1. $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 Back to the Basics Solution Stoichiometry Qualitative Analysis/ Ionic Equation Solubility Curves Acids and Bases/ Titration

2. How many grams of bromine, Br 2, are needed to prepare 0.500 L of a 0.0100M solution in water?

3. Answer: 0.8 grams 1.Find the number of moles of Bromine gas: n=c*v n=0.500L*0.0100M n=0.005 moles of Br 2 2.Covert moles into grams using the following triangle: 3.m=n*M m=0.005*160 m=0.8 grams n M m Note: M Br 2 =2(80) =160 g/mol

4. 1.What is the difference between saturated, unsaturated and supersaturated solution? 2.What is a solute and solvent? 3.Distinguish the difference between concentration and dilution.

5. 1.Saturated solution: A solution that contains the maximum amount of solute that can be dissolved within a given volume. Unsaturated solution: A solution that doesn’t contain the maximum amount of solute, thus requiring more, in order to reach its saturation point. Supersaturated solution: A solution containing more than the maximum amount of solute that can be sustained in a given volume. 2.Solute: The substance that one adds into a solution Solvent: A liquid substance that dissolves the solute 3.Concentration: A solution that contains a lot of solute Dillution: Reducing the amount of solute in a solution that is concentrated. Solution:

6. Classify the following as Polar or Non Polar and identify the type of bond holding it together. 1.C 2 H 2 2.C 2 H 4 3.CH 4 4.NH 3 5.H 2 O 6.HBr 7.HF 8.CCl 3

7. 1.C 2 H 2  Non Polar, London Dispersion 2.C 2 H 4  Non Polar, London Dispersion 3.CH 4  Non Polar, London Dispersion 4.NH 3  Polar, Dipole Dipole 5.H 2 O  Polar, Hydrogen Bonding 6.HBr  Polar, Dipole Dipole 7.HF  Polar, Hydrogen Bonding 8.CCl 3  Polar, Dipole Dipole Solution:

8. A saturated solution of calcium sulfate contains 0.209 grams of solute in 100 mL of solution. Calculate the molar concentration of the calcium sulfate solution.

9. Answer: 0.015 M Calcium sulfate  CaSO 4 1.What is given? c=?, n=?, m=0.209 g, v=0.1 L 2. Find moles. n= m/M n=0.209 g/ M CaSO 4 n=0.209/136 n=0.0015 mol 3. Find concentration. c=n/v c=0.0015 mol/0.1 L c=0.015 mol/L Note: M CaSO 4 =40+32+4(16) =136 g/mol Note: Make sure to ALWAYS convert milliliters into liters for volume because molarity is always in units of moles per liter (mol/L) 1000mL=1L

10. How many milliliters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?

11. Note: Make sure to ALWAYS convert milliliters into liters for volume because molarity is always in units of moles per liter (mol/L). Answer: 65 mL Solution: 1.Use the dillution formula C 1 V 1 = C 2 V 2 2.What is given? C 1 = 5.5 M, V 1 = ?, C 2 = 1.2 M, V 2 = 300 mL 3.Substitute each given value into the equation and solve for V 1 4.C 1 V 1 = C 2 V 2 (5.5 M)*V 1 = (1.2 M)*(0.3L) V 1 = [(1.2 M)*(0.3L)]/5.5M V 1 = 0.065 L 5.Convert 0.065L into mL 6.1L=1000mL 0.065L=x mL 7. Solve for x by cross multiplying. 8. 1L=1000mL 0.065L=x mL x=1000mL*0.065L x=65 mL

12. If you add 500 mL of 0.100 M AgNO 3 solution to a solution containing an excess of chlorine ion. How much moles of AgCl precipitate will you form?

