Chapter 4. Reactions in Aqueous Solutions. Aqueous solutions and their chemistry. Various types of reactions.

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1 Chapter 4 Reactions in Aqueous Solutions Dr. A. AlSaadi 1 Preview Aqueous solutions and their chemistry. Various types of reactions. Precipitation reactions. Acidbase reactions. Oxidationreduction reactions. The concept of molarity. Stoichiometry of reactions in aqueous solutions. Dr. A. AlSaadi 2 1

2 Chapter 4 Section 1 Aqueous Solutions Solution is a solute (for example NaCl, NaOH or ethanol) dissolved in a solvent. (When the solvent is H 2 O, => Aqueous Solutions). What common examples of solutions you can think of? Coffee, Tea, Sea, Can these types of solutions conduct electricity?? Dr. A. AlSaadi Chapter 4 Section 1 General Properties of Aqueous Solutions Aqueous solutions can conduct electric current at different efficiencies (Arrhenius postulate in 1880s). Strong electrolytes. Many ions present in solution (NaCl). Weak electrolytes. Few ions present in solution (Acetic acid). Nonelectrolytes. No ions present in solution (sugar). Lighting a bulb with aqueous solutions. Free ions work as charge carriers in solutions to complete the circuit. Dr. A. AlSaadi 4 2

3 Chapter 4 Section 1 Strong Electrolytes Strong electrolytes are completely dissolved in water to yield a solution that conducts electricity efficiently. Salts (NaCl, KI). Hydration process Strong acids (HCl, H, H 2 SO 4, HClO 4 ). Strong bases (NaOH, KOH). NaCl salt Almost no NaCl units are present. Dr. A. AlSaadi 5 Chapter 4 Section 1 Strong Electrolytes Strong Acid : produces H + ions (protons) and it is completely ionized when dissolved in water. Dr. A. AlSaadi 6

4 Chapter 4 Section 1 Strong Electrolytes When strong acids are put in water, they are completely ionized producing protons (H + ions) and anions. HCl H H 2 SO 4 H 2 O H 2 O H 2 O H + (aq) + Cl (aq) H + (aq) + (aq) H + (aq) + HSO 4 (aq) Strong bases completely dissolve in water to produce OH ions. NaOH (s) H Na + (aq) + OH 2 O (aq) KOH (s) H 2 O K + (aq) + OH (aq) Almost no HCl units are present. Almost no NaOH units are present. Dr. A. AlSaadi 7 Chapter 4 Section 1 Nonelectrolytes Nonelectrolytes can dissolve in water but don t produce ions (no electrical conductivity), like ethanol (C 2 H 5 OH) and sucrose (C 12 H 22 O 11 ). Water is a noneletrolyte and doesn t produce any ions. H 2 O OH (aq) + H + (aq) Dr. A. AlSaadi 8 4

5 Chapter 4 Section 1 Weak Electrolytes Weak electrolytes have a small degree of ionizations and exist predominantly as molecules rater than ions. Weak Acids: they are very slightly ionized in water producing a few number of protons (H + ). HC 2 H O 2 (aq) H H + (aq) + C 2 H O 2 O 2 (aq) Acetic acid has only 1% degree of dissociation. Weak Bases: they very slightly dissolve in water producing a few number of hydroxide ions (OH ). NH (aq) + H 2 O (l) H NH 4+ (aq) + OH 2 O (aq) Dr. A. AlSaadi 9 Chapter 4 Section 1 Electrolytes and Nonelectrolytes Dr. A. AlSaadi 10 5

6 Chapter 4 Section 2 Types of Chemical Reactions in Solutions Types of chemical reactions in solutions are generally: Precipitation reactions. Acidbase reactions. Oxidationreduction reactions. Dr. A. AlSaadi 11 Chapter 4 Section 2 The Hydration Process in Aqueous Solutions Why do salt, sugar, and other solid dissolve in water?? Water is a bent molecule (not linear). OH bonds are covalent (O and H atoms share electrons). Because the oxygen atom has a greater attraction for electrons, shared electrons tend to spend more time closer to the oxygen atom than to either of the hydrogen atoms. In H 2 O, oxygen is partially negative (δ ) and hydrogens are partially positive (δ+), giving rise to a polar molecule. δ means less than one unit of charge. Dr. A. AlSaadi 12 6

