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Chapter 4 Types of Chemical Reactions and Solution Chemistry Topics Water as a solvent Electrolytes and nonelectrolytes Calculations involving molarity of solutions Precipitation reactions Acid base reactions Oxidation reduction reactions 4.1 Water, the common solvent The Water Molecule, Polarity δ+ water meanshas a a Thus, δ- means a partial partial negative partial endpositive (0xygen) and negative charge a partial positive δcharge end (Hydrogen) – O and it is called H H “polar” because of + + δ δ the unequal charge bond angle of water = 105o distribution The Water Molecule bent shape and ability to hydrogen bond gives it many special properties. Water molecules are attracted to one another by dipole interactions This hydrogen bonding gives water: a) its high surface tension, b) its Water’s low vapor pressure. Dissolving ionic salts in water and Hydration Ions have charges and attract the opposite charges on the water molecules. The process of breaking the ions of salts apart is called hydration NH4NO3(s) H2O (l) NH4+(aq) + NO3-(aq) Designates hydration of ions How Ionic solids dissolve in water These ions have been pulled away from the main crystal structure by water’s polarity. H H H H H These ions have been surrounded by water, and are now dissolved! Hydration Solubility in water and Aqueous Solutions Water dissolves ionic compounds (NaCl) and polar covalent molecules (ethanol C2H5OH) The rule is: “like dissolves like” Polar dissolves polar. Nonpolar dissolves nonpolar. Oil is nonpolar. – Oil and water don’t mix. Many Salts are ionic- sea water The Solution Process Called “solvation”. Water breaks the + and - charged pieces apart and surrounds them. Solubility in water depends on the relative attractions of ions for each other and attraction of ions for water molecules In some ionic compounds, the attraction between ions is greater than the attraction exerted by water (slightly soluble slats) – Barium sulfate and calcium carbonate Solids will dissolve if the attractive force of the water molecules is stronger than the attractive force of the crystal. If not, the solids are insoluble. Water doesn’t dissolve nonpolar molecules (like oil) because the water molecules can’t hold onto them. The water molecules hold onto other water molecules, and separate from the nonpolar molecules. Nonpolars? No repulsion between them How does ethanol dissolve in water? Ethanol Molecule Contains a Polar O-H Bond Similar to Those in the Water Molecule The polar water molecule interacts strongly with the polar-O-H bond in ethanol 4.2 The nature of aqueous solutions: strong and weak electrolytes A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount Solution Solvent Solute(s) Soft drink (l) H2O Sugar, CO2 Air (g) N2 O2, Ar, CH4 Soft Solder (s) Pb Sn Electrolytes and Nonelectrolytes Electrolytes- compounds that conduct an electric current in aqueous solution, or in the molten state – all ionic compounds are electrolytes because they dissociate into ions (they are also called “salts”) barium sulfate- will conduct when molten, but is insoluble in water! Nonelectrolytes- Do not conduct an electric current – Most are molecular materials, because they do not have ions Not all electrolytes conduct to the same degree – there are strong electrolytes, and weak electrolytes – Conductivity depends on: degree of dissociation or ionization Electric conductivity of aqueous solutions Nonelectrolyte ______________ Weak electrolyte ______________ Strong electrolyte ______________ 4.1 Dissociation of acids and bases: Strong and weak acids and bases Acids- form H+ ions when dissolved in water (According to Arrhenius) Strong acids dissociate completely into H+ and anions Strong acids- H2SO4 HNO3 HCl HBr HI HClO4 Bases - form OH- ions when dissolved in water Strong bases- KOH, NaOH Weak acids- dissociate partially Acetic acid: HC2H3O2 has 1% dissociation in aqueous solutions The most common weak base is ammonia, NH3 HCl H 2O HNO3 H 2O HC2H3O2(aq) H+ + ClH+ + NO3H+ + C2H3CO2- Strong electrolyte, strong acid Strong electrolyte, strong acid Weak electrolyte, weak acid H2SO4 H+ + HSO4- Strong electrolyte, strong acid HSO4- H+ + SO42- Weak electrolyte, weak acid H3PO4 H2PO4HPO42- H+ + H2PO4H+ + HPO42H+ + PO43- Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid 4.3 H2 O NaOH(s) NH3(aq) + H2O Na+(aq) + OH-(aq) NH4+(aq) + OH-(aq) 4.3 The composition of solutions Molarity (M) Molarity: A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute liters solution Molarity Calculation If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution? Calculating Molarity 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH 40.0 g NaOH 500. mL x 1L_ 1000 mL = 0.500 L 0.10 mole NaOH = 0.20 mole NaOH 0.500 L 1L = 0.20 M NaOH An acid solution is a 0.10 M HCl. How many moles of HCl are in 1500 mL of this acid solution? 