L’eau de Javel et l’eau de Dakin " " : II :
1 : ! "# $ %$ 5 − − .O 2(g) /H 2 O (l) ClO (aq) /Cl (aq) : & R=8,314u.S.I : "( ) * "+ , Ke = 1,00.10-14 : 25°C -./ ( $ 0 "+ ,
Cl −(aq) 6 6() $9 ClO −(aq) 56() 578 $9 :(/ ;7 3 <(7 41 2 % "3 ) - =G H?$ - B ;-/ C D+ E F 56() 578 = >?+ @ A . HO −(aq) - 6- $9 Na +(aq) $9 liqueur de Dakin " L(7 " 2 578 -K 4I # "E J " $ -(( * . ;6 N)?8 "3 ) "7 : L & (1 : < () Q# 3 ) L7 O 56() 578 $ P- 9 OE / N)?
2ClO − ( aq ) → 2Cl − ( aq ) + O2 ( g ) . (S1) L(7 V = 250ml
! :(/ K7. >7 S-! (S0) ;6 L(7 5 R?J$ N)? 0 . & 6 D# V1= 20,0ml TK$ , U6( VW9 X0J$ "Y7 -./ @ *Z$ "+ , <86! "6 $\ D[8 "(! (S1)L(7 t "?(J Y7 -./ H@$ T2 . "2+ T = 300 K . + @ P(t) "(7 * ]*[ + , "(7 ! :@# <( L7 LV^ . "( * #$ P0 : _ t = 0 min -./ 3-# ]*>( & $ . V0 @ L&^1 P- `( $ )a T (1.1 . N)?8 / ?
O 2(g) Cl −(aq) ClO −(aq)
"3-# P :(/ n03 , n02 , n01 S! b 6- 3 ) L7( ?W $ (12
x (mmol)
. -. n 03 n 02 F + (112
1
"1-+ *( "() "3-# P " n0 / #/ (212 0.8
n0 "1-+ n03 / #/ , --78 D O+ L - 89 , / ? 0 x f 3 . c-@ / #/(312
0.6
. "3 . " 7
0.4
T)8 x(t) c-@ @@7 "CV 5#,9 ?W L - L 2 + (13 : )Z :(/
0.2
20
40
60
80
100
120 t (min)
x(t ) =
(P(t ) − P0 ).V0 R.T
(<#$ )1 )Z :.7. ]^ <. x(t) c-@ _ ! d) "/ 68 LV^ ]*> b C (14 t "Y7 -./ X#8 ]/ v(t) "7 "/ A/ (114 . 6 0X / L B 7 -! 4 & LV^ / ? "/2 68 R (214 "C $ # -!. ClO − ( aq )
S0
= 5, 0.10−2 mol.L−1 : 56() 578 $ & (S0) ;6 "@K :(/ 9@$ (314 . NF (/ f #8 . O@7 g. :(/ "#% P6 ha X .n01
" "" L(7 "26 : $ % & .2 cK $+. 6-X L(7 (S2) _ < & $ ;0 R?J :\ " [\ ;7 ;0 0X "26 i@$ . 3,00.10-2 mol.L-1 3,00.10-1 mol.L-1
:(/ <+ $+) $9 $+. 6-X $9 &8 S7+ NaHCO3
− . X#8 ]/9 , " &( K A1 ">7 "+ , A/ . ClO (aq) 56() 578 = O j7( " a "*K T9 (12
[ClO ] / #/ . pK A1 = 7,3 −
− .ClO (aq) L O j7 &8 [HA] S!
[HA]
=9 pH1= 10,3 ; (S2) L(7 pH =9 (/ (22 f k. 8 F
3 / ?8 l-7 . pK A2 = 10,3 . S!K A2 >! "+ ,
HCO −3(aq) /CO 2− 3(aq)
" & j! X $+. 6-X = (32 −
−
. (S2) L(7( ".HO (aq) $9 HCO 3(aq) + −
−
.HO (aq) HCO 3(aq) + W 7 / ? " T9 (132 . X6-C "#86 -! K e K A2 "1-+ K O+ = "+ , / #/ (232 (42 .pH2 = 9,3 <C L(7 pH =9 _ 7+ 5#,9 (142 . (S2) L(7 pH *W9 RY. 0X pH F + (242
[ClO ] P-- "# . T !9 (342 −
.
[HA]
. L-+ "E J " $ i ( RY. L(7 L 2 m -! .@ ##2 ]/9 (442
. ;6 N)?8 "3 ) "7 : L & (1 @ L&^1 P- `( $ )a 1.1 إ ال ClO − + 2 H + + 2e − → Cl − + H 2O أآ ة 2 H 2O → O2 + 4 H + + 4e−
ClO − / Cl − O2 / H 2O (12
: -. n 03 n 02 (112 "3-# P =o <. lV% I$ "Y7 ;9 > ](J =o " " <$9 + n + ClO- =9 N)? . "3 ) "/ 68 "26- VW9 X0J$ "Y7 ;9 -. = $)2 Cl- O2 ) (212
PV 0 0 = n0 RT ⇒ n0 =
PV 0 0 R.T .
