CHEM 101 L5 Aqueous Reactions & Solution Stoichiometry PDF

Aqueous Reactions and Solution Stoichiometry ❑ Aqueous Reactions: Reactions that take place in water (H2O) ❑ Types of aqueous reactions: 1. Precipitation reactions 2. Acid-base reactions 3. Oxidation-reduction reactions ❑ Recap of the general Properties of Aqueous Solutions - A s...

Aqueous Reactions and Solution Stoichiometry ❑ Aqueous Reactions: Reactions that take place in water (H2O) ❑ Types of aqueous reactions: 1. Precipitation reactions 2. Acid-base reactions 3. Oxidation-reduction reactions ❑ Recap of the general Properties of Aqueous Solutions - A solution is a homogeneous mixture of two or more components. – Solute: the component that is dissolved. – Solvent: the component in which the solute is dissolved. – Generally, the component present in the greatest quantity is considered to be the solvent. – The solute can be separated from the solvent by physical methods. - Soluble: A soluble substance readily dissolves in the solvent. - Insoluble: An insoluble substance will NOT dissolve readily in a solvent. - Solubility: describes the amount of solute that will dissolve in a solvent at a specified temperature. For example, 35.7 g of NaCl will dissolve in 100 mL of water at 0 oC , no more. Solubility of ionic compounds When an ionic solid dissolve in water two competing forces come into play: 1. The attractive forces between the oppositely charges ions making up the solid. 2. The attractive forces between water and the ions. ❑ The extent to which solution occurs depends upon a balance between the two forces which are electrical in nature. - If the force attraction between water molecules and the ions in the solid predorminate, the solid dissolves in water. - If the force between oppositely charges ion in the solid dorminate, the solid dissolve to a small extent or does not dissolve at all. ❑ Unfortunately, the relative strength of these two forces cannot be determines from first principles and so we cannot predict water solubilities in advance. Solubility rules for ionic compounds 1. Most alkali metal (Li+, Na+, K+) and ammonium (NH4+) salts are soluble. 2. Nearly all nitrates (NO3-), acetates (C2H3O2-) and perchlorates are soluble. 3. All halides (X- = Cl-, Br-, I-) are soluble, except AgX, Hg2X2 and PbX2. 4. Most sulfates are soluble, except for BaSO4, SrSO4, Hg2SO4 and PbSO4. CaSO4 and Ag2SO4 are slightly soluble.  5. All common acids are soluble. 6. All oxides (O2-) and hydroxides (OH-) are insoluble, except those of alkali metals and alkali earth metals (Ca, Sr, Ba, Ra). 7. All sulfides (S2-) are insoluble, except for those of alkali metals, alkali earth metals, and ammonium sulfide. 8. All phosphates (PO43-), carbonates (CO32-) and chromates (CrO42-) are insoluble except those of alkali metals and the ammonium salts. ❑ EXAMPLE: Classify the following as soluble or insoluble in water – Ba(NO3)2 – AgI – Mg(OH)2  Electrolytes and Non-electrolytes ❑ Electrolyte: substance that dissolves in water to produce a solution that conducts electricity. - Contains ions ❑ Nonelectrolyte: substance that dissolves in water to produce a solution that does not conduct electricity. – Does not contain ions ❑ EXAMPLE: Classify the following as electrolyte or Non-electrolyte – NaOH – CH3OH  Strong and Weak Electrolytes ❑ There two processes by which an electrolyte produces ions in solution: - Dissociation. - Ionization. ❑ Dissociation –process by which ionic compounds separate into constituent ions when dissolved in solution. Example: ❑ Ionization – process by which molecular compounds form/generate ions when dissolved. Example: ❑ Strong Electrolytes: Undergo complete (100%) dissociation/ionization. ❑ Weak Electrolytes: Partially ionized in solution. Exist mostly in their molecular form in solution..  