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pOH = -log [OH - ]

pOH = -log [OH - ]. [H + ][OH - ] = K w = 1.0 x 10 -14. -log [H + ] – log [OH - ] = 14.00. pH + pOH = 14.00. The OH - ion concentration of a blood sample is 2.5 x 10 -7 M . What is the pH of the blood?.

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pOH = -log [OH - ]

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  1. pOH = -log [OH-] [H+][OH-] = Kw = 1.0 x 10-14 -log [H+] – log [OH-] = 14.00 pH + pOH = 14.00

  2. The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood? The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater? pH = -log [H+] = 10-4.82 = 1.5 x 10-5M [H+] = 10-pH pH + pOH = 14.00 pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

  3. تدريب : • احسب قيمة pH و pOH لمحلول يحتوي على تركيز[H+] يساوي 2.0 x 10-3مولار. • احسب [H+] و [HO-] في محلول قيمة pH تساوي 3.70 • احسب pH و pOH لمحلول حجمه 400 مل ويحتوي على 0.31 جم من هيدروكسيد الباريوم Ba(OH)2 • للتأكد توجد الإجابة في كتاب مسائل وحلول بالباب الثاني الفصل7 .

  4. H2O Strong Acids are strong electrolytes NaCl (s)Na+ (aq) + Cl- (aq) HCl (aq) + H2O (l) H3O+(aq) + Cl-(aq) HNO3(aq) + H2O (l) H3O+(aq) + NO3-(aq) CH3COOHCH3COO- (aq) + H+ (aq) HClO4(aq) + H2O (l) H3O+(aq) + ClO4-(aq) H2SO4(aq) + H2O (l) H3O+(aq) + HSO4-(aq) Strong Electrolyte – 100% dissociation Weak Electrolyte – not completely dissociated

  5. Weak Acids are weak electrolytes Weak Bases are weak electrolytes HF (aq) + H2O (l) H3O+(aq) + F-(aq) F-(aq) + H2O (l) OH-(aq) + HF (aq) HNO2(aq) + H2O (l) H3O+(aq) + NO2-(aq) NO2-(aq) + H2O (l) OH-(aq) + HNO2(aq) HSO4-(aq) + H2O (l) H3O+(aq) + SO42-(aq) H2O (l) + H2O (l) H3O+(aq) + OH-(aq) Strong Bases are strong electrolytes H2O NaOH (s) Na+(aq) + OH-(aq) H2O KOH (s) K+(aq) + OH-(aq) H2O Ba(OH)2(s) Ba2+(aq) + 2OH-(aq)

  6. HA (aq) + H2O (l) H3O+(aq) + A-(aq) HA (aq) H+(aq) + A-(aq) [H+][A-] Ka = [HA] Weak Acids (HA) and Acid Ionization Constants Ka is the acid ionization constant

  7. HF (aq) H+(aq) + F-(aq) = 7.1 x 10-4 = 7.1 x 10-4 = 7.1 x 10-4 [H+][F-] x2 x2 Ka Ka = Ka = 0.50 - x 0.50 [HF] HF (aq) H+(aq) + F-(aq) What is the pH of a 0.5M HFsolution (at 250C)? Initial (M) 0.50 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.50 - x x x Ka << 1 0.50 – x 0.50 x2 = 3.55 x 10-4 x = 0.019 M pH = -log [H+] = 1.72 [H+] = [F-] = 0.019 M [HF] = 0.50 – x = 0.48 M

  8. What is the pH of a 0.122M monoprotic acid whose Ka is 5.7 x 10-4? Initial (M) 0.122 0.00 0.00 = 5.7 x 10-4 = 5.7 x 10-4 Change (M) -x +x +x x2 x2 Ka Ka = Equilibrium (M) 0.122 - x x x 0.122 - x 0.122 HA (aq) H+(aq) + A-(aq) Ka << 1 0.122 – x 0.122 x2 = 6.95 x 10-5 x = 0.0083 M pH = -log[H+] = -log[0.0083] = 2.08

  9. x2 Ka = 0.122 - x + = H [ ] K . C a a x2 + 0.00057x – 6.95 x 10-5 = 0 = 5.7 x 10-4  -b ± b2 – 4ac x = ax2 + bx + c =0 2a x = 0.0081 x = - 0.0081 HA (aq) H+(aq) + A-(aq) Initial (M) 0.122 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.122 - x x x pH = -log[H+] = 2.09 [H+] = x = 0.0081 M حل آخر : يتم تطبيق المعادلة التالية مباشرة وهي = 5.7 x 10-4 x 0.122 = 0.0083 M