1500 mL x 1 L = 1000 mL 1.5 L 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1L How many grams of KCl are present in 2.5 L of 0.50 M KCl? 2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl 1L 1 mole KCl How many milliliters of an acid solution, which is 0.10 M HCl, contain 0.15 mole HCl? 0.15 mole HCl x 1 L soln x 1000 mL 0.10 mole HCl 1L = 1500 mL HCl How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 400. mL x 1 L = 0.400 L 1000 mL 0.400 L x 3.0 mole NaOH x 40.0 g NaOH 1L 1 mole NaOH = 48 g NaOH 4.5 A sample of 0.14 M NaCl. What volume of sample contains 1.0 mg NaCl? # mol = M X V (L) V ( L) 1g # mol M # mol V ( L) M 1L 1 mol NaCl x 1.0mg NaCl x = 1.2X10-4 L x 1000 mg 58.5g NaCl 0.14mol NaCl Dilution Adding more solvent to a known solution. The moles of solute stay the same. #moles = M x volume (L) # moles before dilution (1) = # moles after dilution M1 V1 = M2 V2 Stock solution is a solution of known concentration used to make more dilute solutions Preparing a less concentrated solution from a more concentrated solution by dilution Dilution Add Solvent Moles of solute before dilution (i) = Moles of solute after dilution (f) MiVi = MfVf Example How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3? M1 = 4.00 M2 = 0.200 V2 = 60.0 m L V1 = ?mL M1V1 = M2V2 V1 = M2V2 M1 0.200 M X 60.0 mL = 4.00 M = 3.00 mL Dilution process Wash bottle Funnel pipits Volumetric flask 4.4 Types of Chemical Reactions Precipitation reactions Acid-base reactions Oxidation-reduction reactions 4.5 Precipitation Reactions When aqueous solutions of ionic compounds are mixed together a solid forms. A solid that forms from mixed solutions is called precipitate If the substance is not part of the solution, it is a precipitate Precipitation Reactions precipitate Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq) ________________________ Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3- Anions and cations switch partners ________________________ PbI2 Pb2+ + 2I- PbI2 (s) ________________________ Na+ and NO3- are Spectator ions Solubility rules for common ionic compounds in water at 250 C Soluble Compounds Exceptions Compounds containing alkali metal ions and NH4+ NO3-, HCO3-, ClO3Cl-, Br-, I- Halides of Ag+, Hg22+, Pb2+ SO42- Sulfates of Ag+, Ca2+, Sr2+, Ba2+, Hg2+, Pb2+ Slightly soluble Compounds Exceptions CO3 OH- 2-, PO4 3-, CrO4 2-, S2- Compounds containing alkali metal ions and NH4+ Compounds containing alkali metal ions and Ba2+, Ca2+, Mg2+ are marginally soluble Solubility Rules Predicting reaction’s product All nitrates are soluble Alkali metals ions and NH4+ ions are soluble Halides are soluble except Ag+, Pb+2, and Hg2+2 Most sulfates are soluble, except Pb+2, Ba+2, Hg+2,and Ca+2 Solubility Rules Most hydroxides are slightly soluble (insoluble) except NaOH and KOH Sulfides, carbonates, chromates, and phosphates are insoluble Lower number rules supersede so Na2S is soluble 4.6 Describing reactions in solution Types of equations used to represent chemical reactions: – Formula equation – Complete ionic equation – Net ionic equation Writing Net Ionic Equations 1. Write the balanced formula equation. 2. Write the net ionic equation showing the strong electrolytes 3. Determine precipitate from solubility rules 4. Cancel the spectator ions on both sides of the ionic equation Write the net ionic equation for the reaction of silver nitrate with sodium chloride. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) 4.7 Stoichiometry of precipitation reactions Exactly the same, except you may have to figure out what the pieces are. What mass of solid is formed when 100.00 mL of 0.100 M Barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide? What volume of 0.204 M HCl is needed to precipitate the silver from 50.ml of 0.0500 M silver nitrate solution ? Example Calculate mass of solid NaCl that must be to 1.50 L 0f 0.100M AgNO3 solution to precipitate all of the Ag+ ions in the form of AgCl Ag+(aq) + Cl- (aq) AgCl(s) #moles Ag+ = #moles AgNO3 = 1.50L X 0.100 mol Ag+ 1L = 0.15 mol Ag+ (1:1 mole ratio) #mole Cl- required =#mole NaCl = #moles Ag+ 58.45 g NaCl = 0.15 mol NaCl X 1mol NaCl = 8.77 g NaCl 4.8 Acid-Base Reactions For the purpose of this chapter: An acid is a proton donor a base is a proton acceptor usually (donates OH- ions to the solution) base acid acid base Describing acid-base reactions Strong acid HCl (aq) + NaOH (aq) H+ + Cl- + Na+ + OHH+(aq) + OH- (aq) NaCl (aq) + H2O Na+ + Cl- + H2O H2O (l) Weak acid HC2H3O2(aq)+NaOH (aq) HC2H3O2(aq) +Na+ + OHHC2H3O2(aq) + OH- NaC2H3O2(aq) +H2O C2H3O2- +Na+ +H2O C2H3O2- +H2O Neutralization Reactions and Salts • • • • When acid and bases with equal amounts of hydrogen ion H+ and hydroxide ions OH- are