /RC ) ⇐ uC = ‫ـ‬q‫ = ـــ‬Q‫ــــ‬T‫ـــ(ــ‬1‫ـــــ–ـــ‬e‫ــ–ــــ‬t‫ــ‬/‫ـ‬R‫ــ‬C‫ = ـ)ـــ‬U0( 1 – e – t /RC ) : CC i = I0 e – t /RC ⇐ i = ‫ـ‬U‫ـــ‬0‫ـــ–ـــ‬u‫ـــ‬C‫ = ــ‬U‫ــــ‬0‫ــــ–ــ‬U‫ــــ‬0‫ـــ(ـ‬1‫ــــ–ــــ‬e‫ــ–ــــ‬t‫ــ‬/‫ـ‬R‫ــ‬C‫ = ـ)ـــ‬U‫ـــ‬0‫ـ‬e – t /RC : R RR :– . U0 = 10 V R = 2 MΩ C = 4 µF : : (1t (s) 0 4 8 16 24 32 40 . τ = RC = 8 s : (µC)t / τ 0 0,5 1 2 3 4 5e-t/τ 1 0,607 0,368 0,135 0,050 0,018 0,007 q = 40( 1 – e – t /8 )1 – e-t/τ 0 0,393 0,632 0,865 0,950 0,982 0,993 uC = 10( 1 – e – t /8 ) : (V)q (µC) 0 15,72 25,28 34,60 38 39,28 39,72 i = 5 e – t /8 : (µA)uC (V) 0 3,93 6,32 8,65 9,5 9,82 9,93 : (2i (µA) 5 3,04 1,84 0,68 0,25 0,09 0,004 (3 : uC (V) i (µA) . 22 _ : uC (t)10 – 5– . 23 _ : i (t) : 235– 2,5 – i (t) : 22 t (s) uC (t) 0 8 16 24 t (s)0 8 16 24 56

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ : uC (V) : – U0 : (22،23 – ) i(t) = I0 e – t /RC uC(t) = U0( 1 – e – t /RC )0,63U0 ‫ـ‬d‫ــ‬u‫ـــ‬C‫ــ‬ = ‫ـ‬U‫ــــ‬0‫ــ‬ = ‫ـ‬U‫ــــ‬0‫ــ‬ : uC (0) = 0 ⇐ t = 0 : dt RC τ uC (τ) = 0,63 U0 ⇐ t = τ : uC = U0( 1 – e – t /RC ) = U0( 1 – e – t /τ ) : t (s) i (0) = I0 ⇐ t = 0 : 0 τ = RC ًτ : 24 . 24 – i (τ) = 0,37 I0 ⇐ t = τ : . 25 – : τ = RC i (A) uC (t) ... uC (τ) = 0,63 U0 : uC (t) I0 ... i (τ) = 0,37 I0 : – i = I0 e – t /RC t=0 : (...) – = I0 e – t /τ t=τ : (240,37I0 τ = RC t (s) : uC = U0 i (t) ً τ : 25 uC(t) = U0( 1 – e – t /τ ) : 0 ) uC (t) = at : ( )a= ‫ـ‬d‫ــ‬u‫ـــ‬C‫ــ‬ = U‫ــــ‬0‫ـ‬ e – t /τ : dt τ . uC (τ) = U‫ــــ‬0‫ـ‬ τ = U0 : t=τ: ⇐ a= U‫ــــ‬0‫ـ‬ : τ τ :: . Q = 5 mC ( ): C = 5 µF : )τ . R = 0,5 MΩ . (RC : (1 . i (t) (2 (3 : (1 : (RC) τ = 2,5 s ⇐ τ = 5 × 10-6 × 0,5 × 106 = 2,5 s ⇐ τ = RC i(t) = I0 e – t /RC = U‫ــــ‬0‫ ـ‬e – t /τ : (2 R ∴ U‫ــــ‬0‫ـ‬ ‫ــ‬1‫ــ×ـــ‬1‫ـــ‬0‫ــ‬3‫ = ـــ‬2 × 10-3 A = 2 mA ‫ـ‬Q‫ــــ‬ ‫ــ‬5‫ـــ×ــ‬1‫ــ‬0‫ـــ‬-‫ـ‬3‫ــ‬ I0 = R = 5×106 ⇐ U0 = C = 5×10-6 = 1 000 V : i(t) = 2 e – t /2,5 ( mA ) : ( ) i = 1 mA ⇐ uC = uR = U‫ــــ‬0‫ = ـ‬500 V : q = ‫ـ‬Q‫ ــــ‬: t (3 2 2 ∴ t = RC ln2 ⇐ - ‫ــ‬t‫ = ـــ‬ln 0,5 ⇐ e – t /RC = ‫ـ‬1‫ = ــ‬0,5 : RC 2 . t = 1,74 s ⇐ t = 2,5 × 0,69 = 1,74 s 57

‫–א א‬ ‫ א‬ª‫א א‬ 2007 ‫د رة‬ : : ◉ 1 2⊕ K ◉ ◉ ◉ ◉◉ . 26 – ◉ – ◉ C = 1 000 µFE ◉V ◉C L (e = 1,5 V) . ◉⊖ . : 26 . ← –. U = 4,5 V ( 2 200 µF 1 000 µF 100 µF ) ← .. q = CU : UC : AB u D : U C+q –q 27 – . U volts 0u u (t)U ‫ا‬0 Hq : u = vA – vB : OQ . qB = – q < 0 qA = + q > 0 q q+dq B dq A : 27 . dqB = – dq dqA = + dq : u dq dW = u.dq : . u.dq ←) . (27 – ) u = ‫ـ‬q‫ـــ‬ u (q) C .( CU 0 U0 W = ‫ـ‬1‫ ــ‬QU : OHD 2 ‫ـ‬Q‫ ـــ‬U C ‫ـ‬1‫ ــ‬CU2 : CU Q ‫ـ‬1‫ ــ‬Q‫ـــ‬2‫ ـ‬: ← (S.I) 2 Q‫ـــ‬2‫= ـ‬ 2C C Ep = ‫ـ‬1‫ــ‬QU = ‫ـ‬1‫ــ‬ ‫ـ‬1‫ــ‬CU2 : 2 2 2 (C) Q (V) U (F) C Ep (J) . (J)E0− : (t½) . 28 – ) ... Ep = ‫ـ‬1‫ ــ‬C.u2(t) :0,5E0− (t 2 – 0 t½ u(t) =U0 e – t /RC : t – t (s) ‫ـ‬1‫ ــ‬CU02 : 28 Ep = 2 e –2t /RC ⇐ u2(t) =U02 e –2t /RC : 58

‫ א – א א‬ª‫א א‬ 2007 – ‫د رة‬ – Ep = E0 e – 2t /RC ⇐ E0 = ‫ـ‬1‫ــ‬CU02 : e – 2t½ /RC = 1 ⇐ E0 e – 2t½ /RC = E0 ⇐ Ep = E0 : 2 t=0 : τ = RC : 2 t½ = τ ln2 ⇐ –22t½ = – ln2 :2 t = t½ : 2 RC a : ^a (i) (A) i = – dq A = – dN dt dt dN dq = – q dt = – λN0 dt RC q(t) = q0 e – t /RC N(t) = N0 e – λt τ = RC : τ= 1: λ τ ln2 : t½ ln2 : t½ 2 λ K E : . 158 : –1 )①: ( ◉◉ ⊕⊖ C i uR uC R = 320 kΩ E = 12 V .1 . .2 E (V) ◉◉ – .3 : . RC .E i R uC t R E t=0 i0 . . ∞t i du + 1 u – E = 0 : RC .4 dt RC RC . uC(t) = E( 1 – e – t /RC ) :1,5 ‫ـ‬ .C τ .5 । t (s) . t = τ uC .6 0,3 . .7 : ( ) uR = Ri : E = uC(t) + uR : .1 . ( u = E – Ri : ) uC(t) = E – Ri(t) : i(0) = i0 = E : uC(0) = 0 ⇐ t = 0 : (K ) .2 i0 = 3,75 ×R10-5 A ≈ 0,04 mA ⇐ R = 320 kΩ = 32 × 104 Ω E = 12 V : . .i=0: uC → E t → ∞ .3 RC uC(t) = E( 1 – e – t /RC ) = E – Ee – t /RC : .4 : . du = 0 + E e – t /RC : . dt RC 59

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ . E e – t /RC + 1 (E – Ee – t /RC) – E = 0 RC RC RC( ) τ = 0,3 s : uC = E t = τ uC (t) : .5 .6 C ≈ 0,94 × 10-6 F = 0,94 µF ⇐ (R = 32 × 104 Ω τ = 0,3 s) C = τ ⇐ τ = RC : R :t=τ ( ) uC = 7,56 VuC = 0,63 E uC (τ) uC = E uC (t) uC = 0,63 × 12 = 7,56 V : . ( ) uC = E Ep = 1 Cu2 = 1 CE2 : .7 22 Ep ≈ 67,7 µJ ⇐ Ep = 1 × 0,94 × 10-6 × 122 = 67,68 × 10-6 J : . 2 ②: i . C ◉ I . – :( ) t1 t0 = 0 K . K uAB B A◉ V . uAB = It : t . u1 uAB .C C .1 u1 = 6,0 V . t1 = 7,2 s . t1 uAB I = 10 µA : .2 .3 GBF : :( ) GBF R . R = 1 000 Ω 6 V 0 ‫ز را ا ه ازات‬ . (2) (1) τ ( 1 ) ‫ا رة‬ . 63 % τ .1 . RC .2 (2) τ (2) (1) (1) . .3 .C 20 ms/div : .4 2 V/div : ( 2 ) ‫ا رة‬ : 10 ms/div : : (1) 1 V/div : : (2) t :( ) : .( ) q = It : .1 60

‫ א – א א‬ª‫א א‬ 2007 uAB = It : ‫د رة‬ C i C = It1 : q = C.uAB : .2 I = 10 µA : uAB ◉K C= 10×10-6×7,2 ⇐ t1 u1 6,0 V .3 ◉ = 7,2 s u1 =BA 6 .1 q .2 Ep = 1Cu12 : t1 C = 12 × 10-6 F = 12 µF ∴ 2 ⇐ u1 = 6,0 V C = 12 µF : V Ep = 1×12×10-6×62 2 ( 1 ) ‫ا رة‬ Ep = 2,2 × 10-4 J ∴ :( ) τ = RC : RC (2) (1) 1 V/div 2 V/div () .τ↔τ (2) .3 ( 2 ) ‫ا رة‬ u=6V t=0 τ = 10 ms = 10-2 s : (2) .t=τ : ∗ u(τ) = 0,63 × 6 = 3,68 V ⇐ u(τ) = 0,63 umax : τ = 10 ms = 10-2 s : C= τ : τ = RC : .4 R ③: C = 10 × 10-6 F = 10 µF ⇐ C = 10-2 ⇐ R = 1 000 Ω τ = 10-2 s : 1 000 500 V 4,7 µF : . .1 .2 .3 .4 : :∗ Céq = 4,7 + 4,7 = 9,4 µF : .1 ( ) Umax = 500 V : .2 Qtotal = Céq.Umax = 9,4×10-6×500 = 4,7×10-3 C : .3 Ep = 1 QtUm = 1 × 4,7 × 10-3 × 500 = 1,175 J : : .4 22 ∗ Céq = 2,35 µF ⇐ 1= 1+1 =2 .1 Céq 4,7 4,7 4,7 .2 ( ) Umax = 1 000 V : Qtotal = Céq.Umax = 2,35×10-6×1 000 = 2,35×10-3 C : .3 Ep = 1 QtUm = 1 × 2,35 × 10-3 × 1 000 = 1,175 J : .4 22 61

‫–א א‬ ‫ א‬ª‫א א‬ 2007 ‫د رة‬ : (2 (1-2 : . (vernis) .1 2– . i 0 : 01 . : – –KE . L2 : L1 . L1 : L2R=r L1 .( .( )(L , r) L2 : 02 ) ●: ● ℓ : ●R .(Ω) (r)( )N .(H) (L)( ) : 03 ℓ) :L .(N N ) R : ℓ (3 ( LR ) S = πR2 : L= 4π×10-7 .N2.S ℓ Lr : : (L , r) ≈ 0 Ω ) L (L , r) (r● ●: uAB = L di + r i dtAi B uAB (L) ‫ا‬ (r) ‫ا‬uAB = r i : di = 0 : dt A uAB = L di : r=0 : ● ●V1 dt (2 2 L : ( L , r1 ) r1 ο C ● u2 ●V2 {L,R}u(t) ο r2 . -- . – ( r2 ) 4 ● ●M u (t) .T B : 04 (T – t0) ( t0 < T ) 62

u (t) ‫ א – א א‬ª‫א א‬ 20070 t0 5– ‫د رة‬ : . t0 = T : 2 t u (t) = U : 1 ∗ u (t) = 0 : 2 ∗ T : 05 :T (4 ) u(t) .B A u (t) ∗ BC u2 (t) ∗ . 6 – ( u2 (t) i (t) : ) : . r2 . i (t) : . u (t) (L) : 06 R 1K ● U = Cte .7 – A {L,R} ● u(t) i {L,R} U K .K u=0: . K L 2 K' . ● 2 + K' .U R BA ●B .(7– )K : 07 : (diode de roue libre : ) K' ● 8 .( ) A : 21 ● u(t) i u (t) = U : K K L ∗ + D u (t) = 0 : K ∗ R : 9–U i (t) u (t) 6– ●B : :∗ : 08 U0 u (t)u (t) .I=U :R t1 = 2,5 ms . (20 ms) (≈10ns) 0i (t) 0 2,5 5 7,5 10 t (ms) u (t) ( .U 15 :∗ i (t) u (t) : 09 ) .T 0 U 2 63

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ . t2 – T = 2,5 ms : 2. : (3 2 (K ) :1 i e = - L di : 10 – RL :∗ dt ()+U R U – L di – Ri = 0 ⇐ U + e – Ri = 0 Ri dt - :∗ ( ) t = 0+ : 10 i = U ( 1 – e – Rt /L ) : . R ( ) I= U: i = I ( 1 – e – Rt /L ) R » τ=L: « x↦y=(1–e–x) : t↦i: ∴ R . (s) : ∗ . 7– LR Ur2 ( : 4– R . i (t) u2 = r2 i = 1 – e – Rt /L ) : r2 : u1 = U – u2 : : ( L , r1 )t (ms) 0 0,2 0,4 0,8 1,2 1,6 2 u1 = r1 i + L ‫ـ‬d‫ــ‬i‫ـ‬ ∗ dt :t /τ 0 0,5 1 2 3 4 5 U = 7,5 V R = 15 Ω L = 6 mH : e – t/τ 1 0,667 0,368 0,135 0,050 0,018 0,007 τ = L = 6×10-3 = 0,4 × 10-3 s = 0,4 ms1 – e – t/τ 0 0,393 0,632 0,865 0,950 0,982 0,993 R 15 0 0,197 0,316 0,433 0,475 0,491 0,497 I = U = 7,5 = 0,5 A I (A) R 15 i (A) i = 0,5 ( 1 – e – t /0,4 ) ⇐ i (A) t (ms) :0,5 . i = 0,5 A i: t→∞: ( )1 . 11 – i (t)‫ـ‬: . i e – t/τ t (ms) i = 0,497 A : t=5τ0 0,4 0,8 1,2 .%1 .L R (≈5τ) : 11 (K ) :2 K () . ( 12 – ) . ( . . .. )e i :∗ 64

