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‫اﻟﺘﻄﻮّرات اﻟﺮﺗﻴﺒﺔ‬

‫اﻟﻜﺘﺎب اﻷول‬

‫ﺗﻄﻮر ﺟﻤﻠﺔ آﻴﻤﻴﺎﺋﻴﺔ ﻧﺤﻮ ﺣـﺎﻟﺔ اﻟﺘﻮازن‬

‫اﻟﻮﺣﺪة ‪04‬‬

‫ﺣﻠــﻮل ﺗﻤـــﺎرﻳﻦ اﻟﻜﺘﺎب اﻟﻤﺪرﺳﻲ‬

‫‪GUEZOURI Aek – Lycée Maraval - Oran‬‬

‫ﺣﺴﺐ اﻟﻄﺒﻌﺔ اﻟﺠﺪﻳﺪة ‪2011 - 2010‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪01‬‬
‫اﻟﺘﻔﺎﻋﻞ ﺣﻤﺾ – أﺳـﺎس هﻮ اﻟﺘﻔﺎﻋﻞ اﻟﺬي ﻳﺘﻢ ﻓﻴﻪ ﺗﺒﺎدل اﻟﺒﺮوﺗﻮﻧﺎت ‪ H+‬ﺑﻴﻦ اﻟﺤﻤﺾ واﻷﺳــﺎس ‪.‬‬
‫)‪Cu2+ (aq) + 2 OH–(aq) = Cu(OH)2 (s‬‬

‫‪:‬‬

‫ﺗﻔﺎﻋﻞ ﺗﺮﺳﻴﺐ‬

‫)‪ : CH3NH2(aq) + CH3COOH(aq) = CH3NH3+(aq) + CH3COO–(aq‬ﺗﻔﺎﻋﻞ ﺣﻤﺾ – أﺳﺎس ﻷﻧﻪ ﺣﺪث ﺗﺒﺎدل ﺑﺮوﺗﻮن ‪H+‬‬
‫ﺑﻴﻦ ﺣﻤﺾ اﻹﻳﺜﺎﻧﻮﻳﻚ واﻟﻤﻴﺜﺎن أﻣﻴﻦ ‪.‬‬
‫)‪ : CH3COOH(l) + CH3OH(l) = CH3COO-CH3(l) + H2O(l‬ﺗﻔﺎﻋﻞ أﺳﺘﺮة ‪) .‬ﻧﺘﻌﺮّف ﻋﻠﻴﺔ ﻓﻲ وﺣﺪة ﻻﺣﻘﺔ (‬
‫)‪HCl(g) + NH3(g) = NH4Cl(s‬‬

‫)ﻧﺤﺼﻞ ﻓﻲ هﺬا اﻟﺘﻔﺎﻋﻞ ﻋﻠﻰ آﻠﻮر اﻷﻣﻮﻧﻴﻮم ﺻﻠﺐ وﻟﻴﺲ ﻣﺤﻠﻮﻻ ﻷن ‪ HCl‬و ‪ NH3‬ﻏﺎزان( ‪:‬‬

‫ﺗﻔﺎﻋﻞ ﺣﻤﺾ – أﺳﺎس ﻷﻧﻪ ﺣﺪث ﺗﺒﺎدل ﺑﺮوﺗﻮن ‪ H+‬ﺑﻴﻦ ﻏﺎز آﻠﻮر اﻟﻬﻴﺪروﺟﻴﻦ وﻏﺎز اﻟﻨﺸﺎدر ‪.‬‬
‫)‪ : C6H5COOH(l) + H2O(l) = C6H5COO–(aq) + H3O+(aq‬ﺗﻔﺎﻋﻞ ﺣﻤﺾ أﺳﺎس ﻷﻧﻪ ﺣﺪث ﺗﺒﺎدل ﺑﺮوﺗﻮن ‪ H+‬ﺑﻴﻦ ﺣﻤﺾ‬
‫اﻟﺒﻨﺰﻳﻦ واﻟﻤــﺎء ‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪02‬‬
‫‪ – 1‬ﺑﺘﻄﺒﻴﻖ اﻟﻌﻼﻗﺔ ⎦⎤ ‪ pH = − Log ⎡⎣H 3O +‬أو اﻟﻌﻼﻗﺔ اﻟﻌﻜﺴﻴﺔ ﻟﻬﺎ ‪ ⎡⎣H 3O + ⎤⎦ = 10− pH‬ﻧﻤﻸ اﻟﺠﺪول ‪:‬‬
‫‪9,6‬‬

‫‪1,6‬‬

‫‪6,8‬‬

‫‪4,1‬‬

‫‪3,4‬‬

‫‪1,3‬‬

‫‪pH‬‬

‫‪[H3O+] (mol/L) 5,0 × 10–2 4,0 × 10–4 7,4 × 10–5 1,6 × 10–7 2,6 × 10–2 2,5 × 10–10‬‬
‫‪ – 2‬ﻋﻨﺪﻣﺎ ﻳﺘﻨﺎﻗﺺ ]‪ [H3O+‬ﻳﺰداد اﻟـ ‪ ، pH‬وذﻟﻚ ﺣﺴﺐ اﻟﺘﻨﺎﺳﺐ اﻟﻌﻜﺴﻲ ﺑﻴﻨﻬﻤﺎ ﻓﻲ اﻟﻌﻼﻗﺔ ‪. ⎡⎣H 3O + ⎤⎦ = 10− pH‬‬
‫ﻧﺘﺤﻘﻖ ﻣﻦ ذﻟﻚ ﻣﺜﻼ ﻓﻲ اﻟﺨﺎﻧﺘﻴﻦ اﻷوﻟﻰ واﻟﺜﺎﻧﻴﺔ ﻓﻲ اﻟﺠﺪول ‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪03‬‬
‫‪ – 1‬ﻣﻌﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ ‪HCl(g) + H2O(l) = H3O+(aq) + Cl–(aq) :‬‬
‫‪pH = − Log ⎡⎣H 3O + ⎤⎦ - 2‬‬

‫)‪(1‬‬

‫ﺑﻤﺎ أن ﺣﻤﺾ آﻠﻮر اﻟﻬﻴﺪروﺟﻴﻦ ﻳﺘﺸﺮد آﻠﻴﺎ ﻓﻲ اﻟﻤﺎء ‪ ،‬ﻓﺈن ]‪[H3O+] = [HCl‬‬
‫‪nH O+ 0 ,1‬‬
‫ﺗﺮآﻴﺰ ﺷﻮارد اﻟﻬﻴﺪروﻧﻴﻮم ﻓﻲ اﻟﻤﺤﻠﻮل ‪= 0 ,1 mol.L−1 :‬‬
‫= ‪⎡⎣H 3O + ⎤⎦ = 3‬‬
‫‪1‬‬
‫‪V‬‬

‫ﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ اﻟﻌﻼﻗﺔ )‪pH = − Log 0,1 = 1 : (1‬‬
‫ﻣﻼﺣﻈﺔ ‪ :‬اﻟﺘﺮآﻴﺰ ‪ ⎣⎡H 3O + ⎦⎤ = 0 ,1 mol.L−1‬ﻳﻔﻮق ﺣﺪود ﻣﺠﺎل ﺗﻄﺒﻴﻖ اﻟﻌﻼﻗﺔ ⎤⎦ ‪ . pH = − Log ⎣⎡H 3O +‬ﻟﻜﻦ ﺗﻤﺎﺷﻴﺎ ﻣﻊ ﻣﻌﻄﻴﺎت‬
‫اﻟﺘﻤﺎرﻳﻦ ﻧﻌﺘﺒﺮ اﻟﺨﻄﺄ ﻣﻬﻤﻼ وﻧﻮاﺻﻞ ﺣﻠﻮل اﻟﺘﻤﺎرﻳﻦ اﻷﺧﺮى ﻋﻠﻰ أﺳﺎس أن اﻟﻤﺤﺎﻟﻴﻞ ﻣﻤﺪة إذا آﺎن ‪ ، ⎡⎣H 3O + ⎤⎦ = 0 ,1 mol.L−1‬وهﺬا ﻣﺎ‬
‫ﺗﻔﻌﻠﻪ ﻳﻮم اﻻﻣﺘﺤﺎن ‪ ،‬اﺣﺴﺐ ‪ pH‬ﺑﺄﻳﺔ ﻗﻴﻤﺔ ﻟﺘﺮآﻴﺰ ‪ H3O+‬ﺗﻌﻄﻰ ﻟﻚ ‪.‬‬
‫‪ - 3‬ﺑﻤﺎ أن اﻟﺤﻤﺾ ﻗﻮي ﻓﺈن ‪ ، nHCl = nH O+‬وﻟﺪﻳﻨﺎ ‪nH O+ = ⎡⎣H 3O + ⎤⎦ × Vs = 10− pH × Vs = 10−2 × 1 = 10−2 mol‬‬
‫‪3‬‬
‫‪3‬‬
‫وﺑﺎﻟﺘﺎﻟﻲ آﻤﻴﺔ ﻣﺎدة ﻏﺎز ‪ HCl‬اﻟﻤﻨﺤﻠﺔ ﻓﻲ ‪ 1 L‬ﻣﻦ اﻟﻤﺎء هﻲ ‪10–2 mol‬‬
‫‪1‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪4‬‬
‫‪ – 1‬ﻣﻌﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ ‪:‬‬

‫)‪HNO3(l) + H2O(l) = H3O+(aq) + NO3–(aq‬‬

‫‪ – 2‬اﻋﺘﺒﺮﻧﺎ ﺣﻤﺾ اﻵزوت ﻗﻮﻳﺎ ‪ ،‬أي أن ‪C = [H3O+] = 0,1 mol/ L‬‬
‫‪. pH = − Log ⎡⎣H 3O + ⎤⎦ = − Log 0 ,1= 1‬‬
‫‪ – 3‬اﻟﺤﻤﺾ ﻗﻮي ‪ ،‬إذن ‪ nH O+‬ﻻ ﻳﺘﻐﻴﺮ ﻋﻨﺪﻣﺎ ﻧﻤﺪد اﻟﻤﺤﻠﻮل ﺑﺎﻟﻤﺎء ‪.‬‬
‫‪3‬‬

‫ﻟﻴﻜﻦ ‪ [H3O+]1‬هﻮ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد اﻟﻬﻴﺪروﻧﻴﻮم ﻗﺒﻞ اﻟﺘﻤﺪﻳﺪ و ‪ [H3O+]2‬هﻮ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد اﻟﻬﻴﺪروﻧﻴﻮم ﺑﻌﺪ اﻟﺘﻤﺪﻳﺪ ‪.‬‬
‫‪ ، ⎡⎣H 3O + ⎤⎦1 V1 = ⎡⎣H 3O + ⎤⎦ 2 V2‬ﺣﻴﺚ ‪V2 = 90 + 10 = 100 mL ، V1 = 10 mL‬‬
‫‪⎡H 3O + ⎤⎦1 0 ,1‬‬
‫⎣ = ‪ . ⎡⎣H 3O + ⎦⎤ 2‬وﺑﺘﻄﺒﻴﻖ اﻟﻌﻼﻗﺔ ⎦⎤ ‪ pH = − Log ⎡⎣H 3O +‬ﻧﺠﺪ ‪pH = 2‬‬
‫=‬
‫ﻧﺴﺘﻨﺘﺞ ‪= 10−2 mol / L :‬‬
‫‪10‬‬
‫‪10‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪5‬‬
‫‪ – 1‬ﻟﻜﻲ ﻧﺒﻴﻦ إن آﺎن اﻟﺘﻔﺎﻋﻞ ﺗﺎﻣﺎ أو ﻏﻴﺮ ﺗــﺎم ‪ ،‬ﻧﻘـﺎرن ﺑﻴﻦ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﻠﺤﻤﺾ ‪ C‬واﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد ‪H3O+‬‬
‫إذا آﺎن ‪ [H3O+] = C‬ﻓﺈن اﻟﺤﻤﺾ ﻗﻮي ‪.‬‬
‫إذا آﺎن ‪ [H3O+] < C‬ﻓﺈن اﻟﺤﻤﺾ ﺿﻌﻴﻒ ‪.‬‬

‫ ﻣﺤﻠﻮل ﺣﻤﺾ اﻹﻳﺜﺎﻧﻮﻳﻚ ‪[H3O+] = 10– pH = 10–3,9 = 1,26 × 10–4 mol/ L :‬‬‫هﺬﻩ اﻟﻘﻴﻤﺔ أﺻﻐﺮ ﻣﻦ ﺗﺮآﻴﺰ اﻟﺤﻤﺾ ‪ ،‬وﻣﻨﻪ اﻟﺘﻔﺎﻋﻞ ﻏﻴﺮ ﺗـﺎم ‪.‬‬

‫ ﻣﺤﻠﻮل ﺣﻤﺾ آﻠﻮر اﻟﻬﻴﺪروﺟﻴﻦ ‪[H3O+] = 10– pH = 10–3 mol/ L :‬‬‫هﺬﻩ اﻟﻘﻴﻤﺔ ﺗﺴﺎوي ﺗﺮآﻴﺰ اﻟﺤﻤﺾ ‪ ،‬وﻣﻨﻪ اﻟﺘﻔﺎﻋﻞ ﺗـﺎم ‪.‬‬
‫ آﻠﻮر اﻷﻣﻮﻧﻴﻮم هﻮ ﻣﻠﺢ ﺻﻴﻐﺘﻪ ‪ . NH4Cl‬ﻳﺘﺤﻠﻞ ﻓﻲ اﻟﻤﺎء إﻟﻰ ﺷﻮارد اﻷﻣﻮﻧﻴﻮم ‪ NH4+‬وﺷﻮارد اﻟﻜﻠﻮر –‪. Cl‬‬‫ﻳﺘﺤﻠﻞ آﻠﻮر اﻷﻣﻮﻧﻴﻮم ﺗﻤﺎﻣﺎ ﻓﻲ اﻟﻤﺎء ‪ ،‬أي‬

‫‪[ NH 4Cl ] = ⎡⎣ NH 4+ ⎤⎦ = ⎡⎣Cl − ⎤⎦ = C = 10−3 mol.L−1‬‬

‫اﻟﻘﻮة اﻟﺘﻲ ﻧﺘﻜﻠﻢ ﻋﻨﻬﺎ هﻨﺎ هﻲ ﻗﻮة ﺗﻔﺎﻋﻞ ﺷﺎردة اﻷﻣﻮﻧﻴﻮم ‪ NH4+‬ﻣﻊ اﻟﻤــﺎء ‪.‬‬
‫ﻟﻮ ﻟﻢ ﺗﺘﻔﺎﻋﻞ هﺎﺗﺎن اﻟﺸﺎردﺗﺎن ﻣﻊ اﻟﻤﺎء ﻟﻮﺟﺪﻧﺎ ‪ pH‬اﻟﻤﺤﻠﻮل ﻣﺴﺎوﻳﺎ ﻟﻠﻘﻴﻤﺔ ‪ . 7‬ﺳﺒﺐ ﻧﺰول اﻟـ ‪ pH‬إﻟﻰ اﻟﻘﻴﻤﺔ ‪ 6,2‬هﻮ ﺗﻔﺎﻋﻞ اﻟﺤﻤﺾ‬

‫اﻟﻀﻌﻴﻒ ‪ NH4+‬ﻣﻊ اﻟﻤـﺎء ‪NH4+(aq) + H2O(aq) = NH3(aq) + H3O+(aq) :‬‬
‫‪ . [H3O+] = 10– 6,2 = 6,3 × 10–7 mol/ L‬وﺑﻤﻘﺎرﻧﺔ ]‪ [H3O+‬ﻣﻊ ‪ NH4Cl‬ﻧﺤﻜﻢ ﻋﻠﻰ أن اﻟﺘﻔﺎﻋﻞ ﻏﻴﺮ ﺗـﺎم ‪.‬‬
‫ ﻣﺤﻠﻮل ﺣﻤﺾ اﻵزوت ‪ . [H3O+] = 10– pH = 10–3 mol/ L :‬هﺬﻩ اﻟﻘﻴﻤﺔ ﺗﺴﺎوي ﺗﺮآﻴﺰ اﻟﺤﻤﺾ ‪ ،‬وﻣﻨﻪ اﻟﺘﻔﺎﻋﻞ ﺗـﺎم ‪.‬‬‫‪ – 2‬اﻟﺘﻔﺎﻋﻞ ﺗـﺎم ﻣﻌﻨﺎﻩ اﻟﺤﻤﺾ ﻗﻮي ‪ ،‬وﺑﺎﻟﺘﺎﻟﻲ ‪ :‬ﺣﻤﺾ اﻹﻳﺜﺎﻧﻮﻳﻚ ﺿﻌﻴﻒ ‪ ،‬ﺷﺎردة اﻷﻣﻮﻧﻴﻮم ﺣﻤﺾ ﺿﻌﻴﻒ ‪ ،‬ﺣﻤﺾ اﻵزوت ﻗﻮي ‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪06‬‬
‫‪HCl(g) + H2O(l) = H3O+(aq) + Cl–(aq) - 1‬‬
‫اﻟﺜﻨﺎﺋﻴﺘﺎن هﻤﺎ ‪H3O+ / H2O ، HCl / Cl– :‬‬
‫‪-2‬‬

