1. Final exam Exercises

2. Final Exam/ Exercises 1-Which atom of these following atoms will gain two electrons when forming ion. a-Mgb- Sc- Br d- Al 2-alkali metal from period 4 is: a- Na b- Ca c- K d- Li 3-The number of kilograms in one teragram are : a-1 x 10 15 Kg b- 1 x10 9 Kgc- 1 x 10 -9 Kg d- 1 x 10 -15 Kg First comvert from Tg to g : 1X 10 12 m Then from g to Kg 1x 10 12 x 10 -3 = 1 x 10 9

3. Final Exam/ Exercises 1-Which atom of these following atoms will gain two electrons when forming ion. a-Mgb- Sc- Br d- Al 2-alkali metal from period 4 is: a- Na b- Ca c- K d- Li 3-The number of kilograms in one teragram are : a-1 x 10 15 Kg b- 1 x10 9 Kgc- 1 x 10 -9 Kg d- 1 x 10 -15 Kg

4. Final Exam/ Exercises 4- If one mole of calcium hydroxide Ca(OH) 2 is dissolved in enough water, the solution will contain : a- 2 mol of Ca ++ and 2 mol of OH - b- 2 mol of Ca ++ and 1 mol of OH - c- 1 mol of Ca ++ and 2 mol of OH - d- 1 mol of Ca ++ and 1 mol of OH - 5- The mass of one atom of an element is 3.82 x 10 -23 g, what is the molar mass of this element? a- 23 g/mol b- 19 g/mol c- 20 g/mol d- 40 g/mol

5. Final Exam/ Exercises 4 - If one mole of calcium hydroxide Ca(OH) 2 is dissolved in enough water, the solution will contain: a-2 mol of Ca ++ and 2 mol of OH - b- 2 mol of Ca ++ and 1 mol of OH - c- 1 mol of Ca ++ and 2 mol of OH - d- 1 mol of Ca ++ and 1 mol of OH - 5- The mass of one atom of an element is 3.82x10 -23 g, what is the molar mass of this element? a- 23 g/mol b- 19 g/mol c- 20 g/mol d-40 g/mol 6-What volume of a 0.100 M HCl solution is needed to neutralize 25.0 ml of 0.350 M NaOH? a-87.5 ml b- 87.5 L c- 87.5 x 10 -2 ml d- 87.5 x 10 2 L M 1 V 1 = M 2 V 2 0.1 x V 1 =0.35 x 25 V 1 =87.5 mL

6. Final Exam/ Exercises 4 - If one mole of calcium hydroxide Ca(OH) 2 is dissolved in enough water, the solution will contain: a-2 mol of Ca ++ and 2 mol of OH - b- 2 mol of Ca ++ and 1 mol of OH - c- 1 mol of Ca ++ and 2 mol of OH - d- 1 mol of Ca ++ and 1 mol of OH - 5- The mass of one atom of an element is 3.82x10 -23 g, what is the molar mass of this element? a- 23 b- 19 c- 20 d-40 6-What volume of a 0.100 M HCl solution is needed to neutralize 25.0 ml of 0.350 M NaOH? a-87.5 ml b- 87.5 L c- 87.5 x 10 -2 ml d- 87.5 x 10 2 L

7. Final Exam/ Exercises 7- The chemical formula of calcium phosphate is: a- CaPO 4 b-Ca 3 PO 4 c- Ca 3 (PO 4 ) 2 d-Ca(PO 4 ) 2 8- The partial pressure of oxygen was observed to be 130 torr in air with atmospheric pressure of 743 torr. Calculate the mole fraction of O 2 present. a-0.210 b- 0.175 c- 4.76d- 5.72