‫ א – א א‬ª‫א א‬ 2007 : 12 – RL () ‫د رة‬ i – L ‫ـ‬d‫ــ‬i‫ – ـ‬Ri = 0 ⇐ e – Ri = 0 L ‫ـ‬d‫ــ‬i‫ ـ‬+ Ri = 0 dte R Ri dt : . :∗ i = I0 : ( ) : 12 i = I0 e – Rt/L : « » τ=L: x↦y= e–x : t↦i : ∴ R . (s)t (ms) 0 0,2 0,4 0,8 1,2 1,6 ( :∗ t /τ 0 0,5 1 2 3 4 2 ) R = 15 Ω L = 6 mH : τ = 0,4 ms 5 I0 = 0,5 Ae – t/τ 1 0,667 0,368 0,135 0,050 0,018 0,007 i = 0,5 e – t /0,4 ⇐ i (A) t (ms) : . i: t→∞:I (A) 0,5 0,34 0,184 0,068 0,025 0,009 0,004 2 i (A) . 13 – i (t) () :0,5 t=5τ : (i=0)‫ـ‬ : !!! : 12 – ( . . . . ) t (ms) e = R I0 e – Rt/L = e0 e – Rt/L ⇐ e = - L ‫ـ‬d‫ــ‬i‫ = ـ‬Ri dt0 0,4 0,8 1,2 . . .. (t=0: ) e0 = R I0 : : 13 . R I0 = 15 × 0,5 = 7,5 V : e0 !! e0 ! ∆‫ـــ‬i‫ـ‬ : 1A : ∆t × ‫ــــ‬1‫= ـــــ‬Emoy = L = 20 0,001 20 000 V : L = 20 H 1 ms τ R' = 9 R : .. 14 – R' ) τ' = ‫ـــــ‬L‫ــ =ـــــــ‬τ‫ ــ‬:15 – . ( 10 R + R' 10 ●i ● i KL L KD +D + R' R R : 14 : 15 65

KD ‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬●●● ● : 16 – . (4 2A C V 2A .0 L,r . :E● Ep = 1‫ ــ‬L.I2 ● 2 Ep = 1‫ــ‬L.I2 = 1‫ ×ــ‬0,02 × 22 = 0,04 J 22 E=6V ∴L = 20 mH , r = 4,4 Ω – 59 V C = 15,5 µF : 16 :. E'p = 1‫ ــ‬C.u2 = 1‫ ×ــ‬15,5 × 10-6 × ( - 59 )2 = 0,027 J 22 . E'p Ep ∆Ep = 0,040 – 0,027 = 0,013 J : η = 67,5 % ⇐ η = ‫ـ‬E‫ـ'ـــ‬p‫ـ = ـ‬0‫ــ‬,‫ـ‬0‫ـــ‬2‫ــ‬7‫ = ــ‬0,675 : ∴ Ep 0,040 K E : ◉◉ ( . 158 : – 2 )①:A◉ R M◉ .E . ( L , r = 10 Ω ) R YA L,r ◉B t = 0 uBM uMA uMA (V) . uBM (V)1‫ـ‬ .E .1 YB .R L .2 01 R) i .3 (r E L . t = 3 ms .4 t (ms) 1‫ـ‬ t (ms) . .5 . : 01 E = uBA = uBM + uMA ⇐ () : E .1(∗) .......... E = L ‫ـ‬d‫ــ‬i‫ ـ‬+ ( R+r)i ⇐ uBM = L ‫ـ‬d‫ــ‬i‫ ـ‬+ ri : uMA = Ri ⇐ dt dt : ‫ـ‬d‫ــ‬i‫ = ـ‬0 ⇐ i = Cte :E = 9 V ⇐ uBM = 2 V uMA = 7 V : dt ∗ ∗ E = 9 V ⇐ uBM = 4,6 V uMA = 4,4 V : t = 2 ms .2 E = ( R+r)i = 9 V : ‫ـ‬d‫ــ‬i‫ = ـ‬0 :R L ∗ dt ∗R = 35 Ω : R = 7‫ـــ‬r‫ = ـ‬35 Ω : ‫ـ‬u‫ـــ‬M‫ــ‬A‫ـ = ـــ‬R‫ـ = ـــ‬7‫ ـــ‬: uMA = Ri = 7 V : 2 uBM r 2 uBM = ri = 2 V ‫ـ‬d‫ــ‬u‫ـــ‬M‫ـــ‬A‫ = ــ‬R ‫ـ‬d‫ــ‬i‫ ـ‬: uMA = Ri : dt dt uMA = f(t)‫ـ‬d‫ــ‬i‫ـ = ـ‬3‫ـــ‬5‫ـــ‬0‫ــ‬0‫ = ــ‬100 ⇐ ‫ـ‬d‫ــ‬u‫ـــ‬M‫ـــ‬A‫ـــــ = ــ‬7‫ = ــــــ‬3 500 t=0: uBM = f(t)dt R dt 0,002 ‫ـ‬d‫ــ‬i‫ـ‬ L = 0,09 H ⇐ uBM = L dt = 9 V : t=0 66

‫–א א‬ ‫ א‬ª‫א א‬ 2007 .3 : t = 3 ms (r E L R) ‫د رة‬ i : E – L ‫ـ‬d‫ــ‬i‫ ( – ـ‬R+r)i = 0 ⇐ (∗) dt i (t) = ‫ـــ‬E‫ ( ـــــ‬1 – e – ( R+r )t /L ) R+r i = 0,155 A : t = 3 ms = 0,003 s : : .4 Ep = 1,1 mJ ⇐ Ep = 1‫ × ــ‬0,09 × 0,1552 = 1,1 × 10-3 J ⇐ Ep = 1‫ــ‬L.I2 : .5 22 ②: ⇐ τ = ‫ــ‬0‫ـــ‬,‫ـ‬0‫ــ‬9‫ = ــــ‬0,0002 s ⇐ τ = ‫ـــ‬L‫ ـــــ‬: :τ = 2 ms 35+10 R+r ( R+r , L ) . r = 10 Ω L = 0,2 H U = 12 V 0,5 A : i .1 . t1 .2 i .3 t2 = 0,05 s : : i (t) .1 i (t) =I ( 1 – e – r t /L ) : (r,L) ‫ـ‬r‫ــ‬t‫ = ـ‬50t ⇐ τ = 0‫ـــ‬,‫ـ‬2‫ = ــ‬0,02 s ⇐ τ = ‫ـ‬L‫ ــ‬: rL : L 10 r : I = ‫ـ‬U‫ـ = ـــ‬1‫ــ‬2‫ = ـــ‬1,2 A : r 10 i (t) = 1,2 ( 1 – e – 50t ) ( S.I ) . . : t1 .2 50t1 = ln ‫ـ‬1‫ــ‬2‫ = ــ‬0,539 ⇐ e 50t1 = ‫ـ‬1‫ــ‬2‫ ⇐ ــ‬e – 50t1 = ‫ــ‬7‫ ⇐ ـــ‬0,5 = 1,2 ( 1 – e – 50t1 ) : 7 7 12 0‫ــ‬,‫ــ‬5‫ــ‬3‫ــ‬9‫ = ــ‬0,0108 s = 10,8 ms t1 =10,8 ms ⇐ t1 = 50 ∴ : t2 .3 i = 1,1 A ⇐ i = 1,2 ( 1 – e – 2,5 ) = 1,1 A ⇐ r‫ــ‬t‫ـ‬2‫ـ‬ = 50t2 = 50 × 0,05 = 2,5 : L ③: :② E=3V r = 4 Ω L = 0,1 H ( ( S.I ) . . : ) i (t) = 0,75 ( 1 – e – 40t ) .1 t1 = 34,65 ms .2 i = 0,65 A .3 67

‫ א – א א‬ª‫א א‬ 2007 1● ‫د رة‬ ●A : 2● R « » ①: :E ●B E = 5 V R = 10 kΩ C = 100 nF : C ●D : .1 . .1 .β A () ( .1 uBD ) .( uBD = E + A e - βt : t=0 ( t=0 uBD . uBD (t) . ττ() ( . uBD (t) . uBD .2 : .2 ●1 : .3 ●A . (E ●2 .E uBD (t) 2 • R' ( .1 •. uBD (t) ( L ●B : uBD (t) : C .1 ●D R' ( R' . R' = 0 : R' ●A : i (t) 1 R uR E = uAB + uBD :E ● B uAB = Ri (t) : i (t) = C d‫ـــ‬u‫ـــ‬B‫ــ‬D‫ ⇐ ــ‬q (t) = C.uBD i (t) = ‫ـ‬d‫ــ‬q‫ـ(ـــ‬t‫ ـ)ــ‬: q C uC dt dt ●D RC d‫ـــ‬u‫ـــ‬B‫ــ‬D‫ ــ‬+ uBD = E : dt . uBD = E + A e - βt : : d‫ـــ‬u‫ـــ‬B‫ــ‬D‫ = ــ‬- Aβ e – βt ⇐ uBD = E + A e - βt dt . t ( 1 – RCβ ) A e – βt = 0 ⇐ - RCAβ e – βt + E + A e – βt = E β = ‫ـــ‬1‫ ــــ‬: ∴ RC ( . uBD = 0 t=0A=-E ⇐ E + A = 0 ⇐ E + A e0 = 0 : t = 0 uBD = E ( 1 – e – t/RC ) : ∴ 68

‫ א – א א‬ª‫א א‬ 2007 τ = RC : ‫د رة‬ ( ) RC. 63 % : τ = 1 ms ⇐ τ = 104 × 10-7 = 10-3 s = 1 ms ⇐ τ = RC :uBD (t) V : uBD = E ( 1 – e – t/RC ) ( 5 τ = 1 ms : E=5V43210 t (ms) 0 1 2 3 4 5 6 7 8 9 10 . : .2 : – Ep = 1‫ ــ‬C.u2BD : uBD = E : t=0 – 2 R' . R' .3Ep = 1,25 µJ ⇐ Ep = 1‫ × ــ‬100 × 10-9 × 52 = 1,25 × 10-6 J : ( 2 : – – ●A : ( . uBD(0) = E : i (t) :uL L R' uR' t>0: ●B .(1– ) q uC .(3 2– ) ●D uL = L ‫ــ‬d‫ــ‬i‫ ــ‬: dt uL = L ‫ـ‬d‫ــ‬2‫ــ‬q‫ــ‬ = LC ‫ـ‬d‫ـــ‬2‫ـ‬u‫ـــ‬B‫ــ‬D‫ـــ‬ i (t) = ‫ـ‬d‫ـــ‬q‫ــ‬ dt2 dt2 dt R' = 0 uAB = 0 uL + uAB + uBD = 0 : ‫ـ‬d‫ـــ‬2‫ـ‬u‫ـــ‬B‫ــ‬D‫ـــ‬ + ‫ــ‬1‫ــــ‬uBD = 0 : dt2 LCuBD (t) = Ucos(ωt+φ) :. uBD (t) = E cos ωt ⇐ φ = 0 U = E : ( i = 0 uBD = E ⇐ t = 0 ) 1– ‫ا‬ 2– ‫ا‬ 3– ‫ا‬ 69

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ « » ②: E .(1– )C . uC t0 = 0 s (2– .1 ●● R = 100 Ω K Ki t1 . t1 2 – . uC ) uC = f (t) E .2E +q uC . C .2– .3 τ = RC : R τ 1- ‫ا‬ .2– 63 % . .CuC (V) ∆t 2 – .45,00 . τ ∆t .5 R .6 .1– .7 uC1,00 t (ms) : t0 0 0,50 2,50 d‫ـــ‬u‫ـــ‬C‫ـ‬ 2- dt ‫ا‬ E – uC – RC = 0 . i = f (t) . i (t) uC = E ( 1 – e – t/RC ) : .8 t1 = 2,0 ms uC 1– : 2– .1 . ●● ) t1 Ki . E +q uC 1– : C . R( .2 uR i = ‫ـ‬d‫ـــ‬q‫ ــ‬: E = uC + uR dt uR = Ri : E = uC + R ‫ـ‬d‫ـــ‬q‫ ــ‬: dt E = uC : . ‫ـ‬d‫ـــ‬q‫ = ــ‬0 : dt E = 6,00 V : E uC 2 – uC 63 % uC (τ) uC = f (t) τ .3 E(τ) = 0,63 × E = 3,8 V : E = 6,00 V : .(E 63 % ) τ = 0,28 ms : ( 2 – ) C = 2,8 × 10-6 F = 2,8 µF : C = ‫ـ‬τ‫ ـــ‬: C .4 R :. τ = 0,28 ms = 28 × 10-5 s R = 100 Ω 70

‫א א‪ ª‬א – א א‬ ‫‪2007‬‬ ‫) ( ‪t1‬‬ ‫د رة‬ ‫‪.5‬‬ ‫‪∆t = t1 – t0 :‬‬ ‫‪∆t‬‬ ‫‪∆t = 2,0 ms :‬‬ ‫‪t0 = 0 t1 = 2,0 ms :‬‬ ‫‪∆t = 7,1 τ :‬‬ ‫‪ = 7,1 : τ‬ـــ‪0‬ــ‪,‬ــ‪2‬ــ = ـ‪t‬ـــ∆‬ ‫‪∆t‬‬ ‫‪∆t = 5 τ‬‬ ‫‪τ 0,28‬‬ ‫‪99 %‬‬ ‫‪.‬‬ ‫‪R τ = RC‬‬ ‫‪.6‬‬ ‫– ‪ : 2‬ــ‪q‬ـــ‪d‬ـ ‪E = uC + R‬‬ ‫‪.‬‬ ‫‪dt‬‬ ‫‪.7‬‬‫‪: q = C.uC :‬‬ ‫‪q‬‬ ‫‪ = 0‬ـــ‪C‬ـــ‪u‬ــ‪d‬ــ ‪E – uC – RC‬‬ ‫‪:‬‬ ‫ـ)ـــ‪C‬ــ‪u‬ــ‪.‬ـــ‪C‬ــ(ــ‪d‬ـ ‪E = uC + R‬‬ ‫‪dt‬‬ ‫‪dt‬‬ ‫‪: .8‬‬ ‫‪ :‬ــ)ـ‪t‬ــ(ـــ‪C‬ـــ‪u‬ـــ‪d‬ـ ‪i (t) = C‬‬ ‫‪. q (t) = C.uC (t) :‬‬ ‫ـ)ــ‪t‬ـ(ـــ‪q‬ــ‪d‬ـ = )‪i (t‬‬ ‫‪dt‬‬ ‫‪dt‬‬ ‫‪:‬‬ ‫‪i‬‬ ‫‪uC = E ( 1 – e – t/RC ) :‬‬ ‫)‪uC (t‬‬‫‪I0‬‬ ‫‪ e – t/RC :‬ــــ‪E‬ـ = )‪i (t‬‬ ‫‪ :‬ــــ‪E‬ـ = ‪R I0‬‬ ‫‪ e – t/RC‬ـــــ‪E‬ــ = ــ)ـ‪t‬ــ(ـــ‪C‬ـــ‪u‬ـــ‪d‬ـ‬ ‫‪R‬‬ ‫‪dt RC‬‬ ‫‪i (t) = I0 e – t/RC‬‬ ‫‪:‬‬ ‫‪i = f (t) :‬‬ ‫) )‪i (t‬‬ ‫‪0t‬‬ ‫‪t=0‬‬ ‫ــــ‪E‬ـ = ‪I0‬‬ ‫‪(« »+∞ t‬‬ ‫‪R‬‬‫‪ ،‬ن ‪ + … :‬ـــ‪1‬ـ ‪ +‬ـــ‪1‬ـ ‪ +‬ـــ‪1‬ـ = ــــ‪1‬ـــ‬‫‪Céq C1 C2 C3‬‬ ‫)‪(0‬‬ ‫ت ‪ ،‬ن ‪Céq = C1+C2+C3+ … :‬‬ ‫«‬ ‫‪– 160 : 5‬‬ ‫‪» ③:‬‬ ‫دا تا ‪.‬‬ ‫‪ ،‬آ ‪. C1 = 0,1 mF‬‬ ‫ت‬ ‫د ها ت ل‬ ‫‪. 5 mF‬‬ ‫‪.1‬‬ ‫أي‬ ‫‪ .2‬د د ا ت ا ‪.‬‬ ‫‪. u = 40 V‬‬ ‫ا تا‬ ‫‪.3‬‬ ‫اا ؟‬ ‫أ( ه‬ ‫ب( ه آ ؟‬ ‫‪:‬‬ ‫ا ت أن ف ‪:‬‬ ‫‪ .1‬ف‬ ‫‪ Céq‬أ‬ ‫نا ا‬ ‫وا ت ا‬ ‫أي ا‬ ‫‪ Céq‬أآ‬ ‫اع ن ا ا‬ ‫ا‬ ‫ر ا ت ا ع ن ‪Céq > C1 :‬‬ ‫ا ع ن ‪n : ، Céq = n C1 :‬‬ ‫او‬ ‫أن ا ت‬ ‫‪.2‬‬ ‫ـ‪q‬ــ‪é‬ـــ‪C‬ـ = ‪ = 50 ⇐ n‬ــ‪3‬ـ‪-‬ــ‪0‬ـــ‪1‬ــــ×ــــ‪5‬ـ = ‪n = 50 ⇐ n‬‬ ‫∴‬ ‫‪10-4 C1‬‬ ‫‪( .3‬‬ ‫‪qéq = Céq.u = 5 × 10-3 × 40 = 0,2 C :‬‬ ‫= ‪q1‬‬ ‫ــ‪q‬ـ‪é‬ـــ‪q‬ـ‬ ‫‪ = 4 × 10-3 C‬ــ‪2‬ــ‪,‬ــ‪0‬ـ =‬ ‫‪:‬‬ ‫(‬ ‫‪n‬‬ ‫‪50‬‬ ‫‪71‬‬

