1. 1 Chemistry Second Edition Julia Burdge Lecture PowerPoints Jason A. Kautz University of Nebraska-Lincoln Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4 Reactions in Aqueous Solutions

2. 4 4.1General Properties of Aqueous Solutions Electrolytes and Nonelectrolytes Strong Electrolytes and Weak Electrolytes 4.2Precipitation Reactions Solubility Guidelines for Ionic Compounds in Water Molecular Equations Ionic Equations Net Ionic Equations 4.3Acid-Base Reactions Strong Acids and Bases Brønsted Acids and Bases Acid-Base Neutralization 4.4Oxidation-Reduction Reactions Oxidation Numbers Oxidation of Metals in Aqueous Solutions Balancing Simple Redox Equations Other Types of Redox Reactions

3. Reactions in Aqueous Solutions 4 4.5Concentration of Solutions Molarity Dilution Solution Stoichiometry 4.6Aqueous Reactions and Chemical Analysis Gravimetric Analysis Acid-Base Titrations

4. General Properties of Aqueous Solutions A solution is a homogenous mixture of two or more substances. The substance present in the largest amount (moles) is referred to as the solvent. The other substances present are called the solutes. A substance that dissolves in a particular solvent is said to be soluble in that solvent. 4.1

5. General Properties of Aqueous Solutions An electrolyte is a substance that dissolves in water to yield a solution that conducts electricity. An electrolyte undergoes dissociation and breaks apart into its constituent ions. NaCl(s) Na + (aq) + Cl – (aq) H2OH2O

6. General Properties of Aqueous Solutions A non-electrolyte is a substance that dissolves in water to yield a solution that does not conduct electricity. The sucrose molecules remain intact upon dissolving. C 12 H 22 O 11 (s) C 12 H 22 O 11 (aq) H2OH2O

7. General Properties of Aqueous Solutions An electrolyte that dissociates completely is known as a strong electrolyte. Water soluble ionic compounds Strong Acids Strong Bases NaCl(s) Na + (aq) + Cl – (aq) H2OH2O HCl(g) H + (aq) + Cl – (aq) H2OH2O NaOH(s) Na + (aq) + OH – (aq) H2OH2O

8. Aqueous Solutions

9. General Properties of Aqueous Solutions A weak electrolyte is a compound that produces ions upon dissolving but exists in solution predominantly as molecules that are not ionized. Weak Acids Weak Bases HC 2 H 3 O 2 (l) H + (aq) + C 2 H 3 O 2 (aq) – NH 3 (g) + H 2 O(l) NH 4 (aq) + OH – (aq) +

10. General Properties of Aqueous Solutions The double arrow,, denotes a reaction that occurs in both directions. When both the forward and reverse reactions occur at the same rate, the reaction is in a state of dynamic equilibrium. NH 3 (g) + H 2 O(l) NH 4 (aq) + OH – (aq) +

11. General Properties of Aqueous Solutions Classify each of the following as a nonelectrolyte, a weak electrolyte, or a strong electrolyte. Citric acid, H 3 C 6 H 5 O 7 Potassium phosphate, K 3 PO 4 Glucose, C 6 H 12 O 6 Solution: H 3 C 6 H 5 O 7 :Citric acid is a weak acid and therefore a weak electrolyte. K 3 PO 4 : Potassium phosphate is a soluble salt that contains both cations and anions and is therefore a strong electrolyte. C 6 H 12 O 6 : Glucose contains no ions and is therefore a molecular compound and a nonelectrolyte.

12. Precipitation Reactions An insoluble product that separates from a solution is called a precipitate. 2NaI(aq) + Pb(NO 3 ) 2 (aq) PbI 2 (s) + 2NaNO 3 (aq) 4.2

13. Precipitation Reactions A chemical reaction in which a precipitate forms is called a precipitation reaction.

14. Precipitation Reactions Water is a good solvent for ionic compounds because it is a polar molecule. The polarity of water results from electron distributions within the molecule. The oxygen atom has an attraction for the hydrogen atoms’ electrons and is therefore partially negative compared to hydrogen.    –– The oxygen atom is partially negative The hydrogen atoms are partially positive   

15. Precipitation Reactions Hydration occurs when water molecules remove the individual ions from an ionic solid surround them so the substances dissolves.