Method to Distinguish Types of Electrolytes nonelectrolyte weak electrolyte strong electrolyte ❑ The electrical conductivity of a solution depends on two factors: (a) the total concentration of the electrolyte. (b) the extent to which the electrolyte dissociates into ions.  Stoichiometry of Aqueous/Solution Reactions ❑ When dealing with aqueous reactions, the data given is normally for the parent compound and not for the particular ions in the net ionic equation. ❑ Therefore, in solving problems dealing with solution reactions through stoichiometric calculation, one must know: 1. The nature of the reacting species and the product - net ionic equation. 2. The amount of the chemical species present – concentration, e.g. Molarity etc ❑ Example: Determining ions concentration in a solution. Aluminum sulfate, Al2(SO4)3, is a strong electrolyte. Calculate the aluminum and sulfate ion concentrations in 0.0165 M Al2(SO4)3. Analysis: Information given: concentration of Al2(SO4)3; Information implied: The solute is a strong electrolyte and so ionizes completely and stoichiometric ratios of the ions. Information asked for: concentration of Al3+ and SO42- Strategy - Write a balanced chemical equation for the ionization of Al2(SO4)3. - Determine the stoichiometric ratios to relate Al3+ and SO42- to the concentration of Al2(SO4)3. Solution Al2(SO4)3(aq) →2 Al3+ (aq) + 3 SO42-(aq) Use conversion pathway: 0.0165 𝑚𝑜𝑙 Al2(SO4)3 2 𝑚𝑜𝑙 Al3+ 0.0330 𝑚𝑜𝑙 Al3+ [Al3+] = x = = 0.0330 M 1𝐿 1 𝑚𝑜𝑙 Al2(SO4)3 1𝐿 2-] 0.0165 𝑚𝑜𝑙 Al2(SO4)3 3 𝑚𝑜𝑙 SO42- 0.0495 𝑚𝑜𝑙 SO42- [SO4 = x = = 0.0495 M 1𝐿 1 𝑚𝑜𝑙 Al2(SO4)3 1𝐿 ❑ Molecular equation: shows all compounds represented by their chemical formulas plus stoichiometry ❑ Ionic equation: shows all strong electrolytes as ions and all other substances (non- electrolytes, weak electrolytes, gases) by their chemical formulas ❑ Net Ionic equation: shows only the reacting species in the chemical equation. Spectator ions are omitted. ❑ Summary: Steps in writing a net ionic equation – Write the balanced molecular equation. Predict products by exchanging cations and anions in reactants. – Separate strong electrolytes into ions. – Cancel spectator ions. – Use solubility table to determine solubility of the product species. – Use the remaining species to write the net ionic equation. Example: Write molecular, ionic and net ionic equation for reaction that takes place when aqueous solution of sodium sulphate and barium hydroxide are mixed. Strategy: Write the correct chemical formulas for the reactants bearing in mind the stoichiometry. Solution Molecular equation: Ionic equation: Net Ionic equation: shows only the reacting species in the chemical equation – Eliminates spectator ions Net ionic equation: Example 2: Aqueous solutions of silver nitrate and sodium sulfate are mixed. Write the net ionic reaction. PRACTICE AT HOME!! NOTE: Like all equations, net ionic equations must show: 1. Atom balance – there must be the same number of atoms of each element on both sides. 2. Charge balance – there must be the same total charge on both sides. Precipitation Reactions ❑ Precipitation - (formation of a solid from two aqueous solutions) occurs when product(s) is insoluble. Produce insoluble ionic compounds ❑ Precipitate – an insoluble solid compound formed during a chemical reaction in solution. ❑ Application of precipitation reactions: 1. Identification ions present in solution in a lab. 2. Industrial preparation of paints and extraction of metals 3. Separation of products in chemical reactions.  