Chapter 4 Aqueous solutions Types of reactions Parts

Chapter 4 Aqueous solutions Types of reactions

Parts of Solutions n n n n Solution – Homogeneous mixture. Solute – What gets dissolved. Solvent – What does the dissolving. Soluble – Can be dissolved. Insoluble – Can’t be dissolved Miscible – Liquids dissolve in each other. Immiscible – Liquids that don’t dissolve in each other

Aqueous solutions n n Dissolved in water. Water is a good solvent because the molecules are polar. The oxygen atoms have a partial negative charge. The hydrogen atoms have a partial positive charge.

Hydration n Hydration is the process of breaking the ions of salts apart. Ions have charges and attract the opposite charges on the water molecules. An ion is surrounded by other ions of the opposite charge.

Hydration Also check out the animation on the website. H O H O H H H H O O H H H O H H

Solubility n n How much of a substance will dissolve in a given amount of water. Usually g of solute / 100 grams of H 2 O Varies greatly, but if they do dissolve the ions are separated, and they move around. Water can also dissolve non-ionic compounds if they have polar bonds, such as soap.

Electrolytes n n Electricity is moving charges (electrons). The ions that are dissolved can move. Solutions of ionic compounds can conduct electricity. Solutions are classified three ways in their ability to conduct electricity.

Types of solutions n Strong electrolytes - completely dissociate n n Many ions - Conduct well. Weak electrolytes - Partially split into ions. n n n Examples: Na. Cl Examples: Vinegar, ammonia Few ions - Conduct slightly. Non-electrolytes - Don’t split apart. n n Examples: Sand, plastic No ions - Don’t conduct.

Types of solutions - Acidic Acids – form H+1 ions when dissolved. n Strong acids ionize completely. n The following are strong acids. n H 2 SO 4 HNO 3 HCl. O 4 HBr HI HCl. O 3 n Weak acids – partial ionization Example: HC 2 H 3 O 2

Types of Solutions - Basic n n n Bases – form OH-1 ions when dissolved. Strong bases – ionize completely. Any Group 1 hydroxide is considered to be a strong base. n n Examples: Na. OH and KOH Transition metal hydroxides will be weaker bases. n Fe(OH)2 and Cu. OH

Point of Emphasis n n Strong v. Weak & Concentrated v. Dilute These terms often get misapplied or confused. Strong or Weak is a description of how much a species will dissociate into ions and is an intensive property. Concentrated or Dilute is a description of how much solute is used by the experimenter.

Some examples n n Strong and Concentrated means that the substance will dissociate fully and the experimenter used a lot of it – it is in a high concentration. Ex. 8 M HCl Strong and Dilute means that the substance will dissociate fully and the experimenter used a small concentration. Ex. . 001 M HCl

Some more examples n n Weak and concentrated means that the substance doesn’t dissociate fully and the experimenter used a lot of it. Ex. 8 M HC 2 H 3 O 2 (acetic acid) Weak and dilute means that the substance doesn’t dissociate fully and the experimenter used a small concentration. Ex. . 001 M HC 2 H 3 O 2

Measuring Solutions n n n Concentration - how much is dissolved. Molarity = Moles of solute n Liters of solution L abbreviated M (Capital M) 1 M = 1 mole solute / 1 liter solution Calculate the molarity of a solution with 34. 6 g of Na. Cl dissolved in 125 m. L of water.

Solved Answer n n n Calculate the molarity of a solution with 34. 6 g of Na. Cl dissolved in 125 m. L of water. First – find moles of Na. Cl . 592 moles Second – set up formula M = n / L M =. 592 /. 125 L n M = 4. 736 M n

Molarity n n n How many grams of HCl would be required to make 50. 0 m. L of a 2. 7 M solution ? What would the concentration be if you dissolved 27 g of Ca. Cl 2 in 500 m. L of water ? What is the concentration of each ion ?

Solved Answer n n How many grams of HCl would be required to make 50. 0 m. L of a 2. 7 M solution ? Use M = n / L 2. 7 = n /. 050 L n n =. 135 moles n Use n = g / M 4. 921 g of HCl n n This is added with enough water to make 50 m. L of soln.