13. Answer: 0.0500 moles AgCl Solution: 1.Balance the equation: 2AgNO 3(aq) + Cl 2(g) 2AgCl (aq) + 2NO 3(aq) 2.It is given that AgNO 3(g) is the limiting reagent 3.2AgNO 3(aq) + Cl 2(g) 2AgCl (aq) + 2NO 3(aq) c=0.100M v=0.5 L n=? 4. Moles of AgNO 3 : Moles of AgCl 2 : 2 0.05: x x=0.05 moles of AgCl Find the moles of AgNO 3(g) : n=c*v n= 0.100M * 0.5L n=0.05 moles Note: Make sure to ALWAYS convert milliliters into liters for volume because molarity is always in units of moles per liter (mol/L)

14. If you mix 200 mL of 0.100 M Pb(NO 3 ) 2 and 300 mL of 0.200 M MgCl 2, how much moles of PbCl 2 precipitate will you form?

15. Answer: 0.0200 moles PbCl 2 Solution: Pb(NO 3 ) 2 + MgCl 2 PbCl 2 +Mg(NO 3 ) 2 v=0.2 L v=0.3L c=0.100M c=0.200M n=? n=? Find Limiting Reagent Pb(NO 3 ) 2 :PbCl 2 1 : 1 0.02:x x=0.02 mol of PbCl 2 Limiting reagent is Pb(NO 3 ) 2 Since the mole ratio between Pb(NO 3 ) 2 :PbCl 2 is 1:1, the moles of PbCl 2 is 0.02 moles. MgCl 2 :PbCl 2 1 : 1 0.06:x x=0.06 mol of PbCl 2 Note: Make sure to ALWAYS convert milliliters into liters for volume because molarity is always in units of moles per liter (mol/L)

16. Ammonium sulfate is manufactured by reacting sulfuric acid with ammonia. What concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L ammonia solution if 50.0 mL of sulfuric acid is used?

17. Note: Make sure to ALWAYS convert milliliters into liters for volume because molarity is always in units of moles per liter (mol/L) Answer: 0.5368 M 1.Balance the equation: H 2 SO 4(aq) + 2NH 3(aq)  (NH 4 ) 2(aq) + SO 4(aq) 2.What is given? 3. Solve for number of moles in ammonia. 4. Use mole mole ratio to find the number of moles of sulfuric acid 5. Find the concentration of sulfuric acid. c=n/v c=0.02684 moles/0.0500L c=0.5368 M c=? V=0.0500 L n=? c=2.20 M v= 0.0244 L n=? n=c*v n=2.2M*0.0244L n=0.05368 moles H 2 SO 4 : NH 3 1:2 x:0.05368 moles x= 0.05368/2 x=0.02684 moles H 2 SO 4

18. Calcium hydroxide is sometimes used in water treatment plants to clarify water for residential use. Calculate the volume of 0.0250 mol/L calcium hydroxide solution that can be completely reacted with 25.0 mL of 0.125 mol/L aluminum sulfate solution.

19. Solution: 3Ca(OH) 2(aq) + Al 2 (SO 4 ) 3(aq)  2Al(OH) 3(s) + 3CaSO 4(s) Answer: 0.375 L Ca(OH) 2 C=0.025 M v=? n=0.009375 mol c=0.125 M v=0.025 L n=? Find moles for 3Ca(OH) 2 : n=c*v n=0.125M*0.025L n=0.003125 moles Mole Mole Ratio Al 2 (SO 4 ) 3 : Ca(OH) 2 1:3 0.003125 mol: x x= 0.009375 mol of Ca(OH) 2 Volume of 3Ca(OH) 2 : v=n/c =0.009375 mol/0.025M =0.375 L Note : Make sure to ALWAYS convert milliliters into liters for volume.

20. Mr. Krstovic wants 75.0 mL of 0.200 mol/L iron(III) chloride solution to react completely with an excess of 0.250 mol/L sodium carbonate solution. What volume of sodium carbonate solution is needed?