8. Final Exam/ Exercises 7- The chemical formula of calcium phosphate is: a- CaPO 4 b-Ca 3 PO 4 c- Ca 3 (PO 4 ) 2 d-Ca(PO 4 ) 2 8- The partial pressure of oxygen was observed to be 130 torr in air with atmospheric pressure of 743 torr. Calculate the mole fraction of O 2 present. a-0.210 b- 0.175 c- 4.76d- 5.72 9- In a process 500 g of NaCl are treated with an excess of H 2 SO 4 and yield 500 g of Na 2 SO 4. Calculate the percent yield of Na 2 SO 4. 2NaCl + H 2 SO 4  Na 2 SO 4 + 2HCl a- 77.51% b- 90.7% c- 82.39% d- 85.76%

9. Final Exam/ Exercises 9- In a process 500 g of NaCl are treated with an excess of H 2 SO 4 and yield 500 g of Na 2 SO 4. Calculate the percent yield of Na 2 SO 4. 2NaCl + H 2 SO 4  Na 2 SO 4 + 2HCl a- 77.51% b- 90.7% c- 82.39% d- 85.76% Normally we should determine the limiting reagent but in this question it already mention that H 2 SO 4 is excess therefore the limiting regent is NaCl Calculate mole of NaCl Mole= 500/58.5 (molar mass of NaCl)=8.55 From equation 2 mole NaCl ======1 mole Na 2 SO 4 8.55 mole ==========? Mole of Na 2 SO 4 Mole of Na 2 SO 4 = 8.55 /2 = 4.27 mole Then change mole of product to g = 4.275 x 142 (molar mass of Na 2 SO 4 )= 606.84 g X Yield = (500/606.84) x100 = 82.39 %

10. Final Exam/ Exercises 7- The chemical formula of calcium phosphate is: a- CaPO 4 b-Ca 3 PO 4 c- Ca 3 (PO 4 ) 2 d-Ca(PO 4 ) 2 8- The partial pressure of oxygen was observed to be 130 torr in air with atmospheric pressure of 743 torr. Calculate the mole fraction of O 2 present. a-0.210 b- 0.175 c- 4.76d- 5.72 9- In a process 500 g of NaCl are treated with an excess of H 2 SO 4 and yield 500 g of Na 2 SO 4. Calculate the percent yield of Na 2 SO 4. 2NaCl + H 2 SO 4  Na 2 SO 4 + 2HCl a- 77.51% b- 90.7% c- 82.39% d- 85.76%

11. Final Exam/ Exercises 10- The appropriate symbol for an element with Z = 45 and A = 103 is : a-Rhb- Scc- Lrd- Rn 11- What is the molecular weight of CuSO 4 ·5H 2 O? a-249.5 gmol -1 b- 177.5 gmol -1 c- 127 gmol -1 d- 87 gmol -1 12- How many carbon (C) atoms are present in 2 g of nitroglycerin C 3 H 5 N 3 O 9 ? a-1.59 x 10 22 atom b- 2.65 x 10 22 atom c- 4.78 x 10 22 atomd- none

12. Final Exam/ Exercises 12- How many carbon (C) atoms are present in 2 g of nitroglycerin C 3 H 5 N 3 O 9 ? a-1.59 x 10 22 atom b- 2.65 x 10 22 atom c- 4.78 x 10 22 atomd- none First we calculate the number of mole n = 2 / 227 = 0.0088 mole Number of molecules = Avogadro's number x number of mole = 6.022 x 10 23 x 0.0088 = 5.299 x10 21 molecules 1 molecules of nitroglycerin = 3 atom of C 5.299 x10 21 molecules = ? Atom oh C 3 x 5.299 x10 21 = 1.59 x10 22 atoms