More Solved Answers n n What would the concentration be if you dissolved 27 g of Ca. Cl 2 and made 500 m. L of solution ? (FW of Ca. Cl 2 = 110. 98 g) Find moles 27 / 110. 98 =. 243 moles n Use M = n / L n M =. 243 /. 500 L n M =. 487 M (can be read as. 487 Molar)

More Solved Answers n n n The question also included solving the concentration of each ion. Ca. Cl 2 Ca+2 + 2 Cl -1 (bal. eq. ) Notice the 2: 1 ratio of Cl-1 ion to Ca. Cl 2 So the concentration of the Cl-1 ion would be 2 x. 487 . 974 M The concentration of the Ca+2 is. 487 M

Molarity n n n Calculate the concentration of a solution made by dissolving 45. 6 g of Na 2 SO 4 to 475 m. L. (FW of Na 2 SO 4 = 142. 06) What is the concentration of each ion ? A balanced equation is a good place to start Na 2 SO 4 2 Na+1 + SO 4 -2 This shows 2 Na+1 ions for every sulfate.

A Little Stoichiometry n n Calculate the concentration of a solution made by dissolving 45. 6 g of Na 2 SO 4 to 475 m. L. Solve as before: 45. 6 / 142. 06 . 321 moles Use M = n / L. 321 /. 475 =. 676 M Since the Sodium Ions are in a 2 to 1 ratio there would be 1. 352 M Na+1 and there would be. 676 M SO 4 -2

Look at the Stoichiometry n Let’s look at the dissociation n n Na 2 SO 4 2 Na+1 + SO 4 -2 We see the 2 : 1 ratio of the two ions, so that is how we get the values from the slide above.

Making solutions # 1 n n n Describe how to make 100. 0 m. L aliquot (sample) of a 1. 0 M K 2 Cr 2 O 4 solution. The formula wt. of the solute is 246. 2 g Using M = n / L 1. 0 M = n /. 1 L So we need. 1 moles of solute. Using n = g / m . 1 = g / 246. 2 We need 24. 62 grams of solute

Making Solutions # 2 n n n Describe how to make 250. m. L of an 2. 0 M copper (II) sulfate solution. The formula wt. of the solute is 159. 6 g. Using M = n / L 2 = n /. 25 So we need. 5 moles of solute Using n = g / M . 5 = g / 159. 6 We need 79. 8 grams of solute

Dilution n n n Adding more solvent to a known solution. The moles of solute stay the same. moles = M x L from M = n / L moles = moles M 1 V 1 = M 2 V 2 (this is a very handy formula) Stock solution is a solution of known concentration used to make more dilute solutions

Dilution # 1 n n n What volume of a 1. 7 M solution is needed to make 250 m. L of a 0. 50 M solution ? 1. Given info. M 1 = 1. 7 M 2 =. 5 V 2 =. 25 2. Formula M 1 V 1 = M 2 V 2 3. Substitute 1. 7 V 1 =. 5 (. 25) 4. Solve. 0735 L (or 73. 5 m. L)

Dilution # 2 n n n 18. 5 m. L of 2. 3 M HCl is added to 250 m. L of water. What is the concentration of the solution ? 1. Given info. V 1 = 18. 5 M 1 = 2. 3 V 2 =. 25 2. Formula M 1 V 1 = M 2 V 2 3. Substitute 18. 5 (2. 3) = x (. 25) 4. Solve. 1702 M

Dilution n n You have a 4. 0 M stock solution. Describe how to make 1. 0 L of a. 75 M solution. Using M 1 V 1 = M 2 V 2 1. 0 (. 75) = 4. 0 V 2 =. 188 L (or 18. 75 m. L) Take 18. 75 m. L of the 4. 0 M Stock soln. and add enough water to make 1. 0 L

Types of Reactions Precipitation reactions n When aqueous solutions of ionic compounds are poured together a solid forms. n A solid that forms from mixed solutions is called a precipitate n If you’re not a part of the solution, your part of the precipitate (old chem joke)

Precipitation Reactions n n n Only happens if one of the products is insoluble. Otherwise all the ions stay in solutionnothing has happened. Need to memorize the rules for solubility.