‫‪pH = – Log C = – Log 10–3 = 3‬‬

‫‪-3‬‬

‫أ( ‪( H3O+ , Cl–) + ( Na+ , OH–) = (Na+ , Cl–) + 2 H2O‬‬

‫أو اﺧﺘﺼـﺎرا ‪:‬‬

‫)‪= 2 H2O(l‬‬

‫)‪(aq‬‬

‫–‬

‫ب( ]‪pH = – Log [H3O+‬‬

‫‪+ OH‬‬

‫)‪H3O+(aq‬‬

‫)‪(1‬‬

‫)‪(2‬‬

‫ﻧﺤﺴﺐ ﻋﺪد ﻣﻮﻻت –‪ OH‬اﻟﺘﻲ أﺿﻔﻨﺎهﺎ ‪n(OH–) = Cb Vb = 10–3 × 50 × 10–3 = 0,5 × 10–4 mol :‬‬
‫ﻧﺤﺴﺐ ﻋﺪد ﻣﻮﻻت ‪ H3O+‬اﻟﻤﻮﺟﻮدة ﻓﻲ ﻣﺤﻠﻮل ﺣﻤﺾ آﻠﻮر اﻟﻬﻴﺪروﺟﻴﻦ ‪:‬‬

‫‪2‬‬

‫‪n(H3O+) = Ca Va = 10–3 ×0,1 = 10–4 mol‬‬
‫ﺣﺴﺐ اﻟﺘﻔﺎﻋﻞ )‪ ، (1‬ﻓﺈن ﻣﻮﻻ واﺣﺪا ﻣﻦ ‪ H3O+‬ﻳﺘﻔﺎﻋﻞ ﻣﻊ ﻣﻮل واﺣﺪ ﻣﻦ –‪ . OH‬إذن ﻋﺪد ﻣﻮﻻت ﺷﻮارد ‪ H3O+‬اﻟﺒﺎﻗﻴﺔ ﺑﻌﺪ اﻟﺘﻔﺎﻋﻞ‬

‫هﻲ ‪n’ (H3O+) = 10–4 – 0,5 × 10–4 = 0,5 × 10–4 mol :‬‬
‫‪0 ,5 × 10−4‬‬
‫‪= 3,3 × 10−4 mol / L‬‬
‫‪0 ,15‬‬

‫) ‪n' (H 3O +‬‬

‫‪+‬‬

‫=‬
‫= ⎦⎤ ‪⎡⎣H 3O‬‬
‫‪Va + Vb‬‬

‫ﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ اﻟﻌﻼﻗﺔ )‪pH = – Log 3,3 × 10–4 = 3,5 : (2‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪07‬‬
‫‪ – 1‬ﻣﻌـﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ ‪C6H5–COOH(aq) + H2O(l) = C6H5–COO–(aq) + H3O+(aq) :‬‬
‫‪ – 2‬ﺣﺘﻰ ﻧﺘﺄآﺪ أن اﻟﺘﻔﺎﻋﻞ ﻏﻴﺮ ﺗــﺎم ﻧﺤﺴﺐ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد اﻷآﺴﻮﻧﻴﻮم ]‪ [H3O+‬وﻧﻘﺎرﻧﻬﺎ ﻣﻊ اﻟﺘﺮآﻴﺰ ‪. C‬‬
‫ﻟﺪﻳﻨـﺎ‬

‫‪ ، ⎡⎣H 3O + ⎤⎦ = 10− pH = 10−2 ,95 = 1,12 × 10−3 mol / L‬وﻟﺪﻳﻨﺎ ‪. C = 2,0 × 10–2 mol /L‬‬

‫ﺑﻤﺎ أن اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد اﻟﻬﻴﺪروﻧﻴﻮم أﻗﻞ ﻣﻦ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ‪ ، C‬ﻓﺈن ﺗﻔﺎﻋﻞ ﺣﻤﺾ اﻟﺒﻨﺰﻳﻦ ﻣﻊ اﻟﻤﺎء ﻏﻴﺮ ﺗــﺎم ‪.‬‬
‫‪ – 3‬اﻟﻤﻘﺎرﻧﺔ ﺑﻴﻦ ‪ pH‬و ‪: – Log C‬‬

‫‪2,95 3,10 3,25 3,60 3,75 4,25 4,50 5,10‬‬

‫‪pH‬‬

‫‪–Log C 1,70 1,96 2,30 3,00 3,30 4,00 4,30 5,00‬‬
‫ﻣﻦ اﻷﻓﻀﻞ أن ﻳﻜﻮن اﻟﺴﺆال ‪ :‬ﻓﺴّﺮ ﻣﺎ ﺗﺴﺘﻨﺘﺠﻪ ﻣﻦ ﻣﻘﺎرﻧﺘﻚ ‪) .‬ﻟﻴﺲ ‪ :‬ﻋﻠّﻞ(‬
‫ﻧﻼﺣﻆ أن ﻓﻲ آﻞ ﻣﺤﻠﻮل ﻳﻜﻮن ‪ ، pH > – Log C‬وﻧﻌﻠﻢ أن ]‪ ، pH = – Log [H3O+‬وﺑﺎﻟﺘﺎﻟﻲ ‪:‬‬
‫‪ ، – Log [H3O+] > – Log C‬وﻣﻨﻪ ‪ ، Log [H3O+] < Log C‬وهﺬا ﻳﺆدي ﻟﻨﺘﻴﺠﺔ ﺿﻌﻒ اﻟﺤﻤﺾ ‪. [H3O+] < C‬‬

‫‪ – 4‬اﻟﺒﻴــﺎن ‪pH = – Log C‬‬

‫‪pH‬‬

‫‪0,5‬‬

‫‪− LogC‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪8‬‬
‫‪-1‬‬

‫‪0,2‬‬

‫)‪CH2ClCOOH(aq) + H2O(l) = CH2ClCOO–(aq) + H3O+(aq‬‬

‫‪ – 2‬ﺟﺪول اﻟﺘﻘﺪم ‪:‬‬

‫‪H3O+(aq‬‬

‫‪CH2ClCOO–(aq) +‬‬

‫= )‪CH2ClCOOH(aq) + H2O(l‬‬

‫‪0‬‬

‫‪0‬‬

‫زﻳـﺎدة‬

‫‪CV‬‬

‫‪x‬‬

‫‪x‬‬

‫زﻳﺎدة‬

‫‪CV − x‬‬

‫‪xf‬‬

‫‪xf‬‬

‫زﻳﺎدة‬

‫‪CV − x f‬‬

‫‪xm‬‬

‫‪xm‬‬

‫زﻳﺎدة‬

‫‪CV − xm‬‬

‫ﻟﺘﻌﻴﻴﻦ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ ﻧﻀﻊ ‪ CV – xmax = 0‬ﻷن اﻟﺤﻤﺾ هﻮ اﻟﻤﺘﻔﺎﻋﻞ اﻟﻤﺤﺪ ‪ ،‬وﻣﻨﻪ ‪:‬‬

‫‪xmax = CV = 10–2 × 20 × 10–3 = 2,0 × 10–4 mol‬‬
‫‪3‬‬

‫‪ - 3‬ﺗﺼﺤﻴﺢ ‪) pH = 2,4 :‬ﻟﻴﺲ ‪. ( pH = 2,37‬ﻻ ﻧﺒﻘﻲ ﻋﻠﻰ اﻟﺮﻗﻢ اﻟﺜﺎﻧﻲ ﺑﻌﺪ اﻟﻔﺎﺻﻠﺔ ﻓﻲ ﻗﻴﻤﺔ اﻟـ ‪ pH‬إﻻ إذا آﺎن ﻋﺒﺎرة ﻋﻦ ‪5‬‬
‫اﻟﺘﻘﺪّم اﻟﻨﻬـﺎﺋﻲ هﻮ آﻤﻴﺔ ﻣـﺎدة ‪ H3O+‬ﻓﻲ ﻧﻬﺎﻳﺔ اﻟﺘﻔﺎﻋﻞ ‪ ،‬أي ‪:‬‬

‫‪xf = n (H3O+) = [H3O+] × V = 10–pH × V = 10–2,4 × 20 × 10–3 = 7,9 × 10–5 mol‬‬
‫ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ ‪:‬‬

‫‪xf‬‬
‫‪7 ,9 × 10−5‬‬
‫=‬
‫‪= 0 , 40‬‬
‫‪xmax‬‬
‫‪2 × 10−4‬‬

‫=‪τ‬‬

‫‪ ،‬وﻣﻨﻪ اﻟﺘﺤﻮل اﻟﻜﻴﻤﻴﺎﺋﻲ ﻏﻴﺮ ﺗــﺎم ‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪09‬‬
‫‪ – 1‬ﻓﻲ ‪ 100 g‬ﻣﻦ اﻟﻤﺤﻠﻮل )‪ (S0‬ﻳﻮﺟﺪ ‪ 28 g‬ﻣﻦ اﻟﺤﻤﺾ اﻟﻨﻘﻲ ‪.‬‬

‫‪m‬‬
‫‪28‬‬
‫=‬
‫‪= 0 , 22mol‬‬
‫‪M 128‬‬

‫= ) ‪n (HI‬‬

‫)‪ 128 g/mol‬هﻲ اﻟﻜﺘﻠﺔ اﻟﻤﻮﻟﻴﺔ ﻟﻴﻮد اﻟﻬﻴﺪروﺟﻴﻦ ‪(HI‬‬

‫'‪m‬‬
‫‪ 100 g‬ﻣﻦ )‪ (S0‬ﺗﻜﺎﻓﺊ ﺣﺠﻤﺎ ‪ ، V‬ﺣﻴﺚ‬
‫‪V‬‬

‫‪ρ‬‬
‫وﻟﺪﻳﻨﺎ‬
‫‪ρe‬‬

‫=‪، d‬‬

‫ﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ )‪: (1‬‬

‫=‪ρ‬‬

‫)‪(1‬‬

‫‪ ، ρe = 1 g/ cm3‬وهﻲ اﻟﻜﺘﻠﺔ اﻟﺤﺠﻤﻴﺔ ﻟﻠﻤـﺎء ‪ ،‬وﻣﻨﻪ ‪ρ = d × ρe = 1,26 × 1 = 1,26 g/ cm3‬‬

‫‪100‬‬
‫‪= 79 , 4 cm3 = 7 ,94 × 10−2 L‬‬
‫‪1, 26‬‬

‫'‪m‬‬

‫=‬

‫‪ρ‬‬

‫=‪V‬‬

‫) ‪n (HI‬‬
‫‪0 , 22‬‬
‫=‬
‫اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟـ ‪ S0‬هﻮ ]‪ ، [HI‬ﺣﻴﺚ ‪= 2 ,77 mol / L :‬‬
‫‪V‬‬
‫‪7 ,94 × 10−2‬‬

‫= ] ‪[HI‬‬

‫‪ – 2‬ﻋﻨﺪ اﻟﺘﻤﺪﻳﺪ ﻻ ﻳﺘﻐﻴﺮ ﻋﺪد ﻣﻮﻻت ‪ ، HI‬أي ‪ ، n0 (HI) = n (HI) :‬ﺣﻴﺚ ‪:‬‬
‫‪ : n0 (HI) = C0V0‬ﻋﺪد اﻟﻤﻮﻻت ﻗﺒﻞ اﻟﺘﻤﺪﻳﺪ ‪ : n (HI) = CV ،‬ﻋﺪد اﻟﻤﻮﻻت ﺑﻌﺪ اﻟﺘﻤﺪﻳﺪ ‪.‬‬

‫‪CV 0,05 × 0,5‬‬
‫=‬
‫‪ ، C0V0 = CV‬وﻣﻨﻪ ‪≈ 9 × 10−3 L = 9mL :‬‬
‫‪C0‬‬
‫‪2,77‬‬

‫= ‪V0‬‬

‫اﻟﻄﺮﻳﻘﺔ هﻲ ‪ :‬ﻧﺄﺧﺬ ﺣﺠﻤﺎ ‪ V = 9 mL‬ﻣﻦ اﻟﻤﺤﻠﻮل ‪ S0‬وﻧﻀﻴﻒ ﻟﻪ اﻟﻤـﺎء اﻟﻤﻘﻄﺮ إﻟﻰ أن ﻳﺼﺒﺢ ﺣﺠﻢ اﻟﻤﺤﻠﻮل ‪ ، 500 mL‬أي ﻧﻀﻴﻒ‬
‫ج ﻓﻨﺤﺼﻞ ﻋﻠﻰ اﻟﻤﺤﻠﻮل ‪. S1‬‬
‫‪ 491 mL‬ﻣﻦ اﻟﻤـﺎء اﻟﻤﻘﻄﺮ وﻧﺮ ّ‬
‫أ( ﺗﺮآﻴﺰ اﻟﻤﺤﻠﻮل ‪: S2‬‬

‫‪-3‬‬

‫ﻟﺪﻳﻨﺎ )‪ ، n1 (HI) = n2 (HI‬أي ‪ ، C1V1 = C2V2 :‬وﻣﻨﻪ ‪:‬‬

‫‪0 ,05 × 5‬‬
‫‪= 1, 25 × 10−3 mol / L‬‬
‫‪200‬‬

‫= ‪C2‬‬

‫ب( ﺗﻌﺪﻳﻞ ‪ pH :‬اﻟﻤﺤﻠﻮل ‪ S2‬ﻳﺴﺎوي ‪ . 2,90‬اﺣﺴﺐ ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬـﺎﺋﻲ ﻟﻠﺘﻔﺎﻋﻞ ﺑﻴﻦ اﻟﺤﻤﺾ واﻟﻤـﺎء ‪ .‬هﻞ ﻳﻤﻜﻦ اﻋﺘﺒﺎر اﻟﺘﻔﺎﻋﻞ‬
‫ﺗـﺎﻣّﺎ ؟‬

‫⎦⎤ ‪⎡⎣ H 3O + ⎤⎦ × V2 ⎡⎣ H 3O +‬‬
‫=‬
‫=‬
‫اﻟﺠﻮاب ‪ :‬ﻟﺪﻳﻨﺎ ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ‬
‫‪C2V2‬‬
‫‪C2‬‬

‫‪xf‬‬
‫‪xmax‬‬

‫=‪τ‬‬

‫)‪(2‬‬

‫ﻧﻜﺘﺐ ﻣﻌﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ وﻧﻨﺸﺊ ﺟﺪول اﻟﺘﻘﺪّم ﻟﻜﻲ ﻧﺒﻴّﻦ أن )‪ ، xf = n (H3O+‬وﻟﺪﻳﻨﺎ ‪xmax = C2V2‬‬
‫–‪I‬‬

‫‪H3O +‬‬

‫=‬

‫‪H2O‬‬

‫‪0‬‬

‫‪0‬‬

‫زﻳﺎدة‬

‫‪xf‬‬

‫‪xf‬‬

‫زﻳﺎدة‬

‫‪+‬‬

‫‪HI‬‬

‫‪t=0‬‬

‫‪C2V2‬‬

‫‪ C2V2 – xf‬ﻧﻬﺎﻳﺔ اﻟﺘﻔﺎﻋﻞ‬

‫‪4‬‬

‫‪10− pH‬‬
‫‪10−2,9‬‬
‫=‬
‫ﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ اﻟﻌﻼﻗﺔ )‪= 1 ، (2‬‬
‫‪1, 25 × 10−3‬‬
‫‪C2‬‬

