1. 1
‫أﺑــﻮﻇﺒــــ‬ ‫ﻣﺠﻠﺲ‬‫ﻲ‬‫ﻟﻠﺘـﻌـﻠــــﯿﻢ‬
‫اﻟﻜﯿﻤﯿﺎء‬ : ‫اﻟﻤــــــــــﺎدة‬ ‫اﻟﺘﻌﻠﯿﻤﯿﺔ‬ ‫ﻇﺒﻲ‬ ‫أﺑﻮ‬ ‫ﻣﻨﻄﻘﺔ‬ ‫إدارة‬
‫ﺳﺎﻋ‬‫ﺔ‬‫وﻧﺼﻒ‬ : ‫اﻹﺟﺎﺑﺔ‬ ‫زﻣﻦ‬ ‫اﻟﻜﯿﻤﯿﺎء‬ ‫ﺗﻮﺟﯿﮫ‬
‫ﺍﻻ‬ ‫ﳕﻮﺫﺝ‬‫ﻣﺘﺤﺎﻥ‬‫ﺍﻟ‬‫ﺘﺠﺮﻳﺒﻲ‬‫ﺍﻟﻌﻠﻤﻲ‬ ‫ﻋﺸﺮ‬ ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﻟﻠﺼﻒ‬
‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺪﺭﺍﺳﻲ‬ ‫ﺍﻟﻔﺼﻞ‬
‫ﺍﻟﺩﺭﺍﺴﻲ‬ ‫ﻟﻠﻌﺎﻡ‬2009/2010‫ﻡ‬‫ﺍﻷﻭ‬ ‫ﺍﻟﺩﺭﺍﺴﻲ‬ ‫ﺍﻟﻔﺼل‬ ‫ﻤﻨﻬﺞ‬ ‫ﻭﻓﻕ‬ ‫ﺘﻌﺩﻴﻠﻪ‬ ‫ﺘﻡ‬‫ﻟﻠﻌﺎﻡ‬ ‫ل‬2010/2011‫ﻡ‬
‫ﻓﻲ‬ ‫اﻷﺳﺌﻠﺔ‬)5(‫ذﻟﻚ‬ ‫ﻣﻦ‬ ‫اﻟﺘﺄﻛﺪ‬ ‫اﻟﻄﺎﻟﺐ‬ ‫وﻋﻠﻰ‬ ‫ﺻﻔﺤﺎت‬
‫اﻟﻛﻳﻣﻳﺎء‬ ‫ﻣﻌﻠم‬ ‫إﻋداد‬/‫اﻟﻌﻼ‬ ‫أﺑو‬ ‫ﻣﺣﻣد‬ ‫ﺳﺎﻣﻲ‬)‫أﻛﺎدﻳﻣﻳﺔ‬‫اﻟﺟزﻳرة‬‫اﻟﻘدم‬ ‫ﻟﻛرة‬(‫اﻟﻛﻳﻣﻳﺎء‬ ‫ﻣوﺟﻪ‬ ‫ﺑﺈﺷراف‬/‫اﻟﺟﻳﺎر‬ ‫ﻋﺎدل‬
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‫اﻷول‬ ‫اﻟﺴﺆال‬:
ً‫ﻻ‬‫أو‬‫ﻋﻼﻣﺔ‬ ‫ﺿﻊ‬)ü(‫ﯾﺄﺗﻲ‬ ‫ﻣﻤﺎ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺼﺤﯿﺤﺔ‬ ‫اﻟﺘﻜﻤﻠﺔ‬ ‫أﻣﺎم‬:
1-‫اﻟﺘﺮاﻛﯿﺰ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫ﺗﻤﺜﻞ‬ ‫واﻟﺘﻲ‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﺮﺳﻮم‬ ‫ﺗﺄﻣﻞ‬‫ﺑ‬‫اﻟﻤﺎﺋﻲ‬ ‫اﻟﻤﺤﻠﻮل‬ ‫ﻟﻤﻜﻮﻧﺎت‬ ‫ﺎﻟﻤﻮﻻرﯾﺔ‬‫اﻷﺣﻤﺎض‬ ‫ﻣﻦ‬ ‫ﻟﻤﺠﻤﻮﻋﺔ‬
،ً‫ﺎ‬‫ﺻﺤﯿﺤ‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻷﺷﻜﺎل‬ ‫أي‬‫ﻟﻠﺤﻤﺾ‬ ‫اﻟﻤﺎﺋﻲ‬ ‫اﻟﻤﺤﻠﻮل‬ ‫ﻋﻦ‬ ‫ﻟﻠﺘﻌﺒﯿﺮ‬:
* * * *
2-‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬)Ni2+
+ nH2O D [Ni(H2O)n]2+
(‫ﯾﻌﺘﺒﺮ‬H2O:
*‫ﺑﺮوﻧﺸﺘﺪ‬ ‫ﺣﻤﺾ‬–‫ﻟﻮري‬*‫ﺑﺮوﻧﺸﺘﺪ‬ ‫ﻗﺎﻋﺪة‬–‫ﻟﻮري‬*‫ﻟﻮﯾﺲ‬ ‫ﺣﻤﺾ‬*‫ﻟﻮﯾﺲ‬ ‫ﻗﺎﻋﺪة‬
3-‫اﻟ‬ ‫واﻟﺸﻜﻞ‬ ‫اﻟﻜﺎﺗﯿﻮﻧﺎت‬ ‫أﺣﺪ‬ ‫ﻋﻦ‬ ‫ﻟﻠﻜﺸﻒ‬ ‫ﺗﺠﺮﺑﺔ‬ ‫ُﺟﺮﯾﺖ‬‫أ‬‫ﺘﺎﻟﻲ‬‫اﻟﺘﺠﺮﺑﺔ‬ ‫ﯾﺒﯿﻦ‬،
‫اﻟﻤﺮﻓﻖ‬ ‫اﻟﺠﺪول‬ ‫ﻣﻮﻇﻔﺎ‬،
‫ﻋﻨﮫ‬ ‫اﻟﻜﺸﻒ‬ ‫ﺗﻢ‬ ‫اﻟﻜﺎﺗﯿﻮﻧﺎت‬ ‫أي‬ ‫ﺗﻮﻗﻊ‬:
*‫ھﻮ‬ ‫اﻟﻜﺎﺗﯿﻮن‬Pb2+
*‫ھﻮ‬ ‫اﻟﻜﺎﺗﯿﻮن‬Ca2+
*‫ھﻮ‬ ‫اﻟﻜﺎﺗﯿﻮن‬Mg2+
*‫ھﻮ‬ ‫اﻟﻜﺎﺗﯿﻮن‬Ba2+
30
‫ﺑــ‬‫اﻟﺘﺮﻛﯿﺰ‬M
HCl
Cl-
H3O+
‫ﻋﺪا‬ ‫ﻣﺎ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫اﻟﻜﻠﻮرﯾﺪات‬ ‫ﻣﻌﻈﻢ‬Ag+
‫و‬Pb2+
‫ﻋﺪا‬ ‫ﻣﺎ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫ﻻ‬ ‫اﻟﻜﺒﺮﯾﺘﯿﺪات‬ ‫ﻣﻌﻈﻢ‬Ca2+
, Sr2+
, K+
, Na+
‫ﻋﺪا‬ ‫ﻣﺎ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫ﻻ‬ ‫اﻟﻜﺮﺑﻮﻧﺎت‬ ‫ﻣﻌﻈﻢ‬K+
, Na+
, NH4
+
‫ﻋﺪا‬ ‫ﻣﺎ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫اﻟﻜﺒﺮﯾﺘﺎت‬ ‫ﻣﻌﻈﻢ‬Ca2+
, Pb2+
, Ba2+
‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫واﻟﻨﯿﺘﺮات‬ ‫اﻟﻜﻠﻮرات‬
Na2CO3
‫راﺳﺐ‬
K2SO4
‫راﺳﺐ‬ ‫ﯾﺘﻜﻮن‬ ‫ﻻ‬
(NH4)2S
‫راﺳﺐ‬
KNO3
‫راﺳﺐ‬ ‫ﯾﺘﻜﻮن‬ ‫ﻻ‬
‫ﺑــ‬‫اﻟﺘﺮﻛﯿﺰ‬M
HNO3
NO3
-
H3O+
‫ﺑــ‬‫اﻟﺘﺮﻛﯿﺰ‬M
H2SO4
HSO4
-
H3O+
SO4
2-
‫ﺑــ‬‫اﻟﺘﺮﻛﯿﺰ‬M
HF
F-
H3O+

2. 2
4-‫اﻟ‬ ‫اﻟﻤﺤﺎﻟﯿﻞ‬ ‫ﻓﻲ‬‫ﺤﻤﻀ‬‫ﺣﺮارة‬ ‫درﺟﺔ‬ ‫ﻋﻨﺪ‬ ‫ﯿﺔ‬50 ˚C‫ﺗﻜﻮن‬:
*[H3O+
]=[OH-
]*Kw<[H3O+
] . [OH-
]
*Kw >[H3O+
] . [OH-
]*Kw=[H3O+
] . [OH-
]
5-‫أذاب‬ ‫إذا‬‫ﺣﻤﺪان‬0.5 mol‫ﻣﻦ‬Ba(OH)2‫ﻓﻲ‬0.5 L‫اﻟﻤﺤﻠﻮل‬ ‫ﺗﺮﻛﯿﺰ‬ ‫ﻓﯿﻜﻮن‬ ‫اﻟﻤﻘﻄﺮ‬ ‫اﻟﻤﺎء‬ ‫ﻣﻦ‬:
*0.5 M*0.5 m*1.0 M*1.0 m
6-‫اﻟﺤﻤﻮﺿﺔ‬ ‫ﻣﻀﺎدات‬ ‫ﻓﻲ‬ ‫ﯾﺴﺘﺨﺪم‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻤﺮﻛﺒﺎت‬ ‫أﺣﺪ‬:
*KHCO3*KHSO3*NaHCO3*NaHSO3
‫أ‬(ً‫ﺎ‬‫ﺛﺎﻧﯿ‬:‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻌﺒﺎرات‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻋﻠﯿﮫ‬ ‫ﺗﺪل‬ ‫اﻟﺬي‬ ‫اﻟﻌﻠﻤﻲ‬ ‫اﻟﻤﺼﻄﻠﺢ‬ ‫أو‬ ‫اﻻﺳﻢ‬ ‫اﻟﻘﻮﺳﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻛﺘﺐ‬:
7-)(‫ﻣﺎدة‬‫ﻓﻲ‬ ‫وﺗﺴﺘﺨﺪم‬ ‫أﺻﻔﺮ‬ ‫ﻟﻮن‬ ‫اﻟﺒﺮوﺗﯿﻨﺎت‬ ‫ﺗﻜﺴﺐ‬ ‫ھﺎﻣﺔ‬ ‫ﻛﯿﻤﯿﺎﺋﯿﺔ‬‫اﻟﻤﺘﻔﺠﺮات‬ ‫ﺻﻨﺎﻋﺔ‬،
‫واﻟﻤﻄﺎط‬،‫اﻟﺒﻼﺳﺘﯿﻜﯿﺔ‬ ‫واﻟﻤﻮاد‬،‫واﻷﺻﺒﺎغ‬.
8-)(‫ﻛﻘﺎﻋﺪة‬ ‫أو‬ ‫ﻛﺤﻤﺾ‬ ‫ﺗﺘﻔﺎﻋﻞ‬ ‫أن‬ ‫ﯾﻤﻜﻨﮭﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻮاد‬.
9-)(‫ﻣﺎء‬ ‫ﺟﺰﯾﺌﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﺮوﺗﻮن‬ ‫ﺑﺎﻧﺘﻘﺎل‬ ‫ھﯿﺪروﻛﺴﯿﺪ‬ ‫وأﯾﻮن‬ ‫ھﯿﺪروﻧﯿﻮم‬ ‫أﯾﻮن‬ ‫إﻧﺘﺎج‬.
‫راﺑﻌ‬ً‫ﺎ‬:‫ﱠﺎ‬‫ﻤ‬‫ﻋ‬ ‫أﺟﺐ‬‫ﯾﻠﻲ‬:
10-‫اﻟ‬ ‫ُﻀﺮت‬‫ﺣ‬‫ﻤﺤﺎﻟﯿﻞ‬‫اﻟﺘﺎﻟﯿﺔ‬‫ﻣﻮﻻرﯾﺔ‬ ‫ﺑﺘﺮاﻛﯿﺰ‬‫ﻣﺘﺴﺎوﯾﺔ‬)HF , HCl , C6H12O6 , K2SO4.(‫رﺗﺐ‬‫ھﺬه‬‫اﻟﻤﺤﺎﻟﯿﻞ‬
‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫ﻟﻠﺘﯿﺎر‬ ‫ﺗﻮﺻﯿﻠﮭﺎ‬ ‫ﺣﺴﺐ‬ ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﯾ‬.
