Chapter 4, part 2: Reactions in Aqueous Solution Chemical Reactions Chapter 4, part 2: Reactions in Aqueous Solution

The dissolving process When a solid is put into a liquid, solute-solute attractions compete with solute-solvent & solvent-solvent attractions

Solubility Let’s assume our solvent is water . . . If solute-water attractions > solute-solute & water-water attractions, solute particles are pulled out one by one into the water: The solute is SOLUBLE in water

Solubility But if solute-water attractions < solute-solute & water-water attractions, solute particles remain together: The solute is INSOLUBLE in water

Solution conductivity Solution conductivity depends on type of solute particles

Solution conductivity Ionic solutes (salts) made of cations & anions Ions DISSOCIATE (separate) during dissolving Molecular solutes made of molecules each solute particle that moves into solution is identical

Solution conductivity Solutions of ionic solutes contain independent mobile ions Solution conducts electricity Solute is an ELECTROLYTE Solute also conducts when melted, but not when solid (ions can’t move)

Solution conductivity Solutions of most molecular solutes contain independent neutral molecules Solution does not conduct electricity Solute is a NONELECTROLYTE Solute also does not conduct when melted or solid

Solution conductivity Some molecular compounds (acids & bases) react with water to produce ions, as if they dissociated A few acids & bases do this very well, producing lots of ions = strong electrolytes, strong acids & bases Most acids & bases do this weakly, producing a few ions = weak electrolytes, weak acids & bases

All other acids & bases are WEAK Strong acids & bases STRONG ACIDS STRONG BASES HCl HNO3 LiOH Mg(OH)2 HBr H2SO4 NaOH Ca(OH)2 HI HClO4 KOH Sr(OH)2 etc. Ba(OH)2 All other acids & bases are WEAK

Predicting Electrolytes All soluble salts and strong acids & bases are strong electrolytes Weak acids & bases are weak electrolytes All other molecular compounds are nonelectrolytes

Precipitation reactions

Precipitation reactions spectator ions precipitate

Predicting Precipitation Reactions To predict whether a precipitate will form, you need to know which compounds are soluble (no ppt) and which are insoluble (ppt forms) Memorize the guidelines in Table 4.1 on page 154 in your text! Add this guideline: CrO42– acts like SO42–

Solubility guidelines: soluble Compounds of these ions are generally soluble and do NOT form precipitates: Alkali metals (group 1A, except Li1+) Ammonium (NH41+) Nitrates (NO31–) and acetates (C2H3O21–) Chlorides, bromides, iodides, except Pb2+, Ag1+, Hg22+ Sulfates (SO42–) & chromates (CrO42–), except Sr2+, Ba2+, Pb2+, and Hg22+

Solubility guidelines: insoluble Compounds of these ions are generally insoluble and DO form precipitates: Hydroxides (OH1–) and Sufides (S2–) Alkali metals (group 1A) and ammonium (NH41+) are soluble Sulfides of group 2A metals are generally soluble Hydroxides of Ca2+, Sr2+, and Ba2+ are soluble Carbonates (CO32–) and phosphates (PO43–)

Net ionic equations AgNO3 (aq) + NaCl (aq)  AgNO3 (aq) + NaCl (aq)  AgCl + NaNO3 AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq) The (aq) substances are dissociated: Ag1+ (aq) + NO31– (aq) + Na1+ (aq) + Cl1– (aq)  AgCl (s) + Na1+ (aq) + NO31– (aq)

Net ionic equations Na1+ and NO31– are spectator ions Ag1+ (aq) + NO31– (aq) + Na1+ (aq) + Cl1– (aq)  AgCl (s) + Na1+ (aq) + NO31– (aq) The net ionic equation omits spectators: Ag1+ (aq) + Cl1– (aq)  AgCl (s)

Is that really a spectator? An ion is a spectator if and only if it is in exactly the same form in the products and reactants: Na2CO3 (aq) + BaCl2 (aq)  BaCO3 (s) + 2 NaCl (aq) CO32– (aq) + Ba2+ (aq)  BaCO3 (s) Na2CO3 (s) + 2 HCl (aq)  2 NaCl (aq) + H2O + CO2 (g) Na2CO3 (s) + 2 H1+ (aq)  2 Na1+ (aq) + H2O + CO2 (g) Only Cl1– is a spectator Na1+ is not a spectator because it was (s), then (aq)

Examples Indicate whether a ppt forms and if so, complete the reaction as a balanced net ionic equation: AlCl3 (aq) + KOH (aq)  K2SO4 (aq) + FeBr3 (aq)  CaI2 (aq) + Pb(NO3)2 (aq)  Na3PO4 (aq) + AlCl3 (aq)  Al2(SO4)3 (aq) + BaCl2 (aq)  (NH4)2CO3 (aq) + Pb(NO3)2 (aq) 