13. Final Exam/ Exercises 10- The appropriate symbol for an element with Z = 45 and A = 103 is : a-Rhb- Scc- Lrd- Rn 11- What is the molecular weight of CuSO 4 ·5H 2 O? a-249.5 gmol -1 b- 177.5 gmol -1 c- 127 gmol -1 d- 87 gmol -1 12- How many carbon (C) atoms are present in 2 g of nitroglycerin C 3 H 5 N 3 O 9 ? a-1.59 x 10 22 atom b- 2.65 x 10 22 atom c- 4.78 x 10 22 atomd- none 13- The density of benzene (C 6 H 6 ) is 0.8765 g/mL. How many benzene molecules are present in 4.50 mL? a-2.85 x 10 23 molecule b- 3.04 x 10 22 molecule c-1.51 x 10 22 molecule d- 2.18 x 10 23 molecule

14. Final Exam/ Exercises 13- The density of benzene (C 6 H 6 ) is 0.8765 g/mL. How many benzene molecules are present in 4.50 mL? a-2.85 x 10 23 molecule b- 3.04 x 10 22 molecule c-1.51 x 10 22 molecule d- 2.18 x 10 23 molecule d=g/v Mass = 0.8765 x 4.50 = 3.94425 g Mole=mass/molar mass =3.94425 /78=0.05057 mol Number of particle = Avogadro's number x number of moles. = 6.022 x10 23 x 0.05057 =3.04 x 10 22 molecule

15. Final Exam/ Exercises 10- The appropriate symbol for an element with Z = 45 and A = 103 is : a-Rhb- Scc- Lrd- Rn 11- What is the molecular weight of CuSO 4 ·5H 2 O? a-249.5 gmol -1 b- 177.5 gmol -1 c- 127 gmol -1 d- 87 gmol -1 12- How many carbon (C) atoms are present in 2 g of nitroglycerin C 3 H 5 N 3 O 9 ? a-1.59 x 10 22 atom b- 2.65 x 10 22 atom c- 4.78 x 10 22 atomd- none 13- The density of benzene (C 6 H 6 ) is 0.8765 g/mL. How many benzene molecules are present in 4.50 mL? a-2.85 x 10 23 molecule b- 3.04 x 10 22 molecule c-1.51 x 10 22 molecule d- 2.18 x 10 23 molecule

16. Final Exam/ Exercises 14- The empirical formula for a compound that gives the following analysis 75.95% C, 6.33% H, 17.72% N, is : a-C 7 H 6 O b- C 6 H 6 O c- C 5 H 5 N d- C 6 H 7 N 1- change from % to g 75.95 g of C, 6.33 g of H, 17.72 g of N 2- change from g to mole using Divided by the smallest number of mole which is 1.27 75.95 12 = 6.33 mol of C n c = 6.33 1 = 6.33 mol of H n H = 6.33 1.27 = 5 C: 6.33 1.27 = 5 H: 17.72 14 = 1.27 mol of N n N = 1.27 = 1 N: Thus the empirical formula is C 5 H 5 N

17. Final Exam/ Exercises 14- The empirical formula for a compound that gives the following analysis 75.95% C, 6.33% H, 17.72% N, is : a-C 7 H 6 O b- C 6 H 6 O c- C 5 H 5 N d- C 6 H 7 N 15- A 7.691 g sample of MgCl 2 is dissolved in enough water to give 750. mL of solution. What is the chloride ion concentration in this solution? a- 0.216 Mb- 0.188 M c- 0.160 M d- 0.132 M Calculate mole of MgCl 2 Mole= 7.691/95 (molecular mass of MgCl 2 )=0.081 mole Molarity = mole/V (in L) = 0.081 /(750/1000) =0.108M From equation 1 M MgCl 2 ======== 2 M Cl - 0.108 M ========? M [Cl] = 0.108 x 2 = 0.216M

18. Final Exam/ Exercises 14- The empirical formula for a compound that gives the following analysis 75.95% C, 6.33% H, 17.72% N, is : a-C 7 H 6 O b- C 6 H 6 O c- C 5 H 5 N d- C 6 H 7 N 15- A 7.691 g sample of MgCl 2 is dissolved in enough water to give 750. mL of solution. What is the chloride ion concentration in this solution? a- 0.216 Mb- 0.188 M c- 0.160 M d- 0.132 M 16- You have 500 mL of a 0.600 M HCl solution and you want to dilute it to exactly 0.400 M. How much water should you add? a- 250 mLb- 500 mLc- 1000 mLd- 2500 mL M 1 V 1 = M 2 V 2 0.6 X 500 = 0.4 X V 2 V = 750 ml Add water = 750 – 500 =250 mL