7 Chapter 4 Section 2 The Hydration Process When ionic substances dissolve in water, they break up (dissociate) into individual cations and anions. NaCl (s) + H 2 O (l) Na + (aq) + Cl (aq) Hydrations causes the salt to dissociate (fall apart). Dr. A. AlSaadi 1 Chapter 4 Section 2 Precipitation Reactions The abbreviation (aq) means that the ions of the compound are separate and moving around independently in water. K + CrO 4 K + K + CrO 4 K + K + K + K + CrO 4 + Ba ++ Ba ++ Ba ++ CrO 4 Ba ++ K + K + Ba ++ Ba CrO ++ 4 K + K + K + CrO 4 K 2 CrO 4 (aq) Ba( ) 2 (aq) Do these four types of ions remain as ions or some new compound precipitation could form? Dr. A. AlSaadi 14 7

8 Chapter 4 Section 2 Precipitation Reactions When an insoluble substance is produced form mixing two solutions, the reaction is said to be a precipitation p reaction and the insoluble substance is called precipitate. K 2 CrO 4 (aq) Ba( ) 2 (aq)??? 2K + (aq) + CrO 4 2 (aq) + Ba 2+ (aq) + 2 (aq) Products BaCrO 4 OR 2K Dr. A. AlSaadi 15 Chapter 4 Section 2 Precipitation Reactions 2K + (aq) + CrO 4 2 (aq) + Ba 2+ (aq) + 2 (aq) BaCrO 4 (s) + 2K (aq) How can you know which one will precipitate and which one will not? K + CrO 4 K + K + CrO 4 4 K + K + K + CrO 4 + Ba ++ Ba ++ Ba ++ + K + K + K K + K + BaCrO 4 (s) K 2 CrO 4 (aq) Ba( ) 2 (aq) Dr. A. AlSaadi 16 8

9 Chapter 4 Section 2 Precipitation Reactions Another example of precipitation reactions: Ag (aq) +KCl(aq) white solid AgCl(s) or K (s) How can you know which one will precipitate and which one will not? We need to make use of solubility rules of salts in water: Soluble. Insoluble (Not soluble). Dr. A. AlSaadi 17 Chapter 4 Section 2 Solubility Rules for Salts in Water Dr. A. AlSaadi 18 9

10 Chapter 4 Section 2 Solubility Rules for Salts in Water 2K + (aq) + CrO 2 4 (aq) + Ba 2+ (aq) + 2 (aq) BaCrO 4 (s) + 2K (aq) Ag (aq) + KCl(aq) Rule 5 indicates that it is not soluble Rule (exception) indicates that AgCl is not soluble AgCl(s) white solid Rules 1 & 2 indicate that it is soluble K + and are called spectator ions Sample Exercise Using the solubility rules, predict what will happen when the following pairs of solutions are mixed. a. K (aq) & BaCl 2 (aq) b. Na 2 SO 4 (aq) & Pb( ) 2 (aq) c. KOH(aq) & Fe( ) (aq) Dr. A. AlSaadi 19 Chapter 4 Section 2 Solubility Rules (Exercises) Solu uble Compo ounds Exceptions Insoluble Compounds 5 6 Sample Exercise Using the solubility rules above, predict what will happen when the following pairs of solutions are mixed. a. Na 2 SO 4 (aq) & Pb( ) 2 (aq) b. K (aq) & BaCl 2 (aq) c. KOH(aq) & Fe( ) (aq) PbSO 4 solid forms Rule # 4 No precipitation forms. Fe(OH) solid forms Rule # 6 Dr. A. AlSaadi 20 10