21. 2FeCl 3(aq) + 3Na 2 CO 3(aq)  Fe 2 (CO 3 ) 3(s) + 6NaCl (aq) Answer: 0.0900 L Na 2 CO 3 c=0.2M v=0.075L n=? c=0.25M v=? n=? Moles of FeCl 3 n=c*v n=0.2M*0.075L n=0.015 mol of FeCl 3 Mole Mole Ratio 2FeCl 3 : 3Na 2 CO 3 2:3 0.015 mol: x x= 0.0225 mol of Na 2 CO 3 Find volume of Na 2 CO 3 v=n/c v=0.0225 mol/0.25M V=0.09 L of Na 2 CO 3 Note : Make sure to ALWAYS convert milliliters into liters for volume.

22. Describe how you would go about separating a mixture containing solutions of strontium ions, Sr 2+ (aq), and calcium ions, Ca 2+ (aq).

23. CrO 2 -4 Solution: 1.Using the solubility/ insolubility chart, find an element that is insoluble with one of the elements in the mixture requiring separation, and soluble for the other element in the mixture. NOTE: An element that is insoluble for one element is required to be selected in order to create a precipitate. 2.Then, find another element that is insoluble with the remaining element in the solution. By doing this, two solutions will form a precipitate, thus separating the ions in the solution. Sr +2, Ca +2 CO 3 2- Sr 2 CrO 2 (s) Ca +2 (aq) CaCO 3 (s) Note: Answer may vary

24. Using the flow chart method, separate the following cations from the aqueous solution: Ag +, Pb 2+, Ba 2+, Cu+2, Fe 3+

25. Solution: Answers vary, but this is a sample solution. Refer to the solubility/insolubility chart. Pb 2+, Ba 2+, Ag + (aq) C 2 H 3 O 2 -1 (aq) Ag +, Pb 2+, Ba 2+, Fe 3+ Fe(C 2 H 3 O 2 ) 3(s) Ba 2+, Pb +2 (aq) AgBr (s) Br -1 (aq) OH -1 (aq) PbBr 2(s) Ba 2+ (aq) BaCO 3(s) CO 3 -2 (aq)

26. Using the flow chart method, separate the following cations from the aqueous solution: Br -, ClO 3 -2, O -2, NO 3 -

27. Solution: Answers vary, but this is a sample solution. Refer to the solubility/insolubility chart. : Hg +1 (aq) ClO 3 -2, I -1 (aq) Al +3 (aq) ClO 3 -2, O -2, I -1 (aq) Al 2 O 3(s) ClO 3 -2 (aq) PbI 2(s) Pb +1 (aq) HgClO 3(s)

28. Write the total and net ionic equation for the reaction Lead (II) nitrate with hydrochloric acid.

29. Answer: Total Ionic Equation: Pb 2+ (aq) + 2 NO 3 ‾ (aq) + 2 H + (aq) + 2 Cl‾ (aq) PbCl 2 (s) + 2 H + (aq) + 2 NO 3 ‾ (aq) Net Ionic Equation: Pb 2+ (aq) + 2 Cl‾ (aq) PbCl 2 (s) Solution: Finding the total ionic equation: 1.Write a balanced reaction from the information provided.: Pb(NO 3 ) 2 (aq) + 2 HCl (aq) PbCl 2 (s) + 2 HNO 3 (aq) 2.Write the elemental composition for all AQUEOUS compound. For solids, do not dissociate. Pb 2+ (aq) + 2 NO 3 ‾ (aq) + 2 H + (aq) + 2 Cl‾ (aq) PbCl 2 (s) + 2 H + (aq) + 2 NO 3 ‾ (aq) 3. Step 2 is the total ionic equation Finding the net ionic equation: 1.Cancel out elements that have the same charge Pb 2+ (aq) + 2 NO 3 ‾ (aq) + 2 H + (aq) + 2 Cl‾ (aq) PbCl 2 (s) + 2 H + (aq) + 2 NO 3 ‾ (aq) 2.The net ionic equation is the elements from the reactants that do are left over after completing step 1. The remaining reactants should correspond to the product. Pb 2+ (aq) + 2 Cl‾ (aq) PbCl 2 (s)

30. What will be the complete, total ionic and net ionic equations for the reaction of an aqueous solution of nitric acid (HNO 3 ) with one of potassium hydroxide (KOH)?

