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Copy of exercice condu

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 72
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.& 8-KNO3(s)$ A '4 [email protected] I'7 "!
c = 2, 0 ×10 mol ⋅ L−1 b;7 K $ A 'Y f I 0 ",
Y& f I 0 l=1,0 cm]qS=1,0 cm²]'0 > ":
−3
λ NO3−
= 7,1 × 10 −3 S ⋅ m 2 ⋅ mol −1 . λ K + = 7,4 × 10 −3 S ⋅ m 2 ⋅ mol −1 )[email protected]
eau
KNO3(s) 
→ K (+aq ) +NO3(− aq ) ).& 8-KNO3(s)$ A '4 [email protected] "!
σ = (λK + + λNO − ) × c )$ A 'Y f ",
3
c = 2, 0mol ⋅ m −3 )=J_0 mol ⋅ m −3 5 mol ⋅ L−1 # ;7 ' /0=3I
−3
−2
−1
σ == (7, 4 + 7,1).10 × 2 = 2,9.10 S .m )CBA
−4
S
−2 10
−4
G = σ × = 2,9 × 10 × −2 = 2,9 ×10 S. )Y& f ":
l
10
K = ]'' f & 1U L V$ 0,05 mol.L-1b;7 6J 79Y f 1U
0,01 m
G = 10 mS
Y& 9 f I 0 "!
896J f& r& 45 7] & f& l''$ ",
0,005 S.m2.mol
G = K ⋅ σ ) ' [email protected] f & f& [email protected]"!
σ =
G 0,010
=
= 1,0 S .m −1 )4
K 0,010
I'(.& 8-6J 794 [email protected] [ Na + ] = [OH − ] = C 6J 79Y 8-",
NaOH ( S ) → Na(+aq ) + OH (−aq ) )=( σ = Σλi [ X i ] = λ Na × [ Na + ] + λOH × [OH − ] )9Y& f @-8'
Ù σ = λ Na × C + λOH × C
Ù σ = C × (λ Na + λOH ) www.bestcours.net
+
−
+
−
+
λOH =
−
σ
C
−
− λ Na + Ù (λ Na + + λOH − ) =
σ
C
Ù
1,0
− 0,005 = 0,015 S ⋅ m 2 ⋅ mol −1 )CBA
50
−3
mol ⋅ m 5 mol ⋅ L−1 # ;7 ' /0= ΔχϮΤϠϣ
λOH =
−
73
[ =7/0 6 λ & f& c8& ;7 ' σ f& #R 8' [email protected] 74"!
5×10-3 mol·L-1.b;7 (Ag + + NO3- ) GN A 'Y f 25°C60",
13.3 mS.m-189H'f r7 45 GN A 'Y&8& ;7 ' I 0 ":
2
-1
+
2
-1
λ(N O3− ) = 7,14 mS.m .mol λ(Ag ) = 6,19 mS.m .mol )[email protected]
σ = λi ⋅ ci )λ & f& c8& ;7 ' σ f& #R 8' [email protected] "!
i
) 89 f& /0 mol ⋅ m −3 ) 89 ;7 ' /0 S.m2.mol-1 )89 λ & f& /0 g
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S·m-1
σ = (λAg + λNO ) ⋅ c ) GN A 'Y f ",
+
−
3
c = 5 mol·m-3 =J- mol ⋅ m −3 5 mol ⋅ L−1 # ;7 ' /0Y
σ = (6,19 + 7,14) ×10−3 ⋅ 5 = 6, 66 ×10−2 S·m-1 )45
c =
σ
λAg + λCl
+
=
−
0, 0133
= 1mol ⋅ m −3 = 10−3 mol ⋅ L−1 ) GN A '8& ;7 ' ":
−3
(6,19 + 7,14) × 10
Mg2+(aq) + 2 ;h& A '8S Y '@
C0b;7 NO3 –(aq)
H<l' Y&@'$;7 ' K9 1U L V$ ]$U?Y& K]G f &
]''
G = 0,025 S]'&Ur(-K = 0,1 m
Y& K9 f I 0 "!
iY&8& ;7 ' & YXL# l''$ ",
mol.m-3
60λNO3- = 0,00714 S.m2.mol-1 &":
Mg2+X & f& &U
l' &H<& = @& @60"F
λMg2+λNO3- G 0,025
σ = =
= 0,25 S ⋅ m −1 )45 G = Kσ )Y& f "!
