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)ﺏ( ﺇﺫﺍ ﻛﺎﻧﺖ ﻧﺴﺒﺔ ﺍﻟﺘﻔﻜﻚ ﻛﺒﻴﺮﺓ ) أﻛ ﻣ (5 %ﻧﺴﺘﺨﺪﻡ ﻗﺎﻧﻮﻥ ﺍﺳﺘﻔﺎﻟﺪ ﺑﺪﻭﻥ ﺗﻘﺮﻳﺐ ﺣﻴﺚ ﺃﻧﻪ ﻓﻰ ﻫﺬﻩ ﺍﻟﺤﺎﻟﺔ ﻻ
ﻳﻤﻜﻦ ﺇﻫﻤﺎﻝ .............. αﻛﻤﺎ ﻓﻰ ﺍﻷﻣﺜﻠﺔ ﺍﻵﺗﻴﺔ :
ﻋ ـــ ﺿـــ أﺣﺎد اﻟ وﺗ ن ﺗ ـــﺎو 33 %ﻓﻰ ﻣ ﻠ ل ﺗ ــــــــــ ﻩ ) (٨إذا ﺎﻧ ﻧ ـــ ﺔ ﺗﻔ ﺣ
- 0.2 mol/Lإﺣ
Kaﻟﻬ ا اﻟ .
ﺍﻟﺤــﻞ
α = 33 = 0.33
100
= Ka α2 × C = (0.33)2 × 0.2 = 0.0325
) (1 – α
1 – 0.33
ﻗﺎﻋ ة ﺿـ ﻔﺔ ﺗ ـﺎو 25 %ﻓﻰ ﻣ ﻠ ل ﺗ ﻩ ، 0.3 mol/Lاﺣ ـ ﺔ ) (٩إذا ﺎﻧ ﻧ ـ ﺔ ﺗﻔ
Kb
α = 25 = 0.25
ﺍﻟﺤــﻞ
100
= Kb α2 × C = (0.25)2 × 0.3 = 0.025
) (1 – α
1 – 0.25
م اﻟﻌﻼﻗﺔ اﻵﺗ ﺔ : ﻣﻼﺣﻈﺔ ﻟ ﺎب ﻋ د اﻟ ﻻت اﻟ ﻔ ﺔ أو ﻋ دﻫﺎ ﻗ ﻞ اﻟ ﻔ ﺑ ﻻﻟﺔ درﺟﺔ اﻟ ﺄﯾ ) (αﻧ
ﻋﺪد اﻟﻤﻮﻻت اﻟﻤﺘﻔﻜﻜﺔ درﺟﺔ اﻟﺘﻔﻜﻚ )= (α
ﻋﺪد اﻟﻤﻮﻻت اﻟﻜﻠﯿﺔ ﻗﺒﻞ اﻟﺘﻔﻜﻚ
5.1 X 104- Ka )ﺗ ر (
200 ml 0.2 M
.
)اﻟ ﻞ(
= Ka 5.1 × 10-4 = 0.0505
=α Ca 0.2
0.04 mol = 0.2 X 0.2 X
)X (α
2.02 X 103- mol = 0.04 X 0.0505
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