Predicting Precipitation Reactions ❑ To predict whether a precipitate will form when you mix two solutions of ionic compounds, you need to know whether any of the potential products that might form are insoluble or not. ❑ Apply solubility Rules. ❑ Use partner-exchange reactions – The cations from a soluble compound joins with the anion from another soluble compound. The result might be no reaction, one or two precipitates. AR2 A2+ R- Split Into ions B 2X B+ X2-  Solubility Table S = soluble S = slightly soluble I = insoluble - = decomposed by water Example 1: Predicting precipitation What is most likely to be the yellow solid formed in the following reaction? MgCl2 + AgNO3 Strategy Rule 3 – chlorides are soluble with exceptions that does not include MgCl2 Rule 2 - Nitrates are soluble, so AgNO3 is soluble Potential products: AgCl and Mg(NO3)2 Solution MgCl2 (aq) + AgNO3 (aq) Mg(NO3)2 (aq) + AgCl (S) Reaction occurs because AgCl is insoluble and thus it precipitates from the reaction mixture. Write the above precipitation reaction in ionic and net ionic equations. Lets do together!!!! Example 2: What is most likely to be the yellow solid formed in the following reaction? K2CrO4(aq) + Ba(NO3)2(aq) ❑ The possible product combinations are KNO3 and BaCrO4 KNO3 ----- Recall that all group 1 salts are soluble BaCrO4 is therefore the yellow solid Homework: Write molecular, ionic and net ionic equations for the precipitation reaction above.!!!! Practice questions: a) What are the potential products in the reactions between the following aqueous solutions? b) Write balanced equations for each ❑ NaOH + CaCl2 ❑ CuBr2 + (NH4)2CO3 ❑ K2SO4 + Fe(NO3)3 Example 3: Determination of mass of precipitate formed. Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3solution to precipitate all the Ag+ ions in the form of AgCl. Analysis Write a balanced molecular, ionic and net ionic equation using solubility guidelines. From the net ionic equation, calculate number of moles of Cl-ions are required to react with all the Ag+. Calculate mass of NaCl from the number Cl-ions. Solution Molecular equation Ionic equation Net ionic equation  Isolating the precipitate The precipitate from a precipitation reaction can be separated from the reaction mixture by filtration. A Buchner funnel and flask Buchner vacuum can be used to accelerate funnel pump the process. filter paper This apparatus uses a vacuum pump to draw the mixture through the filter. Buchner The filtrate is finally washed flask and dried.  ACID-BASE REACTIONS ❑ Reactions in which an acid and a base react in an aqueous solution to produce a salt and water. General properties of an acid ❑ An acid is a solution that has an excess of H+ ions. ❑ The more H+ ions, the more acidic the solution. ❑ pH less than 7 ❑ Neutralizes bases ❑ Turns blue litmus paper red General properties of a base ❑ A base is a solution that has an excess of OH- ions. ❑ Bases are substances that can accept hydrogen ions ❑ The more OH- ions, the more basic the solution. ❑ pH greater than 7 ❑ Turns red litmus paper blue An acid-base indicator is a dye used to distinguish between acidic and basic solutions by means of color changes it undergoes in these solutions.  Acid-Base Definitions Examples Arrehenius Acids – produce excess H+ ions HCL only in water Bases - produce OH- excess ions NaOH Bronsted-Lowry Acids – donate H+ (proton) H3O+, NH4+, H2O Bases – accept H+ H2O, OH- , NH3 any solvent Lewis Acids – accept e- pair Ag+, BF3 Bases – donate e- pair NH3 Lewis definition is used in organic chemistry for a wider range of substances.  Common Strong and Weak Acids/Bases Strong acids/bases – Undergo complete dissociation into ions in water Strong Acids: HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4 Strong Bases: LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2 Weak acids/bases – Undergo partial dissociation in water forming a mixture of ions and original molecules. Weak Acids: CH3COOH, HF Weak Bases: NH3 NOTE: Dissociation of strong acids/bases is represented by single arrow while that of weak acids/bases is represented by a double arrow implying that the reaction does not go to completion. Example: HCl(aq) H+(aq) + Cl- (aq) strong acid NH3(aq) + H2O(l) NH4+ (aq) + OH-(aq) weak base  Strong/Weak Acid Dissociation Illustration NOTE: H3O+ is oxonium as per IUPAC guideline. Its shorthand notation is H+ H+ is hydrogen ion as per IUPAC guidelines.  