31. 1.Write the reaction as a skeleton equation: 2.HNO 3 (aq) + KOH (aq) KNO 3 (aq) + H 2 O (l) 3.Separate each compound into ions. This is the total ionic equation H + (aq) + NO 3 ¯ (aq) + K + (aq) + OH¯ (aq) K + (aq) + NO 3 ¯ (aq) + H 2 O (l) 4. Cross out the spectator ions. H + (aq) + NO 3 ¯ (aq) + K + (aq) + OH¯ (aq) K + (aq) + NO 3 ¯ (aq) + H 2 O (l) 5. This remainder is your net ionic equation: H + (aq) + OH¯ (aq) H 2 O (l) Answer: Total Ionic Equation: H + (aq) + NO 3 ¯ (aq) + K + (aq) + OH¯ (aq) K + (aq) + NO 3 ¯ (aq) + H 2 O (l) Net Ionic Equation: H + (aq) + OH¯ (aq) H 2 O (l) Solution:

32. 1.What mass of lead (II) nitrate will dissolve to make a saturated solution in 25.0 mL of water at 20°C? 2.What mass of sodium chlorate dissolves 33.3 mL of water at 40°C to make a saturated solution?

33. 1.132.5 grams, assuming 53 g/100 mL of water dissolves 2.38.33 grams assuming 155 g/100 mL dissolves These answers and questions were taken from the following website: http://74.125.93.132/search?q=cache:07X_n6d17SQJ:www.pembinat rails.ca/shaftesbury/mrdeakin/Solubility%2520Curve%2520Questions.052.doc+solubility+curve+questions&cd=1&hl=en&ct=clnk&gl=ca &client=firefox-a http://74.125.93.132/search?q=cache:07X_n6d17SQJ:www.pembinat rails.ca/shaftesbury/mrdeakin/Solubility%2520Curve%2520Questions.052.doc+solubility+curve+questions&cd=1&hl=en&ct=clnk&gl=ca &client=firefox-a Solutions: Still not understanding the concept of solubility curves? Take a look at this site: http://www.thesciencedesk.com/sgsolubilitygraph.htm

34. 1.What mass of potassium chloride precipitates from a saturated solution in 15 g of water when it is cooled from 90°C to 30°C? 2. A 200.0 mL saturated solution of potassium nitrate is made at 90°C. If 15% of the solvent evaporates as the solution is slowly cooled to 30°C, what mass of precipitate is produced?

35. Solutions: 1.1.2 grams with 36 grams/100 mL at 90 degrees Celsius, 28 grams / 100 mL at 30 degrees Celsius 2.325.1 grams with 202.5 grams at 90 degrees Celsius, 47 grams at 30 degrees Celsius. These answers and questions were taken from the following website: http://74.125.93.132/search?q=cache:07X_n6d17SQJ:www.pembinatr ails.ca/shaftesbury/mrdeakin/Solubility%2520Curve%2520Questions. 052.doc+solubility+curve+questions&cd=1&hl=en&ct=clnk&gl=ca& client=firefox-a http://74.125.93.132/search?q=cache:07X_n6d17SQJ:www.pembinatr ails.ca/shaftesbury/mrdeakin/Solubility%2520Curve%2520Questions. 052.doc+solubility+curve+questions&cd=1&hl=en&ct=clnk&gl=ca& client=firefox-a Still not understanding the concept of solubility curves? Take a look at this site: http://www.thesciencedesk.com/sgsolubilitygraph.htm

36. 1.What mass of potassium chromate precipitates from solution if a saturated solution in 60.0 g of water is cooled from 82°C to 40°C? 2.Why is a precipitate formed in the previous question?