19. Final Exam/ Exercises 14- The empirical formula for a compound that gives the following analysis 75.95% C, 6.33% H, 17.72% N, is : a-C 7 H 6 O b- C 6 H 6 O c- C 5 H 5 N d- C 6 H 7 N 15- A 7.691 g sample of MgCl 2 is dissolved in enough water to give 750. mL of solution. What is the chloride ion concentration in this solution? a- 0.216 Mb- 0.188 M c- 0.160 M d- 0.132 M 16- You have 500 mL of a 0.600 M HCl solution and you want to dilute it to exactly 0.400 M. How much water should you add? a- 250 mLb- 500 mLc- 1000 mLd- 2500 mL

20. Final Exam/ Exercises 17- What is the density of NH 3 gas at 100  C and 0.4 atm? a- 0.111 g/L b- 0.222 g/L c- 0.333 g/Ld- 0.445 g/L

21. Final Exam/ Exercises 17- What is the density of NH 3 gas at 100  C and 0.4 atm? a- 0.111 g/L b- 0.222 g/L c- 0.333 g/Ld- 0.445 g/L 18- What is the electronic configuration of Cr? a- [Ar]4s 1 3d 5 b- [Ar]4s 2 3d 4 c- [Kr]5s 1 4d 5 d- [Kr]5s 2 4d 4 19-The correct order of radius in the following is: a- O 2- Cl c- Fe 2+ > Fe d- Fe 2+ < Fe 3+ 20-Which of the following is the correct form of the equilibrium constant expression for the reaction below; Zn(OH) 2(s) Zn 2+ (aq) + 2 OH - (aq) a-K = [Zn 2+ ] [2OH - ] 2 b-K = [Zn 2+ ] [2OH - ] 2 / [Zn(OH) 2 ] c-K = [Zn 2+ ] [OH - ] 2 d-K = [Zn 2+ ] [OH - ] 2 / [Zn(OH) 2 ]

22. Final Exam/ Exercises 21- At equilibrium, the total gas pressure was found to be 0.033 atm. Calculate the equilibrium constant for the following decomposition; NH 4 CO 2 NH 2 (s) 2 NH 3(g) + CO 2(g) a-1.33 x 10 -6 b- 5.3 x 10 -6 c- 3.59 x 10 -5 d- 1.44 x 10 -4 P T = P NH3 + P CO2 0.033 = 2P + P 0.033 = 3P P= 0.011 THEN P NH3 = 2 X 0.011 = 0.022 P CO2 = 0.011 K p =P 2 NH3 X P CO2 = (0.022) 2 X(0.011) =5.3 X 10 -6  

23. Final Exam/ Exercises 21- At equilibrium, the total gas pressure was found to be 0.033 atm. Calculate the equilibrium constant for the following decomposition; NH 4 CO 2 NH 2 (s) 2 NH 3(g) + CO 2(g) a-1.33 x 10 -6 b- 5.3 x 10 -6 c- 3.59 x 10 -5 d- 1.44 x 10 -4 22-The equilibrium constant for the following reaction at 700 K is; H 2(g) + I 2(g) 2 HI (g) K 1 = 10.17 What is the value of the equilibrium constant for the following reaction; HI (g) 1/2 H 2(g) + 1/2 I 2(g) K 2 = ? a-9.668 x 10 -3 b- 3.189 c- 0.314 d- 0.098      