16. Precipitation Reactions Solubility is defined as the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.

17. Classify the following as soluble or insoluble in an aqueous solution. Ba(NO 3 ) 2 AgI Mg(OH) 2 Precipitation Reactions

18. Classify the following as soluble or insoluble in an aqueous solution: Ba(NO 3 ) 2 AgI Mg(OH) 2 Solution: Ba(NO 3 ) 2 :Soluble; compounds containing nitrate are soluble with no exceptions. AgI:Insoluble; although many compounds containing the iodide are soluble; when I ‒ is paired with Ag +, the compound is insoluble. Mg(OH) 2 :Insoluble; most metal hydroxides are insoluble; exceptions include group 1A hydroxides and Ba(OH) 2. Precipitation Reactions

19. Copyright McGraw-Hill 2009 Precipitation Reactions Identify the precipitate in the following reaction: Pb(NO 3 ) 2 (aq) + 2NaI(aq)  2NaNO 3 (?) + PbI 2 (?)

20. Identify the precipitate in the following reaction: Solution: NaNO 3 :Soluble; all nitrates are soluble with no exceptions; all salts that contain group 1A cations are soluble with no exceptions. PbI 2 :Insoluble (precipitates out of solution); although many compounds containing the iodide are soluble; when I ‒ is paired with Ag +, the compound is insoluble. Pb(NO 3 ) 2 (aq) + 2NaI(aq)  2NaNO 3 (?) + PbI 2 (?) Precipitation Reactions Pb(NO 3 ) 2 (aq) + 2NaI(aq)  2NaNO 3 (aq) + PbI 2 (s)

21. Precipitation Reactions In a molecular equation compounds are represented by chemical formulas as though the exist in solution as molecules or formula units Na 2 SO 4 (aq) + Ba(OH) 2 (aq) 2NaOH(aq) + BaSO 4 (s)

22. Precipitation Reactions In the reaction between aqueous Na 2 SO 4 and Ba(OH) 2 the aqueous species are represented as follows: In an ionic equation compounds that exist completely or predominately as ions in solution are represented as those ions. Na 2 SO 4 (aq) → 2Na + (aq) + SO 4 (aq) 2– Ba(OH) 2 (aq) → Ba 2+ (aq) + 2OH – (aq) NaOH(aq) → Na + (aq) + OH – (aq) 2Na + (aq) + SO 4 (aq) + Ba 2+ (aq) + 2OH – (aq) 2Na + (aq) + 2OH – (aq) + BaSO 4 (s) 2– Na 2 SO 4 (aq) + Ba(OH) 2 (aq) 2NaOH(aq) + BaSO 4 (S)

23. Precipitation Reactions An equation that includes only the species that are actually involved in the reaction is called a net ionic equation. Ions that appear on both sides of the equation are called spectator ions. Spectator ions do not participate in the reaction. 2Na + (aq) + SO 4 (aq) + Ba 2+ (aq) +2OH – (aq) 2Na + (aq) + 2OH – (aq) + BaSO 4 (s) 2– Ba 2+ (aq) + SO 4 (aq) BaSO 4 (s) 2–

24. Precipitation Reactions To determine the molecular, ionic and net ionic equations: 1)Write and balance the molecular equation, predicting the products by assuming that the cations trade anions. 2)Write the ionic equation by separating strong electrolytes into their constituent ions. 3)Write the net ionic equation by identifying and canceling spectator ions on both sides of the equation. 4)If both the reactants and products are all strong electrolytes, there is no net ionic equation; no reaction takes place.