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ « – 162 : 12 – » ③ : R = 100 kΩ C = 0,1 µF E=6V E .1 t = 0 .1 +● ●– (D● B● A ● ● 1 uBD = E + a e – b t : uBD = u (t) : C R ( ( ●2 . b ( .τ a=-E : t (s) 0 τ 5 τ : .2 uBD (V) uBD = f (t) : .3 .2 .4 ( E = uAB + uBD : ( uAB = Ri (t) : : ( .1 uAB = u (t) = RC d‫ـــ‬u‫ـــ‬B‫ــ‬D‫⇐ ــ‬ i (t) = C d‫ـــ‬u‫ـــ‬B‫ــ‬D‫⇐ ــ‬ q (t) = C.uBD i (t) = ‫ـ‬d‫ــ‬q‫ـ(ـــ‬t‫ ـ)ــ‬: dt dt dt . ) .......... d‫ــ‬u‫ــ(ــــ‬t‫)ــ‬+ ‫ـــ‬1‫ ـــــ‬u (t) = ‫ـــ‬E‫ ـــــ‬: E = RC d‫ـــ‬u‫ـــ‬B‫ــ‬D‫ ــ‬+ uBD : dt RC RC dt u (t) = E + a e – b t ( d‫ــ‬u‫ــ(ــــ‬t‫ = )ــ‬0 – ab e – bt : d‫ــ‬u‫ــ(ــــ‬t‫)ــ‬ u (t) : dt dt b = ‫ـــ‬1‫ـ = ـــــ‬1‫ ـــ‬: – ab e – bt + ‫ـــ‬1‫ ( ـــــ‬E + a e – b t ) = ‫ـــ‬E‫∴ ـــــ‬ RC τ RC RC b d‫ــ‬u‫ــ(ــــ‬t‫)ــ‬+ ‫ـــ‬1‫ ـــــ‬u (t) = ‫ـــ‬E‫ ـــــ‬: RC u (t) = E + a e – b t : dt RC RC ( . u (0) = 0 t=0a = - E ⇐ E + a = 0 ⇐ E + a e0 = 0 : t = 0 u (t) = E ( 1 – e – t/RC ) : ∴τ = 0,01 s = 10 ms ⇐ ( C = 0,1 µF = 10-7 F R = 100 kΩ = 105 Ω ) τ = RC : uBD (V) t (s) 0 τ 5 τ : .2E uBD (V) 6,00 3,78 0,00 uBD = f (t) : .3 ) ( ( .40 τ 5τ t (s) . uBD = f (t) : ‫ا ن‬ Ep = 1‫ــ‬ × 10-7 × 62 = 1,8 × 10-6 J ⇐ Ep = 1‫ــ‬ C.u2 : ( 2 2 Ep = 1,8 µJ ⇐ 72

q (µC) ‫ א – א א‬ª‫א א‬ 13 – 2007 ‫د رة‬0,2 « – 162 : 20 .(E=5V) » ④: C ++ ++ uC . ( t=0 ) ( R = 100 kΩ ) ---- . q (t) .1 . q (t) = Q0 e – t/τ : .2 : .3 τ = ‫ـــ‬1‫ ــــ‬: t=τ t (ms) . .4 RC i . C .5 . t = 5τ t = 0 .6 . .7 : : RC .1 q uC – Ri = 0 ‫ دارة ا‬R Ri i = – ‫ــ‬d‫ـــ‬q‫ ـــ‬: dq dt uC = ‫ـ‬q‫ـــ‬ : ‫ــ‬d‫ـــ‬q‫ـــ‬ C ‫ـ‬q‫ ـــ‬+ R ‫ــ‬d‫ـــ‬q‫ ـــ‬+ ‫ـــ‬1‫ ــــ‬q (t) = 0 : C dt =0 : dt RC q (t) = Q0 e – t/τ : .2 – Q‫ـــ‬0‫ ـ‬e – t/τ + ‫ـــ‬1‫ ــــ‬Q0 e – t/τ = – Q‫ـــ‬0‫ ـ‬e – t/τ + Q‫ـــ‬0‫ ـ‬e – t/τ = 0 ⇐ ‫ــ‬d‫ـــ‬q‫ – = ـــ‬Q‫ـــ‬0‫ ـ‬e – t/τ τ RC ττ dt τ : ( t = 0 ; q = Q0 ) q = f (t) .3 :( ) q = at + b ( ) b = Q0 ⇐ q = b = Q0 ⇐ t = 0 ∗ a = – Q‫ـــ‬0‫ ⇐ ـ‬t = 0 ‫ ؛‬a = ‫ــ‬d‫ـــ‬q‫ – =ـــ‬Q‫ـــ‬0‫ ـ‬e – t/τ ∗ (« » ) τ dt τ0 = – Q‫ـــ‬0‫ ـ‬t + Q0 : q (t) = – Q‫ـــ‬0‫ ـ‬t + Q0 :∴ τ τ q (t) = 0 : t=τ : τ = 20 ms = 0,02 s : .4 C = 0,2 µF ⇐ C= ‫ـ‬0‫ــ‬,‫ـ‬0‫ـــ‬2‫ = ــ‬2 × 10-7 F ⇐ C = ‫ـ‬τ‫ـــ‬ ⇐ τ = ‫ـــ‬1‫ــــ‬ : .5 105 R RC q (0) = 1 µC ⇐ q (0) = Q0 = 10-6 C ⇐ t = 0 .6q (5τ) = 6,7 nC ⇐ q (5τ) = 10-6 × e – 5 = 6,7 × 10-9 C ⇐ q (5τ) = Q0 e – 5τ / τ ⇐ t = 5τ : i (t) = – ‫ــ‬d‫ـــ‬q‫ = ـــ‬Q‫ـــ‬0‫ـ‬e – t/τ : .7 dt τ ∗ i (0) = Q‫ـــ‬0‫ـ = ـ‬1‫ــ‬0‫ـــ‬-‫ـ‬6‫ = ـ‬50 × 10-6 A i (0) = 50 µA ⇐ τ 0,02 ⇐ t=0i (5τ) = 0,335 µA ⇐ i (5τ) = 50 × 10-6 e – 5 = 0,335 × 10-6 A ⇐ i (5τ) = Q‫ـــ‬0‫ ـ‬e – 5τ / τ ⇐ t = 5τ ∗ τ 73

‫ א – א א‬ª‫א א‬ 2007 – 163 : 16 – ‫د رة‬ « » ⑤: . 12 V 4 mC . .1 .2 . Q0 C t τ : .R .3 .t=τ .4 Ep = 24 mJ ⇐ Ep = 1‫ ×ــ‬4 × 10-3 × 12 = 24 × 10-3 J ⇐ Ep = 1‫ــ‬q.u : : 22 .1 ( q' = C'.u = 2C.u = 2q ) C’ = 2C .2 ( E'p = 1‫ ــ‬q'.u = q.u = 2Ep = 48 mJ ) – 2t/τ ⇐ 2 1‫ ــ‬q‫ـــ‬2‫ـ‬ q (t) = Q0 e – t/τ Ep = 1‫ــ‬ Q‫ــــ‬0‫ـ‬2‫ـ‬ e Ep = 2C : .3 2 C .4 Ep (τ) = 1‫ ــ‬Q‫ــــ‬0‫ـ‬2‫ ـ‬e – 2 2C : t=τ: ( ) C = ‫ـ‬q‫ـ(ــــ‬t‫ =ـ)ــ‬4‫ــــ×ــــ‬1‫ـــ‬0‫ــ‬-‫ـ‬3‫ = ـ‬1‫ ×ــ‬10-3 F ⇐ q (t) = C.u (t) : u (t) 12 3 ⇐ ( C = 1‫ ×ــ‬10-3 F Ep (τ) = 3,24 mJ 3 Q0 = 4 × 10-3 C ) E « – 165 : 22 – » ⑥: R = r = 12 Ω ( L,r ) .i .E 3 V/div :A◉ R M◉ L,r : YA ◉B YB .1 .2 ➀ . .3 . .4 : ➁ : uBM < 0 uAM > 0 : ∗ .1 uBM . uAM :➀ ∗ . . uBM :➁ .2 i = ‫ـ‬u‫ـــ‬A‫ــ‬M‫ـــ‬ ⇐ uAM = Ri : .3 R i = 0,5 A ⇐ ( R = 12 Ω uAM = 2 div × 3 V/div = 6 V ) :. E = uAM – uBM ⇐ E = uAB = uAM + uMB = uAM + ( - uBM ) : .4 E E = 12 V ⇐ E = 6 – (- 6) = 12 V : . K L « – 165 : 24 – » ⑦: ● . uR = 0,1 u0 : t1 t2 ●● .t=0 uR = 0,9 u0 : RM● R = 1 kΩ ● ●B A ● 74

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ .1 : t2 – t1 = 1,65 ms : .2 .τ ( .L ( : () .1 . i (t) = I0 e – t/τ : : uR = Ri = R I0 e – t/τ : ( .2 . R I0 e – t1 /τ = 0,9 u0 ⇐ uR = 0,9 u0 ⇐ t = t1 o . R I0 e – t2 /τ = 0,1 u0 ⇐ uR = 0,1 u0 ⇐ t = t2 o 9 = e t‫ــ‬2‫ــــ–ـــ‬t‫ـ‬1‫ـ‬ ⇐ ‫ــ‬R‫ــــ‬I‫ــ‬0‫ـــ‬e‫ــ–ــــ‬t‫ـ‬1‫ــ‬/‫ـ‬τ‫ــ‬ = ‫ـ‬0‫ـــ‬,‫ـ‬9‫ــــ‬u‫ـــ‬0‫ ــ‬: τ R I0 e – t2 /τ 0,1 u0τ = 0,75 ms ⇐ τ = ‫ــ‬1‫ـــ‬,‫ـ‬6‫ــ‬5‫ = ــــ‬0,75 ms ⇐ τ = ‫ـ‬t‫ـ‬2‫ــــ–ـــ‬t‫ـ‬1‫ ⇐ ــ‬ln 9 = ‫ـ‬t‫ـ‬2‫ــــ–ـــ‬t‫ـ‬1‫ ــ‬: ( ln 9 ln 9 τ L = τ.R ⇐ τ = ‫ـ‬L‫ ـــ‬: RL R L = 0,75 H ⇐ ( τ = 0,75 ms = 7,5 × 10-4 s R = 1 kΩ = 103 Ω ) : .YA Ai K « – 166 : 25 – » ⑧: ●● ● L,r ( L,r )E E = 3,8 V R = 50 Ω M● . R ●B YB . yB t=0 yB . E .1 uR (V) .2 1 t (ms) . I0 .3 20 (R . ‫ــ‬d‫ــ‬i‫ ــ‬i r R L : dt .4 . : .1 ) yB I0 = ‫ـ‬u‫ـــ‬R‫ــ‬ ⇐ uR = RI0 : uR (t) = R.i (t) : .2 R ( ) uR = 3 V : I0 = 60 mA ⇐ ( uR = 3 V R = 50 Ω ) E = ( L ‫ــ‬d‫ــ‬i‫ ــ‬+ ri ) + Ri ⇐ uAM = uAB + uBM : .3 dt E = L ‫ــ‬d‫ــ‬i‫ ــ‬+ ( R+r ) i ∴ r = ‫ـ‬E‫ ـــ‬- R ⇐ E = ( R+r ) i ⇐ ‫ــ‬d‫ــ‬i‫ =ــ‬0 ⇐ i = I0 = Cte : dt I0 dt .4 ( ) ... r = 13,33 Ω ⇐ ( I0 = 0,06 A E = 3,8 V R = 50 Ω ) uR (V) ( ) uR = uR maxuR max τ = 20 ms : 1 t (ms) τ = ‫ـــ‬L‫ ــــــ‬: ( R+r )L R+r τ 20 L = ( 50 + 13,33 ) × 0,02 = 1,266 H ⇐ L = ( R+r )τ ∴ ( ) ... L = 1,266 H ⇐ 75

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ i (A) R = 35 Ω « – 167 : . E = 12 V ( L,r ) 28 – » ⑨ :0,06 L (H)20 t (ms) .L τ t=0 τ (ms) . ( L,r,R ) τ .0,2 2 . .1 .2 .r .3 .4 . .( ( ( E : Ki ( ) : .1 ●●● R ● L,r I0 = ‫ـــ‬E‫ ــــــ‬: .2 R+r ● I0 = 0,24 A ⇐ I0 = 4 div × 0,06 V/div = 0,24 A : r = ‫ـ‬E‫ ـــ‬- R = ‫ــ‬1‫ـــ‬2‫ ــــ‬- 35 = 15 Ω ⇐ ( E = 12 V R = 35 Ω ) I0 0,24 r = 15 Ω ∴( ) i = I max = I0 τ = 20 ms : i = f (t) .3 . τ = ‫ـــ‬L‫ ــــــ‬: ( R+r )L R+r L = 1 H ⇐ L = ( 35 + 15 ) × 0,02 = 1 H ⇐ L = ( R+r )τ ∴ (1) ... L = a.τ : « a » L = f (τ) ( .4 E = ( L ‫ــ‬d‫ــ‬i‫ ــ‬+ ri ) + Ri ⇐ :( dt (2) ... L = ( R+r )τ ⇐ τ= ‫ـــ‬L‫ــــــ‬ : ‫ــ‬d‫ــ‬i‫ ــ‬+ ‫ــ‬1‫ ـــ‬i = ‫ـ‬E‫⇐ ـــ‬ ‫ــ‬d‫ــ‬i‫ ــ‬+ ‫ـ‬R‫ــــ‬+‫ـــ‬r‫ ــ‬i = ‫ــ‬E‫∴ ــــ‬ R+r dt τ L dt L L : ( ) a = R + r : (2) (1) ( : R+r = 35 + 15 = 50 Ω : ∗ a= ‫ـــــ‬0‫ـــ‬,‫ـ‬2‫ــــ×ــــ‬4‫ـــــــ‬ = 50 Ω : ∗ 2 × 8 × 10-3 76