Some Solubility Rules We need to know these. 1 All nitrates and acetates are soluble 2 Alkali metals ions and NH 4+1 ions are soluble 3 Halides are soluble except Ag+1, Pb+2, and Hg 2+2 4 Most sulfates are soluble, except Pb+2, Ba+2, Hg+2, and Ca+2

Solubility Rules 5 Most hydroxides are slightly soluble (insoluble) except Na. OH and KOH 6 Sulfides, carbonates, chromates, and phosphates are generally insoluble

Precipitation reactions n Na. OH(aq) + Fe. Cl 3(aq) ® Na. Cl(aq) + Fe(OH)3(s) n n Na+1(aq)+OH-1(aq) + Fe+3(aq) + Cl-1(aq) ® Na+1(aq) + Cl-1 (aq) + Fe(OH)3 (s) n n n is really So all that really happens is 3 OH-1(aq) + Fe+3(aq) ® Fe(OH)3(s) Essentially a more complex double replacement reaction.

Precipitation reaction n We can predict the products by knowing types of reactions and the rules for solubility Ag. NO 3 (aq) + KCl (aq) ® n Zn(NO 3)2 (aq) + Ba. Cr 2 O 7 (aq) ® n Cd. Cl 2 (aq) + Na 2 S (aq) ® n

Answers n Ag. NO 3 (aq) + KCl (aq) KNO 3 (aq) + Ag. Cl (s) We know the KNO 3 is soluble because all Group 1 metals are soluble. n Zn(NO 3)2 (aq) + Ba. Cr 2 O 7 (aq) Ba(NO 3)2 (aq) + Zn. Cr 2 O 7 (s) The nitrates are soluble. Cd. Cl 2 (aq) + Na 2 S (aq) 2 Na. Cl (aq) + Cd. S (s) The Group 1 Sodium ion is soluble. n

Precipitation Stoichiometry n n 25 m. L of 0. 67 M H 2 SO 4 is added to 35 m. L of 0. 40 M Ca. Cl 2. What mass Ca. SO 4 Is formed ? First we need a balanced equation and moles of each reactant. H 2 SO 4 + Ca. Cl 2 Ca. SO 4 + 2 HCl. 0168. 014

Precipitation Stoichiometry H 2 SO 4 + Ca. Cl 2 Ca. SO 4 + 2 HCl. 0168. 014 The stoichiometry is 1: 1 so Ca. Cl 2 is the limiting reagent. The ratio would be: 1 n Ca. Cl 2 / 1 Ca. SO 4 =. 014 / x x =. 014 n of Ca. SO 4 Convert to grams n = g / M. 014 = g / 136. 14 g = 1. 9 grams

Three Types of Equations n n n Molecular Equation - written as whole formulas, not the ions. K 2 Cr. O 4 (aq) + Ba(NO 3)2 (aq) ® Ba. SO 4 (s) + 2 KNO 3 (aq) Complete Ionic Equation show dissolved electrolytes as the ions. 2 K+1 + Cr. O 4 -2 + Ba+2 + 2 NO 3 -1 ® Ba. Cr. O 4 (s) + 2 K+1 + 2 NO 3 -1 Spectator ions are those that don’t react.

Three Type of Equations n n Net Ionic Equations show only those ions that react, not the spectator ions Ba+2 + Cr. O 4 -2 ® Ba. Cr. O 4 (s) Write three types of equations for the reactions when these solutions are mixed. a. Iron (III) sulfate and potassium sulfide b. Lead (II) nitrate and sulfuric acid.

Three Types of Equations n a. Iron (III) sulfate and potassium sulfide Fe 2(SO 4)3 (aq) + K 2 S (aq) K 2 SO 4 (aq) + Fe 2 S 3 Fe+3 + SO 4 -2 + K+1 + S-2 K+1 + SO 4 -2 + Fe 2 S 3 Fe+3 n (aq) + S-2 (aq) Fe 2 S 3 (s) (s) b. Lead (II) nitrate and sulfuric acid. Pb(NO 3)2 (aq) + H 2 SO 4 (aq) HNO 3 (aq) + Pb. SO 4 (s) Pb+2 + NO 3 -1 + H+1 + SO 4 -2 NO 3 -1 + H+1 + Pb. SO 4 Pb+2 + SO 4 -2 Pb. SO 4 (s)

Stoichiometry of Precipitation n n Exactly the same as regular stoichiometry, except you are finding moles in a different way. Concepts of Limiting Reagent and materials in excess are also the same as before.