Autoionization of pure water ❑ At room temperature water ionizes reversibly to a small extent through a process called autoionization/self-ionization represented as follows; H2O(l) H+(aq) + OH-(aq) ❑ From experimental measurements: [H+]water = [OH-]water = 1.0 x 10-7 M at 25oC ❑ An acidic solution has [H+]  [H+]water ❑ A basic solution has [OH-]  [OH-]water Strength vs. Concentration ❑ The words concentrated and dilute tell how much of an acid or base is dissolved in solution - refers to the number of moles of acid or base in a given volume. ❑ The words strong and weak refer to the extent of ionization of an acid or base. ❑ Is a concentrated, weak acid possible? Common Strong Acids & their Anions Common Weak Acids & their Anions  Types of acids and bases ❑ Acids Monoprotic: one ionizable hydrogen HCl + H2O H3O+ + Cl− Diprotic: two ionizable hydrogens H2SO4 + H2O H3O+ + HSO4− HSO4− + H2O H3O+ + SO42− Triprotic: three ionizable hydrogens H3PO4 + H2O H3O+ + H2PO4− H2PO4− + H2O H3O+ + HPO42− HPO42− + H2O H3O+ + PO43− Polyprotic: generic term meaning more than one ionizable hydrogen ❑ Types of bases Monobasic: One OH− group KOH K+ + OH− Dibasic: Two OH− groups Ba(OH)2 Ba2+ + 2OH− NEUTRALIZATION REACTION ❑ An acidic water solution and basic water solution react together form an aqueous solution of an ionic compound called salt. This reaction is called neutralization. ❑ The nature of the neutralization reaction and the equation written for it depends on whether the acid and base involved are strong or weak. 1. Strong acid – strong base: Molecular equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Ionic equation: H+(aq)+ Cl−(aq) + Na+(aq) + OH−(aq) → Na+(aq) + Cl−(aq) + H2O(l) Net ionic equation: H+(aq) + OH−(aq)→ H2O(l) 2. Weak acid – strong base: When a strong base such as NaOH is added to a solution of a weak acid, HB, a two- step reaction occurs. The first step is the ionization of HB molecule to H+ and B- ions, the second step is the neutralization of the H+ ions by the OH- of the NaOH as shown below: Step 1: HB(aq) H+(aq) + B-(aq) Step 2: H+(aq) + OH-(aq) H2O The overall equation is obtained by adding the two equations and cancelling H+ ions. HB(aq) + OH-(aq) H2O. Example: Solutions of acetic acid and lithium hydroxide are mixed. Write the net ionic reaction. Solution: Molecular equation: CH3COOH(aq) + NaOH(aq) → NaCH3COO (aq) + H2O(l) Ionic equation: CH3COOH aq) + Na+(aq) + OH−(aq) → Na+(aq) + CH3COO-(aq) + H2O(l) Net ionic equation: CH3COOH(aq) + OH−(aq)→ CH3COO-(aq) + H2O(l) 3. Strong acid – weak base: When a strong acid such as HCl is added to a solution of a weak base such as NH3, we consider the reaction to take place in two steps. First step is the reaction of NH3 with H2O to form NH4+ and OH- ions, the second step is the neutralization of OH- by the H+ of the strong acid HCl as show: Step 1: NH3(aq) + H2O NH4+(aq) + OH-(aq) Step 2: H+(aq) + OH-(aq) H2O. The overall equation is obtained by adding the two equations and cancelling OH- and H2O: NH3(aq) + H+(aq) NH4+(aq) Example 1: Writing net ionic equations for acid-base reactions. Write a net ionic equation to represent the reaction of aqueous strontium hydroxide with nitric acid. Analysis: Both the acid and the base are strong, meaning both ionize completely. Strategy: Write molecular and ionic equations, cancel spectator ions to get net ionic equation. Solution Molecular equation: 2HNO3(aq) + Sr(OH)2 →Sr(NO3)2(aq) + 2H2O(l) Ionic equation: 2H+(aq) + 2NO3-(aq) + Sr2+(aq) + 2OH-(aq) →Sr2+ (aq) + 2NO3-(aq) + 2H2O(l) Net ionic equation: 2H+(aq) + 2OH-(aq) → 2H2O(l); Divide by 2 to simply: H+(aq) + OH-(aq) → H2O(l) Example 2: Calculations in acid-base reactions. Calculate the volume of a 0.100 M HCl solution needed to neutralize 25.0 mL of a 0.350 M NaOH solution. Analysis: Reacting species are strong base and acid; write molecular, ionic, and net ionic equations; calculate number of moles for each reactant from their volumes and molarity; use the mole ratios in net ionic equation and number of moles of NaOH to be neutralized to calculate number of moles of HCl required; determine volume of HCL from the number of the moles calculated and Molarity given.  