‫= ‪ ، τ‬وﺑﺎﻟﺘﺎﻟﻲ اﻟﺘﻔﺎﻋﻞ ﺗــﺎم ‪.‬‬

‫ﻟﻠﻤﺰﻳﺪ ‪ :‬ذرات اﻟﻬﺎﻟﻮﺟﻴﻨﺎت )اﻟﻌﻤﻮد اﻟﺴﺎﺑﻊ ﻓﻲ اﻟﺘﺼﻨﻴﻒ اﻟﺪوري اﻟﻤﺨﺘﺼﺮ( ﺗﻜﻮّن ﻣﻊ ذرات اﻟﻬﻴﺪروﺟﻴﻦ ﺣﻤﻮﺿﺎ ﺻﻴﻐﺘﻬﺎ ﻣﻦ اﻟﺸﻜﻞ ‪HA‬‬
‫)‪ . ( HI ، HBr ، HCl ، HF‬إن هﺬﻩ اﻟﺤﻤﻮض ﻟﻴﺴﺖ آﻠﻬﺎ ﻗﻮﻳﺔ ‪ ،‬ﺑﻞ ﺗﺘﻨﺎﻗﺺ ﻗﻮّﺗﻬﺎ ﻣﻦ ‪ HF‬إﻟﻰ ‪ ، HI‬أي أن آﻠﻤﺎ آﺎن ﺣﺠﻢ ذرة‬
‫اﻟﻬﺎﻟﻮﺟﻴﻦ آﺒﻴﺮا آﻠﻤﺎ آﺎن اﻟﺤﻤﺾ أﻗﻮى ‪ .‬أﻗﻮى هﺬﻩ اﻟﺤﻤﻮض هﻮ اﻟﺬي ﻧﺘﺤﺪّث ﻋﻨﻪ ﻓﻲ اﻟﺘﻤﺮﻳﻦ ‪ ، 9‬أي أن ﻣﻦ اﻟﻤﺴﺘﺤﻴﻞ ﺗﻔﺎﻋﻞ ﺷﺎردة‬

‫اﻟﻴﻮد –‪ I‬ﻣﻊ اﻟﻤـﺎء ‪ ،‬ﻓﻬﻲ أﺳﺎس ﺿﻌﻴﻒ ﺟﺪا‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪10‬‬
‫)‪CH3COOH(aq) + H2O(l) = CH3COO–(aq) + H3O+(aq‬‬

‫‪–1‬‬

‫‪xf‬‬
‫‪ – 2‬اﻟﻨﺴﺒﺔ اﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻘﺪم هﻲ‬
‫‪xmax‬‬

‫=‪τ‬‬

‫)‪(1‬‬

‫)ﻳﺠﺐ إﻋﻄﺎء ﻗﻴﻤﺘﻲ اﻟﻨﺎﻗﻠﻴﺘﻴﻦ اﻟﻤﻮﻟﻴﺘﻴﻦ اﻟﺸﺎردﻳﺘﻴﻦ ﻟﻠﺸﺎردﺗﻴﻦ ‪ H3O+‬و –‪ CH3COO‬ﻓﻲ ﻧﺺ اﻟﺘﻤﺮﻳﻦ(‬

‫‪λ1 = λH O = 35 × 10−3 S m 2 mol −1‬‬
‫‪+‬‬

‫‪3‬‬

‫‪λ1 = λCH COO = 4 ,1× 10−3 S m 2 mol −1‬‬

‫‪،‬‬

‫‪−‬‬

‫‪3‬‬

‫ﻟﺪﻳﻨﺎ ⎦⎤ ‪ ، σ 1 = λ1 ⎡⎣H 3O + ⎤⎦ + λ2 ⎡⎣CH 3COO − ⎤⎦ + λOH − ⎡⎣OH −‬وﺑﺈهﻤﺎل ]–‪ [OH‬ﻓﻲ اﻟﻤﺤﻠﻮل ﻳﻜﻮن ﻟﺪﻳﻨﺎ ‪:‬‬

‫) ‪σ 1 = ⎡⎣H 3O + ⎤⎦ (λ1 + λ2‬‬

‫]–‪ ، [H3O+] = [CH3COO‬وﺑﺎﻟﺘﺎﻟﻲ ﻧﻜﺘﺐ ‪:‬‬

‫)‪H3O+(aq‬‬

‫‪+‬‬

‫)‪CH3COO–(aq‬‬

‫‪0‬‬

‫‪0‬‬

‫‪xf‬‬

‫‪xf‬‬

‫=‬

‫)‪H2O(l‬‬

‫‪+‬‬

‫)‪(2‬‬

‫)‪CH3COOH(aq‬‬

‫زﻳﺎدة‬

‫‪CV‬‬

‫زﻳـﺎدة‬

‫‪CV – xf‬‬

‫ﻣﻦ ﺟﺪول اﻟﺘﻘﺪم ﻧﺴﺘﻨﺘﺞ أن )‪ xf = n (H3O+‬و ‪ . xmax = CV‬ﻧﺤﺴﺐ ﻣﻦ اﻟﻌﻼﻗﺔ )‪ (2‬اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ]‪[H3O+‬‬

‫‪σ1‬‬
‫‪4 ,9 × 10−3‬‬
‫=‬
‫‪= 0 ,125 mol / m3 = 1, 25 × 10−4 mol / L‬‬
‫= ⎦⎤ ‪⎡⎣H 3O‬‬
‫‪−3‬‬
‫‪λ1 + λ2 39 ,1× 10‬‬
‫‪+‬‬

‫ﻣﻦ اﻟﻌﻼﻗﺔ )‪(1‬‬
‫‪-3‬‬

‫‪n (H 3O + ) ⎡⎣H 3O + ⎤⎦ × V ⎡⎣H 3O + ⎤⎦ 1, 25 × 10−4‬‬
‫‪xf‬‬
‫=‬
‫=‬
‫=‬
‫=‬
‫‪= 0 ,125‬‬
‫‪xmax‬‬
‫‪10−3‬‬
‫‪CV‬‬
‫‪CV‬‬
‫‪C‬‬

‫=‪τ‬‬

‫أ( اﻟﻤﻄﻠﻮب هﻮ ]‪ [CH3COOH‬وﻟﻴﺲ ]–‪) . [CH3COO‬ﻻ ﻳﻤﻜﻦ ﻣﻌﺮﻓﺔ ]–‪ [CH3COO‬إﻻ ﺑﻤﻌﺮﻓﺔ ‪ pH‬أو ‪( σ‬‬

‫اﻟﺘﻤﺪﻳﺪ ﻳﺆدي إﻟﻰ ‪ ، C2V2 = C1V1 :‬ﺣﻴﺚ ‪ C1V1‬هﻮ ﻋﺪد ﻣﻮﻻت اﻟﺤﻤﺾ ﻗﺒﻞ اﻟﺘﻤﺪﻳﺪ ‪ C2V2 ،‬ﻋﺪد اﻟﻤﻮﻻت ﺑﻌﺪ اﻟﺘﻤﺪﻳﺪ ‪.‬‬
‫ﻣﻊ اﻟﻌﻠﻢ أن ‪C2 = [CH 3COOH ]0‬‬

‫‪C1V1 10−3 × 10‬‬
‫=‬
‫‪= 10−4 mol / L‬‬
‫‪V2‬‬
‫‪100‬‬
‫ب(‬

‫= ‪C2‬‬

‫) ‪ ، σ 2 = λ1 ⎡⎣H 3O + ⎤⎦ + λ2 ⎡⎣CH 3COO − ⎤⎦ = ⎡⎣CH 3COO − ⎤⎦ (λ1 + λ2‬ﻷن ]–‪[H3O+] = [CH3COO‬‬

‫ﺗﺼﺤﻴﺢ ‪ :‬اﻟﻘﻴﻤﺔ اﻟﺼﺤﻴﺤﺔ ﻟـ ‪ σ2‬هﻲ ‪1,55 mS. m-1‬‬

‫)ﻟﻴﺲ ‪ ، 1,2 mS.m-1‬ﻷن هﺬﻩ اﻟﻘﻴﻤﺔ ﻻ ﺗﻮاﻓﻖ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﺑﻌﺪ اﻟﺘﻤﺪﻳﺪ(‬

‫‪σ2‬‬
‫‪1,55 × 10−3‬‬
‫=‬
‫وﻣﻨﻪ ‪= 0 ,038 mol / m3 = 3,8 × 10−5 mol / L :‬‬
‫= ⎦⎤ ‪⎡⎣CH 3COO −‬‬
‫‪−3‬‬
‫‪λ1 + λ2 39 ,1× 10‬‬

‫‪5‬‬

‫‪⎡⎣H 3O + ⎤⎦ 3,8 × 10−5‬‬
‫=‬
‫ﺟـ( اﻟﻨﺴﺒﺔ اﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻘﺪم ‪= 0 ,38 :‬‬
‫= ‪τ2‬‬
‫‪C2‬‬
‫‪10−4‬‬
‫‪ - 4‬آﻠﻤﺎ ﻣﺪدﻧﺎ ﺣﻤﻀﺎ ﺿﻌﻴﻔﺎ ازدادت ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ ‪ ،‬أي ‪τ2 > τ1‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪11‬‬
‫‪-1‬‬
‫‪- 2‬‬

‫‪-3‬‬

‫)‬

‫)‪AgCl(s) = Ag+(aq) + Cl–(aq‬‬
‫)‪Cl–(aq‬‬

‫‪+‬‬

‫)‪Ag+(aq‬‬

‫=‬

‫)‪AgCl(s‬‬

‫‪0‬‬

‫‪0‬‬

‫‪n0‬‬

‫‪xf‬‬

‫‪xf‬‬

‫‪n0 – xf‬‬

‫⎦⎤ ‪ ، σ = λ Ag + ⎡⎣Ag + ⎤⎦ + λCl − ⎡⎣Cl −‬ﺗﺮآﻴﺰا ﺷﺎردﺗﻲ اﻟﻬﻴﺪروﻧﻴﻮم واﻟﻬﻴﺪروآﺴﻴﺪ ﻣﻬﻤﻼن ﻓﻲ هﺬا اﻟﻤﺤﻠﻮل اﻟﻤﻠﺤﻲ ‪.‬‬

‫(‬

‫‪ ، σ = ⎡⎣Ag + ⎤⎦ λ Ag + + λCl −‬ﻷن ]–‪ ، [Ag+] = [Cl‬وﻣﻨﻪ ‪:‬‬

‫‪0 ,19 × 10−3‬‬
‫‪= 0 ,013 mol / m3 = 1,3 × 10−5 mol / L‬‬
‫‪−3‬‬
‫‪(6, 2 + 7 ,6) × 10‬‬

‫=‬

‫)‬

‫‪σ‬‬
‫‪+ λCl −‬‬

‫‪Ag +‬‬

‫‪(λ‬‬

‫= ⎦⎤ ‪⎡⎣Ag +‬‬

‫‪⎡⎣Ag + ⎤⎦ = ⎡⎣Cl − ⎤⎦ = 1,3 × 10−5 mol / L‬‬
‫ﻧﺤﺴﺐ ﺛﺎﺑﺖ اﻟﺘﻮازن ﻟﻬﺬا اﻟﺘﺸﺮّد ‪ . K = ⎡⎣ Ag + ⎤⎦ × ⎡⎣Cl − ⎤⎦ = (1,3 × 10−5 ) = 1, 7 × 10−10 :‬اﻟﻘﻴﻤﺔ اﻟﺼﻐﻴﺮة ﻟـ ‪ K‬ﺗﺒﻴﻦ اﻟﺘﺸﺮّد‬
‫‪2‬‬

‫اﻟﺠﺰﺋﻲ ﻟﻜﻠﻮر اﻟﻔﻀﺔ ‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪12‬‬
‫‪–1‬‬

‫)‪NH3(aq) + H2O(l) = NH4+(aq) + OH–(aq‬‬
‫–‬

‫‪ – 2‬ﻟﻜﻲ ﻧﺒﻴّﻦ أن ﻏـﺎز اﻟﻨﺸﺎدر ﻻ ﻳﺘﻔﺎﻋﻞ آﻠﻴﺎ ﻣﻊ اﻟﻤـﺎء ﻧﻘﺎرن ﺗﺮآﻴﺰ ﺷﻮارد اﻟﻬﻴﺪروآﺴﻴﺪ ‪ OH‬ﻣﻊ ﺗﺮآﻴﺰ اﻷﺳﺎس ‪. C1‬‬

‫‪10−14‬‬
‫‪10−14‬‬
‫= ⎦⎤ ‪ ، ⎡⎣OH‬وﻣﻨﻪ ‪ ، [OH ] < C1‬وﺑﺎﻟﺘﺎﻟﻲ اﻟﺘﻔﺎﻋﻞ ﻏﻴﺮ ﺗـﺎم ‪.‬‬
‫ﻟﺪﻳﻨـﺎ ‪= −11,1 = 1,26 × 10−3 mol / L‬‬
‫‪+‬‬
‫‪⎡⎣H 3O ⎤⎦ 10‬‬
‫‪−‬‬

‫–‬

‫ﻳﻤﻜﻦ أن ﻧﺒﻴّﻦ أن اﻟﺘﻔﺎﻋﻞ ﻏﻴﺮ ﺗﺎم ﺑﺤﺴﺎب ﻗﻴﻤﺔ ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻴﺔ ‪τ1‬‬
‫ﻣﻦ أﺟﻞ هﺬا ﻧﻨﺸﺊ ﺟﺪول اﻟﺘﻘﺪم ‪:‬‬

‫)‪NH3(aq) + H2O(l) = NH4+(aq) + OH–(aq‬‬
‫‪0‬‬

‫‪0‬‬

‫زﻳﺎدة‬

‫‪xf‬‬

‫‪xf‬‬

‫زﻳﺎدة‬

‫‪C1V1‬‬
‫‪C1V1 – xf‬‬

‫‪⎡⎣OH − ⎤⎦ × V1 ⎡⎣OH − ⎤⎦ 1, 26 × 10−3‬‬
‫=‬
‫=‬
‫=‬
‫‪= 1, 26 × 10−2‬‬
‫‪0 ,1‬‬
‫‪C1V1‬‬
‫‪C1‬‬

‫) ‪n (OH −‬‬
‫‪xmax‬‬

‫=‬

‫‪xf‬‬
‫‪xmax‬‬

‫= ‪τ1‬‬

‫‪ ، τ 1 < 1‬إذن اﻟﺘﻔــﺎﻋﻞ ﻏﻴﺮ ﺗـﺎم‬
‫‪ – 3‬ﻋﺪد ﻣﻮﻻت ‪ NH3‬ﻻ ﻳﺘﻐﻴﺮ ﺑﻌﺪ اﻟﺘﻤﺪﻳﺪ ‪ ،‬أي أن ‪ ، C1V1 = C2V2‬ﺣﻴﺚ أن ‪ V1‬هﻮ اﻟﺤﺠﻢ اﻟﺬي ﻧﺄﺧﺬﻩ ﻣﻦ اﻟﻤﺤﻠﻮل اﻷول ‪.‬‬

‫‪C2V2 2 ,5 × 10−2 × 100‬‬
‫= ‪V1‬‬
‫=‬
‫‪= 25 mL‬‬
‫‪0 ,1‬‬
‫‪C1‬‬
‫‪6‬‬

‫اﻟﻄﺮﻳﻘﺔ هﻲ ‪ :‬ﻧﺄﺧﺬ ﺣﺠﻤﺎ ‪ V1 = 25 mL‬وﻧﻀﻌﻪ ﻓﻲ ﻣﺨﺒـﺎر ﺣﺠﻤﻪ ‪ ، 100 mL‬ﺛـﻢ ﻧﻜﻤﻞ اﻟﺤﺠﻢ ﺑﺎﻟﻤﺎء اﻟﻤﻘﻄﺮ )أي ﻧﻀﻴﻒ ‪75 mL‬‬
‫ﻣﻦ اﻟﻤﺎء ( ‪ ،‬ﻓﻨﺤﺼﻞ ﺑﺬﻟﻚ ﻋﻠﻰ ﻣﺤﻠﻮل ‪ S2‬ﺣﺠﻤﻪ ‪ 100 mL‬وﺗﺮآﻴﺰﻩ اﻟﻤﻮﻟﻲ ‪. C2 = 2,5 × 10-2 mol/ L‬‬
‫‪–4‬‬