11 Chapter 4 Section 2 Describing Reactions in Solution Example: Aqueous potassium chloride is added to aqueous silver nitrate. Molecular equation: KCl (aq) + Ag (aq) It shows reactants and products as formula units but not showing the ions. AgCl (s) + K (aq) It shows all substances that are strong Ionic equation: electrolytes in their ionic forms. K + (aq) + Cl (aq) + Ag + (aq) + (aq) AgCl (s) + K + (aq) + (aq) Net ionic equation: Ag + (aq) + Cl (aq) It excludes the spectator ions from the two sides of the equation. AgCl (s) Dr. A. AlSaadi 21 Chapter 4 Section AcidBase Reactions: Introduction Acids Have sour (acidic) taste. Acetic acid in vinegar Citric acid in fruits. Hydrochloric acid of stomach reflux. Carbonic acid in soft drinks. Ascorbic acid is vitamin C. Concentrated acids are very dangerous: can dissolve metals and form hydrogen gas (H 2 ). react with carbonate slat (limestone) to produce carbon dioxide gas (CO 2 ). Dr. A. AlSaadi 22 11

12 Bases Chapter 4 Section AcidBase Reactions: Introduction Have bitter taste. Many soaps, detergents, bleaches, and toothpaste contain NaOH (caustic soda). It can dissolve grease, oil, and fat. (feel slippery). Antacids have Al(OH) or Mg(OH) 2. They neutralize the gastric acid in the stomach. In general, strong bases react with strong acids to give water and salt. Base + Acid H 2 O + salt Dr. A. AlSaadi 2 Chapter 4 Section Definitions of Acids and Bases Strong acids and strong bases: They ionize (dissociate) completely when dissolved in water. They are strong electrolytes. Arrhenius: Acids are substances that produce H + when dissolved in water. Bases are substances that produce OH when dissolved in water. Brønsted: Acids are proton donors. Bases are proton acceptors. Dr. A. AlSaadi 24 12

13 Chapter 4 Section AcidBase Reactions Arrhenius: Acids are substances that produce H + when dissolved in water. Bases are substances that produce OH when dissolved in water. Brønsted: Acids are proton donors. Bases are proton acceptors. NH is a base in the Arrhenius sense and in the Brønsted sense. H 2 O is an acid in the Brønsted sense, but not in the Arrhenius sense Dr. A. AlSaadi 25 Chapter 4 Section Types of Acids Monoprotic acid: The acid has one proton to donate. Most of the strong acids are monoprotic acids. Diprotic acid: The acid has two protons to donate. HCl and H H 2 SO 4 and H 2 CO Only H 2 SO 4 among the polyprotic acids is a strong acid. Triprotic acid: The acid has three protons to donate. Bases can also be monobasic, dibasic, tribasic. NaOH Ba(OH) 2 Al(OH) H PO 4 and H C 6 H 5 O 7 Dr. A. AlSaadi 26 1

14 Chapter 4 Section AcidBase Neutralization Reactions In neutralization reactions an aqueous acid and base produce water and salt. NaOH(aq) + HCl(aq) H 2 O(l) + NaCl(aq) strong base strong acid salt soluble in water Net ionic equation: OH (aq) + H + (aq) H 2 O (l) Other examples of acidbase neutralization reactions: H (aq) + KOH(aq) H 2 SO 4 (aq) + 2NaOH(aq) HCl(aq) + NH (aq) HCl(aq) + NH 4+ (aq) + OH (aq) H 2 O(l) + K (aq) 2H 2 O(l) + Na 2 SO 4 (aq) NH 4 Cl(aq) H 2 O(l) + NH 4 Cl(aq) Dr. A. AlSaadi 27 Chapter 4 Section 4 OxidationReduction Reactions Oxidationreduction reactions (sometimes called redox reactions) ) are reactions involving the transfer of one electron or more from one reactant to another. Redox reaction also involves the change in oxidation states for molecules. These reactions are very common in life: Photosynthesis. (conversion of CO 2 and H 2 O into sugar) Oxidation of sugar and fat in our bodies to produce energy. Combustion that provides humanity with power. Dr. A. AlSaadi 28 14

