ﻓﻨﺤﺼل ﻋﻠﻰ ﻤﺴﺘﻘﻴﻡ ﻴﻤﺭ ﻤﻥ ﺍﻟﻤﺒﺩﺃ ﻤﻴﻠﻪ ﻤﻭﺠﺏ ،ﻤﻌﺎﺩﻟﺘﻪ ﻤﻥ ﺍﻟﺸﻜل : G = a . cﺤﻴﺙ a :ﻤﻘﺩﺍﺭ ﺜﺎﺒﺕ ) ﻤﻌﺎﻤل ﺍﻟﺘﻭﺠﻴﻪ ( ﻨﺘﻴﺠﺔ :ﺍﻟﻨﺎﻗﻠﻴﺔ Gﺘﺘﻨﺎﺴﺏ ﻁﺭﺩﺍ ﻤﻊ ﺍﻟﺘﺭﻜﻴﺯ ﺍﻟﻤﻭﻟﻲ cﻟﻠﻤﺤﻠﻭل ﻤﻼﺤﻅﺔ :ﻫﺫﻩ ﺍﻟﻨﺘﻴﺠﺔ ﻤﺤﻘﻘﺔ ﺒﺸﺭﻁ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﻤﺤﻠﻭل ﺍﻟﺸﺎﺭﺩﻱ ﻤﻤﺩﺩ ﺃﻱ 4c ≤ 10-2 mol /Lـ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ σﻟﻤﺤﻠﻭل ﺸﺎﺭﺩﻱ ) : ( conductivité ﺭﺃﻴﻨﺎ ﺃﻥ ﻨﺎﻗﻠﻴﺔ ﺠﺯﺀ ﻤﻥ ﻤﺤﻠﻭل ﺸﺎﺭﺩﻱ ﻫﻲ ﻤﻘﻠﻭﺏ ﻤﻘﺎﻭﻤﺘﻪﻜﻤﺎ ﺭﺃﻴﻨﺎ ﺃﻥ ﺍﻟﻨﺎﻗﻠﻴﺔ Gﺘﺘﻨﺎﺴﺏ ﻁﺭﺩﺍ ﻤﻊ ﻤﺴﺎﺤﺔ ﺍﻟﺠﺯﺀ ﺍﻟﻤﻐﻤﻭﺭ ﻤﻥ ﺍﻟﻤﺴﺭﻴﻴﻥ Sﻭ ﻋﻜﺴﺎ ﻤﻊ ﺍﻟﻤﺴﺎﻓﺔ Lﺒﻴﻥ ﺍﻟﻤﺴﺭﻴﻴﻥ ﺃﻱ :ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ﺍﻟﻨﺎﻗﻠﻴﺔ Gﺘﺘﻨﺎﺴﺏ ﻁﺭﺩﺍ ﻤﻊ ﺍﻟﺠﺩﺍﺀ ﺃﻱ ﻤﻊG= σ. S ﻭ ﻨﻜﺘﺏ : L ﺤﻴﺙ σﻫﻲ ﺜﺎﺒﺕ ﺍﻟﺘﻨﺎﺴﺏ ﻭ ﺘﺴﻤﻰ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﻟﻠﻤﺤﻠﻭل ﺍﻟﺸﺎﺭﺩﻱﻭ ﺘﺘﻌﻠﻕ ﺒﻁﺒﻴﻌﺔ ﺍﻟﻤﺤﻠﻭل ﺍﻟﺸﺎﺭﺩﻱ ﻭ ﺒﺘﺭﻜﻴﺯﻩ ﻭ ﺘﻘﺎﺱ ﻓﻲ ﺠﻤﻠﺔ ﺍﻟﻭﺤﺩﺍﺕ ﺍﻟﺩﻭﻟﻴﺔ ﺒﻭﺤﺩﺓ ) S . m-1 ( siemens / mètreﺜﺎﺒﺕ ﻴﻤﻴﺯ ﺍﻟﺸﻜل ﺍﻟﻬﻨﺩﺴﻲ ﻟﺨﻠﻴﺔ ﻗﻴﺎﺱ ﺍﻟﻨﺎﻗﻠﻴﺔ ﻭ ﻴﺭﻤﺯ ﺇﻥ ﺍﻟﻤﻘﺩﺍﺭ ﻟﻪ ﺒﺎﻟﺭﻤﺯ Kﻭ ﻭﺤﺩﺘﻪ . m-1ﺍﻟﺠﺩﻭل ﺍﻟﻤﻭﺍﻟﻲ ﻴﻤﺜل ﻗﻴﻡ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﻟﻤﺤﻠﻭل ﻜﻠﻭﺭ ﺍﻟﺒﻭﺘﺎﺴﻴﻭﻡ ﺘﺭﻜﻴﺯﻩ 0.1 mol / Lﻋﻨﺩ ﺩﺭﺠﺎﺕ ﺤﺭﺍﺭﺓ ﻤﺨﺘﻠﻔﺔ .