25. Precipitation Reactions Write the molecular, ionic, and net ionic equations for the combination of Fe(NO 3 ) 2 (aq) with Na 2 CO 3 (aq). Solution: Step 1:Write and balance the molecular equation, predicting the products by assuming that the cations trade anions; checking the solubility of each product. Fe(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) → FeCO 3 (s) + 2NaNO 3 (aq)

26. Precipitation Reactions Write the molecular, ionic, and net ionic equations for the combination of Fe(NO 3 ) 2 (aq) with Na 2 CO 3 (aq). Solution: Step 2:Write the ionic equation by separating strong electrolytes into their constituent ions. Fe 2+ (aq) + 2NO 3 (aq) + 2Na + (aq) + CO 3 (aq) → FeCO 3 (s) + 2Na + (aq) + 2NO 3 (aq) – 2– –

27. Precipitation Reactions Write the molecular, ionic, and net ionic equations for the combination of Fe(NO 3 ) 2 (aq) with Na 2 CO 3 (aq). Solution: Step 3:Write the net ionic equation by identifying and canceling spectator ions on both sides of the equation. Fe 2+ (aq) + 2NO 3 (aq) + 2Na + (aq) + CO 3 (aq) → FeCO 3 (s) + 2Na + (aq) + 2NO 3 (aq) – 2– – Fe 2+ (aq) + CO 3 (aq) → FeCO 3 (s) 2–

28. Acid-Base Reactions Acids can be either strong or weak. A strong acid is a strong electrolyte. 4.3

29. Acid-Base Reactions A weak acid is a weak electrolyte; it does not dissociate completely. Acetic acid, HC 2 H 3 O 2, is an example. Most acids are weak acids. HC 2 H 3 O 2 (l) H + (aq) + C 2 H 3 O 2 (aq) – acidic proton

30. Acid-Base Reactions Strong bases are strong electrolytes. Strong bases are the hydroxides of Group 1A and heavy Group 2A Sodium hydroxide, NaOH, is an example. NaOH(s) Na + (aq) + OH – (aq) H2OH2O

31. Acid-Base Reactions An Arrhenius acid is one that ionizes in water to produce H + ions. An Arrhenius base is one that dissociates in water to produce OH – ions. HCl(g) H + (aq) + Cl – (aq) H2OH2O NaOH(s) Na + (aq) + OH – (aq) H2OH2O

32. Acid-Base Reactions A Brønsted acid is a proton donor. A Brønsted base is a proton acceptor. In these definitions, a proton refers to a hydrogen atom that has lost its electron—also known as a hydrogen ion (H + ). H 2 O is a Brønsted acid: donates a proton to become OH – NH 3 (g) + H 2 O(l) NH 4 (aq) + OH – (aq) + NH 3 is a Brønsted base: accepts a proton to become NH 4 +

33. Acid-Base Reactions Brønsted acids donate protons to water to form the hydronium ion (H 3 O + ). hydrogen ion (H + ) proton hydronium ion (H 3 O + ) HF(aq) + H 2 O(l) H 3 O + (aq) + F – (aq) H2OH2O All refer to the same aqueous species

34. Acid-Base Reactions A monoprotic acid has one proton to donate. Hydrochloric acid is an example: HCl(g) H + (aq) + Cl – (aq) one equivalent of solvated hydrogen ion

35. Acid-Base Reactions A polyprotic acid has more than one acidic hydrogen atom. Sulfuric acid, H 2 SO 4, is an example of a diprotic acid; there are two acidic hydrogen atoms. Polyprotic acids lose protons in a stepwise fashion: H 2 SO 4 (aq) H + (aq) + HSO 4 (aq) – – HSO 4 (aq) H + (aq) + SO 4 (aq) 2– Step 1: Step 2: In H 2 SO 4, the second ionization occurs only to a very small extent. In H 2 SO 4, the first ionization is strong.