37. Solutions: 1.4.8 grams with 74 grams/100 mL at 40 degrees Celsius 2.Solubility of ionic solids decrease with temperatures These answers and questions were taken from the following website: http://74.125.93.132/search?q=cache:07X_n6d17SQJ:www.pembinatra ils.ca/shaftesbury/mrdeakin/Solubility%2520Curve%2520Questions.05 2.doc+solubility+curve+questions&cd=1&hl=en&ct=clnk&gl=ca&clie nt=firefox-a http://74.125.93.132/search?q=cache:07X_n6d17SQJ:www.pembinatra ils.ca/shaftesbury/mrdeakin/Solubility%2520Curve%2520Questions.05 2.doc+solubility+curve+questions&cd=1&hl=en&ct=clnk&gl=ca&clie nt=firefox-a Still not understanding the concept of solubility curves? Take a look at this site: http://www.thesciencedesk.com/sgsolubilitygraph.htm

38. Label the following proportions of the graph on the left.

39. SATURATED UNSATURATED SUPER SATURATED Solution: Still not understanding the concept of solubility curves? Take a look at this site: http://www.thesciencedesk.com/sgsolubilitygraph.htm Still not understanding the concept of solubility curves? Take a look at this site: http://www.thesciencedesk.com/sgsolubilitygraph.htm

40. 1.What mass of lithium sulfate dissolves in 20.0 mL of water at 35°C to make a saturated solution?

41. Still not understanding the concept of solubility curves? Take a look at this site: http://www.thesciencedesk.com/sgsolubilitygraph.htm Solution: 1.7 grams assuming 35 grams/100 mL water dissolves These answers and questions were taken from the following website: http://74.125.93.132/search?q=cache:07X_n6d17SQJ:www.pembina trails.ca/shaftesbury/mrdeakin/Solubility%2520Curve%2520Questio ns.052.doc+solubility+curve+questions&cd=1&hl=en&ct=clnk&gl =ca&client=firefox-a http://74.125.93.132/search?q=cache:07X_n6d17SQJ:www.pembina trails.ca/shaftesbury/mrdeakin/Solubility%2520Curve%2520Questio ns.052.doc+solubility+curve+questions&cd=1&hl=en&ct=clnk&gl =ca&client=firefox-a

42. 20.0 mL of a 3.0M HCl solution are mixed with 20.0 mL of a 5.0M NaOH solution. Is the pH of the resulting solution above 7, below 7, or 7? Explain your answer.

43. Solution found on the following VIDEO (Very DETAILED!!): http://www.youtube.com/watch?v=IX9RHZDVbYM&f eature=related http://www.youtube.com/watch?v=IX9RHZDVbYM&f eature=related

44. The molarity of a hydrochloric acid solution can be determined by titrating a known volume of the solution with a sodium hydroxide solution of known concentration. If 14.7 mL of 0.102 M NaOH is required to titrate 25.00 mL of a hydrochloric acid, HCl, solution, what is the molarity of the hydrochloric acid?

45. Answer: 0.0579 M HCl Solution: 1.Write a balanced equation: HCl +NaOH  H 2 O+ NaCl 2.Write down the given values: 3.Solve for number of moles of NaOH and then use mole-mole ratio to solve for the moles of HCl. 4.Calculate the concentration of HCl. v=0.025L c= ? n=? v=0.0147 L c= 0.102 M n=? Number of Moles NaOH: n=(0.0147 L)(0.102 M) n=0.0014994 moles NaOH Find Number of Moles of HCl: Moles of HCl : Moles of NaOH 1 : 1 x : 0.0014994 Therefore, since mole mole ratio is 1:1, the moles of HCl is also 0.0014994 c=n/v c=(0.0014994 moles)/0.025L c=0.0579 M

46. If 36.2 mL of 0.152 M NaOH is required to neutralize 25.00 mL of an acetic acid, HC 2 H 3 O 2, solution, what is the molarity of the acetic acid?

47. Answer: 0.220 M HC 2 H 3 O 2 Solution: 1.Write a balanced equation: HC 2 H 3 O 2 + NaOH  H 2 O + NaC 2 H 3 O 2 2.Write down the given values: 3.Solve for number of moles of NaOH and then use mole-mole ratio to solve for the moles of HCl. 4.Calculate the concentration of HCl. v=0.025L c= ? n=? v=0.0362 L c= 0.152 M n=? Number of Moles NaOH: n=(0.0362 L)(0.152 M) n=0.0055024 moles NaOH Find Number of Moles of HCl: Moles of C 2 H 3 O 2 : Moles of HCl 1 : 1 x : 0.0055024 Therefore, since mole mole ratio is 1:1, the moles of HCl is also 0.0055024 c=n/v c=(0.0055024 moles)/0.025L c=0.220096 M

48. If 11.3 mL of 0.110 M HCl is required to neutralize 25.00 mL of a barium hydroxide solution, what is the molarity of the barium hydroxide?

49. Answer: 0.0249 M Ba(OH) 2 Solution: 1.Write a balanced equation: Ba(OH) 2 + 2HCl  2H 2 O + BaCl 2 2.Write down the given values: 3.Solve for number of moles of NaOH and then use mole-mole ratio to solve for the moles of HCl. 4.Calculate the concentration of HCl. v=0.025L c= ? n=? v=0.0113 L c= 0.110 M n=? Number of Moles HCl: n=(0.0113 L)(0.110 M) n=0.001243 moles HCl Find Number of Moles of Ba(OH) 2 : Moles of Ba(OH) 2 : Moles of HCl 1 : 2 x : 0.001243 2x=0.001243 x= 0.0006215 Therefore, since mole mole ratio is 1:2, the moles of HCl is 0.0006215 c=n/v c=(0.0006215 moles)/0.025L c=0.02486 M or 0.0249 M Ba(OH) 2

50. Complete equations for the following acids ionizing or dissociating in water: 1.(NH 4 ) 3 PO 4 (s) → ? 2.Mg(OH) 2 (s) → ? 3.H 2 SO 4 (g) + H 2 O (l) → ? 4.HCl (g) + H 2 O (l) → ? Classify the acid, base, conjugate base and conjugate acid 5. NH 3 + H 2 O→ NH 4 + OH - 6. H 2 SO 4 + H 2 O → HSO 4 - + H 3 O + 7. HCl + OH - → Cl - + H 2 O 8. HCl + NH 3 → Cl - + NH 4 +

51. Solutions: 1.(NH 4 ) 3 PO 4 (s) → 3 NH 4 + (aq) + PO 4 3- (aq) 2.Mg(OH) 2 (s) → Mg 2+ (aq) + 2 OH – (aq) 3.H 2 SO 4 (g) + H 2 O (l) → 2 H 3 O + (aq) + SO 4 2- (aq) 4.HCl (g) + H 2 O (l) → H 3 O + (aq) + Cl - (aq) 5.NH 3 + H 2 O→ NH 4 + OH - NH 3 = Base H 2 O= Acid NH 4 = Conjugate Acid OH -1 = Conjugate Base 6. H 2 SO 4 + H 2 O → HSO 4 - + H 3 O + H 2 SO 4 = Acid H 2 O = Base HSO 4 - = Conjugate Acid H 3 O + = Conjugate Base 7. HCl + OH - → Cl - + H 2 O HCl = Acid OH - = Base Cl - = Conjugate Acid H 2 O = Conjugate Base 8. HCl + NH 3 → Cl - + NH 4 + HCl = Acid NH 3 = Base Cl - = Conjugate Acid NH 4 + = Conjugate Base