Stoichiometry of Precipitation – Problem # 1 n n What mass of solid is formed when 100. 00 m. L of 0. 100 M Barium chloride is mixed with 100. 00 m. L of 0. 100 M sodium hydroxide ? Balanced Equation: n n Ba. Cl 2(aq) + 2 Na. OH(aq) Ba(OH)2(s) + 2 Na. Cl(aq) Calculate Moles of each reactant n n Ba. Cl 2 M = n / L. 1 = n /. 1 L n =. 01 n Na. OH same. 01 n

Stoichiometry of Precipitation – Problem # 1 n Look at ratios to determine limiting reagent. n Ba. Cl 2 n n n (aq) + 2 Na. OH (aq) Ba(OH)2 (s) + 2 Na. Cl . 01 2 / 1 =. 01 Na. OH / x x =. 005 moles So Na. OH is the limiting reagent Proportion 2 n Na. OH / 1 n Ba(OH)2 =. 01 / x x =. 005 moles of Ba(OH)2 =. 857 grams (aq)

Stoichiometry of Precipitation – Problem # 2 What volume of 0. 204 M HCl is needed to precipitate all the silver from 50. ml of 0. 0600 M silver nitrate solution ? n HCl (aq) + Ag. NO 3 (aq) Ag. Cl (s) + HNO 3 (aq). 003 n It is a 1: 1 ratio so the moles are the same. Use M = n / L. 204 =. 003 / L L =. 0147 L n

Types of Reactions Acid-Base n For our purposes an acid is a H+1 donor. n n n A base is a OH-1 donor. A neutralization is when there is equal numbers of moles of acid and moles of base. A titration is the lab procedure used to achieve this.

Acid – Base Reactions Always remember the following when dealing with Acid-Base Reactions n n n Acid + Base salt + water Net Ionic Form: H+1 + OH-1 H 2 O Dissociation of Water: H 2 O H+1 + OH-1 General Formula:

Acid - Base Reactions n n n Often called a neutralization reaction Because the acid neutralizes the base. Often titrate to determine concentrations. Solution of known concentration (titrant), Is added to the unknown (analyte), Until the equivalence point is reached where enough titrant has been added to neutralize it.

Titration n n Where the indicator changes color is the endpoint. Not always at the equivalence point. A 50. 00 m. L sample of aqueous Ca(OH)2 requires 34. 66 m. L of 0. 0980 M Nitric acid for neutralization. What is [ Ca(OH)2 ] ? # of [H+1] = # of [OH-1]

Titration Solved Problem n n n 50. 00 m. L sample of aqueous Ca(OH)2 requires 34. 66 m. L of 0. 0980 M Nitric Acid for neutralization. What is [ Ca(OH)2 ] ? Ca(OH)2 + 2 HNO 3 Ca(NO 3)2 + H 2 O. 098 = n /. 03466 n =. 0034 moles H+1 Ca(OH)2 Ca+2 + 2 OH-1. 0017. 0034. 0017 /. 05 = M [ Ca+2] =. 034 M

Acid – Base Reaction This is an important problem n n 75 m. L of 0. 25 M H 2 SO 4 is mixed with 225 m. L of 0. 055 M Na. OH. What is the concentration of the excess H+1 or OH-1 ? Think about what you are being asked to solve and about the process that is going on. In other words, don’t just dive in before you plan the problem.

Acid – Base Reaction First lets get a balanced equation. H 2 SO 4 + 2 Na. OH Na 2 SO 4 + 2 H 2 O n Notice this follows the general form that we talked about a few slides back. Acid + Base Salt + Water H+1 + OH-1 H 2 O

Acid – Base Reaction n We need to determine moles of each reactant. H 2 SO 4: M = n / L. 25 = n /. 075 n =. 0188 n n Na. OH: same format n n . 0124 moles We need to set up a proportion to determine which species is the Limiting Reagent and which is in excess.