36. Acid-Base Reactions Bases that produce only one mole of hydroxide per mole of compound are called monobasic. Sodium hydroxide is an example: NaOH(s) Na + (aq) + OH – (aq) H2OH2O one equivalent of hydroxide

37. Acid-Base Reactions Some strong bases produce more than one hydroxide per mole of compound. Barium hydroxide is an example of a dibasic base. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH – (aq) H2OH2O two equivalents of hydroxide

38. Acid-Base Reactions A neutralization reaction is a reaction between an acid and a base. Generally, a neutralization reaction produces water and a salt. HCl(aq) + NaOH(aq) → H 2 O(l) + NaCl(aq) The net ionic equation of many acid-base reactions is: H + (aq) + OH – (aq) H 2 O(l) acid basewater salt

39. Oxidation-Reduction Reactions An oxidation-reduction (or redox) reaction is a chemical reaction in which electrons are transferred from one reactant to another. Oxidation is the loss of electrons. Reduction is the gain of electrons. Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) Zn metal loses 2 electrons and is oxidized to Zn 2+ Zn 2+ is called the reducing agent Cu 2+ gains 2 electrons and is reduced to Cu metal Cu is called the oxidizing agent 4.4

40. Oxidation-Reduction Reactions Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s)

41. Oxidation-Reduction Reactions A redox reaction is the sum of an oxidation half-reaction and a reduction half-reaction. Oxidation half-reaction: Zn(s)Zn 2+ (aq) + 2e – Cu 2+ (aq) + 2e – Cu(s) Cu 2+ (aq) + Zn(s)Zn 2+ (aq) + Cu(s) Reduction half-reaction: Overall redox reaction: Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) Oxidation (lose 2e – ) Reduction (gain 2e – )

42. Oxidation-Reduction Reactions The oxidation number is the charge an atom would have if electrons were transferred completely. The oxidation number is sometimes called the oxidation state. H 2 (g) + F 2 (g) → 2HF(g) +1–10000+1–1 Oxidation number: Total contribution to charge: N 2 (g) + 3H 2 (g) → 2NH 3 (g) –3–3+10000–3–3+3 Oxidation number: Total contribution to charge:

43. Oxidation-Reduction Reactions To assign oxidation numbers: 1)The oxidation number of an element, in its elemental form, is zero. 2)The oxidation numbers in any chemical species must sum to the overall charge on the species. must sum to zero for any molecule must sum to the charge on any polyatomic ion the oxidation number of a monoatomic ion is equal to the charge on the ion

44. Oxidation-Reduction Reactions To assign oxidation numbers: 3)Know the elements that nearly always have the same oxidation number.

45. Oxidation-Reduction Reactions Assign the oxidation numbers to the elements in the compound KMnO 4. Step 1:Start with the oxidation numbers you know: Step 2:The numbers in the boxes (total contribution to charge) must sum to zero (KMnO 4 is a neutral compound). K Mn O 4 +1 Oxidation number: Total contribution to charge: +7 –2 –8

46. Oxidation-Reduction Reactions Assign the oxidation numbers to the elements in the compound H 2 SO 4. Solution: Step 1:Start with the oxidation numbers you know: Step 2:The numbers in the boxes (total contribution to charge) must sum to zero (the chemical species is neutral). H 2 S O 4 +1 +2 Oxidation number: Total contribution to charge: +6 –2 –8

47. Oxidation-Reduction Reactions Assign the oxidation numbers to the elements in the ion ClO 3. Solution: Step 1:Start with the oxidation numbers you know: Step 2:The numbers in the boxes (total contribution to charge) must sum to negative one (the chemical species is a –1 anion). Cl O 3 +5 Oxidation number: Total contribution to charge: –2 –6–6 – –

48. Oxidation-Reduction Reactions In a displacement reaction, an atom or an ion in a compound is replaced by an atom of another element. Zn(s) + CuCl 2 (aq) → ZnCl 2 (aq) + Cu(s) Zinc displaces, or replaces copper in the dissolved salt. Zn is oxidized to Zn 2+ Cu 2+ is reduced to Cu When a metal is oxidized by an aqueous solution, it becomes an aqueous ion. 0+2–1 +2–1 0 0+2–2+2–2 0

49. Oxidation-Reduction Reactions The activity series is a list of metals (and hydrogen) arranged from top to bottom in order of decreasing ease of oxidation. Cu(s) + ZnCl 2 (aq) → no reaction ElementOxidation Half-Reaction ZincZn → Zn 2+ + 2e – IronFe → Fe 2+ + 2e – NickelNi → Ni 2+ + 2e – HydrogenH 2 → 2H + + 2e – CopperCu → Cu 2+ + 2e – SilverAg → Ag + + e – GoldAu → Au 3+ + 3e – Zn(s) + CuCl 2 (aq) → ZnCl 2 (aq) + Cu(s) Metals listed at the top are called active metals. Metals listed at the bottom are called noble metals. An element in the series will be oxidized by the ions of any element that appears below it in the table. Increasing ease of oxidation