درﺟﺔ اﻟﺤﺮارة) ( ° C )σ (mS.cm-1 17 10.95 18 11.19 19 11.43 20 11.67 21 11.97 22 12.15 23 12.39 24 12.64 25 12.88 5ـ ﺍﻟﻨـﺎﻗـﻠـﻴـﺔ ﺍﻟﻨﻭﻋﻴـﺔ ﺍﻟﻤـﻭﻟﻴـﺔ λiﺍﻟﺸـﺎﺭﺩﻴـﺔ ): (conductivité molaire ionique ﻟﻘﺩ ﺭﺃﻴﻨﺎ ﺃﻥ ﺍﻟﻤﺤﻠﻭل ﺍﻟﺸﺎﺭﺩﻱ ﺒﻪ ﺸﻭﺍﺭﺩ ﻤﻭﺠﺒﺔ ) ﻜﺎﺘﻴﻭﻨﺎﺕ( ﻭ ﺸﻭﺍﺭﺩ ﺴﺎﻟﺒﺔ )ﺃﻨﻴﻭﻨﺎﺕ( ﻭ ﻟﻪ ﻨﺎﻗﻠﻴﺔ ﻨﻭﻋﻴﺔ ، σﺘﺴﺎﻫﻡ ﻓﻴﻬﺎ ﺍﻟﻜﺎﺘﻴﻭﻨﺎﺕ ﻭ ﺍﻷﻨﻴﻭﻨﺎﺕ ﺃﻱ : σ = σ+ + σ- ﺤﻴﺙ :ﺍﻟﻨﺎﻗﻠﻴﺔ σ+ﺘﺘﻨﺎﺴﺏ ﻁﺭﺩﺍ ﻤﻊ ﺘﺭﻜﻴﺯ ﺍﻟﻜﺎﺘﻴﻭﻨﺎﺕ ) ﺍﻟﺸﻭﺍﺭﺩ ﺍﻟﻤﻭﺠﺒﺔ (ﻭﺍﻟﻨﺎﻗﻠﻴﺔ σ-ﺘﺘﻨﺎﺴﺏ ﻁﺭﺩﺍ ﻤﻊ ﺘﺭﻜﻴﺯ ﺍﻷﻨﻴﻭﻨﺎﺕ ) ﺍﻟﺸﻭﺍﺭﺩ ﺍﻟﺴﺎﻟﺒﺔ (ﻓﻤﻥ ﺃﺠل ﻤﺤﻠﻭل ﺸﺎﺭﺩﻱ ﺘﺤﺼﻠﻨﺎ ﻋﻠﻴﻪ ﻤﻥ ﺇﺫﺍﺒﺔ ﻤﺭﻜﺏ ﺸﺎﺭﺩﻱ ﺼﻠﺏ ) AB(Sﻓﻲ ﺍﻟﻤﺎﺀ ﻴﻜﻭﻥAB( S : A+( aq ) + ) B- ( aq) + H2O ] ) σ+ = a . [A+( aq ﺇﺫﻥ : ] ) σ- = b . [B- ( aqﺍﻟﻤﻌﺎﻤل aﻫﻭ ﺜﺎﺒﺕ ﺍﻟﺘﻨﺎﺴﺏ ﻭﻴﺴﻤﻰ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﻟﻴﺔ ﺍﻟﺸﺎﺭﺩﻴﺔ ﻟﺸﻭﺍﺭﺩ ) A+( aqﻭ ﻴﺭﻤﺯ ﻟﻬﺎ λA+ﻭﺍﻟﻤﻌﺎﻤل bﻫﻭ ﺃﻴﻀﺎ ﺜﺎﺒﺕ ﺍﻟﺘﻨﺎﺴﺏ ﻭﻴﺴﻤﻰ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﻟﻴﺔ ﺍﻟﺸﺎﺭﺩﻴﺔ ﻟﺸﻭﺍﺭﺩ ) B- ( aqﻭ ﻴﺭﻤﺯ ﻟﻬﺎ λB-] ) σ+ = λA+ . [A+( aq ﻴﻤﻜﻥ ﺇﺫﻥ ﺃﻥ ﻨﻜﺘﺏ :] ) σ- = λB- . [B- ( aq