Acid – Base Reaction n n 1 n H 2 SO 4 / 2 n Na. OH : x /. 0124 n Na. OH Solving for x gives. 0062 n of H 2 SO 4 When we compare that to the given we see that we have a lot more of the acid than what is needed so we can conclude that H 2 SO 4 is in excess and the Na. OH is the limiting reagent. So what is the next step ?

Acid – Base Reaction Since we are given. 0188 n of H 2 SO 4 and only need. 0062 moles we are left with. 0188 -. 0062 =. 0126 n of H 2 SO 4 in excess n However, each molecule of H 2 SO 4 liberates 2 ions of H+1 so we need to factor that in. H 2 SO 4 2 H+1 + SO 4 -2. 0126. 0252. 0126

Acid – Base Reaction n n This is a limiting reagent problem as well as an acid base problem. After the reaction takes place and accounting for the dissociation of the acid we have the following: . 0252 n of H+1 But we were asked for concentration. Clearly we need to use M = n / L , but there is a catch. Solve for M in your binder.

Acid – Base Reaction n n We need moles and we just solved for that and got. 0252 moles of H+1. What about the volume ? Both reagents are in the solution together, so the volumes need to be combined. The total volume is 75 + 225 m. L. M =. 0252 /. 3 M =. 084 M H+1 Forgetting to add the volumes together is a common mistake in these types of problems.

Types of Reaction - Redox Oxidation-Reduction called Redox n Ionic compounds are formed through the transfer of electrons. n An Oxidation – Reduction reaction involves the transfer of electrons. n We need a way of keeping track.

Oxidation States A way of keeping track of the electrons. n You need to memorize the rules for assigning oxidation states. 1 The oxidation state of elements in their standard states is zero. 2 Oxidation state for monoatomic ions are the same as their charge. n

Oxidation states 3 Oxygen is assigned an oxidation state of -2 in its covalent compounds except as a peroxide. 4 In compounds with nonmetals, hydrogen is assigned the oxidation state +1. 5 In its compounds fluorine is always – 1. 6 The sum of the oxidation states must be zero in compounds or equal the charge of the ion.

Oxidation States n n n Assign the oxidation states to each element in the following. CO 2 NO 3 -1 H 2 SO 4 Fe 2 O 3

Oxidation States n n n Assign the oxidation states to each element in the following. CO 2 C=+4 O = -2 NO 3 -1 N=+5 O = -2 H 2 SO 4 H = + 1 S = + 6 O = -2 Fe 2 O 3 Fe = + 3 O = -2

Oxidation-Reduction n n n Transfer electrons, so the oxidation states change. 2 Na + Cl 2 ® 2 Na. Cl CH 4 + 2 O 2 ® CO 2 + 2 H 2 O Oxidation is the loss of electrons. Reduction is the gain of electrons. OIL RIG or LEO GER or OLE

Oxidation-Reduction n n Oxidation means an increase in oxidation state - lose electrons. Reduction means a decrease in oxidation state - gain electrons. The substance that is oxidized is called the reducing agent. The substance that is reduced is called the oxidizing agent.

Redox Reactions

Redox Vocabulary n n n n Oxidizing agent gets reduced. Gains electrons. More negative oxidation state. Reducing agent gets oxidized. Loses electrons. More positive oxidation state. Both have to happen in a reaction for it to occur.

Identify the following n n n Oxidizing agent & reducing agent Substance oxidized & Substance reduced for the following reactions Fe (s) + O 2(g) ® Fe 2 O 3(s)+ 3 CO(g) ® 2 Fe(s ) + 3 CO 2(g) SO 3 -2 + H+1 + Mn. O 4 -1 ® SO 4 -2 + H 2 O + Mn+2

Answers n Fe (s) + O 2(g) ® Fe 2 O 3(s) Iron is oxidized, Oxygen is reduced Iron is the reducing agent, Oxygen is the oxidizing agent n Fe 2 O 3(s)+ 3 CO(g) ® 2 Fe(s ) + 3 CO 2(g) Fe 2 O 3 is reduced, CO is oxidized Fe 2 O 3 is the oxidizing agent, CO is the reducing agent n SO 3 -2 + H+1 + Mn. O 4 -1 ® SO 4 -2 + H 2 O + Mn+2 S is oxidized, Mn is reduced S is the reducing agent, Mn is the oxidizing agent