50. Oxidation-Reduction Reactions Which of the following reactions will occur? Solution: ?Co(s) + BaI 2 (aq) Cu(s) + Sn 2+ (aq)Sn(s) + CuBr 2 (aq) ? ?Ag(s) + NaCl(aq) No reaction. Cobalt is below barium.Co(s) + BaI 2 (aq) ElementOxidation Half-Reaction SodiumNa → Na + + 1e – BariumBa → Ba 2+ + 2e – CobaltCo → Co 2+ + 2e – TinSn → Sn 2+ + 2e – HydrogenH 2 → 2H + + 2e – CopperCu → Cu 2+ + 2e – SilverAg → Ag + + e – Increasing ease of oxidation No reaction. Silver is below sodium.Ag(s) + NaCl(aq)

51. Oxidation-Reduction Reactions Redox reactions must have both mass balance and charge balance. Before adding half-reactions, the electrons must balance. Cr(s) + Ni 2+ (aq) → Cr 3+ (aq) + Ni(s) Cr(s)Cr 3+ (aq) + 3e – Oxidation half-reaction: Ni(s)Ni 2+ (aq) + 2e – Reduction half-reaction:

52. Oxidation-Reduction Reactions Prior to adding the two half-reactions, balance the electrons. Step 1: Multiply the oxidation half-reaction by 2 Step 3: Multiply the reduction half-reaction by 3 Cr(s)Cr 3+ (aq) + 3e – Oxidation half-reaction: Ni(s)Ni 2+ (aq) + 2e – Reduction half-reaction: 2 3 2Cr(s)2Cr 3+ (aq) + 6e – Oxidation half-reaction: 2Ni(s)3Ni 2+ (aq) + 6e – Reduction half-reaction: 2Ni(s) + 2Cr 3+ (aq)3Ni 2+ (aq) + 2Cr(s)

53. Oxidation-Reduction Reactions Give the overall balanced equation for the following reactions. Solution: ?Cu(s) + H + (aq) 2Ag(s) + Cd 2+ (aq)Cd(s) + 2Ag + (aq) ?Cd(s) + Ag + (aq) No reaction. Copper is below hydrogen.Cu(s) + H + (aq) ElementOxidation Half-Reaction BariumBa → Ba 2+ + 2e – CadmiumCd → Cd 2+ + 2e – TinSn → Sn 2+ + 2e – HydrogenH 2 → 2H + + 2e – CopperCu → Cu 2+ + 2e – SilverAg → Ag + + e – Increasing ease of oxidation

54. Oxidation-Reduction Reactions Combination reactions can involve oxidation and reduction. Hydrogen is oxidized from 0 to +1 Nitrogen is reduced from 0 to –3 N 2 (g) + 3H 2 (g) → 2NH 3 (g) –3–3+100 00–3–3+3

55. Oxidation-Reduction Reactions Decomposition can also be a redox reaction. Na + is reduced to Na H – is oxidized to H 2 NaH(s) → 2Na(s) + 3H 2 (g) –100+1 0–1–10

56. Oxidation-Reduction Reactions Disproportionation reactions occur when one element undergoes both oxidation and reduction. Oxygen in H 2 O 2 (and other peroxides) has an oxidation number of –1. 2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g) –10+1 +2 –2–20 oxidation reduction –2

57. Oxidation-Reduction Reactions Combustion is also a redox process. CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) +1 0 –4 +2+4 0 –2 +4 –2 –4

58. Concentration of Solutions Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution. Other common rearrangements: 4.5

59. Concentration of Solutions Determine the molarity of a 5.00 L solution that contains 235 g of sucrose, C 12 H 22 O 11. Solution: Step 1:Determine the molar mass of sucrose. (342.30 g/mol) Step 2:Determine the moles of sucrose. (0.68653 moles sucrose) Step 3: Determine the molarity of the solution:

60. Concentration of Solutions How many millilitres of 3.50 M NaOH can be prepared from 75.00 grams of solid NaOH? Solution: Step 1:Convert grams to moles. (1.87509 moles NaOH) Step 2:Use the molarity equation to find liters; convert to mL:

61. Concentration of Solutions Dilution is the process of preparing a less concentrated solution from a more concentrated one. moles of solute before dilution = moles of solute after dilution

62. Concentration of Solutions In an experiment, a student needs 250.0 mL of a 0.100 M CuCl 2 solution. A stock solution of 2.00 M CuCl 2 is available. How much of the stock solution is needed? Solution:Use the relationship that moles of solute before dilution = moles of solute after dilution. (2.00 M CuCl 2 )(L c ) = (0.100 M CuCl 2 )(0.2500 L) L c = 0.0125 L or 12.5 mL To make the solution: 1)Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask. 2)Carefully dilute to the calibration mark. M c x L c = M d x L d

63. Concentration of Solutions What volume of 6.0 M H 2 SO 4 is needed to prepare 500.0 mL of a solution that is 0.25 M H 2 SO 4 ? Solution:Use the relationship that moles of solute before dilution = moles of solute after dilution. (6.0 M H 2 SO 4 )(L c ) = (0.25 M H 2 SO 4 )(0.5000 L) L c = 0.021 L or 21 mL M c x L c = M d x L d

64. Concentration of Solutions What are the concentrations of ions in a solution that is 0.750 M in barium nitrate? [Ba(NO 3 ) 2 ] Solution: Step 1:Determine if barium nitrate is a strong electrolyte using solubility rules and, if so, write the equation for the dissociation. Step 2:Determine the concentration of Ba 2+ (aq) ions using the stoichiometric ratio. Square brackets around a chemical species indicates concentration. Ba(NO 3 ) 2 (s) H2OH2O Ba 2+ (aq) + 2NO 3 (aq) ‒

65. Concentration of Solutions What are the concentrations of ions in a solution that is 0.750 M in barium nitrate? [Ba(NO 3 ) 2 ] Solution: Step 3:Determine the concentration of NO 3 – (aq) ions using the stoichiometric ratio. Ba(NO 3 ) 2 (s) H2OH2O Ba 2+ (aq) + 2NO 3 (aq) ‒

66. Aqueous Reactions and Chemical Analysis Gravimetric analysis is an analytical technique based on the measurement of mass. Gravimetric analysis is highly accurate. Applicable only to reactions that go to completion or have nearly 100 % yield. 4.6

67. Aqueous Reactions and Chemical Analysis One common type of gravimetric analysis involves the isolation of a precipitate. Typical steps involve: 1) Mass an unknown solid. 2) Dissolve the unknown in water. 3) React the unknown with an excess amount of a substance that is known to form a precipitate. 4)Filter, dry and weigh the precipitate. 5)Use the formula mass and the mass of the precipitate to find the % mass of the unknown ion.

68. Aqueous Reactions and Chemical Analysis A 0.825 g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with excess silver nitrate. If 1.725 g of AgCl precipitate forms, what is the percent by mass of Cl in the original sample? Solution: Step 1:Find the percent by mass chlorine in AgCl. Step 2:Multiply the percent of Cl by the mass of the precipitate to obtain the mass of Cl in the sample. 0.247 x 1.725 g AgCl = 0.427 g Cl in the sample

69. Aqueous Reactions and Chemical Analysis A 0.825 g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with excess silver nitrate. If 1.725 g of AgCl precipitate forms, what is the percent by mass of Cl in the original sample? Solution: Step 3:Divide the mass of Cl in sample by the total mass of sample; multiply by 100 to determine percent.

70. Aqueous Reactions and Chemical Analysis Quantitative studies of acid-base neutralization reactions are most conveniently carried out using a technique known as a titration. A titration is a volumetric technique that uses burets. The point in the titration where the acid has been neutralized is called the equivalence point.