15 Chapter 4 Section 4 OxidationReduction Reactions Oxidation of zinc in a solution of copper sulfate Dr. A. AlSaadi 29 Chapter 4 Section 4 OxidationReduction Reactions Oxidation is losing electrons: Zn(s) Zn 2+ (aq) + 2e Reduction is gaining electrons: Cu 2+ (aq) + 2e Cu(s) Halfreactions Redox (oxidationreduction) reactions If something in solution gets oxidized, then something else must be reduced (and vice versa). Zn(s) + Cu 2+ (aq)+ 2e Zn 2+ (aq)+ Cu(s) + 2e Zn(s) + Cu 2+ (aq) Zn 2+ (aq)+ Cu(s) Dr. A. AlSaadi 0 15

16 Chapter 4 Section 4 Oxidation States (Oxidation Numbers) Oxidation state is an imaginary charge on an atom if the electrons were transferred completely to that atom. Normally, the shared electrons are completely assigned to the atoms the have stronger attraction for the electrons H 2 O O 2 Dr. A. AlSaadi 1 Chapter 4 Section 4 Oxidation State Rules The oxidation number for any element in its elemental form is zero (O 2, F 2 ). The oxidation number in any chemical species must sum to the overall charge on the species. The oxidation states in ionic compounds are the same as the charge each atom has by its own (PbS, NaCl) Dr. A. AlSaadi 2 16

17 Chapter 4 Section 4 Assigning Oxidation States Dr. A. AlSaadi Chapter 4 Section 4 Assigning Oxidation States Exercise: Assign oxidation states for all atoms in the following: a) CO 2 b) SF 6 c) x CO Total charge = 0 = 2(2) + x x = +4 SF 6 Total charge = 0 = 1(6) + x x 1 6 x = +6 x 2 Total charge = 1 = 2() + x x = +5 Dr. A. AlSaadi 4 17

18 Chapter 4 Section 4 OxidationReduction Process Zn(s) + Cu 2+ (aq) Zn 2+ (aq)+ Cu(s) Zn Cu 2+ 2e + 2e Zn 2+ Cu Oxidized Reduced Losing electron(s) Gaining electron(s) Oxidation state increases Oxidation state decreases Reducing agent Oxidizing agent Dr. A. AlSaadi 5 Chapter 4 Section 4 Oxidation States in Redox Reactions 2Na(s) + Cl 2 (g) 2NaCl(s) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) CH 4 CO 2 + 8e 4 +4 CH 4 is a reducing agent 2O 2 + 8e CO 2 + 2H 2 O O 2 is an oxidizing agent Dr. A. AlSaadi 6 18

19 Chapter 4 Section 4 Exercise For the following two reactions: * determine the oxidation states, * identify the atoms that are oxidized and reduced, and * specify the oxidizing and reducing agents. 2PbS(s) + O 2 (g) 2PbO(s) + 2SO 2 (g) PbO(s) + CO(g) Pb(s) + CO 2 (g) Dr. A. AlSaadi 7 Chapter 4 Section 4 Redox Reactions in Aqueous Solutions Zn(s) + CuCl 2 (aq) ZnCl 2 (aq)+ Cu(s) Displacement reaction What would happen if you place copper metal into a solution of ZnCl 2? Would Cu(s) be oxidized by Zn 2+ (aq) ions the way Zn(s) is oxidized by Cu 2+ (aq) ions? Cu(s) + ZnCl 2 (aq) no reaction Dr. A. AlSaadi 8 19