‫א‬ ‫ א – א‬ª‫א‬ ‫א‬ 2007 ‫د رة‬ « – 167 : RL 30 – » ⑩ : Ep (J) t L: ‫ــ‬d‫ــ‬i‫ ــ‬+ ‫ـ‬R‫ ـــ‬i (t) = 0 : .1 t½ dt L .20,2 ‫ـ‬ . .3 . . R = 100 Ω .4 .5 . I0 τ : . t= ‫ـ‬τ‫ـــ‬ t (s) 20 0,5 ‫ـ‬τ‫ـــ‬ . t½ = 2 ln 2 : i (t) = I0 e – t/τ : : .1 Ep = 12‫ــ‬L.I0 2 e – 2t/τ ⇐ Ep = 1‫ــ‬L.i 2 : .2 2: Ep = at + b : Ep = f (t) ( ) .3 a = - 1‫ــ‬L.I02 ⇐ t = 0 a = ‫ـ‬d‫ــ‬E‫ــــ‬p‫ =ـ‬- 1‫ ــ‬L.I02 e – 2t/τ : ∗ τ dt τ ∗ E0 = 1‫ ــ‬L . I0 2 : t=0 2 Ep = - 1‫ ــ‬L.I02.t + 1‫ ــ‬L . I02 : t=0 ∴ τ 2 t = τ‫ــ‬ ⇐ 1‫ ــ‬t = 1‫ــ‬ ⇐ 0 = - 1‫ ــ‬L.I02.t + 1‫ــ‬L . I02 ⇐ Ep = 0 2 τ2 τ 2 τ = 1 s ⇐ τ‫ = ــ‬0,5 s : .4 2 .5 ( ) Ep = 12‫ــ‬L.I0 2 e – 2t/τ : E0 = 1‫ــ‬L . I0 2 : t=0 ∗ 2 1‫ــ‬ = e – 2 ‫ـ‬t‫ــ½ـ‬ ⇐ E‫ــــ‬0‫=ـ‬ E0 e – 2 ‫ـ‬t‫ــ½ـ‬ ⇐ Ep = E‫ــــ‬0‫=ـ‬ 1‫ــ‬L.I0 2 e – 2 ‫ـ‬t‫ــ½ـ‬ : t = t½ ∗ τ 2 τ τ 2 22 t½ = τ‫ ــ‬ln 2 ⇐ - ln 2 = - 2 ‫ـ‬t‫ــ½ـ‬ : 2 τ t½ = 0,345 s ⇐ t½ = τ‫ــ‬ ln 2 = 0,5 × 0,69 = 0,345 s : . 2 77

‫ א – א א‬ª‫א א‬ : 2007 . ‫د رة‬ .( ) . :4[H3O+] = n‫ــ‬H‫ـــ‬3‫ـ‬O‫ــ‬+ : . pKa Ka V . pH . pH = f (V) S,Soerensen H3O+ ( pH (1 pH :( ) (1-1 : pH ) . 10-14 mol.L-1 < [H3O+] < 10-1 mol.L-1 : H3O+ 10 (Log) 1909 .: pH [H3O+] ≤ 10-1 mol.L-1 () : pH = - Log [H3O+] ⇔ [H3O+] = 10- pH mol.L-1 : [H3O+] pH :➀ . pH [H3O+] . H3O+ pH 10-3 mol.L-1 10-2 mol.L-1 : [H3O+] : pH = 2 pH ⇔ ‫[ أ‬H3O+] pH ⇔ [H3O+] = 10-2 mol.L-1 ‫[ أآ‬H3O+] ⇔ pH = 3 [H3O+] = 10-3 mol.L-1 pH < 7 pH = 7 25 °C :➁ 0 7 14 . pH > 7 | | | pH ‫اا‬ ‫اا‬ pH < 7 pH > 7 ‫ل أو ء‬ ‫و‬ pH = 7 Ke = [H3O+] . [OH-] : Ke :➂ . ( ) [H3O+] = [OH-] = 10-7 mol.L-1 : Ke = 10-14 : 25 °C (2 1 : : pH pH pH ∗ ) 1 .( . – pH ∗ 01 : ‫و‬ 2 pH ‫ورق ا ـ‬ 02 : ‫و‬ pH ‫س ا ـ‬ 78

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ : (2 : (1-2 . pH1 = 2 C1 = 10-2 mol.L-1 : – . pH2 = 3,4 C2 = 10-2 mol.L-1 – : ( S1 ) [H3O+] = C1 : ∗ : ( S2 ) : C [H3O+] [H3O+] = 10- pH1 = 10-2 mol.L-1 : ( S1 ) . HCl HCl ∴CH3COOH [H3O+] < C2 : [H3O+] = 10- pH2 = 10-3,4 = 3,98 × 10-4 mol.L-1 : ( S2 ) ∗ .( .( CH3COOH ∴ .( ) :∗ (« » ( ) HCl (g) + H2O (ℓ) → H3O+ (aq) + Cl- (aq) ) CH3COOH (ℓ) + H2O (ℓ) = H3O+ (aq) + CH3COO- (aq) : ) H3O+ HA HA HA (aq) + H2O (ℓ) → H3O+ (aq) + A- (aq) ) H3O+ HA (aq) + H2O (ℓ) = H3O+ (aq) + A- (aq) : (2-2 CH3NH2 (aq) ( Na+ (aq) + OH- (aq) ) . C = 10-2 mol.L-1 – pH pH 25 °C :C . pH2 = 10,8 pH1 = 12 : [OH-] : 25 °C [H3O+]1 = 10- pH1 = 10-12 mol.L-1 : ∗ [OH-]1 = ‫ـــ‬1‫ــ‬0‫ـــ‬-‫ـ‬1‫ـ‬4‫ــــ‬ = 10-2 mol.L-1 : Ke = [H3O+].[OH-] = 10-14 [H3O+]1 : [OH-]1 = C ∴ . NaOH NaOH (s) H2O Na+ (aq) + OH- (aq) (‫) م‬: [H3O+]2 = 10- pH2 = 10-10,8 = 1,58 × 10-11 mol.L-1 : ∗ CH3NH2 : [OH-]2 < C ⇐ [OH-]2 = ‫ـــ‬1‫ــ‬0‫ـــ‬-‫ـ‬1‫ـ‬4‫ــــ‬ = 6,33 × 10-4 mol.L-1 [H3O+]2 . CH3NH2 (aq) + H2O (ℓ)( ‫ ) ود و= ازن‬CH3NH3+ (aq) + OH- (aq) :. ( ) HO- B B B (aq) + H2O (ℓ) → BH+ (aq) + HO- (aq). ( ) HO- B (aq) + H2O (ℓ) = BH+ (aq) + HO- (aq) 79

‫–א א‬ ‫ א‬ª‫א א‬ 2007 (3 ‫د رة‬ ( : n (mmol) Cl- ‫ و‬H3O+ (1-3 :10 ‫ـ‬ ) 1,0 L5‫ـ‬ 2,0 pH . HCl 240 mL . VM = 24 L.mol-1 : 20 °C0 HCl x (mmol) HCl (g) + H2O (ℓ) → H3O+ (aq) + Cl- (aq) : • : • 5 10 03 : ‫و‬ HCl (g) + H2O (ℓ) → H3O+ (aq) + Cl- (aq) ‫ر آ ا دة‬ n(HCl) n(H2O) n(H3O+) n(Cl-) ‫اع ا‬ 0 n0(HCl) 0 0 ‫ا ا ةا‬ x x x n0(HCl) – x ‫ا ما‬ xmax n0(HCl) – xmax xmax xmax ‫ ام‬: • xmax = 10-2 mol ⇐ xmax = n0 = ‫ـ‬0‫ــ‬,‫ـ‬2‫ـــ‬4‫ = ــ‬0,01 mol ⇐ n0(HCl) – xmax = 0 24 : xf •xf = 10-2 mol ⇐ xf = [H3O+]f .V= 10-2 × 1,0 = 10-2 mol ⇐ [H3O+]f = 10- pH = 10-2 mol.L-1 ⇐ pH = 2 ‫ـــ‬x‫ـــ‬f‫ = ـــ‬100 % : ‫ـــ‬x‫ـــ‬f‫= ـــ‬ ‫ـ‬1‫ـــ‬0‫ــ‬-‫ـ‬2‫ــ‬ = 1 : ‫ـــ‬x‫ـــ‬f‫ـــ‬ • xmax xmax 10-2 xmax n (mol) 3– . HCl :0,05 ‫ـ‬ CH3COO- ‫ و‬H3O+ 2,86 mL ( ) 500 mL . d = 1,05 . 2,9 pH . CH3COOH (ℓ) + H2O (ℓ) = H3O+ (aq) + CH3COO- (aq) : •0 6,3 × 10-4 CH3COOH ρ0 : n0 = ‫ـ‬m‫ = ــــ‬ρ‫ـــ‬V‫ـ = ـــ‬d‫ــ‬ρ‫ـــ‬0‫ــ‬V‫ ـــ‬: n0 x (mol) M M M 04 : 10-2 mol ( ρ0 = 1 g.mL-1 ) xmax 1‫ـــ‬,‫ـ‬0‫ــ‬5‫ــ×ـــ‬1‫ـــ×ـــ‬0‫ــ‬,‫ـ‬5‫ــ‬ n0 = = 5 × ⇐ ‫ و‬60 : • ‫ر آ ا دة‬ ‫اع ا‬ CH3COOH (ℓ) + H2O (ℓ) → H3O+ (aq) + CH3COO- (aq) ‫ا ا ةا‬ ‫ا ما‬ n(CH3COOH) n(H2O) n(H3O+) n(CH3COO -) 0 n0 0 0 ‫ا ازن‬ x x x n0 – x xmax = n0 xmax = n0 xmax n0 – xmax = 0 :• xmax = 5 × 10-2 mol ⇐ xmax = n0 = 5 × 10-2 mol ⇐ n0 – xmax = 0 : xf • • [H3O+]f = 10- pH = 10-2,9 = 1,26 × 10-3 mol.L-1 ⇐ pH = 2,9 xf = 6,3 × 10-4 mol ⇐ xf = nf (H3O+) = [H3O+]f .V= 6,3 × 10-4 mol ∴ ‫ـــ‬x‫ـــ‬f‫ = ـــ‬1,26 % : ‫ـــ‬x‫ـــ‬f‫ـ = ـــ‬6‫ــ‬,‫ـ‬3‫ــــ×ــــ‬1‫ــ‬0‫ـــ‬-‫ـ‬4‫ = ـ‬0,0126 : ‫ـــ‬x‫ـــ‬f‫ـــ‬ xmax xmax 5 × 10-2 xmax CH3COOH : 4– . : . xf . xmax 80

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬(0<τ<1: « » τ ) τ = ‫ــــ‬x‫ ـــــ‬t . τf = ‫ـــ‬x‫ـــ‬f‫ ـــ‬: xmax (2-3 xmax • . τf = 100 % : τf = 1 ∗ . τf < 100 % : τf < 1 25 °C ∗ : ∗ : • . C = 0,10 mol.L-1 50 mL BA . H3O+ . pH = 2,9 pH •. pHA = 2,9 pH .A 0,5 g . pHB = 5 pH . 0,5 g B . H3O+ [H3O+] :A pH ACH3COOH (ℓ) + H2O (ℓ) → H3O+ (aq) + CH3COO- (aq) : :B [H3O+] pHCH3COO- (aq) + H3O+ (aq) → CH3COOH (ℓ) + H2O (ℓ) : ‫ا‬ ‫ا‬:➀ : CH3COOH (ℓ) + H2O (ℓ) H3O+ (aq) + CH3COO- (aq) pH ➁: ‫آ‬ ‫اا‬ pH : [H3O+] ⇔ ‫را ا‬ ➀ ➁ : [H3O+] ⇔ : : () (=) ∗ ∗ CH3COOH (ℓ) + H2O (ℓ) ➀➁= H3O+ (aq) + CH3COO- (aq) .( ) . C = 0,10 mol.L-1 : ‫ا آ ا‬ : • pH = 2,9 : pH ‫ا ـ‬ : [H3O+]f = [CH3COO-]f = 10- pH = 10-2,9 = 1,26 × 10-3 mol.L-1 :CH3COOH [CH3COOH]f = C - [CH3COO-]f = 9,87 × 10-2 mol.L-1 : HO - H2OH3O + CH3COO - ( 55,5 mol.L-1 : ) H2O (ℓ) : ( : OH- (aq) ) H3O+ (aq) CH3COO- (aq) CH3COOH (ℓ) 05 : ‫و‬ ()CH3COOH ‫ل‬ 5– .( ) 81

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ :. : Qr (3-3 . Qr . αA+βB=γC+δD : Qr = ‫ـ[ـ‬C‫ــ]ـــ‬γ‫ـــ‬.‫ــ[ــ‬D‫ــ]ـــ‬δ‫ـ‬ : Qr [A]α . [B]β.( ) Qr [D] [C] [B] [A] : : Qr – . Qr Qrf Qri – . : Qr – . :➀ : CH3COO- HCOOH HCOOH (aq) + CH3COO- (aq) = HCOO- (aq) + CH3COOH (aq) Qr = [‫ــ‬H‫ـــ‬C‫ـــ‬O‫ــــ‬O‫ـــ‬-‫ــ[ـــ×ــ]ـ‬C‫ــــ‬H‫ـــ‬3‫ــ‬C‫ـــ‬O‫ـــ‬O‫ــــ‬H‫ـ]ـــ‬ : [HCOOH]×[ CH3COO-] . :➁ HCOOH (ℓ) + H2O (ℓ) = H3O+ (aq) + HCOO- (aq) : Qr = ‫ـ[ـ‬H‫ــــ‬3‫ـ‬O‫ــــ‬+‫ـــ[ـــ×ـ]ــ‬H‫ـــ‬C‫ــــ‬O‫ـــ‬O‫ــــ‬-‫ـ]ـ‬ : [HCOOH] × 1 . ( [H2O] = 55,5 mol.L-1 : ) [H2O] = 1 : . :➂ Zn (s) + Cu2+ (aq) = Zn2+ (aq) + Cu (s) : II Qr = ‫ـ[ـ‬Z‫ـــ‬n‫ـــ‬2‫ـ‬+‫ــــ×ـــ]ــ‬1‫ــ‬ : [Cu2+] × 1 . [Cu (s)] = 1 [Zn (s)] = 1 : . Qr :➃ : x Qr –CH3COOH (ℓ) + H2O (ℓ) = H3O+ (aq) + CH3COO- (aq) : :V Qr = [‫ــ‬H‫ـــ‬3‫ــ‬O‫ــــ‬+‫ــ[ـــ×ــ]ـ‬C‫ــــ‬H‫ـــ‬3‫ــ‬C‫ـــ‬O‫ـــ‬O‫ــــ‬-‫ـ]ـ‬ [ CH3COOH] = ‫ـ‬n‫ــ‬0‫ــــ–ـــ‬x‫ـــ‬ [ CH3COO-] = [H3O+] = ‫ــ‬x‫ـــــ‬ [CH3COOH] V V xf 0 x ‫ــ‬x‫ــ‬2‫ـــ‬ ‫ـــــــ‬x‫ــ‬2‫ـــــــــ‬ V2 V(n0 – x) Qr = n‫ـــ‬0‫ـــ–ـــ‬x‫ـــ‬ = ∴ V . Qrf Qri Qr 82