‫اﻷﻗﻞ‬----------‫ﺛﻢ‬----------‫ﺛﻢ‬--------------‫ﺛﻢ‬----------‫اﻷﻋﻠﻰ‬‫ﺗﻮﺻﯿﻞ‬
‫اﻟ‬ ‫اﻟﺴﺆال‬‫ﺜﺎﻧﻲ‬:
ً‫ﻻ‬‫أو‬:
11-‫اﻟﻔﻮﺳﻔﻮرﯾﻚ‬ ‫ﺣﻤﺾ‬H3PO4‫ھﯿﺪروﺟﯿﻦ‬ ‫ذرات‬ ‫ﺛﻼث‬ ‫ﻋﻠﻰ‬ ‫ﺻﯿﻐﺘﮫ‬ ‫ﺗﺤﺘﻮي‬،‫ﺣﻤﺾ‬ ‫ﺑﯿﻨﻤﺎ‬ ‫اﻟﺒﺮوﺗﻮن‬ ‫ﺛﻼﺛﻲ‬ ‫ﻛﺤﻤﺾ‬ ‫ﻣﺼﻨﻒ‬ ‫وھﻮ‬
‫اﻷﺳﯿﺘﯿﻚ‬CH3COOH‫ھﯿﺪروﺟﯿﻦ‬ ‫ذرات‬ ‫أرﺑﻊ‬ ‫ﻋﻠﻰ‬ ‫ﺻﯿﻐﺘﮫ‬ ‫ﺗﺤﺘﻮي‬،‫اﻟﺒﺮوﺗﻮن‬ ‫أﺣﺎدي‬ ‫ﻛﺤﻤﺾ‬ ‫ﻣﺼﻨﻒ‬ ‫وھﻮ‬.‫ﺑﺮر‬‫ذﻟ‬‫ﻚ‬‫؟‬
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ً‫ﺎ‬‫ﺛﺎﻧﯿ‬:‫ﻣﺰج‬ ‫ﻋﻨﺪ‬‫ﻣﺤﻠﻮﻟ‬‫ﻲ‬‫اﻟﺮﺻﺎص‬ ‫ﻧﯿﺘﺮات‬IIPb(NO3)2‫و‬‫اﻟﺒﻮﺗﺎﺳﯿﻮم‬ ‫ﻛﻠﻮرﯾﺪ‬KCl‫ﯾ‬‫ﺘﻜﻮ‬‫راﺳﺐ‬ ‫ن‬
‫اﻟﺮﺻﺎص‬ ‫ﻛﻠﻮرﯾﺪ‬ ‫ﻣﻦ‬II:‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻔﻘﺮات‬ ‫ﻋﻦ‬ ‫أﺟﺐ‬:
12-‫ﻛﺘﺎﺑﺔ‬‫اﻟﻌﺎﻣﺔ‬ ‫اﻷﯾﻮﻧﯿﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬--------------------------------------------------------------------------------
13-‫اﻟﻤﺘﻔﺮﺟﺔ‬ ‫اﻷﯾﻮﻧﺎت‬ ‫ﺗﺤﺪﯾﺪ‬------------------------------------------------------------------------------
14-‫اﻟﺼﺮﻓﺔ‬ ‫اﻷﯾﻮﻧﯿﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻛﺘﺎﺑﺔ‬---------------------------------------------------------------------------
15-‫ﺣﺪد‬‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﺘﻔﺎﻋﻼت‬ ‫ﻓﻲ‬‫اﻟﻤﺘﻔﺎﻋﻼت‬‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬‫إﻟﻰ‬‫ﻗﻮاﻋﺪ‬‫أﺣﻤﺎض‬ ‫أو‬‫ارھﯿﻨﯿﻮس‬،‫و‬‫ﺑﺮوﻧﺸﺘﺪ‬-‫ﻟﻮري‬،‫و‬‫ﻟﻮﯾﺲ‬
‫اﻟﻤﻌﺎدﻟﺔ‬‫اﻟﺤﻞ‬
1H+
(aq) + NH3(aq) → NH4
+
(aq)
2HI(aq) + H2O(l) → H3O+
(aq) + I
-
(aq)
3KOH(s) → K+
(aq) + OH
-
(aq)
4C6H5NH2 + H2O D C6H5NH3
+
+OH
-
25

3. 3
ً‫ﺎ‬‫ﺛﺎﻟﺜ‬:‫ﻛﺜﺎﻓﺘﮫ‬ ‫اﻟﻨﯿﺘﺮﯾﻚ‬ ‫ﺣﻤﺾ‬ ‫ﻣﺤﻠﻮل‬1.19 g/cm3
،‫اﻟﻜﺘﻠﻲ‬ ‫اﻟﻤﺌﻮي‬ ‫وﺗﺮﻛﯿﺰه‬32%
‫واﻟﻤﻄﻠﻮب‬‫ﺣﺴ‬‫ﺎ‬‫ب‬:
16-‫ﻓﻲ‬ ‫اﻟﻤﺬﯾﺐ‬ ‫ﻛﺘﻠﺔ‬100g‫اﻟﻤﺤﻠﻮل‬ ‫ھﺬا‬ ‫ﻣﻦ‬
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17-‫؟‬ ‫اﻟﻤﺤﻠﻮل‬ ‫ھﺬا‬ ‫وﻣﻮﻻرﯾﺔ‬ ‫ﻣﻮﻻﻟﯿﺔ‬‫ﺑﺄن‬ ً‫ﺎ‬‫ﻋﻠﻤ‬HNO3 = 63 g/mol
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18-‫ﻣﺠ‬ ‫اﻷﺳﯿﺘﯿﻚ‬ ‫ﺣﻤﺾ‬ ‫ﻣﺤﻠﻮل‬ ‫ﺗﺮﻛﯿﺰ‬ ‫ﻟﺘﻘﺪﯾﺮ‬ ‫ﻣﻌﺎﯾﺮة‬ ‫ﺗﺠﺮﺑﺔ‬ ‫ﻓﻲ‬‫ھﯿﺪروﻛﺴﯿﺪ‬ ‫ﻣﻦ‬ ‫ﻗﯿﺎﺳﻲ‬ ‫ﻣﺤﻠﻮل‬ ‫ﺑﺎﺳﺘﺨﺪام‬ ‫اﻟﺘﺮﻛﯿﺰ‬ ‫ﮭﻮل‬
‫اﻟﺼﻮدﯾﻮم‬0.1 M‫ﻟﻤﻌﺎﯾﺮة‬ ‫اﻟﺘﺠﺮﺑﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺎت‬ ‫إﺣﺪى‬ ‫ﻧﻔﺬت‬30 mL‫اﻟﺤﻤﺾ‬ ‫ﻣﻦ‬‫واﺳﺘﺨﺪﻣﺖ‬ ‫ﻧﻈﯿﻒ‬ ‫ﻛﺄس‬ ‫ﻓﻲ‬
‫ﺑﺎﻟﺮﺳﻢ‬ ‫ﻛﻤﺎ‬ ‫اﻟﻤﻨﺤﻨﻰ‬ ‫ورﺳﻢ‬ ‫اﻟﻨﺘﺎﺋﺞ‬ ‫ﺑﺘﺴﺠﯿﻞ‬ ‫وﻗﺎﻣﺖ‬ ‫واﻟﺴﺤﺎﺣﺔ‬ ‫اﻟﻤﯿﺜﯿﻞ‬ ‫ﺑﺮﺗﻘﺎﻟﻲ‬ ‫ﻛﺎﺷﻒ‬.
‫ﻋﻤﻞ‬ ‫ﻋﻠﻰ‬ ً‫ﺎ‬‫ﺣﻜﻤ‬ ‫اﺻﺪر‬‫وﺟﺪت‬ ‫إن‬ ‫اﻷﺧﻄﺎء‬ ‫ذﻛﺮ‬ ‫ﻣﻊ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬،‫ّﺑﮭﺎ‬‫ﻮ‬‫وﺻ‬.)‫رﻗﻢ‬ ‫اﻟﺼﻔﺤﺔ‬ ‫ﻓﻲ‬ ‫ﺑﺎﻟﻤﺨﻄﻂ‬ ‫اﺳﺘﻌﻦ‬6(
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‫اﻟﺴ‬‫ﺆ‬‫اﻟ‬ ‫ال‬‫ﺜﺎﻟﺚ‬:
ً‫ﻻ‬‫أو‬:‫ﯾﻠﻲ‬ ‫ﻣﺎ‬ ‫ﻓﺴﺮ‬:
19-‫ذوﺑﺎن‬ ‫ﯾﻌﺪ‬‫اﻟﺼﻮدﯾﻮم‬ ‫ﻛﻠﻮرﯾﺪ‬ً‫ﺎ‬‫ﺗﻔﻜﻜ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬‫ذوﺑﺎن‬ ‫ﺑﯿﻨﻤﺎ‬‫اﻟ‬ ‫ﻛﻠﻮرﯾﺪ‬‫ﮭﯿﺪروﺟﯿﻦ‬ً‫ﺎ‬‫ﺗﺄﯾﻨ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬‫؟‬
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20-‫ﺗﻌﺘﺒﺮ‬HS
-
‫ﻣﻦ‬ ‫أﻗﻮى‬ ‫ﻗﺎﻋﺪة‬HSO4
-
‫؟‬
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ً‫ﺎ‬‫ﺛﺎﻧﯿ‬:‫ﻓﻘﺮة‬ ‫ﻟﻜﻞ‬ ‫ﺑﺪاﺋﻞ‬ ‫أرﺑﻌﺔ‬ ‫أﻣﺎﻣﻚ‬–‫اﻟﺒﺪﯾﻞ‬ ‫اﺧﺘﺮ‬‫اﻟﻤﻨﺴﺠﻢ‬ ‫ﻏﯿﺮ‬ً‫ﺎ‬‫ﻋﻠﻤﯿ‬‫ﻣﻊ‬‫ﺗ‬‫ﺒﺮ‬‫ﯾ‬‫إﺟﺎﺑﺘﻚ‬ ‫ﺮ‬:
21-H3PO3،H2SeO3،H3PO4،H2SeO4
‫اﻟﺴﺒﺐ‬:----------------------------------------------------------------------------------------
ً‫ﺎ‬‫ﺛﺎﻟﺜ‬:‫ﺗﻠﯿﮫ‬ ‫اﻟﺘﻲ‬ ‫اﻷﺳﺌﻠﺔ‬ ‫ﻋﻦ‬ ‫وأﺟﺐ‬ ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﺠﺪول‬ ‫ﺗﺄﻣﻞ‬:)‫رﻗﻢ‬ ‫اﻟﺼﻔﺤﺔ‬ ‫ﻓﻲ‬ ‫اﻟﺬوﺑﺎﻧﯿﺔ‬ ‫ﺑﺠﺪول‬ ‫اﺳﺘﻌﻦ‬2(
22-‫رﻗﻢ‬ ‫أي‬‫ﯾﻤﺜﻞ‬‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﻟﻠﺬوﺑﺎن‬ ‫ﻗﺎﺑﻞ‬ ‫أﯾﻮﻧﻲ‬ ‫ﻣﺮﻛﺐ‬‫؟‬----------------------------
23-‫اﻟﻤﺮﻛﺒﯿﻦ‬ ‫ﻣﺤﻠﻮﻟﻲ‬ ‫ﻣﺰج‬ ‫ﻋﻨﺪ‬ ‫اﻟﻤﺘﻔﺮﺟﺔ‬ ‫اﻷﯾﻮﻧﺎت‬ ‫ﻣﺎ‬3‫و‬8‫؟‬-----------------------
24-‫رﻗﻢ‬ ‫أي‬‫ﯾﻤﺜﻞ‬‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫اﻟﺬوﺑﺎن‬ ‫ﺷﺤﯿﺢ‬ ‫أﯾﻮﻧﻲ‬ ‫ﻣﺮﻛﺐ‬‫؟‬----------------------------
25-‫رﻗﻢ‬ ‫أي‬‫ﯾﻤﺜﻞ‬‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺟﺰﺋﯿﺎ‬ ‫ﯾﺘﺄﯾﻦ‬ ‫ﺟﺰﯾﺌﻲ‬ ‫ﻣﺮﻛﺐ‬‫؟‬------------------------------