‫)‪OH–(aq‬‬

‫‪+‬‬

‫)‪= NH4+(aq‬‬

‫)‪H2O(l‬‬

‫‪+‬‬

‫‪0‬‬

‫‪0‬‬

‫زﻳﺎدة‬

‫‪C2V2‬‬

‫‪xf‬‬

‫‪xf‬‬

‫زﻳﺎدة‬

‫‪C2V2 – xf‬‬

‫⎦⎤ ‪⎡OH − ⎤⎦ × V ⎡⎣OH −‬‬
‫اﻟﻨﺴﺒﺔ اﻟﻨﻬﺎﺋﻴﺔ ﻟﺘﻘﺪّم اﻟﺘﻔﺎﻋﻞ هﻲ‬
‫⎣=‬
‫=‬
‫‪C2V‬‬
‫‪C2‬‬
‫ﻟﺪﻳﻨﺎ‬

‫)‪NH3(aq‬‬

‫) ‪n (OH −‬‬
‫‪C2V‬‬

‫=‬

‫‪xf‬‬
‫‪xmax‬‬

‫= ‪τ2‬‬

‫‪6 ,31× 10−4‬‬
‫‪10−14‬‬
‫‪10−14‬‬
‫‪−2‬‬
‫‪−‬‬
‫= ‪τ2‬‬
‫‪ ، ⎡⎣OH ⎤⎦ = − pH = −10 ,8 = 6 ,31× 10−4 mol / L‬وﺑﺎﻟﺘﺎﻟﻲ ‪= 2 ,52 × 10‬‬
‫‪−2‬‬
‫‪2 ,5 × 10‬‬
‫‪10‬‬
‫‪10‬‬

‫وﺟﺪﻧﺎ ‪ ، τ 1 < τ 2‬وﻣﻨﻪ ﻧﺴﺘﺨﻠﺺ أﻧﻪ آﻠﻤﺎ آﺎن اﻷﺳﺎس اﻟﻀﻌﻴﻒ ﻣﻤﺪا ﻳﺘﺸﺮد أآﺜﺮ ‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪13‬‬
‫‪ – 1‬ﻣﻌﺎدﻟﺔ ﺗﻔﺎﻋﻞ ﺣﻤﺾ اﻹﻳﺜﺎﻧﻮﻳﻚ ﻣﻊ اﻟﻤﺎء ‪CH3COOH(aq) + H2O(l) = CH3COO–(aq) + H3O+(aq) :‬‬
‫‪ – 2‬أ( ﺟﺪول اﻟﺘﻘﺪم ‪:‬‬

‫‪CH3COO–(aq) +‬‬

‫)‪H3O+(aq‬‬

‫)‪H2O(l‬‬

‫=‬

‫‪CH3COOH(aq) +‬‬

‫‪0‬‬

‫‪0‬‬

‫زﻳـﺎدة‬

‫‪CV‬‬

‫‪x‬‬

‫‪x‬‬

‫زﻳﺎدة‬

‫‪CV − x‬‬

‫‪xf‬‬

‫‪xf‬‬

‫زﻳﺎدة‬

‫‪CV − x f‬‬

‫‪xm‬‬

‫‪xm‬‬

‫زﻳﺎدة‬

‫‪CV − xm‬‬

‫ﻟﺪﻳﻨﺎ ⎦⎤ ‪ ، σ = λH O + ⎡⎣H 3O + ⎤⎦ + λCH COO − ⎡⎣CH 3COO − ⎤⎦ + λOH − ⎡⎣OH −‬وﺑﺈهﻤﺎل ]–‪ [OH‬ﻓﻲ اﻟﻤﺤﻠﻮل ﻳﻜﻮن ﻟﺪﻳﻨﺎ ‪:‬‬
‫‪3‬‬

‫‪3‬‬

‫)‬

‫]–‪ ، [H3O+] = [CH3COO‬وﺑﺎﻟﺘﺎﻟﻲ ﻧﻜﺘﺐ ‪:‬‬

‫‪−‬‬

‫‪3‬‬

‫‪+‬‬

‫‪3‬‬

‫اﻟﻤﺤﻠﻮل ‪: S1‬‬

‫‪σ1‬‬
‫‪4, 7 ×10−2‬‬
‫=‬
‫‪= 1, 2 mol / m3‬‬
‫‪−3‬‬
‫‪39,1×10‬‬
‫‪+ λCH COO‬‬
‫‪−‬‬

‫ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ ‪:‬‬

‫‪3‬‬

‫‪H 3O +‬‬

‫= ⎦⎤ ‪⎡⎣ H 3O +‬‬
‫‪λ‬‬

‫‪⎡⎣ H 3O + ⎤⎦ 1, 2 ×10−3‬‬
‫= ‪τ1‬‬
‫=‬
‫‪= 1, 2 × 10−2‬‬
‫‪C1‬‬
‫‪0,1‬‬

‫اﻟﻤﺤﻠﻮل ‪: S2‬‬

‫‪σ2‬‬
‫‪1,55 ×10−2‬‬
‫=‬
‫‪= 0,39 mol / m3‬‬
‫‪39,1×10−3‬‬
‫‪+ λCH COO‬‬
‫‪−‬‬

‫‪3‬‬

‫(‬

‫‪σ = ⎡⎣H 3O + ⎤⎦ λH O + λCH COO‬‬

‫‪H 3O +‬‬

‫= ⎦⎤ ‪⎡⎣ H 3O +‬‬
‫‪λ‬‬

‫‪⎡⎣ H 3O + ⎤⎦ 0,39 ×10−3‬‬
‫= ‪τ2‬‬
‫=‬
‫ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ ‪= 3,9 × 10−2 :‬‬
‫‪C1‬‬
‫‪0, 01‬‬

‫‪7‬‬

‫اﻟﻤﺤﻠﻮل ‪: S3‬‬

‫‪σ3‬‬
‫‪4, 6 ×10−3‬‬
‫=‬
‫‪= 0,12 mol / m3‬‬
‫‪−3‬‬
‫‪39,1×10‬‬
‫‪+ λCH COO‬‬
‫‪−‬‬

‫‪3‬‬

‫‪H 3O +‬‬

‫= ⎦⎤ ‪⎡⎣ H 3O +‬‬
‫‪λ‬‬

‫‪⎡ H 3O + ⎤⎦ 0,12 ×10−3‬‬
‫⎣ = ‪τ3‬‬
‫=‬
‫ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ ‪= 0,12 :‬‬
‫‪C1‬‬
‫‪0, 001‬‬
‫ب( آﻠﻤﺎ اﻧﺨﻔﺾ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﻠﺤﻤﺾ ﺗﺰداد ﻧﺴﺒﺔ اﻟﺘﻘﺪّم اﻟﻨﻬﺎﺋﻲ ‪ ،‬ﻷن اﻟﺘﻤﺪﻳﺪ ﻳﺴﺎﻋﺪ ﻋﻠﻰ اﻟﺘﺸﺮد )اﻟﺤﻤﺾ اﻟﻀﻌﻴﻒ ﻳﺘﻔﺎﻋﻞ ﻣﻊ اﻟﻤﺎء( ‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪14‬‬
‫‪ - 1‬ﻣﻌﺎدﻟﺘﺎ اﻟﺘﻔﺎﻋﻠﻴﻦ ‪CH2ClCOOH(aq) + H2O(l) = CH2ClCOO–(aq) + H3O+(aq) :‬‬

‫)‪(1‬‬

‫)‪CHCl2COOH(aq) + H2O(l) = CHCl2COO–(aq) + H3O+(aq‬‬

‫)‪(2‬‬

‫‪) σ2 = 0,33 S m–1‬ﻟﻴﺲ ‪( σ2 = 0,33 mS m–1‬‬

‫ﺗﺼﺤﻴﺢ ‪) σ1 = 0,121 S m–1 :‬ﻟﻴﺲ ‪، ( σ1 = 0,167 mS m–1‬‬
‫‪ – 2‬ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺤﻤﺾ ‪: CH2ClCOOH‬‬

‫⎦⎤ ‪ ، σ 1 = λH O + ⎡⎣H 3O + ⎤⎦ + λCH ClCOO− ⎡⎣CH 2ClCOO − ⎤⎦ + λOH − ⎡⎣OH −‬وﺑﺈهﻤﺎل ]–‪ [OH‬ﻳﻜﻮن ﻟﺪﻳﻨﺎ ‪:‬‬
‫‪2‬‬

‫] ‪ ، [CH ClCOO ] = [H O‬وﺑﺎﻟﺘـﺎﻟﻲ ‪) :‬‬
‫‪+‬‬

‫‪3‬‬

‫–‬

‫‪2‬‬

‫‪−‬‬

‫‪3‬‬

‫(‬

‫‪σ 1 = ⎡⎣H 3O + ⎤⎦ λH O + λCH ClCOO‬‬
‫‪2‬‬

‫‪+‬‬

‫‪3‬‬

‫‪σ1‬‬
‫‪0 ,121‬‬
‫=‬
‫‪= 3,1 mol / m3 = 3,1× 10−3 mol / L‬‬
‫= ⎦⎤ ‪⎡⎣H 3O +‬‬
‫‪−3‬‬
‫‪λH O+ + λCH ClCOO− 39,22 × 10‬‬
‫‪2‬‬

‫‪3‬‬

‫ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺤﻤﺾ ‪: CHCl2COOH‬‬
‫⎦⎤ ‪ ، σ 2 = λH O + ⎡⎣H 3O + ⎤⎦ + λCHCl COO − ⎡⎣CHCl2COO − ⎤⎦ + λOH − ⎡⎣OH −‬وﺑﺈهﻤﺎل ]–‪ [OH‬ﻳﻜﻮن ﻟﺪﻳﻨﺎ ‪:‬‬
‫‪2‬‬

‫] ‪ ، [CHCl COO ] = [H O‬وﺑﺎﻟﺘـﺎﻟﻲ ‪) :‬‬
‫‪+‬‬

‫‪3‬‬

‫–‬

‫‪2‬‬

‫‪−‬‬

‫‪3‬‬

‫(‬

‫‪σ 2 = ⎡⎣H 3O + ⎤⎦ λH O + λCHCl COO‬‬
‫‪2‬‬

‫‪+‬‬

‫‪3‬‬

‫‪σ2‬‬
‫‪0 ,33‬‬
‫=‬
‫‪= 8,5 mol / m3 = 8,5 × 10−3 mol / L‬‬
‫= ⎦⎤ ‪⎡⎣H 3O +‬‬
‫‪−3‬‬
‫‪λH O+ + λCHCl COO− 38,83 × 10‬‬
‫‪2‬‬

‫‪ – 3‬اﻟﻨﺴﺒﺔ اﻟﻨﻬــﺎﺋﻴﺔ ﻟﺘﻘﺪّم ﺗﻔـﺎﻋﻞ ‪: CH2ClCOOH‬‬

‫اﻟﻨﺴﺒﺔ اﻟﻨﻬــﺎﺋﻴﺔ ﻟﺘﻘﺪّم ﺗﻔـﺎﻋﻞ ‪: CHCl2COOH‬‬

‫‪3‬‬

‫⎦⎤ ‪⎡⎣H 3O +‬‬
‫‪3,1 × 10−3‬‬
‫= ‪τ1‬‬
‫=‬
‫‪= 0 ,31‬‬
‫‪10−2‬‬
‫] ‪[CH 2ClCOOH‬‬

‫⎦⎤ ‪⎡⎣H 3O +‬‬
‫‪8,5 × 10−3‬‬
‫=‬
‫‪= 0,85‬‬
‫= ‪τ2‬‬
‫‪10−2‬‬
‫] ‪[CHCl2COOH‬‬

‫‪ – 4‬اﻟﺘﻔــﺎﻋﻞ )‪: (1‬‬

‫‪⎡⎣H 3O + ⎤⎦ f × ⎡⎣CH 2ClCOO − ⎤⎦ f‬‬
‫= ‪K1‬‬
‫‪[CH 2ClCOOH ] f‬‬

‫اﻟﺘﻔــﺎﻋﻞ )‪: (2‬‬

‫‪⎡⎣H 3O + ⎤⎦ f × ⎡⎣CHCl2COO − ⎤⎦ f‬‬
‫= ‪K2‬‬
‫‪[CHCl2COOH ] f‬‬

‫ﻋﻨﺪ اﻟﺘﻮازن ﻳﻜﻮن ﺗﺮآﻴﺰ اﻟﺤﻤﺾ اﻟﺒﺎﻗﻲ ‪ ،‬أي ]‪ [CH2ClCOOH‬أو ]‪ ، [CHCl2COOH‬ﻣﺴﺎوﻳﺎ ﻟﻠﺘﺮآﻴﺰ اﻻﺑﺘﺪاﺋﻲ ﻣﻄﺮوح ﻣﻨﻪ‬
‫ﺗﺮآﻴﺰ ﺷﻮارد اﻟﻬﻴﺪروﻧﻴﻮم ]‪ ، [H3O+‬وﺑﺎﻟﺘﺎﻟﻲ ‪:‬‬

‫‪8‬‬

‫‪2‬‬

‫‪2‬‬
‫‪⎡⎣H 3O + ⎤⎦ f‬‬
‫) ‪3,1 × 10−3‬‬
‫(‬
‫‪= −2‬‬
‫‪= 1, 4 × 10−3‬‬
‫= ‪K1‬‬
‫‪+‬‬
‫‪−3‬‬
‫‪C − ⎡⎣H 3O ⎤⎦ f 10 − 3,1× 10‬‬

‫‪2‬‬

‫‪2‬‬
‫‪⎡⎣H 3O + ⎤⎦ f‬‬
‫) ‪8,5 × 10−3‬‬
‫(‬
‫‪= −2‬‬
‫‪= 4 ,8 × 10−2‬‬
‫= ‪K2‬‬
‫‪+‬‬
‫‪−3‬‬
‫‪C − ⎡⎣H 3O ⎤⎦ f 10 − 8,5 × 10‬‬

‫‪ – 5‬ﻣﻦ أﺟﻞ ﺗﻔﺎﻋﻞ ﻣﻌﻴّﻦ ‪ ،‬ﺛﺎﺑﺖ اﻟﺘﻮازن ﻻ ﻳﺘﻐﻴﺮ ﻣﻬﻤﺎ آﺎن اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ‬
‫إذا أردﻧﺎ اﻟﻤﻘﺎرﻧﺔ ﺑﻴﻦ ﻣﺤﻠﻮﻟﻴﻦ ﻟﺤﻤﻀﻴﻦ ﻣﺨﺘﻠﻔﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ اﻟﺘﺮآﻴﺰ اﻟﻮﻟﻲ ‪ ، C‬ﻓﺈن اﻟﺬي ﻳﻮاﻓﻖ ﺗﻔﺎﻋﻠﻪ ﻣﻊ اﻟﻤﺎء أآﺒﺮ ﻗﻴﻤﺔ ﻟﺜﺎﺑﺖ اﻟﺘﻮازن‬
‫هﻮ اﻟﺬي ﺗﻜﻮن ﻟﻪ أآﺒﺮ ﻗﻴﻤﺔ ﻟﻨﺴﺒﺔ اﻟﺘﻘﺪّم اﻟﻨﻬﺎﺋﻲ ‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪15‬‬
‫‪ – 1‬اﻟﻤﻌﺎدﻟﺔ ‪:‬‬

‫)‪H3O+(aq‬‬

‫‪+‬‬

‫)‪(aq‬‬

‫–‬

‫‪CH3COOH(l) + H2O(l) = CH3COO‬‬

‫ﻣﻦ اﻷﺣﺴﻦ أن ﻳﻜﻮن اﻟﺴﺆال ‪ :‬اآﺘﺐ ﻋﺒﺎرة آﺴﺮ اﻟﺘﻔﺎﻋﻞ ‪ ،‬ﻷن آﺴﺮ اﻟﺘﻔﺎﻋﻞ ﻋﻨﺪ ﺣﺎﻟﺔ اﻟﺘﻮازن هﻮ ﻧﻔﺴﻪ ﺛﺎﺑﺖ اﻟﺘﻮازن ‪ ،‬وهﺬا اﻷﺧﻴﺮ‬
‫ﻣﻄﻠﻮب ﻓﻲ اﻟﺴﺆال ا رﻗﻢ ‪ . 5‬أو ﻣﺜﻼ ‪ :‬أو ﻣﺜﻼ ‪ :‬ﻋﻴّﻦ آﺴﺮ اﻟﺘﻔﺎﻋﻞ اﻻﺑﺘﺪاﺋﻲ‬