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ : K (4-3 :• :C2 = 5 ×10-3 mol.L-1 : ‫( آ ا‬S2) C1 = 10-2 mol.L-1 : ‫آا‬ (S1) pH2 = 3,56 : ‫و‬ pH1 = 3,4 : ‫و‬CH3COOH (ℓ) + H2O (ℓ) = H3O+ (aq) + CH3COO- (aq) : Qrf[H3O+]f = 10- 3,4 = 3,98 ×10-4 mol.L-1 ⇐ pH1 = 3,4 : (S1) Qrf • [CH3COO-]f = [H3O+]f = 3,98 ×10-4 mol.L-1 ∴ [CH3COOH]f = C1 - [CH3COO-]f = 9,6 × 10-3 mol.L-1 ≈ C1 :(1) ... Qrf = 1,65 × 10-5 ⇐ Qrf = ‫ـ[ـ‬H‫ــــ‬3‫ـ‬O‫ــــ‬+‫ـ]ــ‬f‫ــ[ـــ×ــ‬C‫ــــ‬H‫ـــ‬3‫ــ‬C‫ـــ‬O‫ـــ‬O‫ــــ‬-‫ـ]ـ‬f‫ = ـ‬1,65 × 10-5 : [CH3COOH]f[H3O+]f = 10- 3,56 = 2,75 ×10-4 mol.L-1 ⇐ pH2 = 3,56 : (S2) Qrf • [CH3COO-]f = [H3O+]f = 2,75 ×10-4 mol.L-1 ∴ [CH3COOH]f = C2 - [CH3COO-]f = 4,72 × 10-3 mol.L-1 ≈ C2 :(2) ... Qrf = 1,65 × 10-5 ⇐ Qrf = ‫ـ[ـ‬H‫ــــ‬3‫ـ‬O‫ــــ‬+‫ـ]ــ‬f‫ــ[ـــ×ــ‬C‫ــــ‬H‫ـــ‬3‫ــ‬C‫ـــ‬O‫ـــ‬O‫ــــ‬-‫ـ]ـ‬f‫ = ـ‬1,65 × 10-5 : [CH3COOH]f() Qrf (2) (1). Qrf = C te = 1,65 × 10-5 : Qr Qrf . :K » QrfQrf = 1,65 × 10-5 : Qrf • : . :«CH3COOH (ℓ) + H2O (ℓ) = H3O+ (aq) + CH3COO- (aq) . Qrf K. Qrf αA+βB=γC+δD : Qr = K = ‫ـ[ـــ‬C‫ــ]ـــ‬f‫ــ‬γ‫ـــ‬.‫ــ[ــ‬D‫ــ]ـــ‬f‫ــ‬δ‫ـــ‬ [A]f α . [B]f β .: : K .( ) . K : ‫ا ازن‬ ‫آ‬ ‫ا اا‬ (5-3 : • ) C1 = 10-2 mol.L-1 : ( S2 ) ( S1 )σ1 = 143 × 10-4 S.m-1 : . C2 = 10-3 mol.L-1 . σ2 = 43 × 10-4 S.m-1 .( . 83

‫ א – א א‬ª‫א א‬ 2007 . λ C2H5COO - = 35 mS.m2.mol-1 ‫رة‬ ‫د‬ λ H3O+ = 35 mS.m2.mol-1 : C2H5COOH (ℓ) + H2O (ℓ) = H3O+ (aq) + C2H5COO- (aq) : :( ) : C2H5COOH HO- C2H5COO- H3O+ [HO-] [C2H5COO-]f = [H3O+]f [H3O+] τ100 % : ( S1 ) • 0 σ1 = λ H3O+ [H3O+] + λ C2H5COO - [C2H5COO-] ⇐ σ = λ.C : • • ‫ن‬ [H3O+] = λ‫ـــ‬H‫ــ‬3‫ـ‬O‫ــ‬+‫ـــ‬+[‫ـ‬σ‫ـ‬C‫ــ‬λ1‫ـ‬2‫ــ‬HC‫ــ‬2‫ـ‬H5‫ــ‬C5‫ـ‬C‫ــ‬OO‫ــ‬OO- ⇐-]f σ1 = ( λ H3O+ +λ C2H5COO - ) [H3O+] ∴ ‫أ‬ = [H3O+]f = 3,71 × 10-3 mol.L-1 : : [C2H5COOH]f = C1 - [H3O+]f = 9,63 × 10-3 mol.L-1 : ( S2 ) : [C2H5COO-]f = [H3O+]f = 1,11 × 10-4 mol.L-1 [C2H5COOH]f = C2 - [H3O+]f = 8,89 × 10-4 mol.L-1 Ci : 06 : ‫و‬ : τf = ‫ـــ‬x‫ـــ‬f‫ ـــ‬: ( S1 ) ∗ xmax ‫ا ن أن إ ل ا‬ xf = n H3O+ = [H3O+]f .V xmax = n0 = C1.V ∗ ‫أآ آ آ ن ا آ ا ا‬ ‫ـ[ـ‬H‫ــــ‬3‫ـ‬O‫ــــ‬+‫ـ]ــ‬f‫ـ‬ C1 τf = 3,7 % ⇐ τf = ∴ τf = 11 % ⇐ τf = ‫ـ[ـ‬H‫ــــ‬3‫ـ‬O‫ــــ‬+‫ـ]ــ‬f‫ـ‬ : : ( S2 ) ∗ C2 : . τf :• H2O ( AH ) RCOOH HA (aq) + H2O (ℓ) = H3O+ (aq) + A- (aq) :V C HA (aq) + H2O (ℓ) → H3O+ (aq) + A- (aq) τf = ‫ـــ‬x‫ـــ‬f‫ـــ = ـــ‬x‫ـــ‬f‫ـــ‬ xmax = C.V xmax C.V n(H3O+) n(A-) : n(HA) n(H2O) 0 n0 = C.V 0 0 Qrf = ‫ــ[ـ‬H‫ـــ‬3‫ــ‬O‫ـــ‬+‫ــ]ــ‬f‫ـــ[ـــ×ـ‬A‫ـــ‬-‫ــ]ـ‬f‫ ـ‬: x x [AH]f x n0 – x xf = τf .C.V xf = τf .C.V xf n0 – xf = : C.V(1 – τf) τ‫ــ‬f‫ـ‬.‫ــ‬C‫ـــ‬.‫ـ‬V‫ــــ×ـــــ‬τ‫ـ‬f‫ــ‬.‫ـ‬C‫ـــ‬.‫ــ‬V‫= ــ‬ ‫ــ‬τ‫ـ‬2‫ــ‬f‫ــ‬.‫ـ‬C‫ــــ‬ Qrf = C.V(1 – τf) 1 – τf K= ‫ــ‬τ‫ـ‬2‫ــ‬f‫ــ‬.‫ـ‬C‫ــــ‬ : K = Qr : 1 – τf : . K τf . . τf > 99 % K > 104 84

‫ א – א א‬ª‫א א‬ 2007 ⊕⊖ :( / ‫د رة‬ (4 + : ) : (1-42H2O H3O+ HO – 25 °C . ( σ ) σ = 5,5 µS.m-1 : 07 : ‫و‬ 7.: H2O ‫إ ام ل‬ . HO – H3O+ : pH H3O+ ∗ (*) ... 2 H2O (ℓ) = H3O+ (aq) + HO – (aq)HO – ∗ . [H3O+] = [HO –] : (*) –λ HO –= 20 mS.m2.mol-1 λ =H3O+ 35 mS.m2.mol-1 : σ = λ H3O+ [H3O+] + λ HO – [HO –] : – [H3O+] = [HO –] = ‫ــــــــــــ‬σ‫ــــــــــــــ‬ = 10-7 mol.L-1 ⇐ σ = [H3O+] ( λ H3O+ + λ HO –) ⇐ λ H3O+ + λ HO – . 25 °C pH = 7 ⇐ [H3O+] = 10-pH = 10-7 mol.L-1 : :: HO – H3O+ H2O ( ) 2 H2O (ℓ) = H3O+ (aq) + HO – (aq) Ke = [H3O+] . [OH-] : : .8– . 2 H2O (ℓ) = H3O+ (aq) + HO – (aq) :pKe Ke ‫در ا ارة‬ Ke = 10-14 : 25 °C :15 10-15 0 °C Ke = 10 -pKe ⇔ pKe = - Log Ke : pKe14 10-14 25 °C . ( 25 °C ) pKe = 14 :13 10-13 60 °C 08 : ‫و‬ : pH pKe Ke ‫ات‬ pH 14 0 pH0 7 14 9– .| | | pH : : ‫اا‬ ‫اا‬ [H3O+]éq = [HO –]éqpH < 7 pH > 7 Ke = [H3O+]2 : ‫ل أو ء‬ ‫و‬ Log Ke = Log [H3O+]2 = 2 Log [H3O+] ⇐ pH = 7 pH = 7 ⇐ pH = 1‫ــ‬ pKe : – Log [H3O+] =– 1‫ــ‬ Log Ke ∴ 2 2 09 : ‫و‬ :25 °C pH ‫ا ـ‬ : 1‫ــ‬ pH < 7 ⇐ pH < 2 pKe : [H3O+]éq > [HO –]éq :: pH > 7 ⇐ pH > 1‫ــ‬ pKe : [H3O+]éq < [HO –]éq 2 85

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ : : 1‫ــ‬ 25 °C 2 [H3O+] = [HO –] : pH = 7 pH = pKe pH < 7 [H3O+] > [HO –] : pH > 7 pH < 12‫ــ‬pKe [H3O+] < [HO –] : pH > 1‫ــ‬ pKe 2 :( / ) pKa Ka (2-4 ) Ka :( / : H2O HA HA (aq) + H2O (ℓ) = H3O+ (aq) + A- (aq) : ( HA/A- ) Ka K Ka = K = ‫ــ[ـ‬H‫ـــ‬3‫ــ‬O‫ـــ‬+‫ــ]ــ‬f‫ـــ[ـــ×ـ‬A‫ـــ‬-‫ــ]ـ‬f‫ ـ‬:‫اا ة‬ [AH]f ‫ا‬ Ka : ( HA/A- ) pKa‫ا ض أ ى أآ آ‬ Ka = 10 –pKa ⇔ pKa = - Log Ka‫ا أ ى أآ آ‬ : : 10 – pKa Ka A- . A- . pKa Ka . : HA : pKa ⇔ Ka . HA : pKa ⇔ Ka pKa ‫ا ا ة‬ : ‫ا‬ ∗ 10-2 mol.L-1 10 : ‫و‬ . ( HCOOH/HCOO- ) pKa1 = 3,8 Ka1 = 1,58 × 10-4 ‫ةا ضوا‬ ‫ ر‬. ( CH3COOH/CH3COO- ) pKa2 = 4,8 Ka2 = 1,58 × 10-5 pKa Ka pKa1 < pKa2 ⇔ Ka1 > Ka2 ∴ . CH3COOH HCOOH : . CH3COO- HCOO- : 10-2 mol.L-1 ∗ :( ) pKa1 < pKa2 ⇔ Ka1 > Ka2 ( NH4+/NH3 ) pKa1 = 9,21 Ka1 = 6,2 × 10-10 ( CH3NH3+/CH3NH2 ) pKa2 = 10,6 Ka2 = 2,5 × 10-11 . CH3NH3+ NH4+ : . CH3NH2 NH3 : ‫ا ض أ ى أآ آ‬ :∗ 2,5×10-11 6,2×10-10 1,58×10-5 1,58×10-4 Ka | | | | pKa CH3NH3+ NH4+ CH3COOH HCOOH 10,6 9,21 4,8 3,8 | | | | CH3NH2 NH3 CH3COO- HCOO- ‫ا أ ى أآ آ‬ 86

‫א א‪ ª‬א – א א‬ ‫‪2007‬‬ ‫‪:‬‬ ‫د رة‬ ‫ـ‪f‬ــ]ـ‪-‬ـــ‪A‬ـــ[ـــ×ـ‪f‬ــ]ــ‪+‬ـــ‪O‬ــ‪3‬ـــ‪H‬ــ[ـ‬‫⇔ ) ـ‪f‬ــ]ـ‪-‬ـــ‪A‬ـــ[ـــ×ـ‪f‬ــ]ــ‪+‬ـــ‪O‬ــ‪3‬ـــ‪H‬ــ[ـ ( ‪pKa = – Log Ka = – Log‬‬ ‫= ‪Ka‬‬ ‫‪pKa‬‬ ‫‪pH‬‬ ‫‪:‬‬ ‫) ‪( AH/A-‬‬ ‫‪[AH]f‬‬‫ــ‪f‬ـ]ـ‪-‬ــــ‪A‬ــ[ــ ‪pH = pKa + Log‬‬ ‫‪[AH]f‬‬ ‫⇔‬ ‫‪pKa = – Log [H3O+]f‬‬ ‫ــ‪f‬ـ]ـ‪-‬ــــ‪A‬ــ[ــ ‪– Log‬‬ ‫ــ‪f‬ـ]ـ‪-‬ــــ‪A‬ــ[ــ ‪= pH – Log‬‬ ‫∴‬ ‫‪[AH]f‬‬ ‫‪[AH]f‬‬ ‫‪[AH]f‬‬ ‫)‪: ( /‬‬ ‫‪:‬‬ ‫‪pH‬‬ ‫=‬ ‫‪pKa‬‬ ‫‪+‬‬ ‫‪Log‬‬ ‫ـ‪f‬ـ]ــــــــــــــــ[ـ‬ ‫‪ ]f‬ا [‬ ‫‪:‬‬ ‫‪:‬‬ ‫‪ :‬ـ‪f‬ـ]ــــــــــــــــ[ـ ‪pH = pKa + Log‬‬ ‫‪a]f:‬ا‪pK‬‬ ‫[‬ ‫=‬ ‫‪pH‬‬ ‫‪ = 1 :‬ـ‪f‬ـ]ــــــــــــــــ[ـ‬ ‫‪ = 0 :‬ـ‪f‬ـ]ــــــــــــــــ[ـ ‪Log‬‬ ‫‪pH = pKa‬‬ ‫‪ ]f .‬ا [‬ ‫[ ؛‪]f‬وا [‬ ‫[ = ‪ ]f‬ا‬ ‫‪]f‬‬ ‫∴‬ ‫∴‬ ‫‪pH > pKa :‬‬ ‫∴‬ ‫أو ا‬ ‫ـ‪f‬ـ]ــــــــــــــــ[ـ‬ ‫>‬ ‫‪1‬‬ ‫‪:‬‬ ‫‪Log‬‬ ‫ـ‪f‬ـ]ــــــــــــــــ[ـ‬ ‫>‬ ‫‪0‬‬ ‫‪:‬‬ ‫‪pH > pKa‬‬ ‫‪ ]f‬ا [‬ ‫‪ ]f‬ا [‬ ‫ها ‪.‬‬ ‫ا‬ ‫[؛و نا‬ ‫[ < ‪ ]f‬ا‬ ‫‪]f‬‬ ‫‪pH < pKa :‬‬ ‫ـ‪f‬ـ]ــــــــــــــــ[ـ‬ ‫<‬ ‫‪1‬‬ ‫‪:‬‬ ‫‪ < 0‬ـ‪f‬ـ]ــــــــــــــــ[ـ ‪Log‬‬ ‫‪:‬‬ ‫‪pH < pKa‬‬ ‫‪]f‬ا ا‬ ‫ه‬ ‫[ ؛ ‪f‬و] ا‬ ‫[‬ ‫ا‬ ‫[‬ ‫[ > ‪ ]f‬ا‬ ‫‪]f‬‬ ‫‪.‬‬ ‫نا‬ ‫]‪[A] > [B‬‬ ‫]‪[A] = [B‬‬ ‫]‪[A] < [B‬‬ ‫‪pKa‬‬ ‫|‬ ‫و ‪pH 11 :‬‬ ‫‪A‬ه ا‬ ‫ا‬ ‫ا س‪B‬ه ا‬ ‫ل اا‬ ‫‪pH > pKa‬‬ ‫‪pH < pKa‬‬ ‫) ‪( A/B‬‬ ‫ا‬ ‫‪%‬‬ ‫‪pH = pKa‬‬ ‫)‬ ‫‪:‬‬‫‪100‬‬ ‫)‪( /‬‬ ‫– ‪( 12‬‬ ‫‪pH‬‬ ‫‪.‬‬‫‪50‬‬ ‫‪50 % = A-‬‬ ‫‪% = HA‬‬ ‫[ = ‪ ]f‬ا‬ ‫‪%:‬‬ ‫[ ‪. pH = pKa :‬‬ ‫‪]f :‬‬ ‫‪.‬‬‫‪0‬‬ ‫‪ × 100‬ـــــــــــ‪f‬ــ]ــــــــــــــــ[ـــــــــــ = ‪HA %‬‬ ‫[‪ ]f+‬ا س [‬ ‫‪ ]f‬ا‬ ‫])‪pK[HA(aq‬‬ ‫‪pH‬‬ ‫])‪a [A-(aq‬‬ ‫‪ × 100‬ـــــــــــ‪f‬ــ]ــــــــــــــــ[ـــــــــــ = ‪A- %‬‬ ‫ها ها‬ ‫[‪ ]f+‬ا س [‬ ‫‪ ]f‬ا‬ ‫و ‪12 :‬‬ ‫زا ا‬ ‫) ‪( HA/A-‬‬ ‫‪87‬‬