26-‫أي‬‫رﻗﻢ‬‫ﺛﻨﺎﺋﻲ‬ ‫ﺣﻤﺾ‬ ‫ﯾﻤﺜﻞ‬‫؟‬-----------------------------------------------
25
0 5 10 15 20 25 30 35 40 45
‫ﺣﺠﻢ‬NaOH‫ﺑــ‬ ‫اﻟﻤﻀﺎف‬mL
pH
14
12
10
8
6
4
2
0
C2H5OHHClBa(OH)2CH3COOHNaNO3HNO3AgClH2SO4
12345678

4. 4
‫ﺧﺎﻣﺴﺎ‬:‫أﺳﺌﻠﺔ‬ ‫ﻣﻦ‬ ‫ﯾﻠﯿﮫ‬ ‫ﻋﻤﺎ‬ ‫وأﺟﺐ‬ ‫إدرﺳﮫ‬ ‫اﻷدﻟﺔ‬ ‫ﻣﻦ‬ ‫اﻟﻌﺪﯾﺪ‬ ‫ﻣﺪي‬ ‫ﺗﻐﯿﺮ‬ ‫ﯾﺒﯿﻦ‬ ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﻤﺨﻄﻂ‬:
27-‫ﺗ‬ ‫ﻣﺎدة‬‫ﻌﻄﻲ‬‫أﺻﻔ‬ ‫ﻟﻮن‬‫ﺮ‬‫ﻛﺎﺷﻒ‬ ‫ﻣﻊ‬‫أﺣﻤ‬‫اﻟﻔﯿﻨﻮل‬ ‫ﺮ‬‫وﻟﻮن‬‫ﻛﺎﺷﻒ‬ ‫أﺻﻔﺮﻣﻊ‬‫اﻟﻤﯿﺜﯿﻞ‬ ‫ﺑﺮﺗﻘﺎﻟﻲ‬،‫ﻗﯿﻤﺔ‬ ‫ﻓﺘﻜﻮن‬pH‫ﻟﮭﺎ‬:
*3*6*9*11
28-‫ﻣﻦ‬ ‫ﻗﻄﺮة‬ ‫أﺿﯿﻔﺖ‬ ‫إذا‬‫اﻟﺒﺮوﻣﻮﺛﯿﻤﻮل‬ ‫أزرق‬ ‫ﻛﺎﺷﻒ‬‫إﻟﻰ‬25 mL‫ﻣﺤﻠﻮل‬ ‫ﻣﻦ‬0.1 M‫ﺣﻤﺾ‬HCI‫أﺿﯿﻒ‬ ‫ﺛﻢ‬
24.9 mL‫ﻣﻦ‬NaOH0.1 M‫ﻓ‬‫ﺈ‬‫اﻟ‬ ‫ﻟﻮن‬ ‫ن‬‫ﻜﺎﺷﻒ‬:
*‫ﻣﻦ‬ ‫ﯾﺘﻐﯿﺮ‬‫اﻷﺻﻔﺮ‬‫إﻟﻰ‬‫اﻷزرق‬*‫ﻣﻦ‬ ‫ﯾﺘﻐﯿﺮ‬‫اﻷزرق‬‫إﻟﻰ‬‫اﻷﺻﻔﺮ‬*‫ﺗﻐﯿﯿﺮ‬ ‫أي‬ ‫ﻋﻠﯿﮫ‬ ‫ﯾﻄﺮأ‬ ‫ﻻ‬*‫ﯾﺰول‬
29-‫ﻣﻨﮫ‬ ‫ﻗﻄﺮات‬ ‫إﺿﺎﻓﺔ‬ ‫ﻋﻨﺪ‬ ‫اﻟﺰھﺮي‬ ‫اﻷﺣﻤﺮ‬ ‫اﻟﻠﻮن‬ ‫ﻇﮭﻮر‬ ‫ﺗﻔﺴﺮ‬ ‫ﻓﻜﯿﻒ‬ ‫اﻟﺤﻤﻀﯿﺔ‬ ‫اﻟﻜﻮاﺷﻒ‬ ‫ﻣﻦ‬ ‫اﻟﻔﯿﻨﻮﻟﻔﺜﺎﻟﯿﻦ‬ ‫أن‬ ‫ﻋﻠﻤﺖ‬ ‫إذا‬
‫اﻟﺒﻮﺗﺎﺳﯿﻮم‬ ‫ھﯿﺪروﻛﺴﯿﺪ‬ ‫ﻣﺤﻠﻮل‬ ‫إﻟﻰ‬‫؟‬
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30-‫ﻣﻌﺎ‬ ‫ﺗﺠﺮﺑﺔ‬ ‫ﻓﻲ‬‫أﺿﺎﻓ‬ ‫ﯾﺮة‬‫ﺣﻠﯿﻤﺔ‬ ‫ﺖ‬)8(‫اﻟ‬ ‫ﻣﻦ‬ ‫ﻗﻄﺮات‬‫ﻜﺎﺷﻒ‬‫ﺣﺘﻲ‬‫ﺗ‬‫ﺑﻮﺿﻮح‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻧﮭﺎﯾﺔ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺮى‬،‫ﻗﺎﻣ‬ ‫ﻣﺎ‬ ‫ھﻞ‬‫ﺖ‬‫ﺑﮫ‬
‫؟‬ ‫إﺟﺎﺑﺘﻚ‬ ‫ﺑﺮر‬ ‫؟‬ ‫ﺻﺤﯿﺢ‬
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‫ﺳ‬‫ﺎ‬‫د‬‫ﺳ‬ً‫ﺎ‬:‫و‬‫ﺿ‬‫ﺢ‬‫ا‬‫ﻟ‬‫ﻤ‬‫ﻘ‬‫ﺼ‬‫ﻮ‬‫د‬‫ﺑ‬‫ﻜ‬‫ﻞ‬‫ﻣ‬‫ﻦ‬:-
31-‫ﻧ‬‫ﻘ‬‫ﻄ‬‫ﺔ‬‫ا‬‫ﻟ‬‫ﺘ‬‫ﻜ‬‫ﺎ‬‫ﻓ‬‫ﺆ‬
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32-‫ا‬‫ﻟ‬‫ﺘ‬‫ﻔ‬‫ﻜ‬‫ﻚ‬---------------------------------------------------------------------------------------------
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33-‫ا‬‫ﻟ‬‫ﻤ‬‫ﻌ‬‫ﺎﯾ‬‫ﺮ‬‫ة‬
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‫اﻟﻜﺎﺷﻒ‬
‫اﻟﺒﺮوﻣﻮﺛﯿﻤﻮل‬ ‫أزرق‬
‫اﻟﻤﯿﺜﯿﻞ‬ ‫ﺑﺮﺗﻘﺎﻟﻲ‬
‫اﻟﻔﯿﻨﻮل‬ ‫أﺣﻤﺮ‬
‫اﻟﺸﻤﺲ‬ ‫ﺗﺒﺎع‬ ‫ﺻﺒﻐﺔ‬
‫اﻟﻔﯿﻨﻮﻟﻔﺜﺎﻟﯿﻦ‬
‫ﻋﺎم‬ ‫ﻛﺎﺷﻒ‬
‫أزرق‬
‫ﺗﻐﯿﺮ‬
‫أﺻﻔﺮ‬
‫أﺻﻔﺮ‬
‫ﺗﻐﯿﺮ‬
‫أﺣﻤﺮ‬
‫أﺣﻤﺮ‬
‫ﺗﻐﯿﺮ‬
‫أﺻﻔﺮ‬
‫أزرق‬
‫ﺗﻐﯿﺮ‬
‫أﺣﻤﺮ‬
‫زھﺮي‬
‫ﺗﻐﯿﺮ‬
‫اﻟﻠﻮن‬ ‫ﻋﺪﯾﻢ‬
‫ﻏﺎﻣﻖ‬ ‫ﺑﻨﻔﺴﺠﻲ‬
‫ﺑﻨﻔﺴﺠﻲ‬
‫ﻏﺎﻣﻖ‬ ‫أﺣﻤﺮ‬
‫أزرق‬‫أﺧﻀﺮ‬‫أﺻﻔﺮ‬‫ﺑﺮﺗﻘﺎﻟﻲ‬‫أﺣﻤﺮ‬
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
pH

5. 5
‫اﻟ‬ ‫اﻟﺴﺆال‬‫ﺮاﺑﻊ‬:
ً‫ﻻ‬‫أو‬:‫ﯾﻠﻲ‬ ‫ﻋﻤﺎ‬ ‫اﺟﺐ‬:
34-‫ﻗﻮﺗﮭﺎ‬ ‫ﺣﺴﺐ‬ ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﯾ‬ ً‫ﺎ‬‫ﺗﺮﺗﯿﺒ‬ ‫ﻣﺮﺗﺒﺔ‬ ‫اﻟﻘﻮاﻋﺪ‬ ‫ﺑﻌﺾ‬ ‫ﯾﺤﻮي‬ ‫اﻵﺗﻲ‬ ‫اﻟﺠﺪول‬‫اﻟﻨﺴﺒﯿﺔ‬،‫ﺗﻠﯿﮫ‬ ‫اﻟﺘﻲ‬ ‫اﻷﺳﺌﻠﺔ‬ ‫ﻋﻦ‬ ‫وأﺟﺐ‬ ‫أدرﺳﮫ‬:
‫اﻟﺘﺮﺗﯿﺐ‬‫اﻷﻗﻮى‬ ‫اﻷﺿﻌﻒ‬
‫اﻟﻘﺎﻋﺪة‬H2OSO4
2-
CH3COO-
ClO-
NH2
-
*‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﺤﻤﻮض‬ ‫أي‬)NH3 , CH3COOH , H3O+
(‫اﻷﻗﻮى‬ ‫ھﻮ‬-------------
*‫اﻵﺗﻲ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬:NH2
-
+ CH3COOH D CH3COO-
+ NH3
*‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬ ‫اﻻﺗﺰان‬ ‫ﯾﺮﺟﺢ‬ ‫ﺟﮭﺔ‬ ‫أي‬ ‫إﻟﻰ‬-------------------------------------
‫ذﻟﻚ‬ ‫ﺑﺮر‬-----------------------------------------------------------------------------------
*‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺮاﻓﻘﯿﻦ‬ ‫اﻟﺰوﺟﯿﻦ‬ ‫ﺣﺪد‬---------------------------------------
*‫ﻟﻠﻘﺎﻋﺪة‬ ‫اﻟﻤﺮاﻓﻖ‬ ‫اﻟﺤﻤﺾ‬ ‫ﺻﯿﻐﺔ‬ ‫ﻣﺎ‬SO4
2-
‫؟‬------------------------------------
35-‫ﻗﯿﻤﺔ‬ ‫ﻣﺎ‬[ H3O+
]‫ﻣﺤﻠﻮل‬ ‫ﻓﻲ‬Ca(OH)2‫اﻟﮭﯿﺪروﺟﯿﻨﻲ‬ ‫رﻗﻤﮫ‬PH = 8.5‫ﻗﯿﻤﺔ‬ ‫وﻣﺎ‬ ‫؟‬[Ca(OH)2]‫؟‬
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ً‫ﺎ‬‫ﺛﺎﻧﯿ‬:‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻷﺳﺌﻠﺔ‬ ‫ﻋﻦ‬ ‫أﺟﺐ‬:
36-‫اﻟﺮﺧﺎﻣﯿﺔ‬ ‫اﻟﻤﺒﺎﻧﻲ‬ ‫ﺗﻠﻒ‬ ‫ﻓﻲ‬ ‫وﺗﺄﺛﯿﺮھﺎ‬ ‫اﻟﺤﻤﻀﯿﺔ‬ ‫اﻷﻣﻄﺎر‬ ‫ﺗﻜﻮن‬ ‫ﺗﻮﺿﯿﺢ‬ ‫ﻓﻲ‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫وﻇﻒ‬:
)SO3(g) + H2O(l) → H2SO4(aq)(
)CaCO3(s) + 2H3O+
(aq) → Ca2+
(aq) + CO2(g) + 3H2O(l)(
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ً‫ﺎ‬‫راﺑﻌ‬:‫ﯾﻠﻲ‬ ‫ﻣﻤﺎ‬ ً‫ﻼ‬‫ﻛ‬ ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﯾ‬ ‫رﺗﺐ‬:
37-‫اﻟﻤﺘﺴﺎوﯾﺔ‬ ‫اﻟﻤﺤﺎﻟﯿﻞ‬‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﺘﺮاﻛﯿﺰ‬)mol/L(‫ﻟﻘﯿﻢ‬ ً‫ﺎ‬‫ﺗﺒﻌ‬pH.