  10. (1) Find the pH of a 0.135 M aqueous solution of periodic acid (HIO4), for which Ka = 2.3  10-2. A. 1.25 B. 3.28 C. 1.17 D. 1.34 E. 1.64 (2) Find the pH of a 0.183 M aqueous solution of hypobromous acid (HOBr), for which Ka = 2.06  10-9.  A. 4.71 B. 8.69 C. 3.97 D. 4.34 E. 9.28 = 2.3 x 10-2 x 0.135 = 0.0557M pH = - log [0.0557] =1.25 = 2.06 x 10-9 x 0.183 = 1.94 x 10-5 M pH = - log [1.94 x 10-5] =4.71 + + = = H H [ [ ] ] K K . . C C a a a a

  11. NH3(aq) + H2O (l) NH4+(aq) + OH-(aq) [NH4+][OH-] Kb = [NH3] Solve weak base problems like weak acids except solve for [OH-] instead of [H+]. Weak Bases and Base Ionization Constants Kb is the base ionization constant

  12. مثال : احسب التركيز بالمولارية لكل من أيونات (OH)- و Ba2+ في محلول حجمه 200 مل ويحتوي على 0.612 جم من هيدروكسيد الباريوم Ba(OH)2 . الحـل : يتم أولا حساب عدد مولات هيدروكسيد الباريوم بدلالة كتلته 0.612 جم ووزنه الجزيئي (Mol. wt.) = 137.34 + (16 + 1 ) 2 = 171.34 جم/مول إذاً عدد مولات هيدروكسيد الباريوم = مول والتركيز بالمولارية M = ثانياً : بعد معرفة التركيز الحجمي بالمولارية . يتم كتابة معادلة تأين الإلكتروليت Ba(OH)2 الموزونة وذلك لحساب تركيز كل من أيونات (OH)- و Ba2+ في البداية: صفر مولار صفر مولار 0.0179 مولار بعد الذوبانية: 0.0179 + 2 x0.0179 + - 0.0179 مولار قاعدة قوية Ba(OH)2(S) Ba2+(aq) + 2 HO-1(aq) النتيجة: 0.0358 مولار 0.179 مولار صفر مولار تركيز أيونات الهيدروكسيد 2(HO)-1= 2 × 0.0179 = 0.0358 مولار . وتركيز أيونات الباريوم Ba2+ = 0.179 مولاروالآن أحسب قيمة الـ pH للمحلول. الإجابة : pH = 12.55

  13. What is the pH of a 0.40 M Ammonia solution? (if the Kb = 1.8 x 10-5) (a) 11.43 (b) 12.43 (c) 2.57 (d) 3.57 NH3 + H2O NH4+ + HO- 0.4M 1M 0 0 0.4 – x 1M x x Kb = [x][x] ÷ [0.4-x] = 1.8 x 10-5 X2 = 1.8 x 10-5 X 0.4 = 7.2 X 10-6 X = 7.2 X 10-6 = 2.68 X 10-3 pOH = - log 2.68 x 10-3 = 2.57 pH = 14-2.57 = 11.43 حل آخر : يتم تطبيق المعادلة التالية مباشرة وهي = 1.8 x 10-5 X 0.4 = 2.7 x 10-3 pH = -log 2.7 x 10-3 = 2.57 pH = 14-2.57 = 11.43 تهمل قيمة الـ x لان قيمة Kb اصغر من 1

  14. مثال : محلول 0.10 مولار من الأمونيا ثابت تأينه Kb = 1.8 x 10-5. احسب قيم pH و pOH للمحلول . يتم التطبيق المباشر على حيث المعطاة بالمسألة هي : ثابت الاتزان للقاعدة الضعيفة الأمونيا Kb = 1.8 x 10-10 وتركيزها Cb = 0.10 مولار - - = ´ ´ = ´ 10 -3 [ HO ] 1 . 8 10 0 . 1 1 . 342 10 M pOH = - log 1.342 x 10-3 = 2.87 pH = 14 – 2.87 = 11.13

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