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ : (3 4 (2) (1) . (3) 13 – . : 7 pH (1) . ∗ .( ) : 7 pH (2)(2) (1) (3) . ∗ . 13 : : 7 pH (3) ∗ pH = 7 : (1) pH < 7 : (2) ) pH > 7 : (3) . ( : • • . ( /) ( HIn/In- ) : HIn (aq) + H2O (ℓ) = H3O+ (aq) + In- (aq) : Ki = ‫ــ[ـ‬H‫ـــ‬3‫ــ‬O‫ــــ‬+‫ــ]ـ‬f‫ــ[ــ×ــ‬I‫ــ‬n‫ــ‬-‫ــ]ـ‬f‫ ـ‬: Ki : [HIn]f ( HIn/In- ) : pH = pKi + Log ‫ــ[ـ‬I‫ــ‬n‫ــ‬-‫ــ]ـ‬f‫ـ‬ [HIn]f . pH ‫ــ[ـ‬I‫ــ‬n‫ــ‬-‫ــ]ـ‬f‫ـ‬ [HIn]f :. pH ≤ pKi – 1 ⇐ Log ‫ــ[ـ‬I‫ــ‬n‫ــ‬-‫ــ]ـ‬f‫ – ≥ ـ‬1 ⇐ ‫ــ[ـ‬I‫ــ‬n‫ــ‬-‫ــ]ـ‬f‫ ≤ ـ‬0,1 ⇐ [‫ــ‬H‫ـــ‬I‫ــ‬n‫ــ]ــ‬f ≥ 10 : [HIn]f [HIn]f [In-]f . pH ≤ pKi + 1 ⇐ Log ‫ــ[ـ‬I‫ــ‬n‫ــ‬-‫ــ]ـ‬f‫ ≥ ـ‬+ 1 ⇐ ‫ــ[ـ‬I‫ــ‬n‫ــ‬-‫ــ]ـ‬f‫ ≥ ـ‬10 : [HIn]f [HIn]f pH pKi – 1 ≤ pH ≤ pKi + 1 :.« »« »« » pKi – 1 pKi pKi + 1 pH | | | • pH ≤ pKi – 1 pKi – 1 ≤ pH ≤ pKi + 1 pH ≥ pKi + 1 – HIn «» In- – –: : – pH (4 4 .( ) . :( . Va : . pH ) () Ca . Cb 88

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ pH –( .) Vb 14 – . Vb – pH . pH = f (Vb) : . :① • Va = 20 mL ( .) pH . Cb = 2,0 × 10-1 mol.L-1 15 – . pH = f (Vb) : Vb : – 14 : ‫و‬ CH3COOH (aq) + HO- (aq) → H2O (ℓ) + CH3COO- (aq) – pH ‫ز ا ة ا ـ‬ : – . pH 0 < Vb < 12 mL : MNpH pH 12 mL < Vb < 13 mL : NP14 ‫ــ‬ . ( pH )12 ‫ــ‬ P Q pH Vb > 13 mL : NP10 ‫ــ‬ .( ) ● ●8 ‫ــ‬ ()6 ‫ــ‬ ●N : E.4 ‫ــ‬ n0 (CH3COOH) = nE (HO-) ⇔ n0 ( ) = nE ( )2 ●‫ ــ‬M :0      Vb (mL) Ca = Cb ‫ـ‬V‫ـــ‬b‫ــ‬E‫ ⇔ ــ‬Ca.Va = Cb.VbE 0 Va 2 46 8 10 12 14 16 18 20 VbE = 12,5 mL : 15 : ‫و‬ Vb pH ‫ر ا ـ‬ . ( pH Ca = 12,5 × 10-2 mol.L-1 ⇐ Ca = 2,0 × 10-1 × 1‫ـــ‬2‫ــ‬,‫ـ‬5‫ = ــ‬12,5 × 10-2 mol.L-1 : 20 ) pH NP E :pH . :② • Va14 ‫ــ‬ NH3 Vb = 20 mL12 ‫ــ‬ M H3O+(aq) + Cl-(aq)10 ‫ـ●ـ‬ N pH . Ca = 4,0 × 10-2 mol.L-18 ‫ــ‬ ●E VbE = 11 mL . 16 – . pH = f (Va)6 ‫ــ‬ ● pHE = 5,6 :–4 ‫ــ‬ P Q NH3 (aq) + H3O+ (aq) → H2O (ℓ) + NH4+ (aq)2 ‫ــ‬ ● ● :E –0     10 12    Va (mL) Cb = Ca ‫ـ‬V‫ــــ‬a‫ـ‬E‫ ⇔ ــ‬Cb.Vb = Ca.VaE 0 Vb 2 4 6 8 14 16 18 20 Cb = 2,2 × 10-2 mol.L-1 : 16 : ‫و‬ Va pH ‫ر ا ـ‬ . ( pH ) pH NP E : : • : − 17 – . E :− . ( pHE ) 18 – . 89

‫א‬ ‫ א – א‬ª‫א א‬ 2007 ‫د رة‬ pH pH14 ‫ــ‬ Vb (mL)12 ‫ــ‬ 14 ‫ــ‬10 ‫ــ‬ 8 ‫ــ‬ 12 ‫ــ‬ 6 ‫ــ‬ 4 ‫ــ‬ ‫ال‬ 10 10 ‫ــ‬ 2 ‫ــ‬ E● 8,2 < pH < 10 8,2 8 ‫ ــ‬E 0 7,6 6 ‫ــ‬ 0 ‫أزرق ا و ل‬ 6 < pH < 7,6 6 ‫ا‬ 3,1 < pH < 4,4 4,4 4 ‫ــ‬ 3,1 2 ‫ــ‬   6  10 12   Vb (mL) 0 2  8 10 12 14 16  20 0 2 4 8 14 16 18 20 46 18 18 : ‫و‬ 17 : ‫و‬ ‫ ا ةا‬E ‫ ا ا از‬E . g(V) = d‫ـــ‬p‫ــ‬H‫ـــ‬ : ) − dV pH = f (V . g(V) = d‫ـــ‬p‫ــ‬H‫ـــ‬ − dV 19 – : .V : σ VE σ = f (V ) 20 – .σ (mS.cm-1) pH6 ‫ــ‬ 14 ‫ ــ‬d‫ـــ‬p‫ــ‬H‫ـــ‬5 ‫ــ‬ 12 ‫ ــ‬dV4 ‫ــ‬3 ‫ــ‬ 10 ‫ــ‬2 ‫ــ‬ 8 ‫ــ‬1 ‫ــ‬ 6 ‫ــ‬ 4 ‫ــ‬     2 ‫ــ‬ 0 2 4 6 8 10 12 14 16 18 20 Vb (mL) 0 2 4 6  10  14 16 18 20 Vb (mL) VE 8 12 19 : ‫و‬ 20 : ‫و‬ ‫ا ما‬ E ‫س‬E ‫اا ن‬ σ ‫اا‬ d‫ـــ‬p‫ــ‬H‫ـــ‬ dV : ( . 214 : – 1 )①: . HCOOH CH3COO- 500 mL .1 . 0,10 mol 0,10 mol .2 .3 .– .4 . ‫ ً أن‬.5 ‫ آ‬.6 . τf ‫اا‬ ( . Qri ‫ا‬ : . ‫ ه ا‬τf ‫م‬ ‫ ا‬. K = 13 .1 ) Qrf ‫ازن ا ا د ه ا ا‬ ‫ ا ا ؟‬τf HCOOH (aq) + CH3COO- (aq) = HCOO- (aq) + CH3COOH (aq) : 90

‫ א – א א‬ª‫א א‬ 2007 ( ) CH3COO- H+ ‫د رة‬ HCOOH .– : .2 HCOOH (aq) + CH3COO- ( aq) = HCOO- (aq) + CH3COOH(aq) n(HCOOH) n(CH3COO-) n(HCOO-) n(CH3COOH) 0 n0 = 0,10 n0 = 0,10 0 0 x n0 – x n0 – x x x xf n0 – xf n0 – xf xf xf [‫ــ‬H‫ـــ‬C‫ـــ‬O‫ــــ‬O‫ـــ‬-‫ــ[ـــ×ــ]ـ‬C‫ــــ‬H‫ـــ‬3‫ــ‬C‫ـــ‬O‫ـــ‬O‫ــــ‬H‫ـ]ـــ‬ : Qri .3 0 .4 n0 (HCOO-) = n0 (CH3COOH) = 0 : Qri = [HCOOH]×[ CH3COO-] = : : τf Qrf: Qrf = ‫ــ[ـ‬H‫ـــ‬C‫ـــ‬O‫ــــ‬O‫ـــ‬-‫ــ]ـ‬f‫ـــ[ــ×ــ‬C‫ــــ‬H‫ـــ‬3‫ــ‬C‫ـــ‬O‫ـــ‬O‫ــــ‬H‫ــ]ـــ‬f‫ـ‬ : [HCOOH]f ×[ CH3COO-]f ‫ــــــــ‬x‫ــ‬f‫ـــ‬2‫ـــــــــ‬ xf = τf . xmax ⇐ τf = ‫ـــ‬x‫ــ‬f‫ ـــــ‬: Qrf = (0,1 – xf ) 2 ⇐ Qrf = ‫ـــــــــــــــ‬x‫ــ‬f‫ـــ‬.‫ــ‬x‫ــ‬f‫ـــــــــــــــ‬ xmax (0,1 – xf ) ×(0,1 – xf ) Qrf = ‫ــــــ‬1‫ــ‬0‫ـــ‬-‫ـ‬2‫ـــ‬τ‫ــ‬f‫ـ‬2‫ـــــــ‬ ⇐ xmax = 0,10 mol : 10-2 (1 – τf )2 ‫ـــــ‬τ‫ــ‬f‫ـــــ‬ 2 ‫ـــــــ‬τ‫ـ‬f‫ــ‬2‫ــــــ‬ 1 – τf ⇐ (1 – τf )2 : Qrf = Qrf = : ‫ ه ا‬τf ‫م‬ ‫ ا ا‬.5 τf = 78 % : τf = 0,78 ⇐ 0 < τf < 1 : ‫ـــــــ‬τ‫ـ‬f‫ــ‬2‫ = ــــــ‬13 ⇐ Qrf = K = 13 : . (1 – τf )2 .6 ( . 215 : – 2 ) ②: ( C6H5COOH/C6H5COO-) – ( / ) : • . pH1 = 3,1 C1 = 1,0 × 10-2 mol.L-1 pH (01 . Ka ( pH1 ( : ( Na+ (aq) + C6H5COO- (aq) ) pH (02 . pH2 = 8,1 C2 = 1,0 × 10-2 mol.L-1 .( pH2 ( . 5,3 pH () (03 ( . –: ( . ‫ا ازن ا ا د ه ا ا‬ ( . 20 mL 20 mL (04 . –: ( . ‫ا ازن ا ا د ه ا ا‬ ( . pH ( C6H5COOH/C6H5COO-) pKa pH ‫( أو‬ . (05 . ( C6H5COOH/C6H5COO-) . pH . pKa ( C6H5COOH/C6H5COO-) = 4,2 : 91