ﺘﻘﺎﺱ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﻟﻴﺔ ﺍﻟﺸﺎﺭﺩﻴﺔ ﺒﻭﺤﺩﺓ ) ( S . m2 . mol-1 ﺤﻴﺙ ﺍﻟﺘﺭﺍﻜﻴﺯ ﻤﻘﺎﺴﺔ ﺒﻭﺤﺩﺓ mol / m3ﺒﺼﻭﺭﺓ ﻋﺎﻤﺔ ﻴﺭﻤﺯ ﻟﻠﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﻟﻴﺔ ﺍﻟﺸﺎﺭﺩﻴﺔ ﺒﺎﻟﺭﻤﺯ λiﻭﻫﻲ ﺘﻤﺜل ﻤﺴﺎﻫﻤﺔ ﺸﺎﺭﺩﺓ ﻤﺎ ﻓﻲ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﻟﻤﺤﻠﻭل ﻭ ﻫﻲ ﺘﺘﻌﻠﻕ ﺒﺩﺭﺠﺔ ﺍﻟﺤﺭﺍﺭﺓ ) ﺘﻨﺎﺴﺏ ﻁﺭﺩﻱ ( ﻭﻁﺒﻴﻌﺔ ﺍﻟﺸﺎﺭﺩﺓ . ﺍﻟﺠﺩﻭل ﺍﻟﻤﻭﺍﻟﻲ ﻴﻤﺜل ﻗﻴﻡ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﻟﻴﺔ ﺍﻟﺸﺎﺭﺩﻴﺔ ﻟﺒﻌﺽ ﺍﻟﻜﺎﺘﻴﻭﻨﺎﺕ ﻭﺍﻷﻨﻴﻭﻨﺎﺕ ﻋﻨﺩ 25° C λi (S . m2 . mol-1 ).10-3ﺍﻟﻜﺎﺘﻴﻭﻥ ﺍﻷﻨﻴﻭﻥ λi (S . m2 . mol-1 ).10- 3)H3O+(aq 35,0 )OH-(aq 19,9)Li+(aq 3,86 )Cl-(aq 7,63)Na+(aq 5,01 )Br-(aq 7,81)Mg2+(aq 10,6 )F-(aq 5,54 )K+(aq 7,35 )I-(aq 7,70)NH4+(aq 7,34 )NO3-(aq 7,14)Ag+(aq 6,19 )MnO4-(aq 6,10)Ca2+(aq 11,9 )HCO3-(aq 4,45Ba2+(aq 12,7 )CO32-(aq 13,9)Zn2+(aq 10,56 )CH3CO2-(aq 4,09)Fe2+(aq 10,7 )SO42-(aq 16,0
)Al3+(aq 18,9 )H2PO4-(aq 3,3)Fe3+(aq 20,4 )C6H5CO2-(aq 3,23 6ـﺎﻟﻌﻼﻗﺎﺕ ] σ =Σλi[Xiﻭ G = a . cﻓﻲ ﺍﻟﻤﺤﺎﻟﻴل ﺍﻟﺸﺎﺭﺩﻴﺔ ﺍﻟﻤﻤﺩﺩﺓ :ﺍﻟﺠﺩﻭل ﺍﻟﻤﻭﺍﻟﻲ ﻴﺒﻴﻥ ﻗﻴﻡ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﻟﺜﻼﺜﺔ ﻤﺤﺎﻟﻴل ﻤﻥ ﻜﻠﻭﺭ ﺍﻟﺼﻭﺩﻴﻭﻡ ﻤﺘﺴﺎﻭﻴﺔ ﺍﻟﺤﺠﻭﻡ ﻤﺨﺘﻠﻔﺔ) C( mol / L 10-3 2.10-3 ﺍﻟﺘﺭﺍﻜﻴﺯ . )σ(S/m 126 . 10-4 250 . 10-4 3.10-3 380 . 