Balancing Half-Reactions is being deemphasized for the AP Test. (But it is good to know theory behind it) n n n All redox reactions can be thought of as happening in two halves. One produces electrons - Oxidation half. The other requires electrons - Reduction half. Write the half reactions for the following. Na + Cl 2 ® Na+1 + 2 Cl-1 SO 3 -2 + H+1 + Mn. O 4 -1 ® SO 4 -2 + H 2 O + Mn+2

Balancing Half-reactions n n Na + Cl 2 Na+1 + Cl-1 Na Na+1 + e-1 (oxidation ½ reaction) Cl 2 + 2 e-1 2 Cl-1 (reduction ½ reaction) SO 3 -2 + H+1 + Mn. O 4 -1 SO 4 -2 + H 2 O + Mn+2 SO 3 -2 + H 2 O SO 4 -2 + 2 H+1 + 2 e-1 8 H+1 + Mn. O 4 -1 + 5 e-1 Mn+2 + 4 H 2 O

Redox Theory n You need to remember that while these are studied separately, the oxidation halfreaction and the reduction half-reaction must always happen together. You can’t have a species gaining electrons unless you have a species there to lose those same electrons.

Balancing Redox Equations In aqueous solutions the key is the number of electrons produced must be the same as those required. n For reactions in acidic solution an 8 step procedure. 1 Write separate half reactions 2 For each half reaction balance all reactants except H and O 3 Balance O using H 2 O n

Acidic Solution 4 Balance H using H+1 5 Balance charge using e-1 6 Multiply equations to make electrons balance. 7 Add equations and cancel identical species. 8 Check that charges and elements are balanced.

Summary of Rules in Acidic Medium n Balance the atoms in each half-reaction. n n This refers to atoms other than H and O Balance the Oxygen by adding H 2 O where needed. Balance the Hydrogen by adding H+1 where needed. Balance by charge by adding e-1

Balancing in Acidic Medium Problem # 1 n Fe (s) + O 2 (g) ® Fe 2 O 3 (s) Determine the charge on each species n 0 0 +3 -2 Fe (s) + O 2 (g) ® Fe 2 O 3 (s) So: Fe Fe+3 (Fe is oxidized) & O 2 O-2 (O 2 is reduced) n

Balancing in Acidic Medium Problem # 1 Fe Fe+3 (Fe is oxidized) Fe Fe+3 + 3 e-1 O 2 O-2 (O 2 is reduced) O 2 2 O-2 4 e-1 + O 2 2 O-2

Balancing in Acidic Medium Problem # 1 n n The final step is to combine the two ½ reactions. But first we must get rid of the electrons 4 ( Fe Fe+3 + 3 e-1 ) 3 ( 4 e-1 + O 2 2 O-2 ) We get: 4 Fe + 3 O 2 4 Fe+3 + 6 O-2

Balancing in Acidic Medium Problem # 2 n Fe 2 O 3 (s)+ 3 CO +3 n n n -2 (g) +2 -2 Fe 2 O 3 (s)+ 3 CO ® 2 Fe (s ) 0 (g) ® 2 Fe + 3 CO 2 (g) +4 -2 (s ) + 3 CO 2 (g) Fe+3 + 3 e-1 Fe (Fe is reduced) a. CO 2 b. H 2 O + CO 2 c. H 2 O + CO 2 + 2 H+1 d. H 2 O + CO 2 + 2 H+1 + 2 e-1

Balancing in Acidic Medium Problem # 2 n n n Fe+3 + 3 e-1 Fe (Fe is reduced) The CO is oxidized. Here are the steps. Atoms: CO 2 Oxygen (with water) H 2 O + CO 2 Hydrogen (with H+1) H 2 O + CO 2 + 2 H+1 Charge (with e-1) H 2 O + CO 2 + 2 H+1 + 2 e-1

Balancing in Acidic Medium Problem # 2 n Once again, the final step is to factor out the electrons and combine the two ½ reactions. 2 ( Fe+3 + 3 e-1 Fe ) 3 (H 2 O + CO 2 + 2 H+1 + 2 e-1 ) n 2 Fe+3 + 3 H 2 O + 3 CO 2 Fe +3 CO 2 + 6 H+1 n n

Balancing in Acidic Medium Problem # 3 n SO 3 -2 + H+1 + Mn. O 4 -1 SO 4 -2 + H 2 O + Mn+2 n +4 -2 n SO 3 -2 + H+1 + Mn. O 4 -1 SO 4 -2 + H 2 O + Mn+2 n +1 +7 -2 +6 -2 +1 -2 +2 We can see that the S is being oxidized and the Mn is being reduced.