)KOH , CH3COOH , Ba(OH)2 , NH3(‫؟‬
‫اﻷﻗﻞ‬------------‫ﺛﻢ‬------------‫ﺛﻢ‬------------‫ﺛﻢ‬-----------‫اﻷﻋﻠﻰ‬
38-‫ﻗﻮﺗﮭﺎ‬ ‫ﺣﺴﺐ‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻷﺣﻤﺎض‬:
HOIO , HONO2 , HOIO3 , HOI
‫اﻷ‬‫ﺿﻌﻒ‬------------‫ﺛﻢ‬------------‫ﺛﻢ‬------------‫ﺛﻢ‬------------‫اﻷ‬‫ﻗﻮى‬
‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻤﺴﺎﺋﻞ‬ ‫ﺣﻞ‬ ً‫ﺎ‬‫ﺧﺎﻣﺴ‬:
‫أن‬ ‫ﺗﺒﯿﻦ‬ ‫ﻣﻌﺎﯾﺮة‬ ‫ﺗﺠﺮﺑﺔ‬ ‫ﻓﻲ‬25 mL‫ﻣﺤﻠﻮل‬ ‫ﻣﻦ‬Ba(OH)2‫ﻣﻊ‬ ‫ﺗﺘﻌﺎدل‬10 mL،‫ﻣﺤﻠﻮل‬ ‫ﻣﻦ‬
HCl0.0657 M‫اﻟﺘﻜﺎﻓﺆ‬ ‫ﻧﻘﻄﺔ‬ ‫إﻟﻰ‬ ‫ﻟﻠﻮﺻﻮل‬.Ba(OH)2 = 171.35 g/mol
39-‫ﺗﺮﻛﯿﺰ‬ ‫اﺣﺴﺐ‬‫ﺑﺎﻟﻤﻮل‬ ‫اﻟﻘﺎﻋﺪة‬ ‫ﻣﺤﻠﻮل‬/‫ﻟﺘﺮ‬
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40-‫اﺣﺴﺐ‬‫ﺑﺎﻟﺠﺮام‬ ‫اﻟﻘﺎﻋﺪة‬ ‫ﺗﺮﻛﯿﺰ‬/‫؟‬ ‫ﻟﺘﺮ‬
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‫اﻧﺘﮭﺖ‬‫ﺟﻤﯿﻊ‬‫ﻧﻤﺎذج‬‫اﻷﺳﺌﻠﺔ‬
20

6. 6
‫أﺑــﻮﻇﺒــــ‬ ‫ﻣﺠﻠﺲ‬‫ﻲ‬‫ﻟﻠﺘـﻌـﻠــــﯿﻢ‬
‫اﻟﻜﯿﻤﯿﺎء‬ : ‫اﻟﻤــــــــــﺎدة‬ ‫أﺑﻮ‬ ‫ﻣﻨﻄﻘﺔ‬ ‫إدارة‬‫اﻟﺘﻌﻠﯿﻤﯿﺔ‬ ‫ﻇﺒﻲ‬
‫ﺳﺎﻋ‬‫ﺔ‬‫وﻧﺼﻒ‬ : ‫اﻹﺟﺎﺑﺔ‬ ‫زﻣﻦ‬ ‫اﻟﻜﯿﻤﯿﺎء‬ ‫ﺗﻮﺟﯿﮫ‬
‫ﺍﻟﻌﻠﻤﻲ‬ ‫ﻋﺸﺮ‬ ‫ﺍﻟﺜﺎﻧﻲ‬ ‫ﻟﻠﺼﻒ‬ ‫ﺍﻟﺘﺠﺮﻳﺒﻲ‬ ‫ﺍﻻﻣﺘﺤﺎﻥ‬ ‫ﺇﺟﺎﺑﺔ‬ ‫ﳕﻮﺫﺝ‬
‫ﺍﻷﻭﻝ‬ ‫ﺍﻟﺪﺭﺍﺳﻲ‬ ‫ﺍﻟﻔﺼﻞ‬
‫ﺍﻟﺩﺭﺍﺴﻲ‬ ‫ﻟﻠﻌﺎﻡ‬2009/2010‫ﻡ‬‫ﻟﻠﻌﺎﻡ‬ ‫ﺍﻷﻭل‬ ‫ﺍﻟﺩﺭﺍﺴﻲ‬ ‫ﺍﻟﻔﺼل‬ ‫ﻤﻨﻬﺞ‬ ‫ﻭﻓﻕ‬ ‫ﺘﻌﺩﻴﻠﻪ‬ ‫ﺘﻡ‬2010/2011‫ﻡ‬
‫ﻓﻲ‬ ‫ﺍﻷﺴﺌﻠﺔ‬)5(‫ﺫﻟﻙ‬ ‫ﻤﻥ‬ ‫ﺍﻟﺘﺄﻜﺩ‬ ‫ﺍﻟﻁﺎﻟﺏ‬ ‫ﻭﻋﻠﻰ‬ ‫ﺼﻔﺤﺎﺕ‬
‫اﻟﻛﻳﻣﻳﺎء‬ ‫ﻣﻌﻠم‬ ‫إﻋداد‬/‫ﻣﺣﻣد‬ ‫ﺳﺎﻣﻲ‬‫اﻟﻌﻼ‬ ‫أﺑو‬)‫اﻟﻘدم‬ ‫ﻟﻛرة‬ ‫اﻟﺟزﻳرة‬ ‫أﻛﺎدﻳﻣﻳﺔ‬(‫اﻟﻛﻳﻣﻳﺎء‬ ‫ﻣوﺟﻪ‬ ‫ﺑﺈﺷراف‬/‫اﻟﺟﻳﺎر‬ ‫ﻋﺎدل‬
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‫اﻷول‬ ‫اﻟﺴﺆال‬:
ً‫ﻻ‬‫أو‬‫ﻋﻼﻣﺔ‬ ‫ﺿﻊ‬)ü(‫ﯾﺄﺗﻲ‬ ‫ﻣﻤﺎ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺼﺤﯿﺤﺔ‬ ‫اﻟﺘﻜﻤﻠﺔ‬ ‫أﻣﺎم‬:
1-‫اﻷﺣﻤﺎض‬ ‫ﻣﻦ‬ ‫ﻟﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﺎﺋﻲ‬ ‫اﻟﻤﺤﻠﻮل‬ ‫ﻟﻤﻜﻮﻧﺎت‬ ‫ﺑﺎﻟﻤﻮﻻرﯾﺔ‬ ‫اﻟﺘﺮاﻛﯿﺰ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫ﺗﻤﺜﻞ‬ ‫واﻟﺘﻲ‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﺮﺳﻮم‬ ‫ﺗﺄﻣﻞ‬
،‫ﻟﻠﺤﻤﺾ‬ ‫اﻟﻤﺎﺋﻲ‬ ‫اﻟﻤﺤﻠﻮل‬ ‫ﻋﻦ‬ ‫ﻟﻠﺘﻌﺒﯿﺮ‬ ً‫ﺎ‬‫ﺻﺤﯿﺤ‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻷﺷﻜﺎل‬ ‫أي‬:
*** *
2-‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬)Ni2+
+ nH2O D [Ni(H2O)n]2+
(‫ﯾﻌﺘﺒﺮ‬H2O:
*‫ﺑﺮوﻧﺸﺘﺪ‬ ‫ﺣﻤﺾ‬–‫ﻟﻮري‬*‫ﺑﺮوﻧﺸﺘﺪ‬ ‫ﻗﺎﻋﺪة‬–‫ﻟﻮري‬*‫ﻟﻮﯾﺲ‬ ‫ﺣﻤﺾ‬*‫ﻟﻮﯾﺲ‬ ‫ﻗﺎﻋﺪة‬
3-‫اﻟﻜﺎﺗﯿﻮﻧﺎت‬ ‫أﺣﺪ‬ ‫ﻋﻦ‬ ‫ﻟﻠﻜﺸﻒ‬ ‫ﺗﺠﺮﺑﺔ‬ ‫ُﺟﺮﯾﺖ‬‫أ‬‫اﻟ‬ ‫واﻟﺸﻜﻞ‬‫ﺘﺎﻟﻲ‬‫اﻟﺘﺠﺮﺑﺔ‬ ‫ﯾﺒﯿﻦ‬،
‫اﻟﻤﺮﻓﻖ‬ ‫اﻟﺠﺪول‬ ‫ﻣﻮﻇﻔﺎ‬،
‫أي‬ ‫ﺗﻮﻗﻊ‬‫ﻋﻨﮫ‬ ‫اﻟﻜﺸﻒ‬ ‫ﺗﻢ‬ ‫اﻟﻜﺎﺗﯿﻮﻧﺎت‬:
*‫ھﻮ‬ ‫اﻟﻜﺎﺗﯿﻮن‬2+
Pb*‫ھﻮ‬ ‫اﻟﻜﺎﺗﯿﻮن‬2+
Ca*‫ھﻮ‬ ‫اﻟﻜﺎﺗﯿﻮن‬2+
Mg*‫ھﻮ‬ ‫اﻟﻜﺎﺗﯿﻮن‬2+
Ba
30
‫ﺑــ‬‫اﻟﺘﺮﻛﯿﺰ‬M
HCl
Cl-
H3O+
‫ﻋﺪا‬ ‫ﻣﺎ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫اﻟﻜﻠﻮرﯾﺪات‬ ‫ﻣﻌﻈﻢ‬Ag+
‫و‬Pb2+
‫ﻋﺪا‬ ‫ﻣﺎ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫ﻻ‬ ‫اﻟﻜﺒﺮﯾﺘﯿﺪات‬ ‫ﻣﻌﻈﻢ‬Ca2+
, Sr2+
, K+
, Na+
‫ﻋﺪا‬ ‫ﻣﺎ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫ﻻ‬ ‫اﻟﻜﺮﺑﻮﻧﺎت‬ ‫ﻣﻌﻈﻢ‬K+
, Na+
, NH4
+
‫ﻋﺪا‬ ‫ﻣﺎ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫اﻟﻜﺒﺮﯾﺘﺎت‬ ‫ﻣﻌﻈﻢ‬Ca2+
, Pb2+
, Ba2+
‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﺗﺬوب‬ ‫واﻟﻨﯿﺘﺮات‬ ‫اﻟﻜﻠﻮرات‬
Na2CO3
‫راﺳﺐ‬
K2SO4
‫راﺳﺐ‬ ‫ﯾﺘﻜﻮن‬ ‫ﻻ‬
(NH4)2S
‫راﺳﺐ‬
KNO3
‫راﺳﺐ‬ ‫ﯾﺘﻜﻮن‬ ‫ﻻ‬
‫ﺑــ‬‫اﻟﺘﺮﻛﯿﺰ‬M
HNO3
NO3
-
H3O+
‫ﺑــ‬‫اﻟﺘﺮﻛﯿﺰ‬M
H2SO4
HSO4
-
H3O+
SO4
2-
‫ﺑــ‬‫اﻟﺘﺮﻛﯿﺰ‬M
HF
F-
H3O+

7. 7
4-‫اﻟ‬ ‫اﻟﻤﺤﺎﻟﯿﻞ‬ ‫ﻓﻲ‬‫ﺤﻤﻀ‬‫ﺣﺮارة‬ ‫درﺟﺔ‬ ‫ﻋﻨﺪ‬ ‫ﯿﺔ‬50 ˚C‫ﺗﻜﻮن‬:
*[H3O+
]=[OH-
]*Kw<[H3O+
] . [OH-
]
*]-
] . [OH+
O3[H>wK*]-
] . [OH+
O3=[HwK
5-‫أذاب‬ ‫إذا‬‫ﺣﻤﺪان‬0.5 mol‫ﻣﻦ‬Ba(OH)2‫ﻓﻲ‬0.5 L‫اﻟﻤﺤﻠﻮل‬ ‫ﺗﺮﻛﯿﺰ‬ ‫ﻓﯿﻜﻮن‬ ‫اﻟﻤﻘﻄﺮ‬ ‫اﻟﻤﺎء‬ ‫ﻣﻦ‬:
*0.5 M*m0.5*1.0 M*m1.0
6-‫اﻟﺤﻤﻮﺿﺔ‬ ‫ﻣﻀﺎدات‬ ‫ﻓﻲ‬ ‫ﯾﺴﺘﺨﺪم‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻤﺮﻛﺒﺎت‬ ‫أﺣﺪ‬:
*3KHCO*3KHSO*3NaHCO*3NaHSO
‫ب‬(ً‫ﺎ‬‫ﺛﺎﻧﯿ‬:‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻌﺒﺎرات‬ ‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻋﻠﯿﮫ‬ ‫ﺗﺪل‬ ‫اﻟﺬي‬ ‫اﻟﻌﻠﻤﻲ‬ ‫اﻟﻤﺼﻄﻠﺢ‬ ‫أو‬ ‫اﻻﺳﻢ‬ ‫اﻟﻘﻮﺳﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻛﺘﺐ‬:
7-)‫اﻟﻨﯿﺘﺮﯾﻚ‬ ‫ﺣﻤﺾ‬HNO3(‫ﻣﺎدة‬‫ﻓﻲ‬ ‫وﺗﺴﺘﺨﺪم‬ ‫أﺻﻔﺮ‬ ‫ﻟﻮن‬ ‫اﻟﺒﺮوﺗﯿﻨﺎت‬ ‫ﺗﻜﺴﺐ‬ ‫ھﺎﻣﺔ‬ ‫ﻛﯿﻤﯿﺎﺋﯿﺔ‬‫اﻟﻤﺘﻔﺠﺮات‬ ‫ﺻﻨﺎﻋﺔ‬،
‫واﻟﻤﻄﺎط‬،‫اﻟﺒﻼﺳﺘﯿﻜﯿﺔ‬ ‫واﻟﻤﻮاد‬،‫واﻷﺻﺒﺎغ‬.