K
0,1
C0 = 10 3C0 = 0,010 mol.L-1 σ = 0, 25 S ⋅ m −1 f& [- & ;7 ' &U & YXL# ",
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mol.m-3
C09;h& A ';7 &":
2+
–
[NO3 ] = 2 × C0[Mg ] = C0) σ = Σλi [ X i ] = λNO × 2C0 + λMg × C0 )9 f& @-8'
−
2+
3
λMg =
2+
0, 25
σ
− 2 × 0, 00714 = 0, 011 S ⋅ m 2 ⋅ mol −1 )CBA λMg 2+ =
− 2λNO − )45
3
10
C0
x = C y = σ g y = a × xH'[email protected] [email protected]& =f3# &?[' #/ σ = f (C ) Y=E&& & "F
a = (2λNO + λMg
−
3
2+
H<& = @& a
) )9H<& = @& 45 σ = C × (2λNO + λMg ) Ù σ = λNO × 2C + λMg × C )
74
−
2+
−
3
2+
3
)# (& 8' R> 178- G
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0,02 mol/Lb;7 *( P&0Y # 30 mLj
0,01 mol/Lb;7 6J 79Y # 10 mLj
878S&( Y' '@
)[email protected]
ion
Na+
ClH3O+
HO
-3
-3
-3
-3
-1
λ Sm²mol 5,01 10 7,63 10 34,98 10 19,86 10
6J 79*( +&0#=N' S&( [email protected]& I'7 "!
JU [' & =N'& 60=N' [Y<‚3",
S] Y& 8-/< '& A ;7 I 0 ":
S] 8-R>σ f& 60"F
H
3
+
O
+
H
O
-
H 3O + + HO- → 2H 2 O )=N' [email protected] "!
",
=N' [email protected]
2 H 2 O
→
n1 =0,03 × 0,02 = 6 10-4 mol n 2 =0,01× 0,01 = 10-4 mol
6 10-4 -x
10-4 -x
6 10-4 -xmax
10-4 -xmax
oIK n/ -
S / -
/ -
=N' YXL
S] )9JU [' OH 7] 9& =N'& - n1 = 6 10 mol n 2 = 10 mol ) &
xmax=10-4 mol
) S] 8-A ;7 ":
−
n [Na ]f =
+
n f ( Na + )
Vt
-4
-4
n i ( Na + ) 0,01× 10
=
= 2,5.10-3mol L-1 = 2,5mol.m -3 )6J ;7 j
Vt
( 30+10 )
oT N' HkV h'H6 &745=N' 8-ƒ M 6J
n f ( Cl- ) n i ( Cl- ) 0,02 × 30
9( n [Cl ]f =
=
=
= 1,5.10-2 mol L-1 = 15mol.m -3 )( ;7 j
Vt
Vt
40
-
oT N' Lu
o =N' n OH − f =
n f ( OH
Vt
−
) = 10
−4
− xm
= 0mol / L ) 7] ;7 j
Vt
n f ( H 3O + ) 6.10-4 − xm
5.10-4
[H 3O ]f =
=
=
= 1,25.10-2 mol L-1 =12,5mol.m -3 ) 7 ;7 j
-3
Vt
Vt
40.10
+
σ = λCl . Cl − + λNa . Na + + λOH . OH − + λH O . H 3O + )"F
σ =10-3 ( 34,98 × 12,5 + 7,63 ×15+5,01× 2,5) = 0,564 S.m -1 )CBA
−
+
−
+
3
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75
=J-25°C/ <6 V[& .& 8- Fe3 ( PO4 ) 2 ,8H 2 O H&& II AN$-# m '7IK
439 mS.m-1H'f Y
H&& II AN$-4 [email protected] I'7 "!
/< '& AX & AXf& *K7Y&C;7 ' Y& f @R ",
Y&C8& ;7 ' 60":
Y& K9# 1,00 L.YJ cX i K& '7l''$ "F
λ PO = 27,84 mS ⋅ m 2 ⋅ mol −1 λ Fe = 10,70 mS ⋅ m 2 ⋅ mol −1 )[email protected]
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2+
3−
4
Fe3 ( PO4 ) 2 ,8H 2 O( S ) → 3 Fe(2aq+ ) + 2 PO43− ( aq ) + 8 H 2 O( L ) )H&& II AN$-4 [email protected] "!