Balancing in Acidic Medium Problem # 3 Oxidation ½ reaction n n n +4 -2 +1 +7 -2 +6 -2 +1 -2 +2 SO 3 -2 + H+1 + Mn. O 4 -1 SO 4 -2 + H 2 O + Mn+2 a. SO 3 -2 SO 4 -2 b. H 2 O + SO 3 -2 SO 4 -2 c. H 2 O + SO 3 -2 SO 4 -2 + 2 H+1 d. H 2 O + SO 3 -2 SO 4 -2 + 2 H+1 + 2 e-1 Sulfur is oxidized. How can you tell ?

Balancing in Acidic Medium Problem # 3 Reduction ½ reaction n +4 -2 n SO 3 -2 + H+1 + Mn. O 4 -1 SO 4 -2 + H 2 O + Mn+2 n n n +1 +7 -2 +6 -2 +1 -2 +2 a. Mn. O 4 -1 Mn+2 b. Mn. O 4 -1 Mn+2 + 4 H 2 O c. 8 H+1 + Mn. O 4 -1 Mn+2 + 4 H 2 O d. 5 e-1 + 8 H+1 + Mn. O 4 -1 Mn+2 + 4 H 2 O Factor out the electrons and then combine the two half – reactions.

Balancing in Acidic Medium Problem # 3 n n n H 2 O + SO 3 -2 SO 4 -2 + 2 H+1 + 2 e-1 5 e-1 + 8 H+1 + Mn. O 4 -1 Mn+2 + 4 H 2 O Multiply the first ½ rxn. By 5 and the second ½ rxn. by 2 to factor out the e-1. n 5 H 2 O + 5 SO 3 -2 5 SO 4 -2 + 10 H+1 + 10 e-1 + 16 H+1 + 2 Mn. O 4 -1 2 Mn+2 + 8 H 2 O n We can factor out species other than e-1 n

Balancing in Acidic Medium Problem # 3 n n n n 5 H 2 O + 5 SO 3 -2 5 SO 4 -2 + 10 H+1 + 10 e-1 + 16 H+1 + 2 Mn. O 4 -1 2 Mn+2 + 8 H 2 O We can start with the water and electrons. 5 H 2 O + 5 SO 3 -2 5 SO 4 -2 + 10 H+1 + 10 e-1 + 16 H+1 + 2 Mn. O 4 -1 2 Mn+2 + 8 3 H 2 O Next we can factor out the H+1 ions. 5 SO 3 -2 5 SO 4 -2 + 10 H+1 16 6 H+1 + 2 Mn. O 4 -1 2 Mn+2 + 3 H 2 O

Balancing in Acidic Medium Problem # 3 n 5 SO 3 -2 5 SO 4 -2 6 H+1 + 2 Mn. O 4 -1 2 Mn+2 + 3 H 2 O n Combining the two gives us: n n 6 H+1 + 2 Mn. O 4 -1 + 5 SO 3 -2 2 Mn+2 + 3 H 2 O + 5 SO 4 -2

Basic Solution n n Do everything you would with acid, but add one more step. Add enough OH-1 to both sides to neutralize the H+1 Determine the species undergoing redox in the problems below. -1 -1 -1 Cr. I + Cl ® Cr. O + IO + Cl 3 2 4 4

Redox – Basic Solutions n Cr. I 3 + Cl 2 ® Cr. O 4 -1 + IO 4 -1 + Cl-1 Cr+3 Cr+7 (Cr. O 4 -1) Cr is oxidized n I-1 I+7 I is oxidized n Cl 2 Cl-1 Cl is reduced n