OXIDATION-REDUCTION REACTIONS ❑ In aqueous solution, some reactions might involve transfer of electrons between two species. ❑ Such reactions are called oxidation-reduction or simply redox reactions. ❑ Many common reactions fit into this category including the reaction of metals with acids. ❑ In a redox reaction one species loses (i.e donates) electrons and is said to be oxidized. ❑ The other species, which gains (or receives) the electrons is said to be reduced. ❑ To illustrate, consider a reaction between zinc metal and HCl represented by the net equation below: Zn(s) + H+(aq) Zn2+(aq) +H2(g) ❑ This equation can be split into two half-equation, one for oxidation and the other for reduction. ❑ Zn atoms are oxidized to Zn2+ ions by losing electron. Therefore oxidation half- equation is; Oxidation: Zn(s) Zn2+ (aq) + 2e ❑ At the same time, H+ ions are reduced to H2 molecules by gaining electrons; the reduction half-equation is therefore: Reduction: 2H+(aq) + 2e H2(g) ❑ From this example, it is clearer that: 1. Oxidation and reduction occur together in the same reaction i.e. you cant have one without the other. 2. There is no net change in the number of electrons in a redox reaction. i.e. the electron given off by one species in the oxidation half-reaction are taken on by another species in the reduction half-reaction. ❑ Memory Aid: OIL: Oxidation Is Loss; RIG: Reduction Is Gain NOTE: - Oxidizing agent is the chemical species that gains electron and get reduced. - Reducing agent is the chemical species that losses electrons and get oxidized. More examples of a redox reaction: 1. 2Mg(s) + O2(g) 2MgO(s) ---- Burning magnesium ribbon. Oxidized – loss of e-: Mg Mg2+ + 2e Reduced – gain of e-: O + 2e O2- 2. Cu(s) + 2AgNO3(aq) → Cu(NO3 )2(aq) + 2Ag(s) Oxidized – loss of e-: Cu Cu2+ + 2e Reduced – gain of e-: 2Ag2+ + 2e 2Ag(s)  Oxidation Number/State ❑ The concept of oxidation number is used to simplify the electron bookkeeping in redox reactions. ❑ Use of the oxidation number/state in a redox reaction helps in: 1. Identifying the oxidizing and reducing species. 2. Balancing of redox equations. ❑ In oxidation reactions, the oxidation number of an element in the chemical species being oxidized increases. - Meaning that oxidation number of an element in the oxidizing agent (oxidant) reduces. ❑ In reduction reactions, the oxidation number of an element in the chemical species being reduced decreases. - Meaning that oxidation number of an element in the reducing agent (reductant) increase. ❑ In practice, oxidation numbers in all kind of species are assigned according to a set of arbitrary rules given in the table below. Example: Calculating Oxidation Number/State of Chemical Species. 1. Oxidation state of C in CO2? C + 2(-2) = 0; C + -4 = 0 C = +4 ❑ Oxidation state of S in SO42-? S+ 4(-2) = -2; S - 8 = -2; S = -2 + 8 = +6 Example: Identifying oxidizing & reducing agents. For the reaction below, identify whether, H2O2, is an oxidizing or reducing agent. H2O2(aq) + 2 Fe2+(aq) + 2 H+(aq) →2H2O(l) + 2Fe3+(aq) Analysis Select common elements in both the reactants and products (in case its O and Fe) and calculate their O.S. Identify whether the O.S for elements is increasing or decreasing. If OS decrease means the species containing that element is an OXIDANT and vice versa. Solution: The O.S. of O in H2O2 is -1 and -2 in