8-)‫اﻷﻣﻔﻮﺗﯿﺮﯾﺔ‬ ‫اﻟﻤﻮاد‬(‫ﻛﻘﺎﻋﺪة‬ ‫أو‬ ‫ﻛﺤﻤﺾ‬ ‫ﺗﺘﻔﺎﻋﻞ‬ ‫أن‬ ‫ﯾﻤﻜﻨﮭﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻮاد‬.
9-)‫ﻟﻠﻤﺎء‬ ‫اﻟﺬاﺗﻲ‬ ‫اﻟﺘﺄﯾﻦ‬(‫إﻧﺘﺎج‬‫ﻣﺎء‬ ‫ﺟﺰﯾﺌﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﺮوﺗﻮن‬ ‫ﺑﺎﻧﺘﻘﺎل‬ ‫ھﯿﺪروﻛﺴﯿﺪ‬ ‫وأﯾﻮن‬ ‫ھﯿﺪروﻧﯿﻮم‬ ‫أﯾﻮن‬.
ً‫ﺎ‬‫راﺑﻌ‬:‫ﯾﻠﻲ‬ ‫ﱠﺎ‬‫ﻤ‬‫ﻋ‬ ‫أﺟﺐ‬:
10-‫اﻟ‬ ‫ُﻀﺮت‬‫ﺣ‬‫ﻤﺤﺎﻟﯿﻞ‬‫اﻟﺘﺎﻟﯿﺔ‬‫ﻣﻮﻻرﯾﺔ‬ ‫ﺑﺘﺮاﻛﯿﺰ‬‫ﻣﺘﺴﺎوﯾﺔ‬)HF , HCl , C6H12O6 , K2SO4.(‫رﺗﺐ‬‫ھﺬه‬‫اﻟﻤﺤﺎﻟﯿﻞ‬
‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫ﻟﻠﺘﯿﺎر‬ ‫ﺗﻮﺻﯿﻠﮭﺎ‬ ‫ﺣﺴﺐ‬ ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﯾ‬.
‫اﻷﻗﻞ‬C6H12O6‫ﺛﻢ‬HF‫ﺛﻢ‬HCl‫ﺛﻢ‬K2SO4‫اﻷﻋﻠﻰ‬‫ﺗﻮﺻﯿﻞ‬
‫اﻟ‬ ‫اﻟﺴﺆال‬‫ﺜﺎﻧﻲ‬:
ً‫ﻻ‬‫أو‬:
11-‫اﻟﻔﻮﺳﻔﻮرﯾﻚ‬ ‫ﺣﻤﺾ‬H3PO4‫ھﯿﺪروﺟﯿﻦ‬ ‫ذرات‬ ‫ﺛﻼث‬ ‫ﻋﻠﻰ‬ ‫ﺻﯿﻐﺘﮫ‬ ‫ﺗﺤﺘﻮي‬،‫ﺣﻤﺾ‬ ‫ﺑﯿﻨﻤﺎ‬ ‫اﻟﺒﺮوﺗﻮن‬ ‫ﺛﻼﺛﻲ‬ ‫ﻛﺤﻤﺾ‬ ‫ﻣﺼﻨﻒ‬ ‫وھﻮ‬
‫اﻷﺳﯿﺘﯿﻚ‬CH3COOH‫ھﯿﺪروﺟﯿﻦ‬ ‫ذرات‬ ‫أرﺑﻊ‬ ‫ﻋﻠﻰ‬ ‫ﺻﯿﻐﺘﮫ‬ ‫ﺗﺤﺘﻮي‬،‫اﻟﺒﺮوﺗﻮن‬ ‫أﺣﺎدي‬ ‫ﻛﺤﻤﺾ‬ ‫ﻣﺼﻨﻒ‬ ‫وھﻮ‬.‫ﺑﺮر‬‫ذﻟ‬‫ﻚ‬‫؟‬
‫ﻷن‬‫ﻛﻞ‬‫ذر‬‫ة‬‫ھﯿﺪروﺟﯿﻦ‬H‫اﻟﻔﻮﺳﻔﻮرﯾﻚ‬ ‫ﺣﻤﺾ‬ ‫ﺟﺰيء‬ ‫ﻓﻲ‬‫ﻣ‬‫ﺮﺗﺒﻄ‬‫ﺔ‬‫ﺑ‬‫ﺬرة‬‫و‬ ‫أﻛﺴﺠﯿﻦ‬‫ﺗ‬ ‫أن‬ ‫ﯾﻤﻜﻦ‬ ‫ﺑﺎﻟﺘﺎﻟﻲ‬، ‫ﺘﺄﯾﻦ‬
‫ھﯿﺪروﺟﯿﻦ‬ ‫ذرة‬ ‫اﻷﺳﺘﯿﻚ‬ ‫ﺣﻤﺾ‬ ‫ﻓﻲ‬ ‫ﺑﯿﻨﻤﺎ‬H‫واﺣﺪة‬‫ﻣ‬‫ﺑ‬ ‫ﺮﺗﺒﻂ‬‫ﺬرة‬‫و‬ ‫أﻛﺴﺠﯿﻦ‬‫ﺗ‬ ‫أن‬ ‫ﯾﻤﻜﻦ‬‫ﺘﺄﯾﻦ‬‫وذرات‬H‫اﻟﺜﻼث‬
‫اﻟﻜﺮﺑﻮن‬ ‫ﺑﺬرة‬ ‫ﻣﺮﺗﺒﻄﺔ‬ ‫اﻷﺧﺮى‬C‫اﻟﺒﺮوﺗﻮن‬ ‫أﺣﺎدي‬ ‫ﺣﻤﺾ‬ ‫ھﻮ‬ ‫اﻷﺳﯿﺘﯿﻚ‬ ‫ﺣﻤﺾ‬ ‫وﻟﺬﻟﻚ‬ ‫ﺗﺘﺄﯾﻦ‬ ‫وﻻ‬.
ً‫ﺎ‬‫ﺛﺎﻧﯿ‬:‫ﻣﺰج‬ ‫ﻋﻨﺪ‬‫ﻣﺤﻠﻮﻟ‬‫ﻲ‬‫اﻟﺮﺻﺎص‬ ‫ﻧﯿﺘﺮات‬IIPb(NO3)2‫و‬‫اﻟﺒﻮﺗﺎﺳﯿﻮم‬ ‫ﻛﻠﻮرﯾﺪ‬KCl‫ﯾ‬‫ﺘﻜﻮ‬‫راﺳﺐ‬ ‫ن‬
‫اﻟﺮﺻﺎص‬ ‫ﻛﻠﻮرﯾﺪ‬ ‫ﻣﻦ‬II:‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻔﻘﺮات‬ ‫ﻋﻦ‬ ‫أﺟﺐ‬:
12-‫اﻷﯾﻮﻧﯿﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻛﺘﺎﺑﺔ‬‫اﻟﻌﺎﻣﺔ‬2K+
(aq)+ 2Cl
-
(aq)+ Pb2+
(aq) +2NO3
-
(aq) → 2K+
(aq)+ 2NO3
-
(aq) + PbCl2(s)
13-‫اﻟﻤﺘﻔﺮﺟﺔ‬ ‫اﻷﯾﻮﻧﺎت‬ ‫ﺗﺤﺪﯾﺪ‬K+
, NO3
-
14-‫اﻟﺼﺮﻓﺔ‬ ‫اﻷﯾﻮﻧﯿﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻛﺘﺎﺑﺔ‬Pb2+
(aq) +2Cl
-
(aq) → PbCl2(s)
15-‫ﺣﺪد‬‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﺘﻔﺎﻋﻼت‬ ‫ﻓﻲ‬‫اﻟﻤﺘﻔﺎﻋﻼت‬‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬‫إﻟﻰ‬‫ﻗﻮاﻋﺪ‬‫أﺣﻤﺎض‬ ‫أو‬‫ارھﯿﻨﯿﻮس‬،‫و‬‫ﺑﺮوﻧﺸﺘﺪ‬-‫ﻟﻮري‬،‫و‬‫ﻟﻮﯾﺲ‬
‫اﻟﻤﻌﺎدﻟﺔ‬‫اﻟﺤﻞ‬
1H+
(aq) + NH3(aq) → NH4
+
(aq)
H+
‫ﻟﻮﯾﺲ‬ ‫ﺣﻤﺾ‬
NH3‫ﺑﺮوﻧﺸﺘﺪ‬ ‫ﻗﺎﻋﺪة‬–‫وﻟﻮﯾﺲ‬ ‫ﻟﻮري‬
2HI(aq) + H2O(l) → H3O+
(aq) + I-
(aq)
HI‫وﺑﺮوﻧﺸﺘﺪ‬ ‫أرھﯿﻨﯿﻮس‬ ‫ﺣﻤﺾ‬–‫وﻟﻮﯾﺲ‬ ‫ﻟﻮري‬
H2O‫ﺑﺮوﻧﺸﺘﺪ‬ ‫ﻗﺎﻋﺪة‬–‫وﻟﻮﯾﺲ‬ ‫ﻟﻮري‬
3KOH(s) → K+
(aq) + OH-
(aq)NaOH‫أرھﯿﻨﯿﻮس‬ ‫ﻗﺎﻋﺪة‬
4C6H5NH2 + H2O D C6H5NH3
+
+OH-C6H5NH2‫وﺑﺮوﻧﺸﺘﺪ‬ ‫أرھﯿﻨﯿﻮس‬ ‫ﻗﺎﻋﺪة‬–‫وﻟﻮﯾﺲ‬ ‫ﻟﻮري‬
H2O‫ﺑﺮوﻧﺸﺘﺪ‬ ‫ﺣﻤﺾ‬–‫وﻟﻮﯾﺲ‬ ‫ﻟﻮري‬
25

8. 8
ً‫ﺎ‬‫ﺛﺎﻟﺜ‬:‫ﻛﺜﺎﻓﺘﮫ‬ ‫اﻟﻨﯿﺘﺮﯾﻚ‬ ‫ﺣﻤﺾ‬ ‫ﻣﺤﻠﻮل‬1.19 g/cm3
،‫اﻟﻜﺘﻠﻲ‬ ‫اﻟﻤﺌﻮي‬ ‫وﺗﺮﻛﯿﺰه‬32%
‫واﻟﻤﻄﻠﻮب‬‫ﺣﺴ‬‫ﺎ‬‫ب‬:
16-‫ﻓﻲ‬ ‫اﻟﻤﺬﯾﺐ‬ ‫ﻛﺘﻠﺔ‬100g‫اﻟﻤﺤﻠﻮل‬ ‫ھﺬا‬ ‫ﻣﻦ‬
჻‫اﻟﻜﺘﻠﻲ‬ ‫اﻟﻤﺌﻮي‬ ‫اﻟﺘﺮﻛﯿﺰ‬=32%∴‫اﻟﻤﺬاب‬ ‫ﻛﺘﻠﺔ‬=32 g‫اﻟﻤﺬﯾﺐ‬ ‫وﻛﺘﻠﺔ‬68 g
17-‫؟‬ ‫اﻟﻤﺤﻠﻮل‬ ‫ھﺬا‬ ‫وﻣﻮﻻرﯾﺔ‬ ‫ﻣﻮﻻﻟﯿﺔ‬‫ﺑﺄن‬ ً‫ﺎ‬‫ﻋﻠﻤ‬HNO3 = 63 g/mol
‫اﻟﻤﺤﻠﻮل‬ ‫ﺣﺠﻢ‬ = ×
.