20 Chapter 4 Section 4 The Activity Series The activity series shows the order of ease the metal is to be oxidized. Metals at the top of the list are called the active metals. Metals at the bottom of the list are known as noble metals. Dr. A. AlSaadi 9 Chapter 4 Section 4 Balancing OxidationReduction Equations The HalfReaction Method: A half reaction is that reaction that involves either oxidation or reduction. Ce 4+ (aq) + Sn 2+ (aq) Ce + (aq) + Sn 4+ (aq) 2 Ce 4+ (aq) + 2 e 2 Ce + (aq) Sn 2+ (aq) Sn 4+ (aq) + 2e 2Ce 4+ (aq) + Sn 2+ (aq) 2Ce + (aq) + Sn 4+ (aq) Atoms and charges (electrons) must be all balanced. Dr. A. AlSaadi 40 20

21 Chapter 4 Section 5 Concentration of Solutions Concentration is the amount of chemicals (solutes) present (dissolved) in the solution. Moles of solute n Molarity = M = = Liters of solution V Molar Concentration Molarity has the unit of mol/l, mol L 1, M. If you have 0.1 moles of NaOH present in 1L aqueous solution, the solution has 0.1M concentration. Useful web links: Dr. A. AlSaadi 41 Chapter 4 Section 5 Exercise on Molarity Calculations Calculate the molarity of solution prepared by dissolving 1.56g of HCl in water to make 26.8 ml solution. Dr. A. AlSaadi 42 21

22 Chapter 4 Section 5 Preparing a Solution from a Solid A standard solution is a solution whose concentration is accurately known. Steps of preparing a standard solution: Volumetric flask Distilled water Dr. A. AlSaadi 4 Chapter 4 Section 5 Dilution Dilution is the procedure of adding water to stock solutions often are concentrated solutions and kept in the laboratory to achieve the desired concentration. Dr. A. AlSaadi 44 22

23 Chapter 4 Section 5 Dilution Dilution is the procedure of adding water to stock solutions often are concentrated solutions and kept in the laboratory to achieve the desired concentration. It is always true that: Moles of solute before dilution = Moles of solutes after dilution Since no. of moles = M V = moles liters liters Then, M V (before dilution) = M V (after dilution) M c V c = M d V d Dr. A. AlSaadi 45 Chapter 4 Section 5 Dilution What volume of 1.00M KMnO 4 is needed to prepare p 1.00 L of a 0.400M KMnO 4 solution? Moles of solute before dilution = Moles of solutes after dilution M d V d = M c V c V c = M d / M c V d = 0.400M / 1.00M 1.00L = L Answer is 400 ml of the 1.00M KMnO 4 stock solution. Dr. A. AlSaadi 46 2

24 Chapter 4 Section 5 Solution Stoichiometry Soluble ionic compounds are strong electrolytes, i.e. they dissociate completely and exist as ions in aqueous solutions. Examples: H 2 O KMnO 4 (s) K + (aq) + MnO 4 (aq) Concentrations: 0.40M [K + ] = 0.40M [MnO 4 ] = 0.40M H 2 O Na 2 SO 4 (s) 2Na + (aq) + SO 4 2 (aq) Concentrations: 0.40M [Na + ] = 0.80M [SO 4 2 ] = 0.40M Dr. A. AlSaadi 47 Chapter 4 Section 5 Exercises on Solution Stoichiometry Calculations Give the concentration of ClO 4 ions in 1M Fe(ClO 4 ) solution. Dr. A. AlSaadi 48 24

25 Chapter 4 Section 5 Exercises on Solution Stoichiometry Calculations Calculate the number of moles of Cl ions in 1.75L of M ZnCl 2 solution. Dr. A. AlSaadi 49 Chapter 4 Section 5 Exercises on Solution Stoichiometry Calculations 28.0 ml of 0.250M H and 5.0 ml of 0.20M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What are the concentrations of H + and OH ions in excess after the reaction goes to completion? Net ionic equation: H + (aq) + OH (aq) H 2 O (l) From volume and conc. find the moles for H + and OH. moles of H + = mol moles of OH = mol Determine which reactant is the limiting one. Then find the amount of H 2 O formed. Conc. of excess OH Moles of unreacted OH = Total volume of solution Dr. A. AlSaadi 50 25