24. Final Exam/ Exercises 21- At equilibrium, the total gas pressure was found to be 0.033 atm. Calculate the equilibrium constant for the following decomposition; NH 4 CO 2 NH 2 (s) 2 NH 3(g) + CO 2(g) a-1.33 x 10 -6 b- 5.3 x 10 -6 c- 3.59 x 10 -5 d- 1.44 x 10 -4 22-The equilibrium constant for the following reaction at 700 K is; H 2(g) + I 2(g) 2 HI (g) K 1 = 10.17 What is the value of the equilibrium constant for the following reaction; HI (g) 1/2 H 2(g) + 1/2 I 2(g) K 2 = ? a-9.668 x 10 -3 b- 3.189 c- 0.314 d- 0.098 23- When 0.01 mol of N 2 O 4(g) were placed in a 2.0 L flask at 200 °C, it was found that at equilibrium [N 2 O 4 ] = 0.0042 M. What is the value of K c for the following reaction; N 2 O 4(g) 2 NO 2(g) a-0.032 b- 1.5 x 10 -4 c- 6.1 x 10 -4 d- 5.12 x 10 -4        

25. Final Exam/ Exercises 23- When 0.01 mol of N 2 O 4(g) were placed in a 2.0 L flask at 200 °C, it was found that at equilibrium [N 2 O 4 ] = 0.0042 M. What is the value of K c for the following reaction; N 2 O 4(g) 2 NO 2(g) a-0.032 b- 1.5 x 10 -4 c- 6.1 x 10 -4 d- 5.12 x 10 -4 Calculate the initial concentration of N 2 O 4 0.01 / 2 = 0.005M Converted N 2 O 4 [N 2 O 4 ] = initial N 2 O 4 – equilibrium = 0.005-0.0042=0.0008M From equation : 1mole N 2 O 4 --------  2 mol NO 2 0.0008 mol N 2 O 4 ---  ? mol NO 2 [NO 2 ] = 0.0008 x 2 =0.0016 M K c = [NO 2 ] 2 /[N 2 O 4 ] = (0.0016) 2 / 0.0042 = 6.1 x 10 -4  

26. Final Exam/ Exercises 21- At equilibrium, the total gas pressure was found to be 0.033 atm. Calculate the equilibrium constant for the following decomposition; NH 4 CO 2 NH 2 (s) 2 NH 3(g) + CO 2(g) a-1.33 x 10 -6 b- 5.3 x 10 -6 c- 3.59 x 10 -5 d- 1.44 x 10 -4 22-The equilibrium constant for the following reaction at 700 K is; H 2(g) + I 2(g) 2 HI (g) K 1 = 10.17 What is the value of the equilibrium constant for the following reaction; HI (g) 1/2 H 2(g) + 1/2 I 2(g) K 2 = ? a-9.668 x 10 -3 b- 3.189 c- 0.314 d- 0.098 23- When 0.01 mol of N 2 O 4(g) were placed in a 2.0 L flask at 200 °C, it was found that at equilibrium [N 2 O 4 ] = 0.0042 M. What is the value of K c for the following reaction; N 2 O 4(g) 2 NO 2(g) a-0.032 b- 1.5 x 10 -4 c- 6.1 x 10 -4 d- 5.12 x 10 -4        

27. Final Exam/ Exercises 24- For the following equilibrium at 25 °C; Cl 2(g) + 3 F 2(g) 2 ClF 3(g) If [Cl 2 ] = 0.15 M, [F 2 ] = 0.28 M and [ClF 3 ] = 0.01 M at equilibrium, calculate the equilibrium constant K p for the reaction at the same temperature? a-5.1 x 10 -5 b- 0.03 c-18.13 d- 7.2 x 10 -3  