‫‪⎡⎣H 3O + ⎤⎦ × ⎡⎣CH 3COO − ⎤⎦ 0 × 0‬‬
‫=‬
‫آﺴﺮ اﻟﺘﻔـﺎﻋﻞ اﻻﺑﺘﺪاﺋﻲ ‪= 0 :‬‬
‫= ‪Qri‬‬
‫‪C‬‬
‫] ‪[CH 3COOH‬‬
‫‪ ، σ = λH O + ⎡⎣H 3O + ⎤⎦ f + λCH COO − ⎡⎣CH 3COO − ⎤⎦ f + λOH − ⎡⎣OH − ⎤⎦ f - 2‬وﺑﺈهﻤﺎل ]–‪ [OH‬ﻧﻜﺘﺐ ‪:‬‬
‫‪3‬‬

‫‪3‬‬

‫‪σ = λH O ⎡⎣H 3O + ⎤⎦ f + λCH COO ⎡⎣CH 3COO − ⎤⎦ f‬‬
‫‪−‬‬

‫‪+‬‬

‫‪3‬‬

‫‪3‬‬

‫‪ – 3‬ﺟﺪول اﻟﺘﻘﺪم‬

‫)‪H3O+(aq‬‬

‫‪+‬‬

‫))‪(aq‬‬

‫–‬

‫‪CH3COO‬‬

‫= )‪CH3COOH(aq) + H2O(l‬‬

‫‪0‬‬

‫‪0‬‬

‫زﻳـﺎدة‬

‫‪CV‬‬

‫‪t=0‬‬

‫‪x‬‬

‫‪x‬‬

‫زﻳﺎدة‬

‫‪CV – x‬‬

‫اﻟﺤﺎﻟﺔ اﻻﻧﺘﻘﺎﻟﻴﺔ‬

‫‪xéq‬‬

‫‪xéq‬‬

‫زﻳﺎدة‬

‫‪CV – xéq‬‬

‫اﻟﺤﺎﻟﺔ اﻟﻨﻬﺎﺋﻴﺔ‬

‫‪ – 4‬ﻋﺒـﺎرة ‪ : σ‬ﻧﻌﻠﻢ أن ] ‪ ، [CH COO ] = [H O‬وﺑﺎﻟﺘﺎﻟﻲ )‬
‫‪+‬‬

‫–‬

‫‪3‬‬

‫‪3‬‬

‫‪−‬‬

‫‪+‬‬

‫‪3‬‬

‫ﻣﻦ ﺟﺪول اﻟﺘﻘﺪم ﻧﺴﺘﻨﺘﺞ أن ﻋﻨﺪ ﻧﻬـﺎﻳﺔ اﻟﺘﻔﺎﻋﻞ ﻳﻜﻮن ‪xéq = nH O+ = nCH COO−‬‬
‫‪3‬‬

‫)‬

‫‪+ λCH COO−‬‬
‫‪3‬‬

‫‪H 3O +‬‬

‫‪(λ‬‬

‫‪xéq‬‬
‫‪V‬‬

‫(‬

‫‪σ = ⎡⎣H 3O + ⎤⎦ λH O + λCH COO‬‬
‫‪3‬‬

‫)‪(1‬‬

‫‪ ،‬وﺑﺎﺗﺎﻟﻲ ﺗﺼﺒﺢ ﻋﺒـﺎرة ‪ σ‬آﺎﻟﺘـﺎﻟﻲ ‪:‬‬

‫‪3‬‬

‫=‪σ‬‬

‫ﻣﻦ اﻟﻌﻼﻗﺔ )‪ (1‬ﻧﺤﺴﺐ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد اﻟﻬﻴﺪروﻧﻴﻮم ‪.‬‬

‫‪σ‬‬
‫‪1,6 × 10−2‬‬
‫=‬
‫‪= 0 , 41 mol / m3 = 0 , 41× 10−3 mol / L‬‬
‫‪−3‬‬
‫‪+ λCH COO‬‬
‫‪(34,9 + 4,1) × 10‬‬
‫‪−‬‬

‫‪ – 5‬ﻋﻨﺪ ﺣﺎﻟﺔ اﻟﺘﻮازن ﻳﻜﻮن‬

‫‪3‬‬

‫‪+‬‬

‫‪λH O‬‬

‫‪+‬‬
‫‪−‬‬
‫= ⎤⎦ ‪⎣⎡H 3O ⎦⎤ = ⎣⎡CH 3COO‬‬

‫‪3‬‬

‫‪[CH3COOH] = C – [H3O+] = 10–2 – 4 × 10–4 = 9,6 × 10–3 mol/L‬‬

‫‪9‬‬

‫‪2‬‬
‫) ‪⎡⎣H 3O + ⎤⎦ f × ⎡⎣CH 3COO − ⎤⎦ f (4 ,1× 10−4‬‬
‫=‪K‬‬
‫=‬
‫ﺛﺎﺑﺖ اﻟﺘﻮازن ‪= 1,75 × 10−5‬‬
‫‪−3‬‬
‫‪9,6 × 10‬‬
‫‪[CH 3COOH ] f‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪16‬‬
‫)‪NH3(aq) + H2O(l) = NH4+(aq) + OH–(aq‬‬

‫‪-1‬‬

‫‪ - 2‬ﺗﺼﺤﻴﺢ ‪:‬‬
‫ﻗﻴﻢ اﻟﻨﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ اﻟﻤﺴﺠّﻠﺔ ﻓﻲ اﻟﺠﺪول ﺧﺎﻃﺌﺔ ‪.‬‬
‫ﻣﺤﻠﻮل اﻟﻨﺸﺎدر اﻟﺬي ﺗﺮآﻴﺰﻩ ‪ C = 10–2 mol/ L‬ﺗﻜﻮن ﻧﺎﻗﻠﻴﺘﻪ اﻟﻨﻮﻋﻴﺔ ‪ σ = 10,9 mS.m–1‬وﻟﻴﺲ ‪. 100,4 μ S.m–1‬‬
‫ﻟﻤﺎذا اﻟﻘﻴﻢ اﻟﻤﺴﺠﻠﺔ ﻓﻲ اﻟﺠﺪول ﺧﺎﻃﺌﺔ ؟‬

‫ﻳﺠﺐ أن ﻧﻌﻠﻢ أن ﻧﺎﻗﻠﻴﺔ ﻣﺤﻠﻮل )‪ (G‬ﺗﺨﺺ ﻓﻘﻂ ﺟﺰءا ﻣﻦ اﻟﻤﺤﻠﻮل ‪ ،‬أي اﻟﺠﺰء اﻟﻤﺤﺼﻮر ﺑﻴﻦ ﺻﻔﻴﺤﺘﻲ اﻟﺨﻠﻴﺔ ‪G = K σ :‬‬
‫ﺣﻴﺚ ‪ K‬هﻮ ﺛﺎﺑﺖ اﻟﺨﻠﻴﺔ ‪ .‬أﻣﺎ اﻟﻨﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ ﻟﻤﺤﻠﻮل )‪ (σ‬ﺗﺨﺺ اﻟﻤﺤﻠﻮل ‪ ،‬أي أﻧﻬﺎ ﺗﺘﻌﻠﻖ ﺑﻄﺒﻴﻌﺔ اﻟﺸﻮارد اﻟﻤﻮﺟﻮدة ﻓﻲ اﻟﻤﺤﻠﻮل‬
‫وﺗﺮاآﻴﺰهﺎ اﻟﻤﻮﻟﻴﺔ ﻓﻲ هﺬا اﻟﻤﺤﻠﻮل ودرﺟﺔ ﺣﺮارة اﻟﻤﺤﻠـﻮل ‪.‬‬
‫اﻟﻤﻘﺼﻮد ﻣﻦ هﺬا هﻮ ‪ :‬أن ﻣﺤﻠﻮﻻ ﺷﺎردﻳﺎ ﻣﻌﻴﻨﺎ ﺑﺘﺮآﻴﺰ ﻣﻌﻴﻦ ﻓﻲ درﺟﺔ ﺣﺮارة ﻣﻌﻴﻨﺔ ﻻ ﺗﻜﻮن ﻟﻪ إﻻ ﻗﻴﻤﺔ واﺣﺪة ﻟﻠﻨﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ ‪.‬‬
‫اﻟﺠﺪول ﺑﻌﺪ اﻟﺘﺼﺤﻴﺢ ‪:‬‬

‫‪C (mol/L) 1,0 × 10–2 5,0 × 10–3 1,0 × 10–3‬‬
‫‪3,44‬‬

‫‪10,9‬‬

‫‪7,71‬‬

‫‪σ mS.m–1‬‬

‫ﻟﻠﻤﺰﻳﺪ ‪ :‬ﺻﺤّﺤﻨﺎ ﻗﻴﻢ اﻟﻨﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ ﺑﺎﻟﻄﺮﻳﻘﺔ اﻟﺘﺎﻟﻴﺔ ‪ :‬هﻨﺎك ﻋﻼﻗﺔ ﺗﺠﻤﻊ ﺑﻴﻦ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﻸﺳﺎس اﻟﻀﻌﻴﻒ واﻟـ ‪، pH‬‬

‫‪1‬‬
‫)‪pH = (14 + pK A + Log C‬‬
‫‪2‬‬

‫ﺖ ﻣﻄﺎﻟﺒﺎ ﺑﻬﺎ(‬
‫)ﻟﺴ َ‬

‫‪ ،‬ﺣﻴﺚ ‪ C‬هﻮاﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﻸﺳﺎس اﻟﻀﻌﻴﻒ )اﻟﻤﻤﺪد( ‪ pKA ،‬هﻲ‬

‫اﻟﻘﻴﻤﺔ اﻟﺨﺎﺻﺔ ﺑﺎﻟﺜﻨﺎﺋﻴﺔ أﺳـﺎس‪ /‬ﺣﻤﺾ )ﻓﻲ ﻣﺜﺎﻟﻨﺎ اﻟﺜﻨﺎﺋﻴﺔ هﻲ ‪ NH4+ / NH3‬و ‪ pKA = 9,2‬ﻓﻲ اﻟﺪرﺟﺔ ‪. ( 25°C‬‬
‫ﻧﻌﻮّض ﻓﻨﺠﺪ ‪ ، pH = 10,6‬ﺛﻢ ﻧﺴﺘﻨﺘﺞ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﺸﻮارد اﻟﻬﻴﺪروآﺴﻴﺪ ﻓﻲ اﻟﻤﺤﻠﻮل ‪:‬‬

‫‪10−14‬‬
‫‪10−14‬‬
‫‪⎡⎣OH ⎤⎦ = − pH = −10 ,6 = 4 ,0 × 10−4 mol / L‬‬
‫‪10‬‬
‫‪10‬‬
‫‪−‬‬

‫ﻣﻦ ﺟﻬﺔ أﺧﺮى ﻟﺪﻳﻨﺎ ‪× 103 (20 + 7 ,35) × 10−3 = 10 ,9 × 10−3 S :‬‬

‫‪−4‬‬

‫‪) = 4 ×10‬‬

‫‪+‬‬

‫‪4‬‬

‫(‬

‫‪σ = ⎡⎣OH − ⎤⎦ λOH + λNH‬‬
‫‪−‬‬

‫وﺑﻬﺬﻩ اﻟﻄﺮﻳﻘﺔ ﺣﺴﺒﻨﺎ آﻞ اﻟﻘﻴﻢ اﻷﺧﺮى ﻟﻠﻨﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ ‪ .‬وهﻨﺎك اﻟﻄﺮﻳﻘﺔ اﻷﺧﺮى اﻟﺘﻲ ﺗﻌﺘﻤﺪ ﻋﻠﻰ ﻗﻴﻤﺔ ‪ pKA‬اﻟﺜﻨﺎﺋﻴﺔ ‪. NH4+ / NH3‬‬

‫)‪NH3(aq) + H2O(l) = NH4+(aq) + OH–(aq‬‬
‫‪0‬‬

‫‪0‬‬

‫زﻳﺎدة‬

‫‪xf‬‬

‫‪xf‬‬

‫زﻳﺎدة‬

‫‪CV‬‬
‫‪CV – xf‬‬

‫ﻣﻦ ﺟﺪول اﻟﺘﻘﺪم ﻧﺴﺘﻨﺘﺞ أن )–‪ ، xf = n (OH‬وﻟﺪﻳﻨﺎ ]‪ [OH–] = [NH4+‬ﻷن ]‪ [H3O+‬ﻣﻬﻤﻞ ‪.‬‬
‫ﻟﺘﻌﻴﻴﻦ ﺗﺮآﻴﺰي اﻟﺸﺎردﺗﻴﻦ –‪ OH‬و ‪ NH4+‬ﻧﻜﺘﺐ ﻋﺒﺎرة اﻟﻨﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ ﻟﻠﻤﺤﻠﻮل ‪:‬‬

‫⎦⎤ ‪σ = λOH ⎡⎣OH − ⎤⎦ + λNH ⎡⎣NH 4 + ⎤⎦ + λH O ⎡⎣H 3O + ⎤⎦ = λOH ⎡⎣OH − ⎤⎦ + λNH ⎡⎣NH 4 +‬‬
‫‪+‬‬

‫)‬

‫‪+‬‬

‫‪4‬‬

‫‪−‬‬

‫‪4‬‬

‫‪+‬‬

‫‪+‬‬

‫‪3‬‬

‫(‬

‫‪σ = ⎡⎣OH − ⎤⎦ λOH + λNH‬‬
‫‪−‬‬

‫‪10‬‬

‫‪4‬‬

‫‪−‬‬

‫اﻟﻤﺤﻠﻮل اﻷول ‪:‬‬

‫‪σ1‬‬
‫‪10 ,9 × 10−3‬‬
‫=‬
‫‪= 0 , 4 mol / m3 = 4,0 × 10−4 mol / L‬‬
‫‪−3‬‬
‫‪+ λNH‬‬
‫‪27 ,35 × 10‬‬
‫‪+‬‬

‫‪4‬‬

‫‪−‬‬

‫‪λOH‬‬

‫‪−‬‬
‫‪+‬‬
‫= ⎤⎦ ‪⎣⎡OH ⎦⎤ = ⎣⎡NH 4‬‬

‫أﻣﺎ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﻠﺸﺎردة ‪ H3O+‬ﻧﺤﺴﺒﻪ ﻣﻦ اﻟﺠﺪاء اﻟﺸﺎردي ﻟﻠﻤﺎء ‪[H3O+] × [OH–] = 10–14‬‬

‫‪10−14‬‬
‫‪= 2 ,5 × 10−11 mol / L‬‬
‫= ⎦⎤ ‪⎡⎣H 3O +‬‬
‫‪−4‬‬
‫‪4 × 10‬‬

‫‪ ،‬وهﺬا ﻳﺆآّﺪ ﺳﺒﺐ إهﻤﺎﻟﻪ ‪.‬‬

‫اﻟﻤﺤﻠﻮل اﻟﺜﺎﻧﻲ ‪:‬‬

‫‪σ2‬‬
‫‪7 ,71× 10−3‬‬
‫=‬
‫‪= 0 , 28 mol / m3 = 2,8 × 10−4 mol / L‬‬
‫‪−3‬‬
‫‪+ λNH‬‬
‫‪27 ,35 × 10‬‬
‫‪+‬‬

‫‪4‬‬

‫‪−‬‬

‫‪+‬‬

‫= ⎦⎤ ‪⎡⎣OH ⎤⎦ = ⎡⎣NH 4‬‬
‫‪λ‬‬

‫‪OH −‬‬

‫‪10−14‬‬
‫‪= 3,6 × 10−11 mol / L‬‬
‫= ⎦⎤ ‪⎡⎣H 3O +‬‬
‫‪−4‬‬
‫‪2 ,8 × 10‬‬
‫اﻟﻤﺤﻠﻮل اﻟﺜـﺎﻟﺚ ‪:‬‬

‫‪σ3‬‬
‫‪3, 44 × 10−3‬‬
‫=‬
‫‪= 0 ,125 mol / m3 = 1, 25 × 10−4 mol / L‬‬
‫‪−3‬‬
‫‪+ λNH‬‬
‫‪27 ,35 × 10‬‬
‫‪+‬‬