H2O; meaning H2O2 is the oxidant. The O.S of Fe increases from 2 + to +3; meaning Fe is the reductant. Example 2: Using Oxidation Number to Identify Redox Reactions Indicate whether each of the following is an oxidation-reduction reaction. (a) MnO2(s) + 4H+(aq) + 2Cl-(aq) → Mn2+(aq) + 2H2O(l) + Cl2(g) (b) H2PO4-(aq) + OH-(aq) → HPO42-(aq) + H2O(l) Analysis: For each reaction equation, indicate the oxidation states of the elements on both sides of the equation and look for changes in oxidation number. Solution: (a)The O.S. of Mn decreases from +4 in MnO2 to +2 in Mn2+. MnO2 is therefore reduced The O.S. of O remains constant at -2 and that of H at +1. The O.S. of Cl increases from -1 in Cl- to 0 in Cl2. Cl- is therefore oxidized. This is a redox reaction. (a)The O.S. of H is +1 on both sides of the reaction equation. The O.S of O remains constant at -2. The O.S. of P in H2PO4- and HPO42- is +5. Since there is no changes in O.S., this is NOT a redox reaction but an acid-base one.  What is a Half-Reaction ❑ A half-reaction is simply one which shows either reduction OR oxidation, but not both. Consider redox reaction: Ag+ (aq) + Cu(s) Ag(s) + Cu2+ (aq) ❑ It has BOTH a reduction and an oxidation in it. ❑ To identify the two half-reactions, one must first identify the atoms which get reduced and oxidized. ❑ Here are the two half-reactions from the example: Ag+ (aq) Ag(s) Cu(s) Cu2+ (aq) ❑ The silver is being reduced, its oxidation number going from +1 to zero. ❑ The copper's oxidation number went from zero to +2, so it is oxidized in the reaction. NOTE: In order to figure out the half-reactions, you MUST be able to calculate the oxidation number of different chemicals involved.  Balancing Half Reactions ❑ For a half reaction to be said to be balanced, both the mass/atoms and total charge MUST be equal on both sides of the half reaction equation. ❑ The two half-reactions in the preceding slide are already balanced for atoms. So, all we need to do is balance the charge. ❑ To do this you add electrons to the more positive side making sure you add enough to make the total charge on each side become EQUAL. ❑ To the silver half-reaction, we add one electron: Ag+ (aq) + e¯ Ag(s) ❑ To the copper half-reaction, we add two electrons: Cu(s) Cu2+(aq) + 2e¯  Half-Reactions Practice Problems ❑ Balance each half-reaction for atoms and charge: 1) Cl2 ---> Cl¯ 2) Sn ---> Sn2+ 3) Fe2+ ---> Fe3+ 4) I3¯ ---> I¯ 5) ICl2¯ ---> I¯ 6) Sn + NO3¯ ---> SnO2 + NO2 7) HClO + Co ---> Cl2 + Co2+ 8) NO2 ---> NO3¯ + NO  Balancing Half-Reactions in Acid Solution ❑ There are three chemical species available in an acidic solution: 1. H2O - present because the reaction is taking place in solution 2. H+ - available because it is in acid solution. 3. e¯ - available because that's what is transferred in redox reactions Example: MnO4¯ (aq) Mn2+ (aq) in acidic solution Solution 1. Balance the atom being reduced/oxidized.. MnO4¯ (aq) Mn2+ (aq) Mn atoms are balanced. 2. Balance the oxygens. Do this by adding water molecules (as many as are needed) to the side that needs oxygen. MnO4¯(aq) Mn2+ (aq) + 4H2O(l) 3. Balance the hydrogens. Do this by adding hydrogen ions (as many as are needed) to the side that needs hydrogen. 8H+ (aq) + MnO4¯(aq) Mn2+ (aq) + 4H2O(l) 4. Balance the total charge. This will be done adding electrons to the more positive side. It is ALWAYS the last step. 5e¯ + 8H+ (aq) + MnO4¯(aq) Mn2+ (ag) + 4H2O(l) Balancing Half-Reactions in Basic Solution ❑ There are three chemical species available in a basic solution: 1. H2O - present because the reaction is taking place in solution. 2. OH- - available because it is in basic solution 3. e¯ - available because that is what is transferred in redox reactions Example: PbO2 (s) PbO(s) in basic solution Solution 1. Balance the atom being reduced/oxidized. PbO2(s) PbO(s) Pb atoms are balanced. 