‫א א‪ ª‬א – א א‬ ‫‪2007‬‬ ‫د رة‬ ‫‪:‬‬ ‫‪C6H5COOH (aq) + H2O (ℓ) = H3O+ (aq) + C6H5COO- (aq) :‬‬ ‫‪. (01‬‬ ‫‪ :‬ـ‪f‬ــ]ـ‪-‬ـــ‪O‬ــــ‪O‬ـــ‪C‬ــ‪5‬ـــ‪H‬ــ‪6‬ـــ‪C‬ــ[ــ×ــ‪f‬ــ]ــ‪+‬ـــ‪O‬ــ‪3‬ـــ‪H‬ــ[ـ = ‪Ka‬‬ ‫‪[C6H5COOH]f‬‬ ‫⇐ ‪) = C1.V‬ا ( ‪xmax = n0‬‬ ‫‪C6H5COOH‬‬ ‫⇐‬ ‫‪.‬‬ ‫∴ ‪xf = [H3O+]f .V : ، [H3O+]f = 10-3,1 = 7,9 × 10-4 mol.L-1 ⇐ pH1 = 3,1‬‬ ‫م‬ ‫م ‪ = 7,9 × 10-2 :‬ـ‪f‬ــ]ــ‪+‬ـــ‪O‬ــ‪3‬ـــ‪H‬ــ[ = ـــــ‪f‬ــ‪x‬ـــ = ‪ τf‬أي أن ‪ ⇐ τf = 7,9 % :‬ا‬ ‫اا‬ ‫‪xmax‬‬ ‫‪C1‬‬ ‫∴ او ‪.‬‬ ‫)‪C6H5COO- (aq) + H2O (ℓ‬‬ ‫ردة ا وات ا ء ‪= C6H5COOH (aq) + HO- (aq) :‬‬ ‫د‬ ‫‪. (02‬‬ ‫ـ‪f‬ــ]ـ‪-‬ـــ‪O‬ــــ‪H‬ــ[ــ×ــ‪f‬ـ]ــــ‪H‬ـــ‪O‬ــــ‪O‬ـــ‪C‬ــ‪5‬ـــ‪H‬ــ‪6‬ـــ‪C‬ــ[ـ‬ ‫ا ازن ا‬ ‫‪.‬‬ ‫=‪K‬‬ ‫‪[C6H5COO-]f‬‬ ‫‪:‬‬ ‫دا‬ ‫ا‬ ‫⇐ ‪xmax = C2.V‬‬ ‫‪C6H5COO-‬‬ ‫⇐‬ ‫∴ ‪[HO-]f = 10-5,9 = 1,26 × 10-6 mol.L-1 ⇐ [H3O+] = 10-8,1 = 7,9 × 10-9 mol.L-1 ⇐ pH1 = 8,1‬‬ ‫‪xf = [HO-]f .V :‬‬‫م‬ ‫‪ ⇐ τf < 100 %‬ا‬ ‫أي أن ‪:‬‬ ‫ـــــ‪f‬ــ‪x‬ـــ = ‪τf‬‬ ‫=‬ ‫ـــ‪f‬ـ]ـ‪-‬ــــ‪O‬ـــ‪H‬ــ[ـ‬ ‫‪= 1,26 × 10-4‬‬ ‫ا ا م‪:‬‬ ‫‪xmax C2 :‬‬ ‫‪. (03‬‬ ‫‪ C6H5COOH‬ه ا‬ ‫‪pKa = 4,2 pH = 5,3‬‬ ‫ا س ‪ C6H5COO-‬ه ا‬ ‫‪pH‬‬ ‫‪pH < pKa‬‬ ‫||‬ ‫‪pH > pKa‬‬ ‫ا‬ ‫‪pH = pKa‬‬ ‫ردة ا روآ ‪:‬‬ ‫او‬ ‫ب‪ .‬د‬ ‫)‪C6H5COOH (aq) + HO- (aq) = C6H5COO- (aq) + H2O (ℓ‬‬ ‫د ‪ :‬ــــــــــ‪f‬ــ]ـ‪-‬ــــ‪O‬ـــ‪O‬ــــ‪C‬ـ‪5‬ــــ‪H‬ـ‪6‬ـــ‪C‬ــ[ــــــــــ = ‪K‬‬ ‫‪[C6H5COOH]f ×[HO-]f‬‬ ‫ـ‪ .‬ا ازن ا ا‬ ‫‪Ke = [H3O+].[HO-] :‬‬ ‫‪ :‬ـــ‪a‬ـــ‪K‬ـ = ‪K‬‬ ‫⇐ ــــــــ‪f‬ــ]ـ‪+‬ــــ‪O‬ــ‪3‬ـــ‪H‬ــ[ــــ×ــ‪f‬ــ]ـ‪-‬ــــ‪O‬ـــ‪O‬ـــ‪C‬ــ‪5‬ـــ‪H‬ــ‪6‬ـــ‪C‬ــ[ــــــــ = ‪K‬‬ ‫]‪[HO-‬‬ ‫=‬ ‫‪ :‬ـــــ‪e‬ــــ‪K‬ـــ‬ ‫‪Ke‬‬ ‫‪[C6H5COOH]f ×[HO-]f × [H3O+]f‬‬ ‫]‪[H3O+‬‬ ‫‪:‬‬ ‫‪K = 6,3 × 109‬‬ ‫= ــ‪a‬ــ‪K‬ــ‪p‬ـ‪-‬ـــ‪0‬ــ‪⇐ K = 1‬‬ ‫ـــ‪2‬ـ‪,‬ـ‪4‬ـ‪-‬ـــ‪0‬ــ‪1‬ـ‬ ‫‪= 6,3 × 109‬‬ ‫‪e‬ر‪K‬دة ا وات ‪:‬‬ ‫‪10-14‬‬ ‫‪. (04‬‬ ‫)‪C6H5COOH (aq) + C6H5COO- (aq) = C6H5COO- (aq) + C6H5COOH (aq‬‬ ‫د ‪ = 1 :‬ـ‪f‬ــ]ــــ‪H‬ـــ‪O‬ــــ‪O‬ـــ‪C‬ـ‪5‬ــــ‪H‬ـ‪6‬ــــ‪C‬ـ[ــــ×ــ‪f‬ـــ]ـ‪-‬ـــ‪O‬ــــ‪O‬ـــ‪C‬ــ‪5‬ـــ‪H‬ــ‪6‬ـــ‪C‬ــ[ـ = ‪K‬‬ ‫‪ .‬ا ازن ا ا‬ ‫‪[C6H5COOH]f ×[C6H5COO-]f‬‬ ‫‪ pH‬و ‪: pKa‬‬ ‫ـ‪ .‬ا‬ ‫ــ‪f‬ـــ]ـ‪-‬ـــ‪O‬ــــ‪O‬ـــ‪C‬ــ‪5‬ـــ‪H‬ــ‪6‬ـــ‪C‬ــ[ــ ‪pH = pKa + Log‬‬ ‫‪:‬‬ ‫‪[C6H5COOH]f‬‬ ‫‪V = V1 + V2 = 20 + 20 = 40 mL = 4 × 10-2 L‬‬ ‫ب ‪ pH‬ا ‪ ، :‬ا‬ ‫ـــــ‪1‬ــــ‪V‬ـ‪1‬ـــ‪C‬ـــــ = ‪ = 5 × 10-3 mol.L-1 ⇐ [C6H5COOH]i‬ــ‪0‬ـــ‪2‬ــــ×ــ‪2‬ـ‪-‬ـــ‪0‬ــ‪1‬ـ = ‪[C6H5COOH]i‬‬ ‫‪:‬‬ ‫‪40 V1 + V2‬‬ ‫ـــــ‪1‬ــــ‪V‬ـ‪1‬ـــ‪C‬ـــــ = ‪ = 5 × 10-3 mol.L-1 ⇐ [C6H5COO-]i‬ــ‪0‬ـــ‪2‬ــــ×ــ‪2‬ـ‪-‬ـــ‪0‬ــ‪1‬ـ = ‪[C6H5COO-]i‬‬ ‫‪:‬‬ ‫‪40 : V1 + V2‬‬ ‫‪n0(C6H5COOH) = [C6H5COOH]i.V = 5×10-3 × 4×10-2 = 2×10-4 mol‬‬ ‫‪92‬‬

‫ א – א א‬ª‫א א‬ 2007 K = Qrf = ‫ـــــــــــ‬x‫ــ‬f‫ـ‬2‫ = ــــــــــــ‬1 : ‫د رة‬ pH (2×1x0f-4=–1x0f-24) mol : ∴ pH = pKa + Log ‫ــ[ــ‬C‫ـــ‬6‫ــ‬H‫ـــ‬5‫ــ‬C‫ـــ‬O‫ــــ‬O‫ـــ‬-‫ـــ]ـ‬f‫ ــ‬: [C6H5COOH]f :‫ن‬ ، [C6H5COO-] f = [C6H5COOH]f : pH = pKa = 4,2 % % C6H5COOH % C6H5COO- : (C6H5COOH/ C6H5COO-) ‫زا ا‬ . (05 : .100 . ‫ أ‬HA ‫ا و‬ : pH = 3,1 ∗50 HA % = A- % : pH = 4,2 ∗ . ‫ أ‬A- ‫ ردة ا وات‬: pH = 5,3 ∗0 3,1 pKa 5,3 8,1 pH [HA(aq)] 4,2 [A-(aq)] « – 218 : 3 – »③: . . ً H2O ‫ ا رو و ا ء‬HCl ‫آ ر‬ ‫ا‬ ‫ ؟‬pH = 2,0 .‫د ا ا ء‬ ‫ أآ‬.1 ‫ ا ء ا‬1 L HCl ‫ ز‬0,1 mol ‫ ً أ ّ ذا‬، ‫ ا ل ا‬pH ‫ أ‬.2 ‫لل‬ ‫ ا ء ا‬1 L ‫ ا اب‬HCl ‫ آ أن ن آ ا دة ز‬.3 ( HCl (aq) / Cl- (aq) ) ‫ا‬:‫ت‬ ‫ا‬ : HCl (aq) + H2O (ℓ) = H3O+ (aq) + Cl- (aq) : ‫ا ء‬ ‫ آ د ا‬.1 C = ‫ـ‬n‫ـــ‬H‫ـــ‬C‫ــ‬l‫ـ = ـــ‬0‫ــ‬,‫ـ‬1‫ = ــــ‬0,1 mol.L-1 ⇐ nHCl = C.V : ‫ ا ل ا‬pH .2 V1 pH = 1,0 ⇐ [H3O+] = C = 10-pH = 10-1 mol.L-1 ⇐ .3 : ‫ ا اب‬HCl ‫آ ا دة ز‬ nHCl = C.V = [H3O+].V = 10-pH. V = 10-2 × 1 = 0,01 mol : nHCl = 0,01 mol ∴ « – 219 : 8 – » ④: ‫ و‬، C = 10-2 mol.L-1 ‫ آ ا‬، V = 20,0 mL (CH2ClOOH) ‫آ را‬ ً . pH = 2,37 .‫ا ء‬ ‫ أآ د ا‬.1 . ‫ ا ا‬xmax ‫ ّ ا م ا‬.2 ‫م؟‬ ‫ ه ا ل ا‬. τf ‫م‬ ‫ و ا ا‬xf ‫ ّ ا م ا‬.3 : CH2ClCOOH (ℓ) + H2O (ℓ) = H3O+ (aq) + CH2ClCOO- (aq) : ‫ا ء‬ ‫ آ د ا‬.1 : ‫ ا ا‬xmax ‫ا ما‬ .2 CH2ClCOOH ( ℓ ) + H2O ( ℓ ) = H3O+ (aq) + CH2ClCOO-(aq) ‫ و ا‬، ‫ول ا م‬ 0 0 n0 = C.V n0 – xmax = 0 : ‫ن‬ = 2×10-4 mol xmax xmax xmax = n0 = 2 × 10-4 mol ⇐ n0 – xmax xmax = 2 × 10-4 mol : ‫و‬ xf = [H3O+]f.V = 10-pH. V = 10-2,37 × 0,02 = 8,53 × 10-5 mol : xf ‫ا ما‬ .3 ‫اا‬ . : τf < 1 ⇐ τf = ‫ــــ‬x‫ــ‬f‫= ــــــ‬ 8‫ــ‬,‫ــ‬5‫ــ‬3‫ــــ×ــــ‬1‫ــ‬0‫ـــ‬-‫ـ‬5‫ــ‬ = 0,43 : τf ‫م‬ xmax 2 × 10-4 93

‫א א‪ ª‬א – א א‬ ‫‪2007‬‬ ‫د رة‬ ‫«‬ ‫– ‪– 220 : 12‬‬ ‫‪» ⑤:‬‬ ‫ً )‪ (S1‬ز ا در آ ا ‪ ، C1 = 0,10 mol.L-1‬و ا ـ ‪ pH‬ه ‪11,1 :‬‬ ‫ز ا در ‪ NH3‬ا ء ‪.‬‬ ‫‪ .1‬أآ د‬ ‫آً ا ء‪.‬‬ ‫‪ ّ .2‬أن ‪NH3‬‬‫‪C2 = 2,5×10-2 mol.L-1‬‬ ‫ل )‪ (S2‬ز ا در ‪ ، V2 = 100 mL‬و آ ا‬ ‫‪ .3‬أ ح‬ ‫و ه ا إ ً ‪ V2‬و )‪. (S1‬‬ ‫‪ .4‬إذا آ ن ‪ pH‬ا ل )‪ (S2‬وي ‪ ّ . 10,8‬ا ا م ا ا ل )‪. (S2‬‬ ‫‪ NH3‬ا ء ؟‬ ‫‪ .5‬ذا ا ل ا‬ ‫) ‪(NH4+ / NH3‬‬ ‫ا ت‪:‬ا‬ ‫‪:‬‬ ‫ز ا در ‪ NH3‬ا ء ‪NH3 (g) + H2O (ℓ) = NH4+ (aq) + HO- (aq) :‬‬ ‫‪ .1‬د‬ ‫آً ا ء‪:‬‬ ‫‪ .2‬إ ت ن ‪NH3‬‬ ‫‪NH3 ( g ) +‬‬ ‫) ‪H2O ( ℓ‬‬ ‫=‬ ‫‪NH4+‬‬ ‫‪(aq) +‬‬ ‫)‪HO-(aq‬‬ ‫‪n0 – xmax = 0 :‬‬ ‫ا‬‫‪n0 = C1.V1‬‬ ‫‪0‬‬ ‫‪n0 – xmax‬‬ ‫‪1,3 %‬‬ ‫⇐‬ ‫‪0‬‬ ‫‪xmax = 0,1 V1 ⇐ xmax = n0 = C1.V1‬‬ ‫⇐‬ ‫‪xmax‬‬ ‫‪xmax‬‬ ‫أ ى ‪xf = [HO-]f.V1 :‬‬ ‫و‬ ‫≈ ‪τf‬‬ ‫ـــ‪4‬ــ‪1‬ـ‪-‬ــ‪0‬ـــ‪1‬ـ‬ ‫= ‪τf‬‬ ‫‪[HO-]f‬‬ ‫=‬ ‫ـــــ‪e‬ــــ‪K‬ـــ‬ ‫=‬ ‫‪10-pH‬‬ ‫‪:‬‬ ‫‪[H3O+]f‬‬ ‫ـ‪1‬ــــ‪V‬ـ‪.‬ــ‪4‬ـ‪-‬ــ‪0‬ـــ‪1‬ــــ×ـــ‪6‬ـ‪,‬ـــ‪2‬ــ‪1‬ـ = ــــــ‪f‬ــ‪x‬ــــ‬ ‫‪= 0,0126 ⇐ xf = 12,6 × 10-4.V1 :‬‬ ‫‪xmax‬‬ ‫‪0,1 V1‬‬ ‫‪.‬‬ ‫∴ ‪: τf < 100 %‬‬ ‫( ‪nNH3 = C1.V1 = C2.V2 :‬‬ ‫)‬ ‫‪.3‬‬ ‫‪ = 25 mL :‬ــ‪0‬ــ‪0‬ــ‪1‬ــــ×ـــ‪2‬ـ‪-‬ـــ‪0‬ــ‪1‬ـــ×ــ‪5‬ـ‪,‬ـــ‪ = 2‬ــ‪2‬ــــ‪V‬ـ‪.‬ــ‪2‬ـــ‪V1 = 25 mL ⇐ V1 = C‬‬ ‫‪C1 0,1‬‬‫)‪. (S1‬‬ ‫‪25 mL‬‬ ‫‪100 mL‬‬ ‫•‪:‬‬ ‫‪100 mL‬‬ ‫)‪((S1‬‬ ‫)‬ ‫‪25 mL‬‬‫)‪ (S2‬ه ًا ‪.‬‬ ‫‪32‬‬ ‫(‪.‬‬ ‫)‬ ‫م ‪ τf‬ا‬ ‫ا‬ ‫ل )‪: (S2‬‬ ‫‪:‬‬ ‫ا‬ ‫‪.4‬‬ ‫ا‬ ‫أ ى ‪xf = [HO-]f.V2 :‬‬ ‫‪xmax = 2,5×10-2 .V2 ⇐ xmax = C2.V2‬‬ ‫= ‪[HO-]f‬‬ ‫ـــــ‪e‬ــــ‪K‬ـــ‬ ‫=‬ ‫ـــ‪4‬ــ‪1‬ـ‪-‬ــ‪0‬ـــ‪1‬ـ‬ ‫‪:‬‬ ‫⇐ ‪0,0252‬‬ ‫‪[H3O+]f‬‬ ‫‪10-pH‬‬ ‫‪:‬‬ ‫= ــــــ‪f‬ــ‪x‬ــــ‬ ‫ـــ‪2‬ـــ‪V‬ـ‪.‬ــ‪4‬ـ‪-‬ــ‪0‬ـــ‪1‬ــــ×ـــ‪3‬ــ‪,‬ــ‪6‬ــ‬ ‫‪xf = 6,3 × 10-4.V2‬‬ ‫و‬ ‫‪τf ≈ 2,5 %‬‬ ‫⇐‬ ‫‪τf‬‬ ‫=‬ ‫‪xmax‬‬ ‫‪2,5 × 10-2.V2‬‬ ‫=‬ ‫‪.5‬‬ ‫‪τf 1 < τf 2 :‬‬‫داد ا د و ة ا ‪.‬‬ ‫‪ NH3‬ا ء‬ ‫«‬ ‫– ‪– 221 : 18‬‬ ‫‪» ⑥:‬‬‫‪ C1 = 0,10 mol.L-1‬و‬ ‫) ‪( 2Na+ + SO32-‬‬ ‫‪V1 = 30 mL‬‬ ‫‪. C2 = 0,10 mol.L-1‬‬ ‫‪V2 = 30 mL‬‬ ‫ا دث ‪.‬‬ ‫‪ .1‬أآ د ا‬ ‫‪. .2‬‬ ‫‪. Qri‬‬ ‫‪.3‬‬ ‫(‬ ‫‪) τf Qrf‬‬ ‫‪.4‬‬ ‫‪. τ . K = 251‬‬ ‫‪.5‬‬ ‫ا ت ‪ :‬ا ت ) ‪(HSO3- / SO32- ) ، (CH3COOH / CH3COO-‬‬ ‫‪:‬‬ ‫‪ .1‬آ د ا ا دث ‪CH3COOH (aq) + SO32- (aq) = CH3COO- (aq) + HSO3- (aq) :‬‬ ‫‪):‬أ ا ا ا (‬ ‫‪ .2‬ول ا م‬ ‫‪94‬‬