=
‫اﻟﻤﻮﻻرﯾﺔ‬ = × × × = .
‫اﻟﻤﻮﻻﻟﯿﺔ‬ = × × × = .
18-‫ھﯿﺪروﻛﺴﯿﺪ‬ ‫ﻣﻦ‬ ‫ﻗﯿﺎﺳﻲ‬ ‫ﻣﺤﻠﻮل‬ ‫ﺑﺎﺳﺘﺨﺪام‬ ‫اﻟﺘﺮﻛﯿﺰ‬ ‫ﻣﺠﮭﻮل‬ ‫اﻷﺳﯿﺘﯿﻚ‬ ‫ﺣﻤﺾ‬ ‫ﻣﺤﻠﻮل‬ ‫ﺗﺮﻛﯿﺰ‬ ‫ﻟﺘﻘﺪﯾﺮ‬ ‫ﻣﻌﺎﯾﺮة‬ ‫ﺗﺠﺮﺑﺔ‬ ‫ﻓﻲ‬
‫اﻟﺼﻮدﯾﻮم‬0.1 M‫ﻟﻤﻌﺎﯾﺮة‬ ‫اﻟﺘﺠﺮﺑﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺎت‬ ‫إﺣﺪى‬ ‫ﻧﻔﺬت‬30 mL‫اﻟﺤﻤﺾ‬ ‫ﻣﻦ‬‫واﺳﺘﺨﺪﻣﺖ‬ ‫ﻧﻈﯿﻒ‬ ‫ﻛﺄس‬ ‫ﻓﻲ‬
‫ﺑﺎﻟﺮﺳﻢ‬ ‫ﻛﻤﺎ‬ ‫اﻟﻤﻨﺤﻨﻰ‬ ‫ورﺳﻢ‬ ‫اﻟﻨﺘﺎﺋﺞ‬ ‫ﺑﺘﺴﺠﯿﻞ‬ ‫وﻗﺎﻣﺖ‬ ‫واﻟﺴﺤﺎﺣﺔ‬ ‫اﻟﻤﯿﺜﯿﻞ‬ ‫ﺑﺮﺗﻘﺎﻟﻲ‬ ‫ﻛﺎﺷﻒ‬.
‫وﺟﺪت‬ ‫إن‬ ‫اﻷﺧﻄﺎء‬ ‫ذﻛﺮ‬ ‫ﻣﻊ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻋﻤﻞ‬ ‫ﻋﻠﻰ‬ ً‫ﺎ‬‫ﺣﻜﻤ‬ ‫اﺻﺪر‬،‫ّﺑﮭﺎ‬‫ﻮ‬‫وﺻ‬.)‫رﻗﻢ‬ ‫اﻟﺼﻔﺤﺔ‬ ‫ﻓﻲ‬ ‫ﺑﺎﻟﻤﺨﻄﻂ‬ ‫اﺳﺘﻌﻦ‬6(
‫ﻣﻌﺎﯾﺮة‬ ‫ﺑﻌﻤﻠﯿﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺗﻘﻮم‬‫دﻗﯿﻘ‬ ‫ﻏﯿﺮ‬‫ﺔ‬
‫اﻷﺧﻄﺎء‬
ü‫اﻟﻤﻌﺎﯾﺮة‬ ‫ﻋﻤﻠﯿﺔ‬ ‫ﻓﻲ‬ ‫ﻛﺄس‬ ‫اﺳﺘﺨﺪام‬
‫واﻟﺼﻮاب‬:‫ﻧﻈﯿﻒ‬ ‫ﻣﺨﺮوﻃﻲ‬ ‫دورق‬ ‫اﺳﺘﺨﺪام‬
ü‫اﻟﻤﯿﺜﯿﻞ‬ ‫ﺑﺮﺗﻘﺎﻟﻲ‬ ‫ﻛﺎﺷﻒ‬ ‫اﺳﺘﺨﺪام‬
‫واﻟﺼﻮاب‬:‫اﻟﻔﯿﻨﻮﻟﻔﺜﺎﻟﯿﻦ‬ ‫ﻛﺎﺷﻒ‬ ‫اﺳﺘﺨﺪام‬
ü‫ﻗﻮﯾﺔ‬ ‫اﻟﻘﺎﻋﺪة‬ ‫ﻷن‬ ‫ﺧﻄﺄ‬ ‫اﻟﺒﯿﺎﻧﻲ‬ ‫اﻟﺮﺳﻢ‬0.1 M
‫ﻋﻨﺪ‬ ‫اﻟﻤﻨﺤﻨﻲ‬ ‫ﻓﻼﯾﻨﺘﮭﻲ‬11=pH‫ﺑﺎﻟﺮﺳﻢ‬ ‫ﻛﻤﺎ‬
‫واﻟﺼﻮاب‬:‫ﻋﻨﺪ‬ ‫ﯾﻨﺘﮭﻲ‬13=pHً‫ﺎ‬‫ﺗﻘﺮﯾﺒ‬
‫اﻟﺴ‬‫ﺆ‬‫اﻟ‬ ‫ال‬‫ﺜﺎﻟﺚ‬:
ً‫ﻻ‬‫أو‬:‫ﯾﻠﻲ‬ ‫ﻣﺎ‬ ‫ﻓﺴﺮ‬:
19-‫ذوﺑﺎن‬ ‫ﯾﻌﺪ‬‫اﻟﺼﻮدﯾﻮم‬ ‫ﻛﻠﻮرﯾﺪ‬ً‫ﺎ‬‫ﺗﻔﻜﻜ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬‫ذوﺑﺎن‬ ‫ﺑﯿﻨﻤﺎ‬‫اﻟ‬ ‫ﻛﻠﻮرﯾﺪ‬‫ﮭﯿﺪروﺟﯿﻦ‬ً‫ﺎ‬‫ﺗﺄﯾﻨ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬‫؟‬
‫اﻟﺼﻮدﯾﻮم‬ ‫ﻛﻠﻮرﯾﺪ‬‫ﺗﻨﻔﺼ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ذوﺑﺎﻧﮫ‬ ‫وﻋﻨﺪ‬ ‫وﺳﺎﻟﺒﺔ‬ ‫ﻣﻮﺟﺒﺔ‬ ‫أﯾﻮﻧﺎت‬ ‫ﻣﻦ‬ ‫ﯾﺘﻜﻮن‬ ‫أﯾﻮﻧﻲ‬ ‫ﻣﺮﻛﺐ‬‫ﻞ‬‫اﻷﯾﻮﻧﺎت‬
ً‫ﻼ‬‫أﺻ‬ ‫اﻟﻤﻮﺟﻮدة‬‫ﺑﻌﻀﮭﺎ‬ ‫ﻋﻦ‬،‫ﺑﯿﻨﻤﺎ‬‫اﻟﮭﯿﺪروﺟﯿﻦ‬ ‫ﻛﻠﻮرﯾﺪ‬‫ﺑﻔﻌﻞ‬ ‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫إذاﺑﺘﮫ‬ ‫ﻋﻨﺪ‬ ‫ﯾﺘﺄﯾﻦ‬ ‫ﻗﻄﺒﻲ‬ ‫ﺟﺰﯾﺌﻲ‬ ‫ﻣﺮﻛﺐ‬
‫اﻟﺬاﺋﺐ‬ ‫ﻏﯿﺮ‬ ‫اﻟﻤﺮﻛﺐ‬ ‫ﻓﻲ‬ ً‫ﻼ‬‫أﺻ‬ ‫ﻣﻮﺟﻮدة‬ ‫ﺗﻜﻦ‬ ‫ﻟﻢ‬ ‫اﻟﺘﻲ‬ ‫اﻷﯾﻮﻧﺎت‬ ‫وﺗﺘﻜﻮن‬ ‫اﻟﻘﻄﺒﻲ‬ ‫اﻟﻤﺎء‬ ‫ﺟﺰﯾﺌﺎت‬
20-‫ﺗﻌﺘﺒﺮ‬HS
-
‫ﻣﻦ‬ ‫أﻗﻮى‬ ‫ﻗﺎﻋﺪة‬HSO4
-
‫؟‬
‫ﻟﻠﻘﺎﻋﺪة‬ ‫اﻟﻤﺮاﻓﻖ‬ ‫اﻟﺤﻤﺾ‬ ‫ﻷن‬HS
-
‫وھﻮ‬H2S‫ﻟﻠﻘﺎﻋﺪة‬ ‫اﻟﻤﺮاﻓﻖ‬ ‫اﻟﺤﻤﺾ‬ ‫ﻣﻦ‬ ‫أﺿﻌﻒ‬HSO4
-
‫وھﻮ‬H2SO4
‫اﻟﻘﺎﻋﺪة‬ ‫ﻗﺪرة‬ ‫ﺗﻜﻮن‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬HS
-
‫اﻟﻘﺎﻋﺪة‬ ‫ﻗﺪرة‬ ‫ﻣﻦ‬ ‫أﻛﺒﺮ‬ ‫ﺑﮫ‬ ‫واﻻﺣﺘﻔﺎظ‬ ‫اﻟﺒﺮوﺗﻮن‬ ‫ﺟﺬب‬ ‫ﻋﻠﻰ‬HSO4
-
ً‫ﺎ‬‫ﺛﺎﻧﯿ‬:‫ﻓﻘﺮة‬ ‫ﻟﻜﻞ‬ ‫ﺑﺪاﺋﻞ‬ ‫أرﺑﻌﺔ‬ ‫أﻣﺎﻣﻚ‬–‫اﻟﺒﺪﯾﻞ‬ ‫اﺧﺘﺮ‬‫اﻟﻤﻨﺴﺠﻢ‬ ‫ﻏﯿﺮ‬ً‫ﺎ‬‫ﻋﻠﻤﯿ‬‫ﻣﻊ‬‫ﺗ‬‫ﺒﺮ‬‫ﯾ‬‫إﺟﺎﺑﺘﻚ‬ ‫ﺮ‬:
21-3PO3H،3SeO2H،4PO3H،4SeO2H
‫اﻟﺴﺒﺐ‬:‫اﻟﺒﺮوﺗﻮن‬ ‫ﺛﻨﺎﺋﯿﺔ‬ ‫أﺣﻤﺎض‬ ‫واﻟﺒﻘﯿﺔ‬ ‫اﻟﺒﺮوﺗﻮن‬ ‫ﺛﻼﺛﻲ‬ ‫ﺣﻤﺾ‬ ‫ﻷﻧﮫ‬
25
0 5 10 15 20 25 30 35 40 45
‫ﺣﺠﻢ‬NaOH‫ﺑــ‬ ‫اﻟﻤﻀﺎف‬mL
pH
14
12
10
8
6
4
2
0

9. 9
ً‫ﺎ‬‫ﺛﺎﻟﺜ‬:‫ﺗﻠﯿﮫ‬ ‫اﻟﺘﻲ‬ ‫اﻷﺳﺌﻠﺔ‬ ‫ﻋﻦ‬ ‫وأﺟﺐ‬ ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﺠﺪول‬ ‫ﺗﺄﻣﻞ‬:)‫رﻗﻢ‬ ‫اﻟﺼﻔﺤﺔ‬ ‫ﻓﻲ‬ ‫اﻟﺬوﺑﺎﻧﯿﺔ‬ ‫ﺑﺠﺪول‬ ‫اﺳﺘﻌﻦ‬2(
22-‫رﻗﻢ‬ ‫أي‬‫ﯾﻤﺜﻞ‬‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫ﻟﻠﺬوﺑﺎن‬ ‫ﻗﺎﺑﻞ‬ ‫أﯾﻮﻧﻲ‬ ‫ﻣﺮﻛﺐ‬‫؟‬5