Practice n n n Cr(OH)3 + OCl-1 + OH-1 Cr. O 4 -2 + Cl-1 + H 2 O Mn. O -1 + Fe+2 ® Mn+2 + Fe+3 4 Cu + NO 3 -1 ® Cu+2 + NO(g) Pb + Pb. O 2 + SO 4 -2 ® Pb. SO 4 Mn+2 + Na. Bi. O 3 ® Bi+3 + Mn. O 4 -1

Balancing in Basic Medium Problem # 1 n n n Mn+2 + Na. Bi. O 3 ® Bi+3 + Mn. O 4 -1 You must know the procedure to do this or any problem. Get the oxidation states. Mn+2 + Na. Bi. O 3 ® Bi+3 + Mn. O 4 -1 +2 +1 +5 -2 +3 +7 -2

Balancing in Basic Medium Problem # 1 n Mn+2 + Na. Bi. O 3 ® Bi+3 + Mn. O 4 -1 n +2 n n n +1 +5 -2 +3 +7 -2 Do the ½ reactions Mn+2 Mn. O 4 -1 4 H 2 O + Mn+2 Mn. O 4 -1 + 8 H+1 8 OH-1 + 4 H 2 O + Mn+2 Mn. O 4 -1 + 8 H 2 O Notice how the OH-1 has been added above 8 OH-1 + 4 H 2 O + Mn+2 Mn. O 4 -1 + 8 H 2 O + 5 e-1

Balancing in Basic Medium Problem # 1 n Bi. O 3 -1 Bi+3 6 H+1 + Bi. O 3 -1 Bi+3 + 3 H 2 O 6 H 2 O + Bi. O 3 -1 Bi+3 + 3 H 2 O + 6 OH-1 n 2 e-1 + 6 H 2 O + Bi. O 3 -1 Bi+3 + 3 H 2 O + 6 OH-1 n n

Balancing in Basic Medium Problem # 1 n n n We now combine the two ½ reactions. 8 OH-1 + 4 H 2 O + Mn+2 Mn. O 4 -1 + 8 H 2 O + 5 e-1 2 e-1 + 6 H 2 O + Bi. O 3 -1 Bi+3 + 3 H 2 O + 6 OH-1 We need to take care of the electrons. 16 OH-1 + 8 H 2 O + 2 Mn+2 2 Mn. O 4 -1 + 16 H 2 O 30 H 2 O + 5 Bi. O 3 -1 5 Bi+3 + 15 H 2 O + 30 OH-1

Balancing in Basic Medium Problem # 1 n n 16 OH-1 + 8 H 2 O + 2 Mn+2 2 Mn. O 4 -1 + 16 H 2 O 30 H 2 O + 5 Bi. O 3 -1 5 Bi+3 + 15 H 2 O + 30 OH-1 Combine n 16 OH-1 + 38 H 2 O + 2 Mn+2 + 5 Bi. O 3 -1 2 Mn. O 4 -1 + 31 H 2 O + 5 Bi+3 30 OH-1 Reduce n 7 H 2 O + 2 Mn+2 + 5 Bi. O 3 -1 2 Mn. O 4 -1 + 5 Bi+3 + 14 OH-1

Redox Titrations n n n Same as any other titration. The permanganate ion is used often because it is its own indicator. Mn. O 4 -1 is purple, Mn+2 is colorless. When reaction solution remains clear, Mn. O 4 -1 is gone. Chromate ion is also useful, but color change, orange yellow to green, is harder to detect.

Example n The iron content of iron ore can be determined by titration with standardized KMn. O 4. The iron ore is dissolved in HCl, and the iron reduced to Fe+2 ions. This solution is then titrated with KMn. O 4 soln. , producing Fe+3 and Mn+2 ions in acidic solution. If it requires 41. 95 m. L of 0. 205 M KMn. O 4 to titrate a solution made with 0. 6128 g of iron ore, what percent of the ore was iron ?

Solved answer n n n We need an equation. Fe+2 + Mn+7 Fe+3 + Mn+2. 205 M x 41. 95 m. L 8. 6 x 10 -3 moles Mn+7 1: 1 ratio so 8. 6 x 10 -3 moles Fe+2 is needed. 8. 6 x 10 -3 moles = g / 55. 85 g/n . 480 g Fe. 480 /. 6128 = 78. 38 % Fe in the sample.