‫‪4‬‬

‫‪+‬‬

‫‪−‬‬

‫= ⎦⎤ ‪⎡⎣OH ⎤⎦ = ⎡⎣NH 4‬‬
‫‪λOH −‬‬

‫‪10−14‬‬
‫‪= 8,0 × 10−11 mol / L‬‬
‫= ⎦⎤ ‪⎡⎣H 3O +‬‬
‫‪−4‬‬
‫‪1, 25 × 10‬‬
‫اﻟﻨﺴﺒﺔ اﻟﻨﻬـﺎﺋﻴﺔ ﻟﻠﺘﻘﺪم ﻓﻲ آﻞ ﻣﺤﻠﻮل ‪:‬‬
‫ﻟﺪﻳﻨﺎ‬

‫⎦⎤ ‪⎡⎣OH −‬‬
‫=‬
‫‪C‬‬

‫‪xf‬‬
‫‪xmax‬‬

‫=‪τ‬‬

‫‪4 × 10−4‬‬
‫= ‪. τ1‬‬
‫اﻟﻤﺤﻠﻮل اﻷول ‪= 0 ,04 :‬‬
‫‪10−2‬‬
‫‪2 ,8 × 10−4‬‬
‫اﻟﻤﺤﻠﻮل اﻟﺜﺎﻧﻲ ‪= 0 ,056 :‬‬
‫‪5 × 10−3‬‬

‫= ‪τ2‬‬

‫‪1, 25 × 10−4‬‬
‫اﻟﻤﺤﻠﻮل اﻟﺜﺎﻟﺚ ‪= 0 ,125 :‬‬
‫= ‪τ3‬‬
‫‪10−3‬‬

‫‪ .‬ﻧﻌﻢ ﺗﺘﻌﻠﻖ اﻟﻨﺴﺒﺔ اﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻘﺪم ﺑﺎﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﻸﺳﺎس ‪ ،‬ﺣﻴﺚ آﻠﻤﺎ آﺎن اﻷﺳﺎس‬

‫ﻣﻤﺪا ﺗﻜﻮن ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻴﺔ أآﺒﺮ ‪.‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪17‬‬
‫ﻣﻌـﺎدﻟﺔ اﻟﺘﻔـﺎﻋﻞ ‪2 (Ag+ , NO3–) + Cu = (Cu2+ , 2 NO3–) + 2 Ag :‬‬
‫أو اﺧﺘﺼﺎرا ‪:‬‬

‫‪2 Ag+ + Cu = Cu2+ + 2 Ag‬‬

‫) –‪ NO3‬ﺷﺎردة ﻏﻴﺮ ﻓﻌـﺎﻟﺔ( ‪.‬‬

‫ﻟﻠﻤﺰﻳﺪ ‪ :‬ذرة اﻟﻨﺤﺎس ﺑﺈﻣﻜﺎﻧﻬﺎ ﺗﻘﺪﻳﻢ اﻹﻟﻜﺘﺮوﻧﺎت ﻟﺸﻮارد اﻟﻔﻀﺔ ﺣﺴﺐ اﻟﻜﻤﻮن اﻟﻨﻈﺎﻣﻲ اﻟﻤﻌﻄﻰ ﻓﻲ ﺟﺪول اﻟﻜﻤﻮﻧﺎت اﻟﻨﻈﺎﻣﻴﺔ ﻓﻲ‬
‫اﻟﻮﺣﺪة اﻷوﻟﻰ ‪.‬‬
‫‪2+‬‬

‫⎦⎤ ‪⎡Cu‬‬
‫⎣ = ‪ ، Qr‬ﺗﺮآﻴﺰا ‪ Cu‬و ‪ Ag‬ﻻ ﻳﻈﻬﺮان ﻓﻲ ﻋﺒﺎرة ‪ Qr‬ﻷﻧﻬﻤﺎ ﺻﻠﺒﺎن ‪.‬‬
‫‪ - 1‬آﺴﺮ اﻟﺘﻔـﺎﻋﻞ ‪:‬‬
‫‪2‬‬
‫⎦⎤ ‪⎡⎣Ag +‬‬

‫‪11‬‬

‫‪m 6 ,35‬‬
‫=‬
‫‪ - 2‬آﻤﻴﺔ ﻣـﺎدة اﻟﻨﺤـﺎس ‪= 0 ,1 mol‬‬
‫‪M 63,5‬‬
‫ﺟﺪول اﻟﺘﻘ ّﺪم ‪:‬‬

‫‪Cu2+ + 2 Ag‬‬

‫=‬

‫= )‪n (Cu‬‬
‫‪+‬‬

‫‪Cu‬‬

‫‪2 Ag+‬‬

‫‪0‬‬

‫‪0‬‬

‫‪n0‬‬

‫‪CV‬‬

‫‪xq‬‬

‫‪xéq‬‬

‫‪n0 – xéq‬‬

‫‪CV – 2 xéq‬‬

‫‪ - 3‬ﺛﺎﺑﺖ اﻟﺘﻮازن‬

‫‪⎡⎣Cu 2+ ⎤⎦ f‬‬
‫‪2‬‬

‫‪+‬‬
‫‪⎣⎡Ag ⎦⎤ f‬‬

‫= ‪K = Qr , f‬‬

‫)‪(1‬‬

‫‪xéq‬‬
‫وﻟﺪﻳﻨﺎ ﻣﻦ ﺟﺪول اﻟﺘﻘﺪم ‪ ، n (Cu2+) = xéq‬وﺑﺎﻟﺘﺎﻟﻲ ‪:‬‬
‫= ⎦⎤ ‪⎡⎣Cu 2+‬‬
‫‪V‬‬
‫‪CV − 2 xéq‬‬

‫وﻟﺪﻳﻨﺎ آﺬﻟﻚ ﻣﻦ ﺟﺪول اﻟﺘﻘﺪم ‪ ، n ( Ag + ) = CV − 2 xéq‬وﺑﺎﻟﺘﺎﻟﻲ‬

‫‪V‬‬

‫‪xéq‬‬
‫‪xéq V‬‬
‫‪V‬‬
‫=‪K‬‬
‫=‬
‫وﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ اﻟﻌﻼﻗﺔ )‪: (1‬‬
‫‪2‬‬
‫‪2‬‬
‫) ‪(CV − 2 xéq‬‬
‫⎞ ‪⎛ CV − 2 xéq‬‬
‫⎜‬
‫⎟‬
‫‪V‬‬
‫⎝‬
‫⎠‬

‫= ⎦⎤ ‪⎡⎣Ag +‬‬

‫)‪(2‬‬

‫‪ - 4‬ﻟﻜﻲ ﻧﺘﺄآﺪ ﻣﻦ ذﻟﻚ ﻧﻌﻮّض ‪ xéq = 1,0 × 10–3 – 4,8 × 10–11 mol‬ﻓﻲ اﻟﻤﻌﺎدﻟﺔ )‪ ، (2‬وذﻟﻚ ﻟﻜﻲ ﻧﺠﺪ اﻟﻨﺘﻴﺠﺔ‬

‫‪(10−3 − 4,8 × 10−11) × 0,02‬‬
‫‪2 × 10 −5 − 9 ,6 × 10−13‬‬
‫=‪K‬‬
‫=‬
‫=‬
‫‪2‬‬
‫‪2‬‬
‫‪2‬‬
‫])‪(CV − 2 xéq ) [0,1× 0,02 − 2 (10−3 − 4,8 × 10−11‬‬
‫)‪(9,6 × 10−11‬‬
‫‪xéq V‬‬

‫ﻧﻬﻤﻞ ﻓﻲ اﻟﺒﺴﻂ اﻟﻘﻴﻤﺔ ‪ 9,6 × 10–13‬أﻣـﺎم اﻟﻘﻴﻤﺔ ‪ ، 2 × 10–5‬ﻓﻨﺠﺪ ‪. K ≈ 2, 2 × 1015‬‬
‫‪ - 5‬ﻋﻨﺪ اﻟﺘﻮازن ﻳﻜﻮن ‪) n (Ag+) = CV – 2 xéq = 9,6 × 10–11 mol‬اﻧﻈﺮ ﻟﻤﻘﺎم ﻋﺒﺎرة ‪ ، (K‬وﺑﺎﻟﺘﺎﻟﻲ ‪:‬‬

‫‪9 ,6 × 10−11‬‬
‫=‬
‫‪= 4 ,8 × 10−9 mol / L‬‬
‫‪0 ,02‬‬

‫‪CV − 2 xéq‬‬
‫‪V‬‬

‫‪+‬‬

‫= ⎦⎤ ‪⎡⎣Ag‬‬

‫ﻟﻜﻲ ﻧﺤﺴﺐ ‪ xmax‬ﻳﺠﺐ ﺗﺤﺪﻳﺪ اﻟﻤﺘﻔﺎﻋﻞ اﻟﻤﺤﺪ أوﻻ ‪ ،‬ﻣﻦ أﺟﻞ ذﻟﻚ ﻧﻜﺘﺐ ‪:‬‬

‫‪ ، n0 – x = 0‬وﻧﺴﺘﻨﺘﺞ ‪x = 0,1 mol‬‬
‫‪CV 0 ,1 × 0 ,02‬‬
‫=‬
‫‪ CV – 2 x = 0‬وﻧﺴﺘﻨﺘﺞ ‪= 1,0 × 10−3 mol‬‬
‫‪2‬‬
‫‪2‬‬

‫=‪x‬‬

‫اﻟﻘﻴﻤﺔ اﻟﺼﻐﻴﺮة ﻟﻠﺘﻘﺪم ‪ x‬هﻲ اﻟﻤﻮاﻓﻘﺔ ﻟﺸﻮارد اﻟﻔﻀﺔ ‪ ،‬وﺑﺎﻟﺘﺎﻟﻲ ﺷﻮارد اﻟﻔﻀﺔ هﻲ اﻟﻤﺘﻔﺎﻋﻞ اﻟﻤﺤﺪ ‪ ،‬وﻣﻨﻪ ﻳﻜﻮن ‪xmax = 10–3 mol‬‬
‫ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬـﺎﺋﻲ ﻟﻠﺘﻔﺎﻋﻞ هﻲ‬

‫‪10−3 − 4 ,8 × 10−11‬‬
‫=‬
‫‪≈1‬‬
‫‪10−3‬‬

‫‪xf‬‬
‫‪xmax‬‬

‫‪12‬‬

‫= ‪ ، τ‬ﻳﻤﻜﻦ اﻋﺘﺒﺎر اﻟﺘﻔﺎﻋﻞ ﺗﺎﻣــﺎ ‪.‬‬

18 ‫اﻟﺘﻤﺮﻳﻦ‬
(2 Na+ , SO32–)(aq) + CH3COOH(aq) = CH3COO–(aq) + (Na+ , HSO3–)(aq) : ‫ ﻣﻌﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ‬- 1
SO32–(aq) + CH3COOH(aq) = CH3COO–(aq) + HSO3–

: ‫أو اﺧﺘﺼﺎرا‬
-2

SO3

2–

(aq)

+ CH3COOH(aq)

=

HSO3

–

(aq)

–

+

CH3COO

C1V1

C2V2

0

0

C1V1 – x

C2V2 – x

x

x

C1V1 – xf

C2V2 – xf

xf

xf

(aq)

CH 3COO − ][HSO3− ]
0×0
[
Qr ,i =
=
2−
[SO3 ][CH 3COOH ] C1 × C2

(1)

Qr , f

[CH 3COO −] f [HSO3−] f
=
[SO32−] f [CH 3COOH ] f

=

=0

x 2f

(C1V1 − x f ) × (C2V2 − x f )

(‫ وﻣﻦ اﻟﻤﻌﻄﻴﺎت ﻟﺪﻳﻨﺎ ﻋﺪدا ﻣﻮﻻت اﻟﻤﺘﻔﺎﻋﻠﻴﻦ ﻣﺘﺴﺎوﻳﺎن )ﻧﻔﺲ اﻟﺘﺮآﻴﺰ وﻧﻔﺲ اﻟﺤﺠﻢ‬، τ =

xf
xmax

-3

-4

‫ﻟﺪﻳﻨﺎ اﻟﻨﺴﺒﺔ اﻟﻨﻬﺎﺋﻴﺔ ﻟﻠﺘﻘﺪم‬

: ‫( ﻧﻜﺘﺐ‬1) ‫ وﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ اﻟﻌﻼﻗﺔ‬، xmax = C1V1 = C2V2

‫إذن‬

2

2
⎛
⎞ ⎛
⎞
2
⎟ ⎜ 1 ⎟
xf
xf
xf
⎞ ⎜
=
= ⎜⎛
=
⎟
⎟ =⎜
2
( xmax − x f )
⎝ xmax − x f ⎠ ⎜ x ⎛ xmax − 1⎞ ⎟ ⎜⎜ 1 − 1⎟⎟
⎟⎟ ⎝τ
⎜ f⎜ x
⎠
⎠⎠
⎝ ⎝ f
2

Qr , f

Qr , f =
(2)

τ=

K
251
=
= 0,94
1 + K 1 + 251

13

τ2
(1 − τ )2
K=

τ2
(1 − τ )2

‫ ﻧﺴﺘﻨﺘﺞ‬،

K =

‫ وﻣﻨﻪ‬، Qr,f = K ‫ﻧﻌﻠﻢ أن‬

τ
1−τ

-5

: ‫( ﻧﻜﺘﺐ‬2) ‫ﺑﺠﺬر اﻟﻤﻌﺎدﻟﺔ‬

‫اﻟﺘﻤﺮﻳﻦ ‪19‬‬
‫)‪HCO3–(aq) + NH3(aq) = NH4+(aq) + CO32–(aq‬‬

‫‪ – 1‬ﻣﻌﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ ‪:‬‬
‫‪ – 2‬ﺟﺪول اﻟﺘﻘﺪّم ‪:‬‬

‫‪-3‬‬

‫ﻟﺪﻳﻨﺎ‬

‫=‬

‫‪0‬‬

‫‪0‬‬

‫‪C2V2‬‬

‫‪C1V1‬‬

‫‪x‬‬

‫‪x‬‬

‫‪C2V2 – x‬‬

‫‪C1V1 – x‬‬

‫‪xf‬‬

‫‪xf‬‬

‫‪C2V2 – xf‬‬

‫‪C1V1 – xf‬‬

‫‪xm‬‬

‫‪xm‬‬

‫‪C2V2 − xm‬‬

‫‪C1V1 − xm‬‬

‫] ‪NH 4 + ][CO32−‬‬
‫[‬
‫= ‪Qr‬‬
‫]‪[HCO3−][NH 3‬‬
‫‪=0‬‬

‫‪-4‬‬

‫)‪CO32–(aq‬‬

‫‪NH4+(aq) +‬‬

‫)‪NH3(g‬‬

‫‪HCO3–(aq) +‬‬

‫‪[NH 4 + ][CO32−] = 0 × 0‬‬
‫‪[HCO3−][NH 3] C1 × C2‬‬
‫‪x f2‬‬
‫) ‪(C1V1 − x f ) (C2V2 − x f‬‬

‫‪xf‬‬
‫‪xmax‬‬

‫= ‪، τ‬‬

‫= ‪Qr ,i‬‬

‫= ‪Qr , f = K‬‬

‫)‪(1‬‬

‫و ﻟﻜﻲ ﻧﺤﺪّد اﻟﺘﻘﺪّم اﻷﻋﻈﻤﻲ ‪ xmax‬ﻳﺠﺐ ﺗﺤﺪﻳﺪ اﻟﻤﺘﻔﺎﻋﻞ اﻟﻤﺤﺪ ﻓﻲ ﺣﺎﻟﺔ ﻓﺮض أن اﻟﺘﻔـﺎﻋﻞ ﺗﺎم ‪.‬‬

‫‪ ، C1V1 – x = 0‬وﻣﻨﻪ ‪x = C1V1 = 0,15 × 0,03 = 4,5 × 10–3 mol‬‬
‫‪ ، C2V2 – x = 0‬وﻣﻨﻪ ‪x = C2V2 = 0,1 × 0,02 = 2 × 10–3 mol‬‬
‫ﻧﺴﺘﻨﺘﺞ أن اﻟﻤﺘﻔﺎﻋﻞ اﻟﻤﺤﺪ هﻮ ﻣﺤﻠﻮل اﻟﻨﺸﺎدر ‪ ،‬وﺑﺎﻟﺘﺎﻟﻲ ‪xmax = C2V2‬‬
‫ﻣﻦ ﺟﻬﺔ أﺧﺮى ﻟﺪﻳﻨﺎ ‪ ، C1V1 = 2,25 C2V2‬أي ‪C1V1 = 2,25 xmax‬‬
‫ﻧﻌﻮّض ﻓﻲ اﻟﻌﻼﻗﺔ )‪: (1‬‬