10-4 NaCl(s ) + H2O ﺤﻴﺙ ﻜﻠﻭﺭ ﺍﻟﺼﻭﺩﻴﻭﻡ ﻴﺘﻔﻜﻙ ﻓﻲ ﺍﻟﻤﺎﺀ ﻭﻓﻕ : ) Na+(aq) + Cl-(aqﻤﻥ ﺨﻼل ﻫﺫﺍ ﺍﻟﺠﺩﻭل ﻴﻤﻜﻥ ﺍﻟﻘﻭل ﺃﻥ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﻟﻠﻤﺤﻠﻭل ﺘﺘﻨﺎﺴﺏ ﻁﺭﺩﺍ ﻤﻊ ﺍﻟﺘﺭﻜﻴﺯ c ≤ ) c ( 10-2 mol /L )σ = Λ . c ………..(1 ﺃﻱ: ﺤﻴﺙ Λﺜﺎﺒﺕ ﺍﻟﺘﻨﺎﺴﺏ ﻭ ﻴﻌﺭﻑ ﺒﺎﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﻟﻴﺔ ﻟﻠﻤﺫﺍﺏ ﻓﻲ ﺍﻟﻤﺤﻠﻭل ﻭﻴﻘﺎﺱ ﺒﻭﺤﺩﺓ )( S . m2 . mol-1 σ = σ+ + σ- ﻭ ﻗﺩ ﺭﺃﻴﻨﺎ ﺃﻥ ] ) σ = λA+ . [A+( aq ) ] + λB- . [B- ( aq ﺃﻱ : ] σ = Σ λi [ Xi ﺒﺼﻭﺭﺓ ﻋﺎﻤﺔ : ﻭ ﻟﻜﻥ [A+( aq ) ] = c : [B- ( aq ) ] = c σ = ( λA+ + λB- ) . c )………….. (2 ﻭ ﻤﻨﻪ : ﻤﻥ ﺍﻟﻌﻼﻗﺘﻴﻥ ) (1ﻭ ) (2ﻨﺠﺩ : Λ = λA+ + λB- Λ = Σ λi ﺒﺼﻭﺭﺓ ﻋﺎﻤﺔ : ﻤﻼﺤﻅﺔ :ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻤﺫﺍﺏ ﻤﻥ ﺍﻟﺸﻜل ) AB2(sﻓﺈﻨﻪ ﻴﺘﻔﻜﻙ ﺤﺴﺏ )AB2(s )A2+(aq) + 2 B-(aq
ﻭ ﻴﻜﻭﻥ [A+( aq ) ] = c : [B- ( aq ) ] = 2 cﺇﺫﻥ :ﺒﺘﻁﺒﻴﻕ ﺍﻟﻌﻼﻗﺔ ] σ = Σ λi [ Xi) σ = (λA2+ . c + λB- . 2c ﻴﻜﻭﻥσ = (λA2+ + 2 λB- ) . c ﺃﻱﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻟﻨﺎﻗﻠﻴﺔ ﻭ ﺍﻟﺘﺭﻜﻴﺯ ) : G = f ( c ﻓﻲ ﺍﻟﻔﻘﺭﺓ 3ـ 4ﺭﺃﻴﻨﺎ ﻤﻥ ﺨﻼل ﺍﻟﺩﺭﺍﺴﺔ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ ﻟﻠﻌﻭﺍﻤل ﺍﻟﻤﺅﺜﺭﺓ ﻓﻲ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻤﺤﺎﻟﻴلﻤﻤﺩﺩﺓ ﺃﻱ ﺘﺭﻜﻴﺯﻫﺎ c ≤ 10-2 mol /Lﺃﻥ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺘﺘﻨﺎﺴﺏ ﻁﺭﺩﺍ ﻤﻊ ﺍﻟﺘﺭﻜﻴﺯ ﻭ ﻫﺫﺍ ﻤﺎ ﺘﺜﺒﺘﻪ ﺍﻟﺩﺭﺍﺴﺔ ﺍﻟﻨﻅﺭﻴﺔ ﺍﻟﻤﻭﺍﻟﻴﺔ :G= σ. ﻨﻌﻠﻡ ﺃﻥ : SL ﻭ=K L S ﺤﻴﺙ K :ﻫﻭ ﺜﺎﺒﺕ ﺍﻟﺨﻠﻴﺔG= σ ﻭﻤﻨﻪ : ﺇﺫﻥ K σ=K.Gσ= Λ.c ﻭ ﻟﻜﻥ ﺭﺃﻴﻨﺎ ﺃﻥ : ﺒﺎﻟﻤﻁﺎﺒﻘﺔ ﺒﻴﻥ ﺍﻟﻌﻼﻗﺘﻴﻥ ﻨﺠﺩ :K.G =Λ.cG = KΛ . c ﻭ ﺒﺎﻟﺘﺎﻟﻲ : ﺇﺫﻥ : G =a.cﺃﻱ ﻓﻌﻼ ﺍﻟﻨﺎﻗﻠﻴﺔ Gﻟﻠﻤﺤﺎﻟﻴل ﺍﻟﺸﺎﺭﺩﻴﺔ ﺍﻟﻤﻤﺩﺩﺓ ﺘﺘﻨﺎﺴﺏ ﻁﺭﺩﺍ ﻤﻊ ﺍﻟﺘﺭﻜﻴﺯ .ﺒﻌﺩ ﻫﺫﺍ ﺍﻟﻌﺭﺽ ﻫل ﺘﻤﻜﻨﺕ ﻤﻥ ﻓﻬﻡ ﻜﻴﻔﻴﺔ ﺍﻟﺘﺄﻜﺩ ﻤﻥ ﺘﺭﻜﻴﺯ ﺍﻟﻤﺼل ﺍﻟﻤﻘﺘﺭﺡ
ﻓﻲ ﺍﻟﻔﻘﺭﺓ 2؟؟؟ ﺒﺎﻟﻁﺒﻊ ﻨﻌﻡ ﻴﺒﻘﻰ ﻓﻘﻁ ﺃﻥ ﺘﺴﺘﻐل ﺍﻟﻤﻨﺤﻨﻰ ) G = f ( cﻭ ﺘﺘﺒﻊ ﺍﻟﺨﻁﻭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ : 1ـ ﺘﻤﺩﻴﺩ ﻤﺤﻠﻭل ﺍﻟﻤﺼل ) ( sérumﻨﺄﺨﺫ ﺤﺠﻡ 10 mLﻤﺜﻼ ﻤﻥ ﺍﻟﻤﺼل ) ( sérumﻭ ﻨﻤﺩﺩﻩ 10ﻤﺭﺍﺕ ﺃﻱ ﻨﻜﻤل ﺍﻟﺤﺠﻡ ﻓﻲ ﻤﺨﺒﺎﺭ ﻤﺩﺭﺝ ﺇﻟﻰ 100 mLﺒﺎﻟﻤﺎﺀ ﺍﻟﻤﻘﻁﺭ . 2ـ ﻗﻴﺎﺱ ﻨﺎﻗﻠﻴﺔ ﺍﻟﻤﺤﻠﻭل ﺍﻟﻤﻤﺩﺩ ﻟﻠﻤﺼل ) ﻗﻴﺎﺱ Uﻭ ( Iﺃﻭ ﻤﺒﺎﺸﺭﺓ ﺒﺎﺴﺘﻌﻤﺎل ﺠﻬﺎﺯ : conductimètre ﻓﻲ ﺍﻟﻤﺜﺎل ﺍﻟﻤﺩﺭﻭﺱ ﻨﺠﺩ G = 10.9 mS 3ـ ﺒﺎﻹﺴﻘﺎﻁ ﻋﻠﻰ ﺍﻟﻤﻨﺤﻨﻰ ﻨﺠﺩ ﺍﻟﺘﺭﻜﻴﺯ ﺍﻟﻤﻭﻟﻲ ﻟﻠﻤﺤﻠﻭل ﺍﻟﻤﻤﺩﺩ ) c1 = 15,6 .10-3 mol /Lﺃﻨﻅﺭ ﺍﻟﺸﻜل ﺍﻟﻤﻭﺍﻟﻲ (