28. Final Exam/ Exercises 24- For the following equilibrium at 25 °C; Cl 2(g) + 3 F 2(g) 2 ClF 3(g) If [Cl 2 ] = 0.15 M, [F 2 ] = 0.28 M and [ClF 3 ] = 0.01 M at equilibrium, calculate the equilibrium constant K p for the reaction at the same temperature? a-5.1 x 10 -5 b- 0.03 c-18.13 d- 7.2 x 10 -3 25- The compound CH 3 OCH 3 is classified as: a-Alcohol b- Ketone c- An esterd- An ether 26-From the following, which compound has the double bond? a-C 8 H 18 b- C 8 H 16 c- C 8 H 14 d- C 9 H 20  

29. Final Exam/ Exercises 27- How many hydrogen atoms in this following organic compound? a-10 b- 14c- 16d- 18 28-How many single (σ) bonds in the following molecule? a-24 b- 10 c- 25d- 22

30. Final Exam/ Exercises 29-From question (28), What is the hybridization for carbon atom number 4 (C4)? a-sp b- sp 2 c- sp 3 d- sp 3 d 2 30- The IUPAC name of the following compound is: a-4,6-methyl-3-hepteneb- 4,6-dimethyl-4-heptene c- 4,6-methyl-3-heptened- 4,6-dimethyl-3-heptene 31- The IUPAC name of the following compound is: a-1,2,4-trichlorobenzeneb- 1,3,4-trichlorobenzene c- 1,2,3-trichlorobenzened- 1,4,5-trichlorobenzene

31. Final Exam/ Exercises 32- What is the concentration of H + in a 2.5 M HCl solution? a- 0.75 M b- 1.0 M c- 2.5 M d- 5.0 M 33- What is the solubility of HgCl 2 (Ksp = 1.8 x 10 -18 )? a-7.66x10 -7 b- 1.89x10 -3 M c- 7.8x10 -5 M d- 1.3 x 1 K sp = [Hg +2 ][Cl - ] 2 K sp = (s)(2s) 2 K sp = 4s 3 4s 3 = 1.8 x 10 -18 s 3 = 4.5 x 10 -19 s= 7.66 x 10 -7 M [Hg +2 ]= s [Cl - ]= 2s

32. Final Exam/ Exercises 32- What is the concentration of H + in a 2.5 M HCl solution? a- 0.75 M b- 1.0 M c- 2.5 M d- 5.0 M 33- What is the solubility of HgCl 2 (Ksp = 1.8 x 10 -18 )? a-7.66x10 -7 b- 1.89x10 -3 M c- 7.8x10 -5 M d- 1.3 x 10 -3 34- Calculate the pH of 0.1 M KOH solution. a- 0.1 b- 1.5 c- 13.3 d- 13.0

33. Final Exam/ Exercises 32- What is the concentration of H + in a 2.5 M HCl solution? a- 0.75 M b- 1.0 M c- 2.5 M d- 5.0 M 33- What is the solubility of HgCl 2 (Ksp = 1.8 x 10 -18 )? a-7.66x10 -7 b- 1.89x10 -3 M c- 7.8x10 -5 M d- 1.3 x 10 -3 34- Calculate the pH of 0.1 M KOH solution. a- 0.1 b- 1.5 c- 13.3 d- 13.0 35- Calculate the H + ion concentration in lemon juice having a pH 2.0. a- 0.01 M b- 2.5x10 -4 M c- 10.0 M d- 3.0 M

34. Final Exam/ Exercises 32- What is the concentration of H + in a 2.5 M HCl solution? a- 0.75 M b- 1.0 M c- 2.5 M d- 5.0 M 33- What is the solubility of HgCl 2 (Ksp = 1.8 x 10 -18 )? a-7.66x10 -7 b- 1.89x10 -3 M c- 7.8x10 -5 M d- 1.3 x 10 -3 34- Calculate the pH of 0.1 M KOH solution. a- 0.1 b- 1.5 c- 13.3 d- 13.0 35- Calculate the H + ion concentration in lemon juice having a pH 2.0. a- 0.01 M b- 2.5x10 -4 M c- 10.0 M d- 3.0 M