26 Chapter 4 Section 5 Exercises on Solution Stoichiometry Calculations When aqueous solutions of Na 2 SO 4 and Pb( ) 2 are mixed, PbSO 4 precipitates. p Calculate the mass of PbSO 4 formed when 1.25L of M Pb( ) 2 and 2.00L of M Na 2 SO 4 are mixed. How many ions of Pb 2+ will remain unreacted in the solution? 1. Identify the ions and possible solid product. 2Gi 2. Give net ionic i equation.. Find numbers of moles for Pb 2+ and SO Which one is limiting? 5. Calculate moles (then grams) of PbSO 4 based on limiting reactant. Dr. A. AlSaadi 51 Chapter 4 Section 5 Exercises on Solution Stoichiometry Calculations What mass of Na 2 CrO 4 is required to precipitate p all of the silver ions from 75.0 ml of a M solution of Ag? Dr. A. AlSaadi 52 26

27 Chapter 4 Section 6 Aqueous Reactions and Chemical Analysis Many aqueous reactions are very useful for determining how much of a particular substance is present in a sample. Gravimetric analysis. Acidbase titration. Dr. A. AlSaadi 5 Chapter 4 Section 6 Gravimetric Analysis It is an analytical technique that is based on the measurement of mass. The precipitate formed out of a precipitation reaction is isolated and measured. The reaction must have 100% yield. The precipitate must be completely insoluble. Dr. A. AlSaadi 54 27

28 Chapter 4 Section 6 Gravimetric Analysis A 0.86g sample of an ionic compound MCl x is dissolved in water and treated with an excess Ag. if g of AgCl precipitate forms, what is the % by mass of Cl in MCl x? % mass of Cl in AgCl = mass of fcli in AgCl ppt = % mass of Cl in MCl x = Dr. A. AlSaadi 55 Chapter 4 Section 6 AcidBase Titrations Titration (or standardization) is used to characterize aqueous solutions (acidic or basic) of an unknown concentration. It is done by gradually adding a strong acid (or strong base) solution (titrant) of known concentration to a base (or acid) solution (analyte) for which the concentration is needed to be determined, with the presence of an indicator. Base (titrant) Acid (analyte) + indicator Dr. A. AlSaadi 56 28

29 Chapter 4 Section 6 AcidBase Titrations OH OH OH Titrant OH OH OH OH OH OH # mol OH = VxM OH OH OH OH (dispensed) d) (Known concentration) OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH Equivalence point: Change in color (indicator) #mol OH = # mol H + Analyte (known volume, + H H + H + + H OH H + OH H + OH Unknown H + H + H + H + H + H + H + H + OH OH OH H + OH H + OH concentration) Dr. A. AlSaadi H + (aq) + OH (aq) H 2 O (l) 57 Chapter 4 Section 6 AcidBase Titrations In this example, when the reaction is completed, the base titrant neutralizes the acid analyte. The point of neutralization (end point or equivalence point) The end point can be visually located by using indicators, that change their colors when an access of the titrant is present in the solution. This experiment has to be done very carefully. Known volume and concentration Known volume Dr. A. AlSaadi 58 29

30 Chapter 4 Section 6 Standardization of NaOH Solution Using KHP Standardization is the process to accurately determine the concentration of a solution before using it as a titratnt in a titration experiment. KHP is potassium hydrogen phthalate. K It is a monoprotic acid that is used to standardize + NaOH solutions of unknown concentrations. H( NaOH(aq) )+KHP( KHP(aq) Na + (aq)+oh (aq)+k + (aq)+hp (aq) OH (aq) + HP (aq) P 2 (aq) + H 2 O(l) (Net ionic equation) In this case, solution of KHP which has a known mass is titrated with NaOH of unknown concentration. Dr. A. AlSaadi 59 Chapter 4 Section 6 Standardization of NaOH Solution Using KHP K + In a titration experiment, it was found that ml of NaOH solution was needed to neutralize g of KHP. What is the concentration of the NaOH solution? OH (aq) + HP (aq) P 2 (aq) + H 2 O(l) Dr. A. AlSaadi 60 0

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