Acids Acids produce H3O1+ in aqueous solution: Strong acids are molecular compounds that react completely with water to produce H3O1+: HCl (g) + H2O  H3O1+ (aq) + Cl1– (aq) For convenience, we often show acids as simply dissociating to produce H1+: HCl (g)  H1+ (aq) + Cl1– (aq) There are only 6 strong acids (memorize them, pg 161)

Acids Acids produce H3O1+ in aqueous solution: Weak acids are molecular compounds that react incompletely with water: HC2H3O2 (aq) + H2O  H3O1+ (aq) + C2H3O21– (aq) HC2H3O2 (aq)  H1+ (aq) + C2H3O21– (aq) All acids that are not strong are weak

Acids H1+ is a proton The reaction with water is proton transfer: HCl (g) + H2O  H3O1+ (aq) + Cl1– (aq) HC2H3O2 (aq) + H2O  H3O1+ (aq) + C2H3O21– (aq) Acids with one H1+ to transfer are monoprotic acids HCl HNO3 HC2H3O2 Acids with more than one H1+ to transfer are polyprotic acids H2SO4 H3PO4 H2C3H2O4

Acids Polyprotic acids produce H3O1+ in steps: H2SO4 (aq) + H2O  H3O1+ (aq) + HSO41– (aq) HSO41– (aq) + H2O  H3O1+ (aq) + SO42– (aq) For H2SO4 the first step is strong and the second weak H2C2O4 (aq) + H2O  H3O1+ (aq) + HC2O41– (aq) HC2O41– (aq) + H2O  H3O1+ (aq) + C2O42– (aq) For all other polyprotic acids, the first step is weak and the second step is weaker

Bases Bases produce OH1– in aqueous solution: Strong bases are ionic hydroxide compounds that are completely dissociated in water: NaOH (s)  Na1+ (aq) + OH1– (aq) The strong bases are the hydroxides of group 1A and 2A metals (memorize them) Weak bases are molecular compounds that react incompletely with water to produce OH1–: NH3 (aq) + H2O (l)  NH41+ (aq) + OH1– (aq) Amines (–NH2 ) are weak bases

Neutralization Acids and bases neutralize each other The H1+ from the acid is transferred to the base: HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq) H1+ (aq) + OH1– (aq)  HOH (l) The base is not always an OH1– compound: HCl (aq) + NH3 (aq)  NH41+ (aq) + Cl1– (aq) H1+ (aq) + NH3 (aq)  NH41+ (aq)

Neutralization and net ionic equations It is important to recognize strong acids and bases when writing net ionic equations HCl (aq) + NaOH (aq)  H2O (l) + NaCl (aq) H1+ (aq) + OH1– (aq)  H2O (l) HC2H3O2 (aq) + NaOH (aq)  H2O (l) + NaC2H3O2 (aq) HC2H3O2 (aq) + OH1– (aq)  H2O (l) + C2H3O21– (aq) The weak acid is not significantly ionized, so acetate is not a spectator (even if it is aq)

Examples Write the molecular and net ionic equations for HNO3 + NaOH  HF + KOH  HC2H3O2 + NH3  H2SO4 + Ba(OH)2  H2C2O4 + NaOH  HCHO2 + Ca(OH)2 

Gas forming reactions Some neutralizations produce a gas: CO32– + 2 H1+  H2CO3 H2CO3 is unstable and decomposes immediately H2CO3  CO2 (g) + H2O The overall reaction is CO32– + 2 H1+  CO2 (g) + H2O

Gas forming reactions Memorize these gas-formers (pg 166): SO32– + 2 H1+  SO2 (g) + H2O HSO31– + H1+  SO2 (g) + H2O CO32– + 2 H1+  CO2 (g) + H2O HCO31– + H1+  CO2 (g) + H2O S2– + 2 H1+  H2S (g) NH41+ + OH1–  NH3 (g) + H2O burning sulfur smell rotten egg smell ammonia

Redox Redox (oxidation-reduction) reactions are those in which electrons are transferred The loss of electrons is oxidation The gain of electrons is reduction LEO says GER

Oxidation states The oxidation state (O.S.) or oxidation number is a convenient but artificial way to describe the electron environment around an atom It is related to the number of electrons gained, lost, or apparently used in forming compounds Oxidation states are assigned using the rules on page 169 of your text (memorize these in order)

Assigning oxidation states 1. The O.S. of each atom in an element is zero 2. The O.S. of a monoatomic ion is equal to its charge 3. The total of the O.S. of all atoms in any species (formula unit, molecule or ion) equals the charge on that species 4. In compounds, metals always have a positive O.S. • Group 1A metals are always O.S. +1 and Group 2A metals are always O.S. +2 5. For nonmetals in compounds, • the O.S. of fluorine is –1. • the O.S. of hydrogen is +1. • the O.S. of oxygen is –2. 6. In binary compounds, the O.S. of a Group 7A element is –1, Group 6A element –2, and Group 5A element –3.