2. Balance the oxygens. Do this by adding water molecules (as many as are needed) to the side that needs oxygen. PbO2(s) PbO(s) + H2O(l) 3. Balance the hydrogens. Do this by adding hydrogen ions (as many as are needed) to the side that needs hydrogen. 2H+(aq) + PbO2(s) PbO(s) + H2O(l) 4. Balance the total charge. This will be done adding electrons to the more positive side. It is ALWAYS the last step. 2e + 2H+(aq) + PbO2(s) PbO(s) + H2O(l) 5. Convert all H+ to H2O. Do this by adding OH¯ ions to both sides. The side with the H+ will determine how many hydroxide to add. In our case, the left side has 2 hydrogen ions, while the right side has none, so: 2e + 2H+(aq) + 2OH-(aq) + PbO2(s) PbO(s) + H2O(l) + 2OH-(aq) 2H2O 6: Cancel any duplicate molecules or ions. This means one water molecule may be cancelled from each side to reduce the equation to simplest terms: 2e¯ + H2O(l) + PbO2(s) PbO(s) + 2OH¯ (aq) The half-reaction is now correctly balanced. Example: Balancing the equation for a redox reaction in an acidic solution The redox reaction below is used to determine the sulfite ion concentration present in wastewater from papermaking plant. Balance the equation in an acidic solution. SO32-(aq) + MnO4-(aq) →SO42-(aq) + Mn2+(aq) Analysis and Strategy The medium is acidic; split the equation into half equations; identify which equation represents oxidation and reduction using oxidation number. Solution Step 1: Write half-equations based on the species undergoing oxidation & reduction: Oxidation: SO32-(aq) →SO42-(aq) Reduction: MnO4-(aq) →Mn2+(aq) Step 2: Balance each half-equation for numbers of atoms, in this order: - balance atoms other than H and O - balance O atoms by adding H2O - then balance H atoms by adding H+ For the 2 half equations S and Mn atoms are already balanced. So, balance O atoms: Oxidation: SO32-(aq) + H2O(l) →SO42-(aq) Reduction: MnO4-(aq) →Mn2+(aq) + 4H2O(l) Then balance H atoms: Oxidation: SO32-(aq) + H2O(l) →SO42-(aq) + 2H+(aq) Reduction: MnO4-(aq) + 8H+(aq) →Mn2+(aq) + 4H2O(l) Step 3: Balance each half-equation for electric charge by adding the number of electrons necessary to get the same electric charge on both sides of each half- equation: Oxidation: SO32-(aq) + H2O(l) →SO42-(aq) + 2H+(aq) + 2e- Reduction: MnO4-(aq) + 8H+(aq)+ 5e-→Mn2+(aq) + 4H2O(l) Step 4: Combine the half-equations to get the overall redox equation. Multiply the oxidation half-equation by 5 and the reduction half-equation by 2, such that there will be 10e- on each half equation and that the 10e- electrons cancels when the two half equations are added to get the overall equation. Electrons must NOT appear in the final equation: Oxidation: 5SO32-(aq) + 5H2O(l) → 5SO42-(aq) + 10H+(aq) + 10 e- Reduction: 2MnO4-(aq) + 16H+(aq)+ 10 e- → 2Mn2+(aq) + 8H2O(l) Overall: 5SO32-(aq) + 2MnO4-(aq) + 5H2O(l) + 16H+(aq) → 5SO42-(aq) + 2Mn2+(aq) + 8H2O(l) + 10H+(aq) Step 5: Simplify the equation by cancelling species that occur on both sides and reduce the coefficients to the smallest whole numbers. This is done for H2O and H+ species: Overall: 5 SO32-(aq) + 2MnO4-(aq) + 6H+(aq) →5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) Example: Balancing a redox reaction equation in a basic solution Balance the redox equation for the reaction below between cyanide ions cyanate and permanganate ions in a basic solution. CN-(aq) + MnO4-(aq) →OCN-(aq) + MnO2(s) Analysis and Strategy The medium is acidic; split the equation into half equations; identify which equation represents oxidation and reduction using oxidation number. Solution Step 1: Write half-equations based on the species undergoing oxidation & reduction: Oxidation: CN-(aq) →OCN-(aq) Reduction: MnO4-(aq) →MnO2(s) Step 2: Balance each half-equation for numbers of atoms, in this order: - balance atoms other than H and O - balance O atoms by adding H2O - then balance H atoms by