23-‫اﻟﻤﺮﻛﺒﯿﻦ‬ ‫ﻣﺤﻠﻮﻟﻲ‬ ‫ﻣﺰج‬ ‫ﻋﻨﺪ‬ ‫اﻟﻤﺘﻔﺮﺟﺔ‬ ‫اﻷﯾﻮﻧﺎت‬ ‫ﻣﺎ‬3‫و‬8‫؟‬‫ﻣﺘﻔﺮﺟﺔ‬ ‫أﯾﻮﻧﺎت‬ ‫ﺗﻮﺟﺪ‬ ‫ﻻ‬
24-‫رﻗﻢ‬ ‫أي‬‫ﯾﻤﺜﻞ‬‫اﻟﻤﺎء‬ ‫ﻓﻲ‬ ‫اﻟﺬوﺑﺎن‬ ‫ﺷﺤﯿﺢ‬ ‫أﯾﻮﻧﻲ‬ ‫ﻣﺮﻛﺐ‬‫؟‬AgCl
25-‫رﻗﻢ‬ ‫أي‬‫ﯾﻤﺜﻞ‬‫ﻓﻲ‬ ‫ﺟﺰﺋﯿﺎ‬ ‫ﯾﺘﺄﯾﻦ‬ ‫ﺟﺰﯾﺌﻲ‬ ‫ﻣﺮﻛﺐ‬‫اﻟﻤﺎء‬‫؟‬COOH3CH
26-‫أي‬‫رﻗﻢ‬‫ﺛﻨﺎﺋﻲ‬ ‫ﺣﻤﺾ‬ ‫ﯾﻤﺜﻞ‬‫؟‬HCl
‫ﺧﺎﻣﺴﺎ‬:‫أﺳﺌﻠﺔ‬ ‫ﻣﻦ‬ ‫ﯾﻠﯿﮫ‬ ‫ﻋﻤﺎ‬ ‫وأﺟﺐ‬ ‫إدرﺳﮫ‬ ‫اﻷدﻟﺔ‬ ‫ﻣﻦ‬ ‫اﻟﻌﺪﯾﺪ‬ ‫ﻣﺪي‬ ‫ﺗﻐﯿﺮ‬ ‫ﯾﺒﯿﻦ‬ ‫اﻟﺘﺎﻟﻲ‬ ‫اﻟﻤﺨﻄﻂ‬:
27-‫ﺗ‬ ‫ﻣﺎدة‬‫ﻌﻄﻲ‬‫أﺻﻔﺮ‬ ‫ﻟﻮن‬‫ﻛﺎﺷﻒ‬ ‫ﻣﻊ‬‫اﻟﻔﯿﻨﻮل‬ ‫أﺣﻤﺮ‬‫وﻟﻮن‬‫ﻛﺎﺷﻒ‬ ‫أﺻﻔﺮﻣﻊ‬‫اﻟﻤﯿﺜﯿﻞ‬ ‫ﺑﺮﺗﻘﺎﻟﻲ‬،‫ﻗﯿﻤﺔ‬ ‫ﻓﺘﻜﻮن‬pH‫ﻟﮭﺎ‬:
*3*6*9*11
28-‫ﻣﻦ‬ ‫ﻗﻄﺮة‬ ‫أﺿﯿﻔﺖ‬ ‫إذا‬‫اﻟﺒﺮوﻣﻮﺛﯿﻤﻮل‬ ‫أزرق‬ ‫ﻛﺎﺷﻒ‬‫إﻟﻰ‬25 mL‫ﻣﺤﻠﻮل‬ ‫ﻣﻦ‬0.1 M‫ﺣﻤﺾ‬HCI‫أﺿﯿﻒ‬ ‫ﺛﻢ‬
24.9 mL‫ﻣﻦ‬NaOH0.1 M‫ﻓ‬‫ﺈ‬‫اﻟ‬ ‫ﻟﻮن‬ ‫ن‬‫ﻜﺎﺷﻒ‬:
*‫ﻣﻦ‬ ‫ﯾﺘﻐﯿﺮ‬‫اﻷﺻﻔﺮ‬‫إﻟﻰ‬‫اﻷزرق‬*‫ﻣﻦ‬ ‫ﯾﺘﻐﯿﺮ‬‫اﻷزرق‬‫إﻟﻰ‬‫اﻷﺻﻔﺮ‬*‫ﺗﻐﯿﯿﺮ‬ ‫أي‬ ‫ﻋﻠﯿﮫ‬ ‫ﯾﻄﺮأ‬ ‫ﻻ‬*‫ﯾﺰول‬
29-‫ﻣﻨﮫ‬ ‫ﻗﻄﺮات‬ ‫إﺿﺎﻓﺔ‬ ‫ﻋﻨﺪ‬ ‫اﻟﺰھﺮي‬ ‫اﻷﺣﻤﺮ‬ ‫اﻟﻠﻮن‬ ‫ﻇﮭﻮر‬ ‫ﺗﻔﺴﺮ‬ ‫ﻓﻜﯿﻒ‬ ‫اﻟﺤﻤﻀﯿﺔ‬ ‫اﻟﻜﻮاﺷﻒ‬ ‫ﻣﻦ‬ ‫اﻟﻔﯿﻨﻮﻟﻔﺜﺎﻟﯿﻦ‬ ‫أن‬ ‫ﻋﻠﻤﺖ‬ ‫إذا‬
‫اﻟﺒﻮﺗﺎﺳﯿﻮم‬ ‫ھﯿﺪروﻛﺴﯿﺪ‬ ‫ﻣﺤﻠﻮل‬ ‫إﻟﻰ‬‫؟‬
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫اﻟﻤﺤﻠﻮل‬ ‫ﻓﻲ‬ ‫ﻟﻠﻜﺎﺷﻒ‬ ‫اﺗﺰان‬ ‫ﯾﺤﺪث‬:HIn +H2O D H+
3O + In-
‫ﻣﺤﻠﻮل‬ ‫وﻓﻲ‬KOH
‫أﯾﻮﻧﺎت‬ ‫ﺗﺘﺤﺪ‬OH-
‫أﯾﻮﻧﺎت‬ ‫اﻟﻘﺎﻋﺪة‬ ‫ﻣﻦ‬H3O+
‫ﻧﻘﺺ‬ ‫ﻟﺘﻌﻮﯾﺾ‬ ‫اﻟﻜﺎﺷﻒ‬ ‫ﺟﺰﯾﺌﺎت‬ ‫ﺗﺄﯾﻦ‬ ‫ﻓﯿﺰداد‬ ‫اﻟﻜﺎﺷﻒ‬ ‫ﻣﻦ‬H3O+
‫وﻟﺬﻟﻚ‬
‫ﻧﺴﺒﺔ‬ ‫ﻓﺘﺰداد‬ ‫اﻷﻣﺎﻣﻲ‬ ‫اﻻﺗﺠﺎه‬ ‫ﻓﻲ‬ ‫اﻻﺗﺰان‬ ‫ﯾﺰاح‬‫اﻟﻜﺎﺷﻒ‬ ‫أﯾﻮﻧﺎت‬In-
‫اﻷﺣﻤﺮ‬ ‫وھﻮ‬ ‫اﻟﻤﺤﻠﻮل‬ ‫ﻓﻲ‬ ‫ﻟﻮﻧﮭﺎ‬ ‫ﯾﻈﮭﺮ‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬
30-‫ﻣﻌﺎ‬ ‫ﺗﺠﺮﺑﺔ‬ ‫ﻓﻲ‬‫أﺿﺎﻓ‬ ‫ﯾﺮة‬‫ﺣﻠﯿﻤﺔ‬ ‫ﺖ‬)8(‫اﻟ‬ ‫ﻣﻦ‬ ‫ﻗﻄﺮات‬‫ﻜﺎﺷﻒ‬‫ﺣﺘﻲ‬‫ﺗ‬‫ﺑﻮﺿﻮح‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻧﮭﺎﯾﺔ‬ ‫ﻧﻘﻄﺔ‬ ‫ﺮى‬،‫ﻗﺎﻣ‬ ‫ﻣﺎ‬ ‫ھﻞ‬‫ﺖ‬‫؟‬ ‫إﺟﺎﺑﺘﻚ‬ ‫ﺑﺮر‬ ‫؟‬ ‫ﺻﺤﯿﺢ‬ ‫ﺑﮫ‬
‫ﺧﺎﻃﺊ‬:‫ﻣﻦ‬ ‫ﺟﺰء‬ ‫ﯾﺴﺘﮭﻠﻚ‬ ‫ﻣﻨﮫ‬ ‫اﻟﻤﺰﯾﺪ‬ ‫إﺿﺎﻓﺔ‬ ‫ﻟﺬا‬ ‫ﺿﻌﯿﻔﺔ‬ ‫ﻗﺎﻋﺪة‬ ‫أو‬ ‫ﺿﻌﯿﻒ‬ ‫ﺣﻤﺾ‬ ‫ﻋﻦ‬ ‫ﻋﺒﺎرة‬ ‫اﻟﻜﺎﺷﻒ‬ ‫ﻷن‬
‫اﻟﻤﻮاد‬‫اﻟﻜﺎﺷﻒ‬ ‫ﻣﻦ‬ ‫أﺛﻨﺘﯿﻦ‬ ‫أو‬ ‫ﻗﻄﺮة‬ ‫ﺗﻀﺎف‬ ‫ﻟﺬا‬ ‫اﻟﻤﻌﺎﯾﺮة‬ ‫ﺣﺴﺎب‬ ‫ﻓﻲ‬ ‫ﺧﻄﺄ‬ ‫ھﻨﺎك‬ ‫ﯾﻜﻮن‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ‫اﻟﻤﺘﻔﺎﻋﻠﺔ‬
‫ﺳ‬‫ﺎ‬‫د‬‫ﺳ‬ً‫ﺎ‬:‫و‬‫ﺿ‬‫ﺢ‬‫ا‬‫ﻟ‬‫ﻤ‬‫ﻘ‬‫ﺼ‬‫ﻮ‬‫د‬‫ﺑ‬‫ﻜ‬‫ﻞ‬‫ﻣ‬‫ﻦ‬:-
31-‫ﻧ‬‫ﻘ‬‫ﻄ‬‫ﺔ‬‫ا‬‫ﻟ‬‫ﺘ‬‫ﻜ‬‫ﺎ‬‫ﻓ‬‫ﺆ‬ً‫ﺎ‬‫ﻛﻤﯿﺎﺋﯿ‬ ‫ﻣﺘﻜﺎﻓﺌﺔ‬ ‫ﺑﻜﻤﯿﺎت‬ ‫اﻟﻤﻌﺎﯾﺮة‬ ‫ﻋﻤﻠﯿﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺴﺘﺨﺪﻣﺎن‬ ‫اﻟﻤﺤﻠﻮﻻن‬ ‫ﻓﯿﮭﺎ‬ ‫ﯾﻜﻮن‬ ‫اﻟﺘﻲ‬ ‫اﻟﻨﻘﻄﺔ‬ ‫ھﻲ‬.
32-‫ا‬‫ﻟ‬‫ﺘ‬‫ﻔ‬‫ﻜ‬‫ﻚ‬‫اﻷﯾﻮﻧﻲ‬ ‫اﻟﻤﺮﻛﺐ‬ ‫ذوﺑﺎن‬ ‫ﻟﺪى‬ ‫ﺑﻌﺾ‬ ‫ﻋﻦ‬ ‫ﺑﻌﻀﮭﺎ‬ ‫اﻷﯾﻮﻧﺎت‬ ‫اﻧﻔﺼﺎل‬ ‫ﻋﻤﻠﯿﺔ‬
33-‫ا‬‫ﻟ‬‫ﻤ‬‫ﻌ‬‫ﺎﯾ‬‫ﺮ‬‫ة‬‫ﻣﻦ‬ ‫ﻣﻌﯿﻨﺔ‬ ‫ﻛﻤﯿﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻹﺗﻤﺎم‬ ‫وﻻزﻣﺔ‬ ‫اﻟﺘﺮﻛﯿﺰ‬ ‫ﻣﻌﻠﻮم‬ ‫ﻣﺤﻠﻮل‬ ‫ﻣﻦ‬ ‫ﻗﯿﺎﺳﮭﺎ‬ ‫ﯾﺘﻢ‬ ‫ﻟﻜﻤﯿﺎت‬ ‫ﻓﯿﮭﺎ‬ ‫اﻟﻤﺘﺤﻜﻢ‬ ‫اﻹﺿﺎﻓﺔ‬ ‫ﻋﻤﻠﯿﺔ‬ ‫ھﻲ‬
‫اﻟﺘﺮﻛﯿﺰ‬ ‫ﻣﺠﮭﻮل‬ ‫ﻣﺤﻠﻮل‬.