35. Final Exam/ Exercises 36- What is the pH of a 0.01 M HA (weak acid) solution that is 10 % ionized? a- 1.5 b- 2.5 c- 3.0 d- 4.5

36. Final Exam/ Exercises 36- What is the pH of a 0.01 M HA (weak acid) solution that is 10 % ionized? a- 1.5 b- 2.5 c- 3.0 d- 4.5 37- Find the pH of a 0.1 M aqueous solution of hypobromous acid (HOBr), for which K a = 2.06  10 -9. a- 4.84 b- 2.55 c- 3.0 d- 4.5 pH = -log[H + ] = 4.84

37. Final Exam/ Exercises 36- What is the pH of a 0.01 M HA (weak acid) solution that is 10 % ionized? a- 1.5 b- 2.5 c- 3.0 d- 4.5 37- Find the pH of a 0.1 M aqueous solution of hypobromous acid (HOBr), for which K a = 2.06  10 -9. a- 4.84 b- 2.55 c- 3.0 d- 4.5 38- Calculate the pH of a buffer solution containing 0.1 M CH 3 COOH (Ka = 1.8x10 -5 ) and 0.2 M CH 3 COONa. a-4.96 b- 5.04 c- 4.75 d- 4.35

38. Final Exam/ Exercises 38- Calculate the pH of a buffer solution containing 0.1 M CH 3 COOH (Ka = 1.8x10 -5 ) and 0.2 M CH 3 COONa. a-4.96 b- 5.04 c- 4.75 d- 4.35 pH = pK a + log [A - ] [HA] pH = -log 1.8 x 10 -5 +log 0.2 0.1 pH = 5.04

39. Final Exam/ Exercises 36- What is the pH of a 0.01 M HA (weak acid) solution that is 10 % ionized? a- 1.5 b- 2.5 c- 3.0 d- 4.5 37- Find the pH of a 0.1 M aqueous solution of hypobromous acid (HOBr), for which K a = 2.06  10 -9. a- 4.84 b- 2.55 c- 3.0 d- 4.5 38- Calculate the pH of a buffer solution containing 0.1 M CH 3 COOH (K a = 1.8x10 -5 ) and 0.2 M CH 3 COONa. a-4.96 b- 5.04 c- 4.75 d- 4.35 39-From question (38), By adding 0.01 M of NaOH to the above buffer solution, the resultant pH is: a-5.11 b- 5.18 c- 4.92 d- 5.04

40. Final Exam/ Exercises 39-From question (38), By adding 0.01 M of NaOH to the above buffer solution, the resultant pH is: a-5.11 b- 5.18 c- 4.92 d- 5.04 This mean that the buffer will shift toward product to resist the addition of base. That mean the acid concentration will be 0.1-0.01=0.09 That mean the base concentration will be 0.2+0.01=0.21 pH = 5.11

41. Final Exam/ Exercises 36- What is the pH of a 0.01 M HA (weak acid) solution that is 10 % ionized? a- 1.5 b- 2.5 c- 3.0 d- 4.5 37- Find the pH of a 0.1 M aqueous solution of hypobromous acid (HOBr), for which K a = 2.06  10 -9. a- 4.84 b- 2.55 c- 3.0 d- 4.5 38- Calculate the pH of a buffer solution containing 0.1 M CH 3 COOH (Ka = 1.8x10 -5 ) and 0.2 M CH 3 COONa. a-4.96 b- 5.04 c- 4.75 d- 4.35 39-From question (38), By adding 0.01 M of NaOH to the above buffer solution, the resultant pH is: a-5.11 b- 5.18 c- 4.92 d- 5.04

42. Final Exam/ Exercises 40- What is the smallest solubility of these salts. a-BaCO 3 (Ksp = 5.1x10 -9 ) b- AgI (Ksp = 8.5x10 -17 ) c- PbI 2 (Ksp = 8.8x10 -9 ) d- HgCl 2 (Ksp = 1.8 x 10 -18 )