Examples What is the oxidation state of each element in S8 Cr2O72– Cl2O KO2 S2O32– Hg2Cl2 KMnO4 H2CO

Redox In a redox reaction, atoms change (O.S.) The element oxidized loses electrons and its O.S. becomes more positive The element reduced gains electrons and its O.S. becomes more negative

Examples Which of these reactions is a redox reaction? Identify the species oxidized and reduced. HCl (aq) + NaOH (aq)  H2O + NaCl (aq) 2 Pb(NO3)2 (s)  2 PbO (s) + 4 NO2 (g) + O2 (g) NH4Cl (s) + NaOH (aq)  NH3 (g) + H2O + NaCl (aq) Identify the species oxidized and reduced: 5 VO2+ (aq) + MnO41– (aq) + H2O  5 VO21+ (aq) + Mn2+ (aq) + 2 H1+ (aq)

Agents of Oxidation and Reduction An agent makes something happen An oxidizing agent makes oxidation happen by being reduced A reducing agent makes reduction happen by being oxidized The “agent” is the entire species in which the oxidized or reduced atom appears

Examples Identify the elements oxidized & reduced and the oxidizing & reducing agents in 2 NO2 (g) + 7 H2 (g)  2 NH3 (g) + 4 H2O (g) 5 H2O2 (aq) + 2 MnO41– (aq) + 6 H1+ (aq)  8 H2O + 2 Mn2+ (aq) + 5 O2 (g) S2O32– (aq) + 4 Cl2 (aq) + 5 H2O  2 HSO41– (aq) + 8 H1+ (aq) + 8 Cl1– (aq) 6 Fe2+ (aq) + 14 H1+ (aq) + Cr2O72– (aq)  6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O S2O32– (aq) + 2 H1+ (aq)  S (s) + SO2 (g) + H2O

Reactions You now know how to write & balance 4 types of reactions combustion precipitation acid-base neutralization gas-forming and how to recognize redox reactions

Reaction quizzes Reaction quizzes (RQ) will replace NQ I give you the names of the reactants You write the formulas of the reactant and the formulas of the products Cross out spectator ions Write the balanced net ionic equation You need not include state symbols such as (aq) or (g)

Example Solutions of silver nitrate and potassium chloride are mixed AgNO3 (aq) + KCl (aq)  AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq) cross out K1+ and NO31– as spectators Ag1+ + Cl1–  AgCl check: it is already balanced

Example Hydrochloric acid solution is added to solid sodium hydrogen carbonate HCl (aq) + NaHCO3 (s)  HCl (aq) + NaHCO3 (s)  H2O + CO2 (g) + NaCl (aq) cross out Cl1– as a spectator do NOT cross out Na1+ because NaHCO3 is solid H1+ + NaHCO3  H2O + CO2 + Na1+ check: it is already balanced

Example Ethoxyethane burns in air C2H5OC2H5 + O2  C2H5OC2H5 + O2  CO2 + H2O there are no spectators in combustion balance C2H5OC2H5 + 6 O2  4 CO2 + 5 H2O

Example Solutions of nitrous acid and sodium hydroxide are mixed HNO2 (aq) + NaOH (aq)  HNO2 (aq) + NaOH (aq)  H2O + NaNO2 (aq) cross out Na1+ as a spectator do NOT cross out NO21– because HNO2 is a weak acid and is not significantly dissociated in solution! HNO2 + OH1–  H2O + NO21– check: it is balanced

Titrations The text includes examples of acid-base titration and redox titration These are just stoichiometry problems, based on a particular type of reaction Do examples 5-9 and 5-10 to practice these They will be on the test, and you may have to write and/or balance your own equations

Exercise 26 Both NH3 (aq) and HC2H3O2 (aq) conduct weakly, but when these solutions are mixed, the resulting solution conducts electricity very well. Explain.

Exercise 32 Assuming volumes are additive, what is [NO31–] in a solution obtained by mixing 275 mL 0.283 M KNO3, 328 mL 0.421 M Mg(NO3)2, and 784 mL H2O?