σ = Σλi [ X i ] ) [email protected]",
2+
3−
PO4 Fe )&9Y&#< '& #
σ = λ Fe × [ Fe 2+ ] + λ PO × [ PO43− ] )9 f& @45
2+
σ = C (3λ Fe
2+
3−
4
[ PO43− ] = 2 × C [ Fe 2+ ] = 3 × C )- ' 7' AX @& '
+ 2λ PO ) Ù σ = λ Fe × 3C + λ PO × 2C )^J f& @45
3−
4
2+
3−
4
σ = C (3λ Fe + 2λ PO ) )":
2+
C =
3−
4
σ
(3λ Fe 2 + + 2λ PO 3− )
)4
4
C =
0,439
= 5,00 mol ⋅ m −3 )CBA
−3
(3 × 10,70 ⋅ 10 + 2 × 27,84 ⋅ 10 )
−3
5,00 × 10-3 mol.L-1)-mol.L-1/0 5Y
)Y& # 1,00 L.YJ cX i K& '7"F
M = 3 × M Fe + 2 × M P + 8 × M O + 8 × M H O )i K& & '( 3I M = 3 × 55,8 + 2 × 31,0 + 8 × 16,0 + 8 × 18,0 = 501 g ⋅ mol −1 )CBA
n = C × V)=( I'(/6& &7C = 5,00 × 10-3 mol.L-1)9i K&8& ;7 ' &
m = C × V × MÙm = n × M)89i K& '7m = 5,00 × 10-3 × 1,00 × 501 = 2,51 g)CBA
2
)[email protected]
)25°C / <6 & Af& +
-3
2
λ(Na (aq)) = 5,01.10 S.m .mol-1;λ(HCOO-(aq)) = 5,46.10-3 S.m2.mol-1
M(C) = 12 g.mol-1 ; M(H) = 1 g.mol-1 ; M(O) = 16 g.mol-1 ; M(Na) = 23 g.mol-1 ) & ='(
8-HCOONa(s)6J A E # m = 68,0 mg '7 4~8S Y&V = 100,0 mL&0 Gj
V[& .&
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.& 8-6J A E 4 [email protected] I'7 "!
c;7 ' I 0 M(HCOONa(s)) V ,m Y&c8& ;7 ' @R ",
Y& 8-/< '& [email protected] 8& ;7 ' I 0 ":
]'&UI 0 ?c;7 ' Y&σ1 f& @R "F
#S ' #6 '( # (' f L &h?k Y& 5 V[& .& # V’&0QGj
U = 1,00 =J-S = 3,21 cm2&h& .; 0 L = 1,00 cm - &]=JN#c ' @-/V
#'NJ #&h& Y& 8-/ I = 2,47 mA
k > '?G f & I 0 "x
Y&σ2 f& l''$ "y
QN>& Y&c’8& ;7 ' I 0 "„
* <[email protected] V[& .&V’? l''$ "
76
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HCOONa(s) → HCOO-(aq) + Na+(aq)).& 8-6J A E 4 [email protected] "!
n
m
)Y&c8& ;7 ' @",
c = 0,010 mol.L-1)CBAc = =
V
(M × V)
)Y& 8-/< '& [email protected] 8& ;7 ' ":
+
n ( Na (aq) ) = n ( HCOO (aq) ) = n ( HCOONa (s ) ) =c × V )=N' [email protected] YXL# [Na+(aq)] = [HCOO-(aq)] = 0,010 mol.L-1 = 10 )CBA[Na+(aq)] = [HCOO-(aq)] = c)45
mol.m-3
σ1 = λ(Na+(aq)) × [Na+(aq)] + λ(HCOO-(aq)) × [HCOO-(aq)])Y&σ1 f& @"F
σ1 = (λ(Na+(aq)) + λ(HCOO-(aq))) × c)-[Na+(aq)] = [HCOO-(aq)] = c) &
σ1 = (5,46.10-3 + 5,01.10-3) × 10 = 1,05.10-1 S.m-1)CBA
I 2,47.10-3
G = =
= 2,47.10-3 S)G f & i 0"x
U
1,00
L 1,00.10-2
= 31,2 m-1) k > '
k= =
S 3,21.10-4
σ2 = k × G = 2,47.10-3 × 31,2 = 7,69.10-2 S.m-1)Y&σ2 f& "y
)QN>& Y&c’8& ;7 ' "„
+
σ2 = (λ(Na (aq)) + λ(HCOO (aq))) × c’)45c’;7 'c;7 ' R[- h fFY a UX[
c’ = σ2/(λ(Na+(aq)) + λ(HCOO-(aq))))-8'
c’ = 7,69.10-2/(5,46.10-3 + 5,01.10-3) = 7,35 mol.m-3)C"A
n = c’ × (V+ V’))-8'i K& ? /6 &7 h'QN>' &YXL) V[& .&V’? "
-3
-3
-1
-2
V’ = 1,0.10 /7,35.10 - 1,0.10 = 3,6.10 L = 36 mL)CBAV’ = n/c’ – V)45
# &h& .; f (_C1b;7 6J (S1Y # 100 mL8- f L &h
G1 = 1,22 mS89#6 '( #Y&
.; f (_C2b;7 6J (S2Y # 100 mL8- f & L`N &h?