‫ א – א א‬ª‫א‬ ‫א‬ CH3COOH (aq) + SO32- (aq) = HSO3- (aq) + CH3COO-(aq) 2007 .3 n2 = C2.V2 n1 = C1.V1 0 0 ‫د رة‬ n2 – xf n1 – xf xf xf : Qri : Qri =[[CC‫ــــ‬HH‫ــــ‬33‫ـ‬CC‫ــــ‬OO‫ــــ‬OO‫ــــ‬H-‫ــ]ـ‬i‫×ـ]ـ‬i‫ـ[ـ[ــ×ــ‬H‫ـ‬S‫ــ‬O‫ـ‬S‫ــ‬O‫ـ‬3‫ــ‬2‫ـ‬3‫ـ‬-‫]ـ‬-‫]ـ‬ii Qri = 0 : ، [CH3COO-]i = [HSO3-]i = 0 : K = Qrf = ‫ـ[ـ[ـ‬C‫ـ‬C‫ـــ‬HH‫ـــ‬3‫ـ‬3‫ـ‬C‫ـ‬C‫ـــ‬OO‫ــــ‬OO‫ــــ‬H-‫ــ]ـ‬f‫ـ]ـ‬f×‫[ــ[ـ×ــ‬H‫ــ‬S‫ــ‬SO‫ـــ‬O‫ـ‬3‫ــ‬2‫ـ‬3‫ـ‬--‫ــ]]ـ‬ff‫= ــ‬ ‫ـــــــــــــــــ‬x‫ــ‬f‫ــ‬2‫ـــــــــــــــــــ‬ : τf Qrf .4 ( n1 - xf) ( n2 - xf) : n1 = n2 = xmax : ‫ــ‬x‫ـــ‬f‫ ــــ‬2 ⇐ n1 = n2 : ‫ـــــ‬τ‫ــ‬f‫ــ‬2‫ــــــــ‬ K = Qrf = xmax =2 (1 – τf)2 : (xmax)2 Qrf ‫ــ‬n‫ــ‬1‫ـــ‬-‫ـــ‬x‫ـــ‬f‫ـــ‬ xmax ‫ـــــ‬τ‫ــ‬f‫ــ‬2‫ــــــــ‬ K = Qrf = (1 – τf)2 : : τ .5 : K = Qrf K = 251 τf = 0,94 = 94 % ⇐ 250 τf 2 – 502 τf – 251 = 0 « – 223 : 25 – » ⑦: . ( Corrosif ) . pH [ ( HCOOH/HCOO- ) ] % % % .100 ‫ــ‬ % ‫أ س‬% pH = pKa .1 . pKa 80 ‫ــ‬ .% % pH = 5 ‫ أ‬.260 ‫ــ‬ [HCOOH]éq = 2 [HCOO-]éq : pH ّ .340 ‫ــ‬ pH ‫إ د ا ـ‬ ‫ه‬20 ‫ــ‬ ‫ ؟‬pKa pH 0       pH = pKa + Log ‫ـ[ــ‬A‫ــــ‬-‫ ـــ]ـ‬: : 0 2 4 6 8 10 12 14 pH [AH] .1 . pH = pKa : ‫ ن‬، [AH] = [A-] : pKa ( HCOOH/HCOO- ) = 3,8 ⇐ pH = pKa = 3,8 : : pH = 5 ‫أ‬ % % .2 % AH = 6 % : HCOOH : % A- = 94 % : HCOO- : [HCOOH]éq = 2 [HCOO-]éq : pH .3 ‫ــ[ـــــــــــــ‬H‫ــــ‬C‫ـــ‬O‫ــــ‬O‫ــــ‬H‫ــ]ــــ‬é‫ـ‬q‫ـ = ــــــــــــــ‬2‫ = ـــ‬67 % : : [HCOOH]éq + [HCOO-]éq 3 % AH = 67 % : HCOOH % A- = 33 % : HCOO- pH ≈ 3, 5 : pH 95

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ « – 224 : : 26 – » ⑧ : : CBA pH 3,1 – 4,4 A∗ . 4,8 – 6,4 . ) . C B∗ 5,2 – 6,8 6,8 – 8,0 ( pH C∗ 8,2 – 10,0 .1 .2 11,6 – 14,0 pH : pH : pHB = 6,8 . pHA ≤ 4,8 : A .1 pHB ≥ 6,8 . pHA ≥ 4,4 pHB ≤ 6,8 A 4,4 ≤ pHA ≤ 4,8 : B: ‫ و‬pHC ≥ 10,0 : C 10,0 ≤ pHC ≤ 11,6 : ‫ و‬، pHC ≤ 11,6 ( ) C ‫ إ ر‬.2 . C pH < 10,0 . pHCpH « – 226 : 30 – » ⑨:14 ‫ ــ‬. CB VB = 20 mL12 ‫ــ‬10 ‫ــ‬ CA = 10-2 mol.L-1 VA 8 ‫ــ‬ pH 6 ‫ــ‬ pH = f (VA) 4 ‫ــ‬ . .1 .2 ‫ــ‬ .K .20   E (VAE , pHE) .30 4 8 12 16 20 24 28 VA (mL) pH = 2 .4 pH = 9,2 pH = 5,2 . pKa3 (H2O/HO-) = 14 pKa2 (H3O+/H2O) = 0 pKa1 (NH4+/NH3) = 9,2 : : NH3 (aq) + H3O+ (aq) = NH4+ (aq) + H2O (ℓ) : ‫ة‬ ‫ا‬ ‫ آ د‬.1 .2 :K K= ‫ـ[ــــــــــ‬N‫ــــ‬H‫ـــ‬4‫ــ‬+‫ــ]ــ‬é‫ـ‬q‫ــــــــــــ‬ = ‫ــــــ‬1‫= ــــــ‬ ‫ــــــ‬1‫= ــــــ‬ ‫ــــــ‬1‫ــــــ‬ = 1,58 × 109 [NH3]éq × [H3O+]éq Ka1 10-pKa1 10- 9,2 K = 1,58 × 109 ∴ : E (VAE , pHE) .3 E (VAE = 18 mL , pHE = 5,6) : .4 – : – – . ‫ ه ا‬NH4+ pH < pKa1 ⇐ pKa1 = 9,2 : pH = 2 . ‫ ه ا‬NH4+ pH < pKa1 ⇐ pH = 5,2 .‫ا ل‬ ‫ ا ان ا‬NH4+/NH3 pH = pKa1 ⇐ pH = 9,2 96

‫ א – א א‬ª‫א א‬ 2007 « – 227 : 33 – ‫د رة‬ » ⑩: . ρ = 1,23 kg.L-1 : ( déboucheur ) •:. 20 % ‫روآ ا د م‬ ‫اا ا‬ . M ( NaOH ) = 40 g.mol-1 • . 20 mL 10 mL 5 mL : • . 1000 mL 500 mL 100 mL : . 25 mL : . pH (S)‫ا ل‬ . .1. CA = 0,10 mol.L-1 ‫أ ح‬.(S) 100 .2 ( S ) VB = 20 mL .3 : pHVA (mL) 2 4 6 8 10 11 12 12,5 13,5 14 15 16 17 20 22 25 pH 12,7 12,6 12,5 12,3 12,0 11,6 10,8 10,0 2,9 2,4 2,1 1,9 1,7 1,5 1,4 1,3 .( . pH = f (VA) ( . ( . (S) ( .1 ( : : .1ρ = 1,23 kg.L-1 = 1230 g/L NaOH (s) 20 % Cm(NaOH) = m/V = 0,20 × ρ = 0,2 × 1230 = 246 g/L : : C = Cm/M = 246/40 = 6,2 mol.L-1 : : .2 100 mL ( hotte )10 mL 10,0 mL . 100 mL . 250 mL 1000 mL . 32 . 32pH : ( .314 ‫ــ‬ H3O+ (aq) + HO- (aq) = 2 H2O (ℓ)12 ‫ــ‬ ( ) pH = f (VA) (10 ‫ ــ‬: (8 ‫ــ‬ •E :6 ‫ــ‬ E ( VAE = 13 mL , pHE = 7 )4 ‫ــ‬ : (S) (2 ‫ ــ‬: () CA.Véq = CB.VB ⇐ xéq = nHO- = nH3O+0   0 4 8 12 16 20 24 28 VA (mL) pH = f (VA) : CS =C‫ـــ‬A‫ـــ‬.‫ـ‬V‫ـــ‬é‫ــ‬q‫ = ــ‬0‫ــ‬,‫ــ‬1‫ــ‬0‫ــــ×ــــ‬1‫ــ‬3‫ـــ‬ = 6,5 × 10-2 mol/L ⇐ CB = CS : VB 20 (S) 100 CS0 = 100 CS = 6,5 mol/L : H3O+ (aq) + HO- (aq) = 2 H2O (aq) n(H3O+)éq n(HO-)éq ( n – xéq = 0 n – xéq = 0 :( )1 ∆‫ـــ‬C‫ـ|ـ = ــــ‬6‫ـــ‬,‫ـ‬5‫ــــ–ـــ‬6‫ــ‬,‫ــ‬2‫ = ـــ|ــ‬6,5 % C 6,2 97

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ . :5 : . . . . : (1 :( ) (1 1 ( –– ) . . . (1– ) .. . .( ) : 01 : (2 1(1727 – 1642) : .1.2.1 ) . z (C) ‫ ر‬، 2. : . ( z . ‫ ( – ا‬O, i→ , j→, k→) : ∗ M yy . ‫نو أ‬ ‫أ دا آ‬ • (t=0) :∗ → :- k→ () →j . iO ()xx . M : 02 :- z y x: ( héliocentrique ) : x → = → → z → : «» ) (S) ‫ و ه‬، « Repère de Copernic : ‫اا‬ OM i+ yj+ k ( géocentrique ) .( » 3– ‫ا‬. ‫ا آ د‬ : S . () 4– ‫ا‬. : T . : 03 () 98

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬ `[ : .2.2.1 W. :- () :- : ( ) : . : 04 : .3.2.1 () :C y .C ( ) •M2 •G •Mn ( )C : •M1 M• 3 (5– )G . ( M1 , M2 , M3 , … , Mn ) →j Z O→G.∑n mi = m1. → + → + → + … + → OM1 m2.OM2 m3.OM3 mn.OMn → i=1 x k O → : (3 1 i z r→ : .1.3.1 G : 05 Oxyz :- ( M1 , M2 , M3 , … , Mn ) r→= x → y →j + z → : (x,y,z) i+ k P1 t r→1 P1 r→2 : y P1(t) ∆→r (6– ) (t + ∆t) • • P2(t+∆t) ∆→r = r→2 – r→1 = ∆x →i + ∆y →j + ∆z k→ v→moy → ⊕^ :- r1 r→2 (C) : ^ → ‫ـ→ــ∆ــ‬r‫ــ‬ : ∆t x vmoy = Oz : 06 . ∆→r →vmoy ∆r→ : - → vmoy y P(t) →v b ∆t ∆t → 0 : • t : (C) ) →vmoy → v→ : →r ⊕^ .(7– O (C) : ^ →v =∆lti→m0→vmoy =∆lti→m0‫ـ→ــ∆∆ــ‬rt‫ــ‬ ‫ــ‬d‫ــ‬r‫ــ→ـ‬ : x dt =z →v = vx →i + vy →j + vz →k : : 07 vz = ‫ــ‬d‫ــ‬z‫ـــ‬ ‫ــ‬d‫ــ‬y‫ـــ‬ ‫ــ‬d‫ــ‬x‫ـــ‬ : v→ dt dt dt vy = vx = 7– . →v 99

‫ א – א א‬ª‫א א‬ 2007 ‫د رة‬. m.s-1 m/s : (S.I) v→moy v = √ vx2 + vy2 + vz2 : v→ : ‫→ـــ∆ـ‬v‫ـــ‬ ∆t - ∆t ∆t → 0 : →amoy = :« » t(C) →aG : a→moy → a→G : . →aG =∆lti→m0→amoy =∆lti→m0∆‫ـ∆ــ‬v→‫ــ‬tG‫ــ‬ = ‫ـ‬d‫ــ‬v‫ـ→ـ‬G‫= ــ‬ ‫ـ‬d‫ــ‬2‫ــ‬r→‫ــ‬ : dt dt2 a→G = ax →i + ay →j + az →k : ‫ـ‬d‫ــ‬v‫ـــ‬z‫ـ‬ ‫ـ‬d‫ــ‬2‫ــ‬z‫ــ‬ ‫ـ‬d‫ــ‬v‫ـــ‬y‫ـ‬ ‫ـ‬d‫ــ‬2‫ــ‬y‫ــ‬ ‫ـ‬d‫ــ‬v‫ـــ‬x‫ـ‬ ‫ـ‬d‫ــ‬2‫ــ‬x‫ــ‬ : dt dt2 dt dt2 dt dt2 az = = ay = = ax = =. m.s-2 m/s2 : (S.I) aG = √ ax2 + ay2 + az2 : : ( ‫ ا ن ا ول ) أ ا‬.2.3.1 :« »“ : ”() : ) .( .3.3.1 ) :( . () 8– . aG‫ـ→ـ‬a‫ـــ‬ ‫ــ‬a→‫ـــ‬ -2 3 → → 8-ⓐ . m 2m 3m aG - F F ⓐ 8-ⓑ . F 2a→ 3a→ F = k.maG : F/m aG . k: → m ⓑ → 1 kg 1 N 2F : 08m 3F F = maG : k = 1 : 1 m/s2 1 N = 1 kg.m/s2 : ∑ → = m a→G : Fext : (G ) ∑ → ” → a→G Fext Fext “ ∑ = m : → a→G : • F (9– .G →vG • v→G ) G .G ∆v→ • a→G ⊕^ ^ 100 → : 09 F