Exercise 34 Predict whether a reaction occurs, and if so, write a net ionic equation: a. AgNO3 (aq) + CuCl2 (aq)  b. Na2S (aq) + FeCl2 (aq)  c. Na2CO3 (aq) + AgNO3 (aq) 

Exercise 39 Write net ionic equations to represent these neutralizations of stomach acid (HCl): a. sodium bicarbonate b. calcium carbonate c. magnesium hydroxide

Exercise 42 Give net ionic equations for the preparation of each salt by acid-base neutralization: (NH4)2HPO4, NH4NO3, (NH4)2SO4

Exercise 47 Balance these redox reactions in acid solution: a. MnO41– + I1–  Mn2+ + I2 (s) b. BrO31– + N2H4  Br1– + N2 c. VO43– + Fe2+  U  UO22+ + NO (g)

Exercise 49 Balance these redox reactions in basic solution: a. CN1– + MnO41–  MnO2 (s) + CNO1– b. [Fe(CN)6]3– + N2H4  [Fe(CN)6]4– + N2 c. Fe(OH)2 (s) + O2 (g) (OH)3 (s) C2H5OH + MnO41–  C2H3O21– + MnO2 (s)

Exercise 52 Write a balanced equation for each redox reaction: a. Oxidation of nitrite ion to nitrate ion by permanganate ion in acid solution, producing Mn2+ b. Reduction of H2S with SO2. Both are converted to elemental sulfur and water. c. Reaction of sodium metal with hydroiodic acid. d. Reduction of VO2+ to V3+ by zinc metal, in acidic solution.

Exercise 53 Identify the oxidizing and reducing agents in these reactions: a. 5 SO32– + 2 MnO41– + 6 H+  5 SO42– + 2 Mn2+ + 3 H2O b. 2 NO2 + 7 H2  2 NH3 + 4 H2O c. 2 [Fe(CN)6]4– + H2O2 + 2 H+  2 [Fe(CN)6]3– + 2 H2O

Exercise 60 a. How many mL of concentrated HCl (d = 1.19 g/mL, 38% HCl by mass) must be diluted to 20.0 L to prepare 0.10 M HCl? b. 25.00 mL of the 0.1 M HCl requires 20.93 mL of 0.1186 M NaOH for titration. What is the molarity of the HCl? c. Why could you not prepare 0.1000 M HCl simply by diluting concentrated HCl?

Exercise 64 5.00 mL battery acid (H2SO4) requires 49.74 Ml 0.935 M NaOH for complete neutralization. Is the concentration of the battery acid within the desired range (4.8 – 5.3 M)?

Exercise 67 Iron ore weighing 0.9132 g is dissolved in HCl to produce Fe2+, which is then titrated with 28.72 mL 0.05051 M K2Cr2O7. What is the % Fe by mass in the ore? 6 Fe2+ + 14 H+ + Cr2O72–  6 Fe3+ + 2 Cr3+ + 7 H2O

Half-reactions A half reaction describes either the oxidation or the reduction part of a redox reaction It includes the number of electrons lost or gained The overall reaction is the sum of two half-reactions, one oxidation and one reduction

Examples 5-5A & 5-5B Represent reaction 4.2 with half-reactions and the overall net reaction: 2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g) Represent the reaction of chlorine gas with aqueous sodium bromide to produce liquid bromine and aqueous sodium chloride with half-reactions and a net equation

Balancing Redox Equations Redox equations require electrical balance as well as material (atom) balance Write balanced half reactions for oxidation & reduction First balance atoms, adding H1+ and H2O as needed When atoms are balanced, add e– to balance charge. Multiply each half-reaction by a factor such that the number of electrons lost and gained are the same Combine the half reactions and cancel species that are identical on both sides

Examples 5-6A & 5-6B Balance these equations in acid solution Fe2+ (aq) + MnO41– (aq)  Fe3+ (aq) + Mn2+ (aq) UO2+ (aq) + Cr2O72– (aq)  UO22+ (aq) + Cr3+ (aq)

Redox in basic solution To balance a redox equation in basic solution, Balance it as if it is in acid solution After combining the half-reactions, add equal moles of OH1– to both sides such that all H1+ is neutralized to form H2O Simplify

Examples 5-7A & 5-7B Balance these equations in basic solution S (s) + OCl1– (aq)  SO32– (aq) + Cl1– (aq) MnO41– (aq) + SO32– (aq)  MnO2 (s) + SO42– (aq)

Disproportionation In disproportionation reactions, the same element is both oxidized and reduced 2 H2O2 (aq)  2 H2O + O2 (g) S2O32– (aq)  S (s) + SO2 (g)

Common Redox agents Common oxidizing agents include MnO41– (permanganate) in acid solution Cr2O72– (dichromate) in acid solution O2 or O3 Common reducing agents include S2O32– (thiosulfate) H2