‫‪x f2‬‬
‫⎞ ‪2, 25 xmax ⎞ ⎛ xmax‬‬
‫⎜⎛ ‪x f x f‬‬
‫⎜ ⎟‪− 1‬‬
‫⎟‪− 1‬‬
‫‪⎝ xf‬‬
‫‪⎠⎝ xf‬‬
‫⎠‬

‫=‬

‫‪x f2‬‬
‫) ‪(2, 25 xmax − x f ) ( xmax − x f‬‬

‫‪τ2‬‬
‫=‬
‫=‬
‫‪2, 25‬‬
‫‪1‬‬
‫) ‪− 1 τ − 1 (2, 25 − τ ) (1 − τ‬‬
‫‪τ‬‬

‫) ()‬

‫‪τ2‬‬
‫‪ - 5‬ﻧﺤﻞ اﻟﻤﻌﺎدﻟﺔ‬
‫) ‪(2, 25 − τ )(1 − τ‬‬

‫‪1‬‬

‫(‬

‫‪Qr , f‬‬

‫= ‪ Qr , f‬ذات اﻟﻤﺠﻬﻮل ‪. τ‬‬

‫)‪τ2 = Qr,f ( 2,25 – 3,25 τ + τ2‬‬
‫‪(Qr,f – 1) τ2 – 3,25 Qr,f τ + 2,25 Qr,f = 0‬‬

‫‪ ،‬وﻟﺪﻳﻨﺎ ‪Qr,f = K = 7,9 × 10–2‬‬

‫ﺣﻞ اﻟﻤﻌﺎدﻟﺔ ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﻴﺔ ﻳﻌﻄﻴﻨﺎ ﺟﺬرﻳﻦ هﻤﺎ ‪ ) τ2 = – 0,59 ، τ1 = 0,32‬ﻣﺮﻓﻮض(‬
‫ﻧﺴﺒﺔ اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ هﻲ ‪. 32 %‬‬

‫‪14‬‬

‫= ‪Qr , f‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪20‬‬

‫‪ – 1‬ﻣﻌـﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ ‪+ Ag(s) :‬‬
‫‪-2‬‬

‫] ‪[Fe3+‬‬
‫] ‪[Fe2+ ][Ag +‬‬

‫= ‪Qr‬‬

‫)‪Fe3+(aq‬‬

‫=‬

‫)‪Ag+(aq‬‬

‫‪+‬‬

‫)‪Fe2+(aq‬‬

‫)‪(1‬‬

‫ﺛﺎﺑﺖ اﻟﺘﻮازن اﻟﻤﻌﻄﻰ ﻓﻲ اﻟﺘﻤﺮﻳﻦ ‪ K = 3,2‬ﺧﺎص ﺑﺎﻟﺘﻔﺎﻋﻞ اﻟﻤﺒﺎﺷﺮ ‪ ،‬أي ﺗﻔﺎﻋﻞ ﺷﻮارد اﻟﺤﺪﻳﺪ اﻟﺜﻨﺎﺋﻲ ﻣﻊ ﺷﻮارد اﻟﻔﻀﺔ ‪.‬‬

‫‪10−2‬‬
‫‪Qr = −2‬‬
‫اﻟﺤﺎﻟﺔ اﻷوﻟﻰ ‪= 100 :‬‬
‫‪10 × 10−2‬‬
‫‪5 ×10−3‬‬
‫‪Qr = −1‬‬
‫اﻟﺤﺎﻟﺔ اﻟﺜـﺎﻧﻴﺔ ‪= 0,5 :‬‬
‫‪10 ×10−1‬‬
‫‪ - 3‬ﻟﻮ وﺟﺪﻧﺎ ﻓﻲ إﺣﺪى اﻟﺤﺎﻟﺘﻴﻦ ﻣﺜﻼ ‪ ، Qr = 3,2‬ﻓﻬﺬا ﻣﻌﻨﺎﻩ أن اﻟﺠﻤﻠﺔ ﻓﻲ ﺣـﺎﻟﺔ اﻟﺘﻮازن ‪ ،‬أي ﻻ ﺗﻨﻤﻮ ‪.‬‬
‫اﻟﺤـﺎﻟﺔ ‪ : 1‬وﺟﺪﻧﺎ ‪ ، Qr > K‬إذن اﻟﺠﻤﻠﺔ ﻏﻴﺮ ﻣﺘﻮازﻧﺔ ‪ ،‬ﻓﻠﻜﻲ ﻳﺼﺒﺢ ‪ Qr = K‬ﻳﺠﺐ أن ﺗﻨﻤﻮ ﻟﻜﻲ ﻳﺘﻨﺎﻗﺺ ‪ ، Qr‬ﻓﻤﻦ أﺟﻞ هﺬا‬
‫اﻟﻐﺮض ﻳﺠﺐ أن ﻳﻨﻘﺺ اﻟﺒﺴﻂ ﻓﻲ اﻟﻌﻼﻗﺔ )‪ (1‬وﻳﺰداد اﻟﻤﻘـﺎم ‪ .‬ﻣﻌﻨﻰ هﺬا ﻳﺠﺐ أن ﻧﻀﻴﻒ اﻟﺘﻘﺪم )‪ (x‬ﻟـ ‪ Fe2+‬و ‪ Ag+‬وﻧﻨﻘﺼﻪ ﻣﻦ‬
‫‪ ، Fe3+‬وﺑﺎﻟﺘﺎﻟﻲ ﺗﻨﻤﻮ اﻟﺠﻤﻠﺔ ﻧﺤﻮ اﻟﻴﺴـﺎر ‪.‬‬
‫اﻟﺤـﺎﻟﺔ ‪ : 2‬وﺟﺪﻧﺎ ‪ ، Qr < K‬إذن اﻟﺠﻤﻠﺔ ﻏﻴﺮ ﻣﺘﻮازﻧﺔ ‪ ،‬ﻓﻠﻜﻲ ﻳﺼﺒﺢ ‪ Qr = K‬ﻳﺠﺐ أن ﺗﻨﻤﻮ ﻟﻜﻲ ﻳﺰداد ‪ ، Qr‬ﻓﻤﻦ أﺟﻞ هﺬا اﻟﻐﺮض‬

‫ﻳﺠﺐ أن ﻳﺰداد اﻟﺒﺴﻂ ﻓﻲ اﻟﻌﻼﻗﺔ )‪ (1‬وﻳﻨﻘﺺ اﻟﻤﻘـﺎم ‪ .‬ﻣﻌﻨﻰ هﺬا ﻳﺠﺐ أن ﻧﻀﻴﻒ اﻟﺘﻘﺪم )‪ (x‬ﻟـ ‪ Fe3+‬وﻧﻨﻘﺼﻪ ﻣﻦ ‪ Fe2+‬و ‪Ag+‬‬
‫وﺑﺎﻟﺘﺎﻟﻲ ﺗﻨﻤﻮ اﻟﺠﻤﻠﺔ ﻧﺤﻮ اﻟﻴﻤﻴﻦ ‪.‬‬
‫‪ – 4‬ﻓﻲ اﻟﺘﺤﻮل اﻷول )اﻟﺤﺎﻟﺔ ‪ (1‬اﻟﺘﻔﺎﻋﻞ اﻟﻐﺎﻟﺐ هﻮ اﻟﺘﻔﺎﻋﻞ ﻏﻴﺮ اﻟﻤﺒﺎﺷﺮ ‪ ،‬ﻟﺬﻟﻚ ﻳﻜﻮن ﺛﺎﺑﺖ اﻟﺘﻮازن ﻟﻬﺬا اﻟﺘﻔﺎﻋﻞ هﻮ ‪:‬‬

‫‪1‬‬
‫‪1‬‬
‫=‬
‫‪= 0,31‬‬
‫‪K 3, 2‬‬

‫='‪K‬‬

‫ﻓﻲ اﻟﺘﺤﻮّل اﻟﺜﺎﻧﻲ )اﻟﺤﺎﻟﺔ ‪ (2‬اﻟﺘﻔﺎﻋﻞ اﻟﻐﺎﻟﺐ هﻮ اﻟﺘﻔﺎﻋﻞ اﻟﻤﺒﺎﺷﺮ ‪ ،‬أي ‪K = 3,2‬‬
‫‪ – 5‬اﻟﺤـﺎﻟﺔ اﻷوﻟﻰ ‪:‬‬

‫)‪Ag(s‬‬

‫‪+‬‬

‫)‪Fe3+(aq‬‬

‫‪10–1‬‬
‫‪10–1 – x‬‬

‫)‪Ag+(aq‬‬

‫=‬

‫‪10–2‬‬

‫‪10–2‬‬

‫‪10–2 + x‬‬

‫ﻋﻨﺪ اﻟﺘﻮازن ﻳﻜﻮن آﺴﺮ اﻟﺘﻔﺎﻋﻞ ﻣﺴﺎوﻳﺎ ﻟﺜﺎﺑﺖ اﻟﺘﻮازن ‪ ،‬وﺑﺎﻟﺘﺎﻟﻲ ‪= 3, 2 :‬‬
‫‪10–2 – x = 3,2 (10–2 + x)2‬‬

‫‪+‬‬

‫‪10–2‬‬

‫‪10–2 – x‬‬

‫أو‬

‫)‪Fe2+(aq‬‬
‫‪10–2 + x‬‬

‫‪10−2 − x‬‬
‫‪2‬‬

‫)‪(10−2 + x‬‬

‫=‪K‬‬

‫‪3,2 x2 + 1,06 x – 9,7 × 10–3 = 0‬‬

‫ﺑﺤﻞ هﺬﻩ اﻟﻤﻌﺎدﻟﺔ ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﻴﺔ ﻧﺠﺪ ﺟﺬرﻳﻦ هﻤﺎ ‪ x1 = 8,75 × 10–3 :‬و ‪) x2 = – 0,34‬ﻣﺮﻓﻮض ﻷﻧﻪ ﺳﺎﻟﺐ(‬

‫اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ هﻮ ‪xf = 8,75 × 10–3 mol‬‬
‫اﻟﺘﺮآﻴﺐ اﻟﻨﻬﺎﺋﻲ ﻟﻠﻮﺳﻂ ‪:‬‬

‫‪Ag‬‬

‫‪3+‬‬

‫‪+‬‬

‫‪Ag‬‬

‫‪Fe‬‬

‫‪2+‬‬

‫‪Fe‬‬

‫= ‪10–2 + 8,75 × 10–3 = 10–2 + 8,75 × 10–3 = 10–2 – 8,75 × 10–3 = 10–1 – 8,75 × 10–3‬‬
‫‪9,1 ×10–2 mol‬‬

‫‪1,87 ×10–2 mol‬‬

‫‪1,25 ×10–3 mol‬‬

‫‪15‬‬

‫‪1,87 ×10–2 mol‬‬

‫‪،‬‬

‫اﻟﺤـﺎﻟﺔاﻟﺜﺎﻧﻴﺔ ‪:‬‬

‫)‪Ag(s‬‬

‫)‪Fe3+(aq‬‬

‫‪+‬‬

‫‪10–1‬‬

‫)‪Ag+(aq‬‬

‫=‬

‫‪5 ×10–3‬‬
‫‪5 ×10–3 + x‬‬

‫‪10–1 + x‬‬

‫‪+‬‬

‫‪10–1‬‬

‫‪10–1‬‬

‫‪10–1 – x‬‬

‫‪10–1 – x‬‬

‫ﻋﻨﺪ اﻟﺘﻮازن ﻳﻜﻮن آﺴﺮ اﻟﺘﻔﺎﻋﻞ ﻣﺴﺎوﻳﺎ ﻟﺜﺎﺑﺖ اﻟﺘﻮازن ‪ ،‬وﺑﺎﻟﺘﺎﻟﻲ ‪= 3, 2 :‬‬
‫‪5 × 10–2 + x = 3,2 (10–1 – x)2‬‬

‫أو‬

‫)‪Fe2+(aq‬‬

‫‪5 × 10−3 + x‬‬
‫‪2‬‬

‫)‪(10−1 − x‬‬

‫=‪K‬‬

‫‪3,2 x2 – 1,64 x + 0,027 = 0‬‬

‫ﺑﺤﻞ هﺬﻩ اﻟﻤﻌﺎدﻟﺔ ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﻴﺔ ﻧﺠﺪ ﺟﺬرﻳﻦ هﻤﺎ ‪ x1 = 1,71 × 10–2 :‬و ‪) x2 = 0,49‬ﻣﺮﻓﻮض ﻷﻧﻪ أآﺒﺮ ﻣﻦ ‪(10–1‬‬

‫اﻟﺘﻘﺪم اﻟﻨﻬﺎﺋﻲ هﻮ ‪xf = 8,75 × 10–3 mol‬‬
‫اﻟﺘﺮآﻴﺐ اﻟﻨﻬﺎﺋﻲ ﻟﻠﻮﺳﻂ ‪:‬‬

‫‪Fe3+‬‬

‫‪Ag‬‬

‫‪Fe2+‬‬

‫‪Ag+‬‬

‫= ‪10–1 – 1,71 × 10–2 = 10–1 – 1,71 × 10–2 = 5 × 10–3 + 1,71 × 10–2 = 10–1 + 1,71 × 10–2‬‬
‫‪11,7 ×10–2 mol‬‬

‫‪8,3 ×10–2 mol‬‬

‫‪2,21 ×10–2 mol‬‬

‫‪8,3 ×10–2 mol‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪21‬‬
‫‪ – 1‬ﻣﻌـﺎدﻟﺔ اﻟﺘﻔﺎﻋﻞ ‪CH3COOH(aq) + H2O(l) = CH3COO–(aq) + H3O+(aq) :‬‬
‫‪ - 2‬ﺟﺪول اﻟﺘﻘﺪم ‪:‬‬

‫)‪+ H3O+(aq‬‬

‫)‪CH3COO–(aq‬‬

‫)‪H2O(l‬‬

‫=‬

‫‪+‬‬

‫)‪CH3COOH(aq‬‬

‫‪0‬‬

‫‪0‬‬

‫زﻳﺎدة‬

‫‪C0V0‬‬

‫‪x‬‬

‫‪x‬‬

‫زﻳﺎدة‬

‫‪C0V0 – x‬‬

‫‪xf‬‬

‫‪xf‬‬

‫زﻳﺎدة‬

‫‪C0V0 – xf‬‬

‫‪xm‬‬

‫‪xm‬‬

‫زﻳﺎدة‬

‫‪C0V0 − xm‬‬

‫ﻣﻦ ﺟﺪول اﻟﺘﻘﺪم ﻧﺴﺘﻨﺘﺞ )‪ ، xf = n (H3O+‬وﻟﺪﻳﻨﺎ ‪xmax = C0V0‬‬
‫] ‪n0 ( H 3O + ) [H 3O + ] × V0 [H 3O +‬‬
‫=‬
‫=‬
‫اﻟﻨﺴﺒﺔ اﻟﻨﻬـﺎﺋﻴﺔ ﻟﻠﺘﻘﺪّم‬
‫‪C0V0‬‬
‫‪C0V0‬‬
‫‪C0‬‬

‫=‬

‫‪xf‬‬
‫‪xmax‬‬

‫= ‪ ، τ‬وﻣﻨﻪ ‪[H 3O + ] = [CH 3COO − ] = C0 ×τ :‬‬

‫‪-3‬‬
‫) ‪[CH 3COOH ] f = C0 − [H 3O + ] = C0 − C0 ×τ = C0 (1 − τ‬‬

‫‪ - 4‬ﺛﺎﺑﺖ اﻟﺤﻤـﻮﺿﺔ‬

‫‪⎡⎣H 3O + ⎤⎦ f × ⎡⎣CH 3COO − ⎤⎦ f‬‬
‫= ‪KA‬‬
‫‪[CH 3COOH ] f‬‬
‫‪C0 2 × τ 2‬‬
‫‪τ2‬‬
‫‪= C0‬‬
‫‪1−τ‬‬
‫) ‪C0 (1 − τ‬‬