‫اﻟﻜﺎﺷﻒ‬
‫اﻟﺒﺮوﻣﻮﺛﯿﻤﻮل‬ ‫أزرق‬
‫اﻟﻤﯿﺜﯿﻞ‬ ‫ﺑﺮﺗﻘﺎﻟﻲ‬
‫اﻟﻔﯿﻨﻮل‬ ‫أﺣﻤﺮ‬
‫اﻟﺸﻤﺲ‬ ‫ﺗﺒﺎع‬ ‫ﺻﺒﻐﺔ‬
‫اﻟﻔﯿﻨﻮﻟﻔﺜﺎﻟﯿﻦ‬
‫ﻋﺎم‬ ‫ﻛﺎﺷﻒ‬
‫أزرق‬
‫ﺗﻐﯿﺮ‬
‫أﺻﻔﺮ‬
‫أﺻﻔﺮ‬
‫ﺗﻐﯿﺮ‬
‫أﺣﻤﺮ‬
‫أﺣﻤﺮ‬
‫ﺗﻐﯿﺮ‬
‫أﺻﻔﺮ‬
‫أزرق‬
‫ﺗﻐﯿﺮ‬
‫أﺣﻤﺮ‬
‫زھﺮي‬
‫ﺗﻐﯿﺮ‬
‫اﻟﻠﻮن‬ ‫ﻋﺪﯾﻢ‬
‫ﻏﺎﻣﻖ‬ ‫ﺑﻨﻔﺴﺠﻲ‬
‫ﺑﻨﻔﺴﺠﻲ‬
‫ﻏﺎﻣﻖ‬ ‫أﺣﻤﺮ‬
‫أزرق‬‫أﺧﻀﺮ‬‫أﺻﻔﺮ‬‫ﺑﺮﺗﻘﺎﻟﻲ‬‫أﺣﻤﺮ‬
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
pH
C2H5OHHClBa(OH)2CH3COOHNaNO3HNO3AgClH2SO4
12345678

10. 10
‫اﻟ‬ ‫اﻟﺴﺆال‬‫ﺮاﺑﻊ‬:
ً‫ﻻ‬‫أو‬:‫ﯾﻠﻲ‬ ‫ﻋﻤﺎ‬ ‫اﺟﺐ‬:
34-‫ﻗﻮﺗﮭﺎ‬ ‫ﺣﺴﺐ‬ ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﯾ‬ ً‫ﺎ‬‫ﺗﺮﺗﯿﺒ‬ ‫ﻣﺮﺗﺒﺔ‬ ‫اﻟﻘﻮاﻋﺪ‬ ‫ﺑﻌﺾ‬ ‫ﯾﺤﻮي‬ ‫اﻵﺗﻲ‬ ‫اﻟﺠﺪول‬‫اﻟﻨﺴﺒﯿﺔ‬،‫ﺗﻠﯿﮫ‬ ‫اﻟﺘﻲ‬ ‫اﻷﺳﺌﻠﺔ‬ ‫ﻋﻦ‬ ‫وأﺟﺐ‬ ‫أدرﺳﮫ‬:
‫اﻟﺘﺮﺗﯿﺐ‬‫اﻷﻗﻮى‬ ‫اﻷﺿﻌﻒ‬
‫اﻟﻘﺎﻋﺪة‬H2OSO4
2-
CH3COO-
ClO-
NH2
-
*‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﺤﻤﻮض‬ ‫أي‬)+
O3COOH , H3, CH3NH(‫اﻷﻗﻮى‬ ‫ھﻮ‬+
O3H
*‫اﻵﺗﻲ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬:NH2
-
+ CH3COOH D CH3COO-
+ NH3
*‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬ ‫اﻻﺗﺰان‬ ‫ﯾﺮﺟﺢ‬ ‫ﺟﮭﺔ‬ ‫أي‬ ‫إﻟﻰ‬‫اﻷﻣﺎﻣﻲ‬ ‫اﻻﺗﺠﺎه‬
‫ذﻟﻚ‬ ‫ﺑﺮر‬‫اﻷﺿﻌﻒ‬ ‫اﻟﻘﺎﻋﺪة‬ ‫و‬ ‫اﻷﺿﻌﻒ‬ ‫اﻟﺤﻤﺾ‬ ‫إﻧﺘﺎج‬ ‫ﺗﻔﻀﻞ‬ ‫اﻟﺒﺮوﺗﻮن‬ ‫اﻧﺘﻘﺎل‬ ‫ﻋﻠﻰ‬ ‫اﻟﻘﺎﺋﻤﺔ‬ ‫اﻟﺘﻔﺎﻋﻼت‬ ‫ﻷن‬
*‫اﻟﺴﺎﺑﻖ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺮاﻓﻘﯿﻦ‬ ‫اﻟﺰوﺟﯿﻦ‬ ‫ﺣﺪد‬-3/ NH
-
2NH،-
COO3COOH /CH3CH
*‫ﻟﻠﻘﺎﻋﺪة‬ ‫اﻟﻤﺮاﻓﻖ‬ ‫اﻟﺤﻤﺾ‬ ‫ﺻﯿﻐﺔ‬ ‫ﻣﺎ‬-2
4SO‫؟‬
-
4SOH
35-‫ﻗﯿﻤﺔ‬ ‫ﻣﺎ‬[ H3O+
]‫ﻣﺤﻠﻮل‬ ‫ﻓﻲ‬Ca(OH)2‫اﻟﮭﯿﺪروﺟﯿﻨﻲ‬ ‫رﻗﻤﮫ‬PH = 8.5‫ﻗﯿﻤﺔ‬ ‫وﻣﺎ‬ ‫؟‬[Ca(OH)2]‫؟‬
[ ] = . ×
∴ [ ] = [ ]
=
. ×
= . ×
[ ( ) ] =
. ×
×
( )
= . ×
ً‫ﺎ‬‫ﺛﺎﻧﯿ‬:‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻷﺳﺌﻠﺔ‬ ‫ﻋﻦ‬ ‫أﺟﺐ‬:
36-‫اﻟﺮﺧﺎﻣﯿﺔ‬ ‫اﻟﻤﺒﺎﻧﻲ‬ ‫ﺗﻠﻒ‬ ‫ﻓﻲ‬ ‫وﺗﺄﺛﯿﺮھﺎ‬ ‫اﻟﺤﻤﻀﯿﺔ‬ ‫اﻷﻣﻄﺎر‬ ‫ﺗﻜﻮن‬ ‫ﺗﻮﺿﯿﺢ‬ ‫ﻓﻲ‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫وﻇﻒ‬:
)SO3(g) + H2O(l) → H2SO4(aq)(
)CaCO3(s) + 2H3O+
(aq) → Ca2+
(aq) + CO2(g) + 3H2O(l)(
•‫اﻟﺠﻮ‬ ‫ﻓﻲ‬ ‫وﯾﻨﺘﺸﺮ‬ ‫اﻟﺼﻨﺎﻋﯿﺔ‬ ‫اﻟﻌﻤﻠﯿﺎت‬ ‫ﻓﻲ‬ ‫اﻟﻜﺒﺮﯾﺖ‬ ‫أﻛﺴﯿﺪ‬ ‫ﺛﺎﻟﺚ‬ ‫ﻏﺎز‬ ‫ﯾﻨﺘﺞ‬،‫ﻟﯿﻨﺘﺞ‬ ‫اﻟﻐﯿﻮم‬ ‫ﻣﺎء‬ ‫ﻓﻲ‬ ‫ﯾﺬوب‬ ‫وھﻮ‬
‫ﺛﻠﻮج‬ ‫أو‬ ‫أﻣﻄﺎر‬ ‫ﺷﻜﻞ‬ ‫ﻋﻠﻰ‬ ‫ﯾﺘﺴﺎﻗﻂ‬ ‫اﻟﺬي‬ ‫اﻟﻜﺒﺮﯾﺘﯿﻚ‬ ‫ﺣﻤﺾ‬.
•‫اﻟﻜﺎﻟﺴﯿﻮم‬ ‫ﻛﺮﺑﻮﻧﺎت‬ ‫ﻣﻦ‬ ‫اﻟﺮﺧﺎﻣﯿﺔ‬ ‫اﻟﻤﺒﺎﻧﻲ‬ ‫ﺗﺘﻜﻮن‬CaCO3،‫اﻟﺤﻤﻀﻲ‬ ‫اﻟﻤﻄﺮ‬ ‫ﯾﺴﻘﻂ‬ ‫وﻋﻨﺪﻣﺎ‬2H3O+
‫ﻋﻠﻰ‬
‫اﻟﺮﺧﺎﻣﯿﺔ‬ ‫اﻟﻤﺒﺎﻧﻲ‬،‫اﻟﺮﺧﺎم‬ ‫وﯾﺬﯾﺐ‬ ‫اﻟﻜﺎﻟﺴﯿﻮم‬ ‫ﻛﺮﺑﻮﻧﺎت‬ ‫ﻣﻊ‬ ‫ﯾﺘﻔﺎﻋﻞ‬.
ً‫ﺎ‬‫راﺑﻌ‬:‫ﯾﻠﻲ‬ ‫ﻣﻤﺎ‬ ً‫ﻼ‬‫ﻛ‬ ً‫ﺎ‬‫ﺗﺼﺎﻋﺪﯾ‬ ‫رﺗﺐ‬:
37-‫اﻟﻤﺘﺴﺎوﯾﺔ‬ ‫اﻟﻤﺤﺎﻟﯿﻞ‬‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﺘﺮاﻛﯿﺰ‬)mol/L(‫ﻟﻘﯿﻢ‬ ً‫ﺎ‬‫ﺗﺒﻌ‬pH.
)KOH , CH3COOH , Ba(OH)2 , NH3(‫؟‬
‫اﻷﻗﻞ‬CH3COOH‫ﺛﻢ‬NH3‫ﺛﻢ‬KOH‫ﺛﻢ‬Ba(OH)2‫اﻷﻋﻠﻰ‬
38-‫ﻗﻮﺗﮭﺎ‬ ‫ﺣﺴﺐ‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻷﺣﻤﺎض‬:
HOIO , HONO2 , HOIO3 , HOI
‫اﻷ‬‫ﺿﻌﻒ‬HOI‫ﺛﻢ‬HOIO‫ﺛﻢ‬HONO2‫ﺛﻢ‬HOIO3‫اﻷ‬‫ﻗﻮى‬
‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻤﺴﺎﺋﻞ‬ ‫ﺣﻞ‬ ً‫ﺎ‬‫ﺧﺎﻣﺴ‬:
‫أن‬ ‫ﺗﺒﯿﻦ‬ ‫ﻣﻌﺎﯾﺮة‬ ‫ﺗﺠﺮﺑﺔ‬ ‫ﻓﻲ‬25 mL‫ﻣﺤﻠﻮل‬ ‫ﻣﻦ‬Ba(OH)2‫ﻣﻊ‬ ‫ﺗﺘﻌﺎدل‬10 mL،‫ﻣﺤﻠﻮل‬ ‫ﻣﻦ‬HCl0.0657 M‫إﻟﻰ‬ ‫ﻟﻠﻮﺻﻮل‬
‫اﻟﺘﻜﺎﻓﺆ‬ ‫ﻧﻘﻄﺔ‬.Ba(OH)2 = 171.35 g/mol
39-‫ﺑﺎﻟﻤﻮل‬ ‫اﻟﻘﺎﻋﺪة‬ ‫ﻣﺤﻠﻮل‬ ‫ﺗﺮﻛﯿﺰ‬ ‫اﺣﺴﺐ‬/‫ﻟﺘﺮ‬
2HCl + Ba(OH)2 → BaCl2 + 2H2O
‫اﻟﺤﻤﺾ‬ ‫ﻣﻮﻻت‬ =
.
× × = . ×
[ ( ) ] = . × ×
( )
× × = . ( )
40-‫اﺣﺴﺐ‬‫ﺑﺎﻟﺠﺮام‬ ‫اﻟﻘﺎﻋﺪة‬ ‫ﺗﺮﻛﯿﺰ‬/‫؟‬ ‫ﻟﺘﺮ‬
[ ( ) ] =
. ( )
×
.
( )
= . / ( )
‫اﻧﺘﮭﺖ‬‫اﻹﺟﺎﺑﺔ‬ ‫ﻧﻤﺎذج‬ ‫ﺟﻤﯿﻊ‬
20