4ـ ﺤﺴﺎﺏ ﺍﻟﺘﺭﻜﻴﺯ ﺍﻟﻤﻭﻟﻲ cﻟﻠﻤﺼل ) ﺍﻟﻤﺤﻠﻭل ﺍﻟﻤﺭﻜﺯ ( ﺤﻴﺙ :c = 10 . c1c = 0,156 mol / L ﺇﺫﻥ : 5ـ ﺍﺴﺘﻨﺘﺎﺝ ﺍﻟﺘﺭﻜﻴﺯ ﺍﻟﻜﺘﻠﻲ ﻟﻠﻤﺼل t = c . M :ﺤﻴﺙ Mﻫﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﻭﻟﻴﺔ ﻟﻜﻠﻭﺭ ﺍﻟﺼﻭﺩﻴﻭﻡ ) ( Na+ , Cl- ﺇﺫﻥ M = 58,5 g / mol :t = 0,156 x 58,5 ﻭ ﻤﻨﻪ :t = 9,13 g / L ﺇﺫﻥ : 6ـ ﺍﻻﺴﺘﻨﺘﺎﺝ ) ﺍﻟﺤﻜﻡ ( ﻓﻴﻤﺎ ﻴﺨﺹ ﺼﺤﺔ ﻤﺎ ﻜﺘﺏ ﻋﻠﻰ ﺍﻟﺒﻁﺎﻗﺔ : ﻗﻴﻤﺔ ﺍﻟﺘﺭﻜﻴﺯ ﺍﻟﺫﻱ ﻜﺘﺒﺕ ﻋﻠﻰ ﺍﻟﺒﻁﺎﻗﺔ 9 g / Lﺘﺘﻭﺍﻓﻕ ﻤﻊ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺘﻲ ﻭﺠﺩﻨﺎﻫﺎ ﻤﻥ ﺨﻼل ﺍﻟﺩﺭﺍﺴﺔ ﺍﻟﺘﺠﺭﻴﺒﻴﺔ . ﺨﻼﺼﺔ :ﻤﻥ ﺨﻼل ﻗﻴﺎﺱ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺘﻌﺭﻓﻨﺎ ﻋﻠﻰ ﻁﺭﻴﻘﺔ ﻓﻴﺯﻴﺎﺌﻴﺔ ﻟﺤﺴﺎﺏ ﺘﺭﻜﻴﺯ ﻤﺤﻠﻭل ﻭ ﻜﻤﻴﺔ ﻤﺎﺩﺘﻪ . -7ﺃﺴﺌﻠﺔ ﺍﻟﺘﺼﺤﻴﺢ ﺍﻟﺫﺍﺘﻲ : ( 1ﺨﻠﻴﺔ ﻗﻴﺎﺱ ﺍﻟﻨﺎﻗﻠﻴﺔ ﻤﺅﻟﻔﺔ ﻤﻥ ﻤﺴﺭﻴﻴﻥ ﻤﺴﺘﻭﻴﻴﻥ ﻤﺘﻭﺍﺯﻴﻴﻥ ﻤﺴﺎﺤﺔ ﺍﻟﺴﻁﺢ ﺍﻟﻤﻐﻤﻭﺭ ﻟﻜل ﻤﻨﻬﻤﺎ S = 1 cm2ﻭ ﺍﻟﺒﻌﺩ ﺒﻴﻨﻬﻤﺎ . L = 1 cm ﺘﻐﻤﺭ ﺍﻟﺨﻠﻴﺔ ﻓﻲ ﻤﺤﻠﻭل ﺸﺎﺭﺩﻱ ﻭ ﻨﻁﺒﻕ ﺒﻴﻥ ﻁﺭﻓﻴﻬﺎ ﺘﻭﺘﺭ ﻤﺘﻨﺎﻭﺏ ﺠﻴﺒﻲﻓﻨﻘﺭﺃ ﻋﻠﻰ ﺍﻟﻔﻭﻟﻁﻤﺘﺭ ﺍﻟﻘﻴﻤﺔ U = 2,5 vﻭﻋﻠﻰ ﺍﻷﻤﺒﻴﺭﻤﺘﺭ ﺍﻟﻘﻴﻤﺔ I = 5,0 mAﺃ ـ ﺍﺤﺴﺏ ﻨﺎﻗﻠﻴﺔ ﺠﺯﺀ ﺍﻟﻤﺤﻠﻭل ﺍﻟﺸﺎﺭﺩﻱ ﺍﻟﻤﺤﺼﻭﺭ ﺒﻴﻥ ﺍﻟﻤﺴﺭﻴﻴﻥ . ﺏ ـ ﺍﺴﺘﻨﺘﺞ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﺍﻓﻘﺔ . ( 2ﺍﺤﺴﺏ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﻟﻴﺔ ﺍﻟﺸﺎﺭﺩﻴﺔ ﻟـ : Cl- , K+ , Ag+ , NO3 - ﻭ ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﻟﻴﺔ ﻟـ AgNO3 ﻴﻌﻁﻰ ﻋﻨﺩ : 25 ° C Λ( HCl = 42,6.10-3 s.m2./mol λ( H3O+) = 35.10-3 s.m2/mol Λ(KCl) = 15.10-3 s.m2/mol Λ(AgCl) = 12,7.10-3 s.m2/mol Λ(KNO3) = 14,5 .10-3 s.m2/mol