adding H+ For the 2 half equations C and Mn atoms are already balanced. So, balance for O and H: Oxidation: CN-(aq) + H2O(l) →OCN-(aq) + 2H+(aq) Reduction: MnO4-(aq) + 4H+(aq) →MnO2(s) + 2H2O(l) Step 3: Balance each half-equation for electric charge by adding the number of electrons necessary to get the same electric charge on both sides of each half-equation: Oxidation: CN-(aq) + H2O(l) →OCN-(aq) + 2 H+(aq) + 2 e- Reduction: MnO4-(aq) + 4H+(aq) + 3e-→MnO2(s) + 2H2O(l) Step 4: Combine the half-equations to get the overall redox equation. Multiply the oxidation half-equation by 3 and the reduction half-equation by 2 to get 6 e- on each side, then cancel them to get the overall equation. Oxidation: 3CN-(aq) + 3H2O(l) →3OCN-(aq) + 6H+(aq) + 6 e- Reduction:2MnO4-(aq) + 8H+(aq) + 6 e-→2MnO2(s) + 4H2O(l) Overall: 3CN-(aq) + 3H2O(l) + 2MnO4-(aq) + 8H+(aq) →3OCN-(aq) + 6H+(aq) + 2 MnO2(s) + 4H2O(l) Step 5: Simplify the equation by cancelling species that occur on both sides and reduce the coefficients to the smallest whole numbers. This is done for H2O and H+ species. Overall: 3CN-(aq) + 2MnO4-(aq) + 2H+(aq) →3OCN-(aq) + 2MnO2(s) + H2O(l) Step 6: Change from an acidic to a basic medium by noting the number of H+ and adding the same number of OH- ions both sides of the overall equation. In our case 2OH- ions. Combine 2H+ and 2OH- to form 2H2O and simplify. Overall: 3CN-(aq) + 2MnO4-(aq) + 2H+(aq) + 2OH-(aq) →3OCN-(aq) + 2MnO2(s) + H2O(l) + 2OH-(aq) 3CN-(aq) + 2MnO4-(aq) + 2H2O(l) →3OCN-(aq) + 2MnO2(s) + H2O(l) + 2OH-(aq) Lastly simplify for H2O molecule: Overall: 3CN-(aq) + 2MnO4-(aq) + H2O(l) →3OCN-(aq) + 2MnO2(s) + 2OH-(aq)  Solution Stoichiometry (recap) ❑ The quantitative study of the relative amounts of reactants and products in a chemical reaction is referred to as stoichiometry. ❑ Therefore, solution stoichiometry deals with relative quantities of reactants and products for chemical reactions occurring in solutions. ❑ In solution stoichiometry, we use mole ratio/stoichiometric coefficients in a balanced chemical equation, especially net ionic equation to calculate; concentration (e.g molarity, volume, and/or mass...etc of reactants and products in any given reaction problem. ❑ Note that in most stoichiometric problems, data given are those of the parent compounds and not of particular ions in the net equation. After all, you do not have reagents bottles labelled 3 M OH- but rather 3 M NaOH. ❑ Therefore, mole ratios are used to relate ions to their parent compounds (reactant or products) in terms of moles and from this, the volume, mass, concentration of the parent compounds can be determined. Example: Application of solution stoichiometry in precipitation reaction. When aqueous solution of sodium hydroxide and iron (III) nitrate are mixed, a red precipitate is formed. a) Write ionic equation for the reaction b) What volume of 0.136 M iron (III) nitrate is required to produce 0.886 of precipitate? c) How many grams of precipitate are formed when 50.00 mL of 0.200 M NaOH and 30.00 mL of 0.125 M Fe(NO3)3 are mixed? Strategy a) Identify the ions in the solution from the two reaction; come up the with possible precipitate and use solubility to decide which one is insoluble; write ionic equation. b) From mass of precipitate, determine # of moles of ppt then determine # of moles of ion, then # of mole of parent compound then volume of parent compound. c) Determine # of moles of NaOH and Fe(NO3)3 using V x M, then determine # of moles of Fe3+ and OH- ions using mole ratio; then use the stoichiometric coefficients in the net ionic equation and calculated moles to determine the limiting reactant – recall limiting reactant is the one with less # moles than required from stoichiometric ratio; convert the # of moles the limiting reactant to # of moles of precipitate then to mass using the molar mass of the precipitate.