G2 = 1,44 mS89#6 '( #Y& # &h&
C2;7 ' # hf 3 7 C1;7 ' 7 4 #i 0"!
Y& 8- [ f & L &h?ƒ S1Y # 100 mL5 V[& .& # 100 mLQG",
Y& K]G3 f & &Ul''$ C1 H=J& Y&C;7 ' 9 H=J&
Cb;7 ‚-(H=J& Y& #S2Y # 100 mLS1Y # 100 mLT;&":
Y&G4 f & &UI 0 ?C2C1#;7 ' V$' &U=E&
Y&C2;7 ' # hf S1Y&C1;7 ' )-Y& ;7 g 6 q I$'G f & &"!
S2
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G3. =
77
G1
C
)89Y& K9 f -8' C = 1 b;7 -QJ 5QNLY& &",
2
2
G3 = 0,61 Ms)CBA
n n +n
C ×V + C ×V
C = = 1 2 = 1 1 2 2 )H=J& Y& ;7 ":
Vt
Vt
Vt
V
C1 + C2
)- V1 = V2 = t ) &
C2C1#;7 ' V$' &U=E&C;7 ' -8' C =
2
2
G + G2
G4 = 1
= 1,33 mS )-Y& ;7 g 6 q I$'G f & &
2
+
-3
2
-l
λ(K ) = 7,35. 10 S. m .mol ; λ(Cl-) = 7,63. 10-3 S. m2.mol-l) & f& )[email protected]
8S Y # .;< f =< # =&@' & 8 ' I7 ' =E&=(c?U"!
-3
c = 5,00.10 b;7 (K+ + Cl-)$ ((S1)8S Y # .;<1 f L V$ ",
mol.L-1
I1 = 3,52 89Y& 8-& 8S ]( ' /-U1 = 0,800 V#'NJ 8V # ' ( mA
Y& # &h& .;G1 f & I 0 !",
U2 = 0,500 V 08-]$U#(&& I2' /89 ,",
(S1)Y&σ1 f& I 0 [email protected]& & f&*'@'$:",
'E bK] V & = @ 89 k > ' [ O$ # l''$ F",
-3
-1
c = 5,00.10 mol.L b;7 ' Rb+ + Cl- ((S2)Y g [ > `N=&@' ":
G2 = 4,53 .10-3 SH'f f& l''$ Cl-Rb+# # =( & f& (S2)Y&σ2 f& @R
λ(Rb+) &
=J& (S3) Y&σ3 f& @< (S2)Y& # 100 mL(S1)Y& # 100 mLT;&"F
]'&UI 0 ?σ2σ1 H
)=&@' & 8 ' I7 ' "!
# '&VN R 8 ' / 8- ' R _0
'/ & 6;#08- f & 1U L8V
i'
)Y& # &h& .;G1 f& !",
G1 =I1/U1 = 3,52.10-3/0,800 = 4,40.10-3 S
)- h'Y&G f & &)I2' /,",
I2 = G1 × U2 = 2,2.10-3 A)4 G1 = I2/U2
)(S1)Y&σ1 f& :",
+
+
+
σ1 = λ(K ) × [K ] + λ(Cl ) × [Cl ] = c × (λ(K ) + λ(Cl-)) = 0,0749 S.m-1
k = σ1/G1 = 17,0 m-14 σ1 = k × G1) [email protected])k > 'F",
σ2 = λ(Rb+) × [Rb+] + λ(Cl-) × [Cl-] = c × (λ(Rb+) + λ(Cl-)(S2)Y&σ2 f& @":
))
+
) λ(Rb ) & f&
k × G2
σ2 = k × G2 = 0,0770 S.m-1λ(Rb+) =
- λ(Cl-) =7,79.10-3 S.m2.mol-1 c
)(S3) Y&σ3 f& @"F
78
+
+
+
+
σ3 = λ(Rb ) × [Rb ]’ + λ(Cl ) × [Cl ]’ + λ(K ) × [K ]’ )
)(S3)Y& 8-A ;7 I [Rb ]'=
+
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n ( Rb + )
V1 + V2
[K ]'=
+
[Cl ]'=
-
=
n (K+ )
V1 + V2
n ( Cl- )
V1 + V2
=
c × V1
c
+
= ) Rb A ;7 j
V1 + V2 2
=
c × V2
c
+
= )K A ;7 j
V1 + V2 2
c × V1 + c × V2
= c )Cl A ;7 j
V1 + V2
+
λ(Rb )
λ(K+) σ1 σ2
σ3 = c × (
+ λ(Cl ) +
) = + = 0,0760 S.m-1)45
2
2
2 2
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