71. Aqueous Reactions and Chemical Analysis The equivalence point is usually signalled by a color change. The color change is brought about by the use of an indicator. Indicators have distinctly different colors in acidic and basic media. The indicator is chosen so that the color change, or endpoint, is very close to the equivalence point. Phenolphthalein is a common indicator.

72. Aqueous Reactions and Chemical Analysis Typical steps of a titration include: 1)Prepare an unknown solution 2)Add an appropriate indicator to visualize the endpoint. 3)Carefully add a standard solution with a buret until the endpoint is reached. 4)Using solution stoichiometry, calculate the molarity of the unknown solution.

73. Aqueous Reactions and Chemical Analysis Sodium hydroxide solutions are commonly used in titrations. NaOH solutions must be standardized as the concentrations change over time. (NaOH reacts with CO 2 that slowly dissolves into the solution forming carbonic acid.) The acid potassium hydrogen phthalate (KHP) is frequently used to standardize NaOH solutions. acidic proton of KHP; KHP is a monoprotic acid

74. Aqueous Reactions and Chemical Analysis A sample of 1.00 g of KHP was used to standardize a sodium hydroxide solution. The standardization required 15.00 mL of NaOH to reach the endpoint. What is the concentration of the NaOH solution? Solution: Step 1:Use the molar mass of KHP to determine the moles of KHP.

75. Aqueous Reactions and Chemical Analysis A sample of 1.00 g of KHP was used to standardize a sodium hydroxide solution. The standardization required 15.00 mL of NaOH to reach the endpoint. What is the concentration of the NaOH solution? Solution: Step 2:Using the net ionic equation, convert to moles of NaOH KHC 8 H 4 O 4 (aq) + NaOH(aq) → KNaC 8 H 4 O 4 (aq) + H 2 O(l)

76. Aqueous Reactions and Chemical Analysis A sample of 1.00 g of KHP was used to standardize a sodium hydroxide solution. The standardization required 15.00 mL of NaOH to reach the endpoint. What is the concentration of the NaOH solution? Solution: Step 3:Use the molarity equation to calculate molarity of the sodium hydroxide solution.

77. Aqueous Reactions and Chemical Analysis How many milliliters of a 1.42 M H 2 SO 4 solution are needed to neutralize 95.5 mL of a 0.336 M KOH solution? Solution: Step 1:Write and balance the chemical equations that corresponds to the neutralization reaction: H 2 SO 4 (aq) + 2KOH(aq) → K 2 SO 4 (aq) + 2H 2 O(l) Step 2:Use the molarity and volume given to determine the moles of KOH. moles of KOH = 0.336 M KOH x 0.0955 L = 0.032088 moles KOH

78. Aqueous Reactions and Chemical Analysis How many milliliters of a 1.42 M H 2 SO 4 solution are needed to neutralize 95.5 mL of a 0.336 M KOH solution? Solution: Step 3:Using the moles of KOH and the stoichiometric ratios from the balanced equation, convert to moles of H 2 SO 4. H 2 SO 4 (aq) + 2KOH(aq) → K 2 SO 4 (aq) + 2H 2 O(l)

79. Aqueous Reactions and Chemical Analysis How many millilitres of a 1.42 M H 2 SO 4 solution are needed to neutralize 95.5 mL of a 0.336 M KOH solution? Solution: Step 4:Using the moles of H 2 SO 4 and the concentration given, calculate the millilitres of solution.

80. Chapter Summary: Key Points 4 Properties of Aqueous Solutions Electrolytes and Nonelectrolytes Strong Electrolytes and Weak Electrolytes Precipitation Reactions Solubility of Ionic Compounds in Water Molecular Equations Ionic Equations Net Ionic Equations Acid-Base Reactions Strong Acids and Bases Brønsted Acids and Bases Acid-Base Neutralization Oxidation-Reduction Reactions Oxidation Numbers Oxidation of Metals in Aqueous Solutions Balancing Simple Redox Equations Other Types of Redox Reactions Concentration of Solutions Molarity Dilution Solution Stoichiometry Aqueous Reactions and Chemical Analysis Gravimetric Analysis Acid-Base Titrations