‫= ‪KA‬‬
‫‪16‬‬

‫)‪(1‬‬

‫‪ - 5‬أ( إﺗﻤﺎم اﻟﺠﺪول ‪:‬‬

‫‪5,0 × 10–4‬‬

‫‪1,0 × 10–3‬‬

‫‪5,0 × 10–3‬‬

‫‪1,0 × 10–2‬‬

‫)‪C0 (mol/L‬‬

‫‪16,0‬‬

‫‪12,5‬‬

‫‪5,6‬‬

‫‪4,0‬‬

‫‪τ × 10–2‬‬

‫‪2000‬‬

‫‪1000‬‬

‫‪200‬‬

‫‪100‬‬

‫‪1‬‬
‫)‪(L / mol‬‬
‫‪C0‬‬

‫‪16,7 × 10–4 33,2 × 10–4 1,78 × 10–2 3,04 × 10–2‬‬

‫‪τ2‬‬
‫‪1‬‬
‫‪= KA‬‬
‫ﻣﻦ اﻟﻌﻼﻗﺔ )‪ (1‬ﻧﺴﺘﻨﺘﺞ‬
‫‪C0‬‬
‫‪1−τ‬‬

‫‪τ2‬‬
‫‪1−τ‬‬

‫=‪X‬‬

‫=‪y‬‬

‫‪ ،‬ﻣﻌﺎدﻟﺔ ﻣﺴﺘﻘﻴﻢ ﺷﻜﻠﻬﺎ ‪ ، Y = a X‬ﺣﻴﺚ ‪a = K A‬‬
‫‪τ2‬‬
‫‪1−τ‬‬

‫ب( رﺳﻢ اﻟﺒﻴﺎن‬

‫ﻧﻈﺮي‬
‫ﺗﺠﺮﻳﺒﻲ‬

‫‪0,05‬‬
‫•‬

‫‪1‬‬
‫)‪(L / mol‬‬
‫‪C0‬‬

‫‪200‬‬

‫ﻣﻦ أﺟﻞ ﺣﺴﺎب ﺛﺎﺑﺖ اﻟﺤﻤﻮﺿﺔ ‪ KA‬ﻧﺄﺧﺬ ﻣﺜﻼ اﻟﻨﻘﻄﺘﻴﻦ ‪:‬‬
‫)‪ (y1 = 16,7 × 10–4 ، X1 = 100 L/mol‬و )‪(y2 = 33,2 × 10–4 ، X2 = 200 L/mol‬‬
‫‪16,5 × 10−4‬‬
‫‪= 1,65 × 10−5‬‬
‫‪100‬‬

‫= ‪KA‬‬

‫اﻟﺘﻤﺮﻳﻦ ‪22‬‬
‫‪–1‬‬

‫‪V0 = 500 mL‬‬
‫‪V=1L‬‬

‫‪C0 = 0,2 mol/ L‬‬

‫اﻟﻤﺤﻠﻮل ‪S0‬‬

‫‪ C = 2 × 10–3 mol/ L‬اﻟﻤﺤﻠـﻮل ‪S‬‬

‫ﻣﻼﺣﻈﺔ ‪ :‬آﺎن ﻣﻦ اﻷﻓﻀﻞ ﺗﻮﻓﻴﺮ ﻣﺎﺻﺎت ﻋﻴﺎرﻳﺔ ‪50 mL ، 20 mL ، 10 mL ، 5 mL :‬‬
‫ﻋﺪد ﻣﻮﻻت ﺣﻤﺾ اﻟﺒﺮوﺑﺎﻧﻮﻳﻚ )‪ n0 (C2H5COOH‬ﻻ ﻳﺘﻐﻴﺮ ﻋﻨﺪﻣﺎ ﻧﻀﻴﻒ اﻟﻤـﺎء ‪ ،‬أي ’‪CV = C0V0‬‬
‫ﺣﻴﺚ ‪ V’0‬هﻮ اﻟﺤﺠﻢ اﻟﺬي ﻧﺄﺧﺬﻩ ﻣﻦ اﻟﻤﺤﻠﻮل ‪ S0‬وﻧﻀﻴﻒ ﻟﻪ اﻟﻤـﺎء ‪.‬‬

‫‪17‬‬

‫‪n0 0,1‬‬
‫=‬
‫ﻟﺪﻳﻨﺎ ‪= 0, 2 mol / L‬‬
‫‪V0 0,5‬‬

‫‪CV 2 × 10−3 × 1‬‬
‫=‬
‫‪= 10 mL‬‬
‫‪C0‬‬
‫‪0, 2‬‬

‫= ‪ ، C0‬وﺑﺎﻟﺘﺎﻟﻲ‬

‫= ' ‪V0‬‬

‫ﻣﻼﺣﻈﺔ ‪ :‬ﻧﻘﻮل ‪ :‬اﻗﺘﺮح ﺑﺮوﺗﻮآﻮﻻ ﺗﺠﺮﻳﺒﻴﺎ ‪ ،‬ﻻ ﻧﻘﻮل ‪ :‬اﻗﺘﺮح ﺑﺮوﺗﻮآﻮل ﺗﺠﺮﻳﺒﻲ ‪.‬‬
‫اﻟﻄﺮﻳﻘﺔ ‪ :‬ﻧﺄﺧﺬ ﺑﻮاﺳﻄﺔ اﻟﻤـﺎﺻﺔ اﻟﺘﻲ ﺳﻌﺘﻬﺎ ‪ 10 mL‬اﻟﺤﺠﻢ ’‪ V0‬ﻣﻦ اﻟﻤﺤﻠﻮل ‪ S0‬وﻧﻀﻌﻪ ﻓﻲ ﻣﺨﺒـﺎر ﺳﻌﺘﻪ ‪ 1 L‬ﺛﻢ ﻧﻜﻤﻞ اﻟﺤﺠﻢ ﺑﺎﻟﻤﺎء‬
‫اﻟﻤﻘﻄّﺮ ‪ ،‬وﻧﺤﺼﻞ ﺑﺬﻟﻚ ﻋﻠﻰ اﻟﻤﺤﻠﻮل ‪. S‬‬
‫‪–2‬‬
‫ﺟﺪول اﻟﺘﻘﺪّم ‪:‬‬

‫)‪H3O+(aq‬‬

‫‪-3‬‬

‫‪+‬‬

‫)‪C2H5COO–(aq‬‬

‫)‪H2O(l‬‬

‫=‬

‫‪+‬‬

‫)‪C2H5COOH(aq‬‬

‫‪0‬‬

‫‪0‬‬

‫زﻳﺎدة‬

‫‪2 × 10–3‬‬

‫‪x‬‬

‫‪x‬‬

‫زﻳﺎدة‬

‫‪2 × 10–3 – x‬‬

‫‪xéq‬‬

‫‪xéq‬‬

‫زﻳﺎدة‬

‫‪2 × 10–3 – xéq‬‬

‫⎦⎤ ‪ ، σ = λH O + ⎣⎡H 3O + ⎦⎤ + λC H COO − ⎣⎡C2 H 5COO − ⎦⎤ + λOH − ⎣⎡OH −‬وﺑﺎهﻤﺎل ]–‪ [OH‬ﻧﻜﺘﺐ ‪:‬‬
‫‪3‬‬
‫‪2 5‬‬
‫⎦⎤ ‪ ، σ = λH O + ⎡⎣H 3O + ⎤⎦ + λC H COO − ⎡⎣C2 H 5COO −‬وﺑﻤﺎ أن ]–‪ [H3O+] = [C2H5COO‬ﻧﻜﺘﺐ ‪:‬‬
‫‪5‬‬

‫)‬

‫‪−‬‬

‫‪3‬‬

‫‪2‬‬

‫(‬

‫‪σ = ⎡⎣H 3O + ⎦⎤ λH O + λC H COO‬‬
‫‪5‬‬

‫‪2‬‬

‫‪+‬‬

‫‪3‬‬

‫)‪(1‬‬

‫ﻣﻦ ﺟﺪول اﻟﺘﻘﺪّم ﻟﺪﻳﻨﺎ ‪ ، n (H3O+) = xéq‬وﺑﺎﻟﺘﺎﻟﻲ‬

‫)‬

‫ﺑﺎﻟﺘﻌﻮﻳﺾ ﻓﻲ اﻟﻌﻼﻗﺔ )‪+ λC H COO − : (1‬‬
‫‪5‬‬

‫‪2‬‬

‫‪H 3O +‬‬

‫‪(λ‬‬

‫‪xéq‬‬
‫‪V‬‬

‫‪xéq‬‬
‫‪V‬‬

‫= ] ‪[H 3O +‬‬

‫=‪σ‬‬

‫)‪(1‬‬

‫‪ - 4‬ﺗﺼﺤﻴﺢ ‪) σ = 6,2 × 10–3 S.m–1 :‬ﻟﻴﺲ ‪(σ = 6,2 × 10–5 S.m–1‬‬
‫أ( آﻴﻔﻴﺔ ﻗﻴـﺎس اﻟﻨﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ ‪:‬‬
‫ اﻟﻄﺮﻳﻘﺔ اﻷوﻟﻰ ‪ :‬ﻳﻮﺟﺪ ﺟﻬـﺎز ﻳﺴﻤﻰ ﻣﻘﻴﺎس اﻟﻨﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ ‪ ،‬ﻳﺘﺄﻟﻒ ﻣﻦ ﻣﺴﺒـﺎر ﻣﻮﺻﻮل ﻟﺠﻬﺎز ﻋﺮض رﻗﻤﻲ ‪.‬‬‫ﻟﻤﺎ ﻧﻐﻤﺮ اﻟﻤﺴﺒﺎر ﻓﻲ اﻟﻤﺤﻠﻮل اﻟﻤﺮاد ﻗﻴﺎس ﻧﺎﻗﻠﻴﺘﻪ اﻟﻨﻮﻋﻴﺔ ﻧﻘﺮأ ﻋﻠﻰ ﺷﺎﺷﺔ اﻟﺠﻬﺎز ﻗﻴﻤﺔ اﻟﻨﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ ﻣﻘﺪرة ﺑـ ‪. S.m–1‬‬
‫ اﻟﻄﺮﻳﻘﺔ اﻟﺜﺎﻧﻴﺔ ‪ :‬ﻧﺴﺘﻌﻤﻞ ﺧﻠﻴﺔ ﻗﻴـﺎس اﻟﻨﺎﻗﻠﻴﺔ ﻟﻘﻴﺎس ﻧﺎﻗﻠﻴﺔ اﻟﻤﺤﻠﻮل )‪ . (G‬ﻧﻀﺒﻂ ﺗﻮﺗﺮا آﻬﺮﺑﺎﺋﻴﺎ ﻣﺘﻨﺎوﺑﺎ ﺑﻴﻦ اﻟﺼﻔﻴﺤﺘﻴﻦ ﻗﻴﻤﺘﻪ‬‫اﻟﻤﻨﺘﺠﺔ ‪) Ueff‬ﻻ ﻧﺴﺘﻌﻤﻞ ﺗﻮﺗﺮا ﻣﺴﺘﻤﺮا ‪ ،‬ﻷن ﻣﺮور اﻟﺘﻴﺎر اﻟﻤﺴﺘﻤﺮ ﻳﻤﻜﻦ أن ﻳﺴﺒّﺐ ﺗﺤﻠﻴﻼ آﻬﺮﺑﺎﺋﻴﺎ ﻟﻠﻤﺤﻠﻮل ﻣﻤﺎ ﻳﺠﻌﻞ ﻗﻴﺎس ﻧﺎﻗﻠﻴﺘﻪ ﻏﻴﺮ‬
‫دﻗﻴﻖ ( ‪.‬‬
‫ﻧﻘﺮأ ﺷﺪة اﻟﺘﻴﺎر اﻟﻤﻨﺘﺠﺔ ﻋﻠﻰ ﻣﻘﻴﺎس اﻷﻣﺒﻴﺮ ‪ ،‬ﺛﻢ ﻧﺤﺴﺐ اﻟﻨﺎﻗﻠﻴﺔ‬

‫‪I‬‬
‫‪G‬‬
‫‪ ، G = eff‬وﻣﻦ اﻟﻌﻼﻗﺔ‬
‫‪K‬‬
‫‪U eff‬‬

‫= ‪ σ‬ﻧﺴﺘﻨﺘﺞ اﻟﺘﺎﻗﻠﻴﺔ اﻟﻨﻮﻋﻴﺔ ‪ ،‬ﻣﻊ اﻟﻌﻠﻢ‬

‫أن ‪ K‬هﻮ ﺛﺎﺑﺖ اﻟﺨﻠﻴﺔ وﻗﻴﻤﺘﻪ ﻣﺴﺠّﻠﺔ ﻋﻠﻰ اﻟﺠﻬـﺎز ‪.‬‬
‫‪–2‬‬

‫‪–3‬‬

‫ب( ﻳﺠﺐ أن ﻧﺴﺘﻌﻤﻞ ﻣﺤﺎﻟﻴﻞ ﻣﻤﺪّدة ) ﻣﻦ اﻷﻓﻀﻞ أن ﻳﻜﻮن ﺗﺮآﻴﺰهﺎ ﻣﺤﺼﻮرا ﺑﻴﻦ ‪ 10 mol/L‬و ‪ ، ( 10 mol/L‬ﻷن إذا آﺎن‬
‫اﻟﻤﺤﻠﻮل ﻣﺮآﺰا ﻓﺈن اﻟﻨﺎﻗﻠﻴﺔ اﻟﻤﻮﻟﻴﺔ اﻟﺸﺎردﻳﺔ )‪ (λ‬ﺗﺼﺒﺢ ﺗﺘﻌﻠﻖ ﺑﺘﺮآﻴﺰ اﻟﻤﺤﻠﻮل ‪ ،‬وﺑﺎﻟﺘﺎﻟﻲ ﺗﺼﺒﺢ اﻟﻌﻼﻗﺔ اﻟﺘﻲ ﻧﻄﺒﻘﻬﺎ ‪ σ = λ C‬ﻏﻴﺮ‬
‫دﻗﻴﻘﺔ ‪ ،‬أﻣﺎ إذا آﺎن اﻟﻤﺤﻠﻮل ﻣﻤﺪدا ﻓﺈن ‪ λ‬ﺗﻜﻮن ﻣﺴﺘﻘﻠﺔ ﻋﻦ اﻟﺘﺮآﻴﺰ اﻟﻤﻮﻟﻲ ﻟﻠﻤﺤﻠﻮل ‪.‬‬
‫ﻣﺜﻼ ‪ λH O+ = 35 × 10−3 S .m 2 .mol −1‬ﻣﻦ أﺟﻞ ﻣﺤﻠﻮل ﻣﻤﺪد ﻓﻲ اﻟﺪرﺟﺔ ‪ ، 25°C‬أﻣﺎ إذا آﺎن ﻣﺮآﺰا ﻓﺈن هﺬﻩ اﻟﻘﻴﻤﺔ ﻏﻴﺮ ﺛﺎﺑﺘﺔ ‪.‬‬
‫‪3‬‬

‫‪18‬‬

V = 1 L = 10–3 m3 ‫ ﻣﻊ اﻟﻌﻠﻢ أن‬، xéq =

λH O
3

+

Vσ
‫( ﻧﺴﺘﻨﺘﺞ اﻟﺘﻘﺪم ﻋﻨﺪ اﻟﺘﻮازن‬1) ‫ﺟـ( ﻣﻦ اﻟﻌﻼﻗﺔ‬
+ λC H COO−
2

5

xéq =

1×10−3 × 6, 2 × 10−3
= 1, 6 × 10−4 mol
38,58 × 10−3

[H 3O + ] = [C2 H 5COO − ] =

[OH − ] =

xf
V

=

1, 6 × 10−4
= 1, 6 × 10−4 mol / L
1

10−14
10−14
=
= 6, 25 ×10−11 mol / L
+
−4
[H 3O ] 1, 6 ×10

[¨C2 H 5COOH ] f = C − [H 3O + ] f = 2 × 10−3 − 1, 6 × 10−4 = 1,84 × 10−3 mol / L : ‫ – ﻋﻨﺪ ﺣﺎﻟﺔ اﻟﺘﻮازن ﻳﻜﻮن‬5

K=

[C2 H 5COO − ] f × [H 3O + ] f
[C2 H 5COOH ] f

19

=

(1, 6 ×10−4 )
1,84 ×10

2

−3

= 1, 4 × 10−5

: ‫ ﺛﺎﺑﺖ اﻟﺘﻮازن‬- 6



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