"() "3-# P " n0 / #. :( / ) *( : <. 1/5 X O2 "# $ T !
n03 =
n0 PV = 0 0 5 5R.T
: "() * P " T)$ ?W L - (13
nT (t ) = n0 + x(t ) ⇒ x(t ) = nT (t ) − n0 x (t ) = x(t ) =
P(t )V PV − 0 0 وT ( =T 0 ), ( ٍوV =V 0 ) RT RT0
( P(t ) − P0 )V0 و RT
x(t) c-@ _ ! d) "/ 68 LV^ ]*> b C (14 1)Z :.7. ]^ <. "7 "/ (114 : " "CV + / ? "/2 A8
.6 0X / L B 7 4 & LV^ / ? "/2 68 (214 pC .8 "/ =o <. & LV^ pC . :.7.( 2 ( < ( "@( "@ . & LV^ . * ]*[ X 3 ) 6 0X / L B 7
: 56() 578 $ & (S0) ;6 "@K :(/ 9@$ (314
ClO − ( aq ) = 5, 0.10−2 mol.L−1 S0
: $ #
• -4
n0 = 0,2 mmol =2.10 mol : • ClO−( aq) 5,0.10−2 S0 n0 = ClO ( aq) .V0 = .V0 = .20.10−3 = 2.10−4 mol S1 5 5 −
" "" L(7 "26 : $ % & .2 − P-/ @ X ClO (aq) 56() 578 = O j7( " a "*K (12
HClO j7( "@
→ H 3O + + ClO − HClO + H 2O ← H 3O + aq . ClO − aq éq éq K A1 = HClO aq éq
: .- (22 ClO − ( aq ) ClO − ( aq ) éq éq pH = pK A1 + log ⇔ log = pH − pK A1 HClO( aq ) HClO( aq ) éq
éq
ClO ( aq ) éq pH − pK A1 ) = 10( HClO( aq ) −
⇒
éq
ClO ( aq ) éq = 10(10,3− 7,3) = 103 ⇔ ClO − ( aq ) = 103. HClO( aq ) éq éq HClO( aq ) −
éq
( I. ) P X P-/ @
(32
→ H 2O( aq ) + CO32− ( aq ) HCO3− ( aq ) + HO − ( aq ) ←
:
−
−
:HO (aq) HCO 3(aq) + W 7 / ? " (132
X6-C "#86 K e K A2 "1-+ K O+ = "+ , 232
K=
. CO32 − aq HCO3
−
éq
. HO − ( aq ) éq
aq éq
1 − H 3O + aq . CO32 − aq K HO ( aq ) éq 1 1 éq éq ⇒ = = K A2 = . − + − K A2 H 3O aq HCO3 aq HO ( aq ) H 3O + aq éq éq éq éq 1 −14 −10,3 −24,3 −25 ⇒ K = K A2 .K e = 10 .10 = 10 = 5.10 (42 L(7 pH _ ! (142
pH 2 = pK A 2
CO 3 2 − ( aq ) éq + log − HCO 3 ( aq )
éq
CO3
2−
= 0, 3 mol . L ; HCO 3 − ( aq ) = 3.10 − 2 mol .L− 1 éq éq −1
( aq )
⇒ pH 2 = 10, 3 + log
0, 3 = 10, 3 − 1 = 9, 3 0, 03
. ( )(S2) L(7 pH *W9 ( L7 )RY. 0X pH F =9 . (242 -
HO (aq) $9 " & n0X j! + / ?8 l-7 HClO / ClO − : j!/P-/ C " & :(/ <3! + L(7 . n-! *W9 L(7 pH=9 ;9 pH pC .8 T#
[ClO ] P-- "# . T 7. (342 −
.
[HA]
ClO ( aq ) ClO ( aq ) éq éq ⇔ log = pH 2 − pK A1 pH 2 = pK A1 + log HClO( aq ) HClO( aq ) −
−
éq
éq
ClO − ( aq ) éq pH − pK ⇒ = 10( 2 A1 ) HClO( aq ) éq
ClO ( aq ) ClO − ( aq ) éq éq ( 9,3−7,3) 2 = 10 = 10 ⇔ = 102. HClO( aq ) HClO( aq ) éq éq −
"$- PH "C X L-+ "E J " $ i ( RY. L(7 L 2 m ;0 D.@ T# (442 . "E J " $ 9 -( :(/ ,B8 =9 ) 1 $ )1 -# 9 DC H?$ #/ L K81 I[ 0X :(/ O(8 9 "Y!V )
[email protected]