U=R.I -8ﺃﺠﻭﺒﺔ ﺍﻟﺘﺼﺤﻴﺢ ﺍﻟﺫﺍﺘﻲ R = UI (1 R = 500 Ω ﺃ – ﻨﻌﻠﻡ ﺃﻥ : = 5.21,05-3 ﺇﺫﻥ :=G 1 ﻟﻜﻥ ﺍﻟﻨﺎﻗﻠﻴﺔ Gﻫﻲ ﻤﻘﻠﻭﺏ ﺍﻟﻤﻘﺎﻭﻤﺔ R=G 1 500G = 2.10-3 s ﺏ – ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ σ = K . G : =K L = ﺤﻴﺙ Kﻫﻭ ﺜﺎﺒﺕ ﺍﻟﺨﻠﻴﺔ S 1 10-2 K = 100 m-1 σ = 100 x 2.10-3 ﺇﺫﻥ : σ = 0,2 s . m-1 ﻭ ﻤﻨﻪ : = Λ( HCl λ( H3O+) + (λ - ) (2 Cl ﺇﺫﻥ : (λ - ) = Λ( HCl )- λ( H3O+ Cl (λ - ) = 42,6 . 10-3 - 35 .10-3 Cl
λ( - ) = 7,6 . 10-3 s.m2./mol Cl Λ(KCl) = λ (K+) + λ( - ) : ﺒﻨﻔﺱ ﺍﻟﻁﺭﻴﻘﺔ Cl λ (K+) = Λ(KCl) - λ( - ) Cl λ (K+) = 7,4 .10-3 s.m2./mol : ﺒﺎﻟﺘﻌﻭﻴﺽ ﻨﺠﺩ Λ(AgCl) = λ (Ag+) + λ( - ) : ﺃﻴﻀﺎ Cl : ﺒﺎﻟﺘﻌﻭﻴﺽ ﻨﺠﺩ λ (Ag+) = Λ(AgCl) - λ( - ) Cl λ (Ag+) = 5,1 .10-3 s.m2./molΛ(KNO3) = λ (K+) + λ - : ﻜﺫﻟﻙ (NO3 ) : ﺒﺎﻟﺘﻌﻭﻴﺽ ﻨﺠﺩ λ - = Λ(KNO3) - λ (K+) (NO3 ) λ - = 7,1 .10-3 s.m2./mol (NO3 ) AgNO3 ﺍﻟﻨﺎﻗﻠﻴﺔ ﺍﻟﻨﻭﻋﻴﺔ ﺍﻟﻤﻭﻟﻴﺔ ﻟﻤﺤﻠﻭل Λ(AgNO3) = λ (Ag+) + λ - (NO3 )Λ(AgNO3) = 12,2 .10-3 s.m2./mol : ﺒﺎﻟﺘﻌﻭﻴﺽ ﻨﺠﺩ
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