mixed, the resulting solution is neutral. This reaction is called Neutralization reaction • HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) H2SO4(aq) + 2KOH (aq) K2SO4(aq) + 2H2O(l) Notice salt (NaCl) and water are the products Salt = ionic compound whose cation comes from a base and anion from an acid. Neutralization between acid and metal hydroxide produces water and a salt. Example How many mL of 2.00 M H2SO4 are required to neutralize 50.0 mL of 1.00 M KOH? H2SO4 + 2KOH K2SO4 + 2H2O 0.0500 L x 1.00 mole KOH x 1 mole H2SO4 x 1L 1L 2 mole H2SO4 x 2 mole KOH 1000 mL = 1L 12.5 mL Another solution H2SO4 + 2KOH K2SO4 + 2H2O #mole KOH = (M) KOH X (V) KOH #mole KOH = 0.05 L X1M = 0.05 mol KOH 1 mol H2SO4 #mole H2SO4 = # mole KOH X 2mol KOH = 0.025 mol H2SO4 #mole H2SO4 = M H2SO4 X vol (L) H2SO4 0.025 mol H2SO4 = 2.00 M X V V = 0.0125 L = 12.5 mL Acid - Base Titrations Often called a neutralization titration Because the acid neutralizes the base Often called volumetric analysis since titration is made to determine concentrations It involves delivery of a measured volume of solution of known concentration (titrant) Titrant is added to the unknown (analyte) until the equivalence (stoichiometric) point is reached where enough titrant has been added to neutralize it. Titration Equivalence point is marked by using an indicator Where the indicator changes color is the endpoint End point does not match always the equivalence point. A successful titration requires: – A rapid known exact reaction – Endpoint is very close to the equivalence point – Accurate Measurement of volume of titrant Accurate determination of a solution concnetration is called Titration Equivalence point is marked by using an indicator Where the indicator changes color is the endpoint End point does not match always the equivalence point. A 50.00 mL sample of aqueous Ca(OH)2 requires 34.66 mL of 0.0980 M Nitric acid for neutralization. What is [Ca(OH)2 ]? # of H+ x MA x VA = # of OH- x MB x VB Titration A solution of accurately known concentration is gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point– the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Endpoint –the point at which the color of indicator changes Slowly add base to unknown acid UNTIL The indicator changes color (pink) Example In the titration of KHP (molar mass = 204.22g/mol), 34.67 mL of NaOH solution was required to react with 0.1082 g KHP. Calculate the molarity of NaOH solution. Solution NaOH(aq) + KHP(aq) → NaKP(aq) + H2O(l) # mol NaOH = 0.1082 g KHP × 1 mol KHP 1 mol NaOH 204.22 g KHP mol KHP = 5.298X10-4 mol NaOH There are 5.298X10-4 mol of sodium hydroxide in 34.67 mL of solution. #mole NaOH = MNaOH X VNaOH MNaOH = # molNaOH V = 5.298 10 4 mol 34.67 10 3 L NaOH = 1.528 M NaOH 4.9 Oxidation - Reduction (Redox) Reactions 0 1 0 1 2 Na ( s) Cl ( g ) 2 Na Cl ( s) 2 Each sodium atom loses one electron: 1 0 Na Na e Each chlorine atom gains one electron: 0 1 Cl e Cl Definitions Loss of Electrons = Oxidation 1 0 Na Na e Sodium is oxidized Increase of the oxidation state Gain of Electrons = Reduction 0 1 Cl e Cl Chlorine is reduced Decrease of the oxidation state Definitions - The substance that loses electrons is called reducing agent (Electron donor) - The substance that gains electrons is called oxidizing agent (Electron acceptor) +2 0 Mg is the reducing agent -2 0 Mg(s) + S(s) → MgS(s) Mg is oxidized – loses e- S is reduced – gains eS is the oxidizing agent Not All Reactions are Redox Reactions - Reactions in which there has been no change in oxidation number are not redox reactions. Examples: 1 5 2 1 1 1 1 1 5 2 Ag N O3 (aq) Na Cl (aq) Ag Cl ( s) Na N O3 (aq) 1 2 1 1 6 2 1 6 2 1 2 2 Na O H (aq) H 2 S O 4 (aq) Na 2 S O 4 (aq) H 2 O(l ) Assigning Oxidation States • An “oxidation state” (oxidation number) is a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction. • Generally, a bonded atom’s oxidation number is the charge it would have if the electrons in the bond were assigned to the atom of the more electronegative element Rules for Assigning Oxidation States 1) The oxidation State of any uncombined element is zero. 2) The oxidation State of a monatomic ion equals its charge. 0 0 1 1 2 Na Cl 2 2 Na Cl Rules for Assigning Oxidation States 3) The oxidation state of oxygen in compounds is -2, except in peroxides, such as H2O2 where it is -1. 4) The oxidation state of hydrogen in compounds is +1, except in metal hydrides, like NaH, where it is -1. 1 2 H2O Rules for Assigning Oxidation states 5) The sum of the oxidation states of the atoms in the compound must equal 0. 1 2 H2O 2(+1) + (-2) = 0 H O 2 2 1 Ca(O H ) 2 (+2) + 2(-2) + 2(+1) = 0 Ca O H Rules for Assigning Oxidation States 6) The sum of the oxidation States in the formula of a polyatomic ion is equal to its ionic charge. ? 2 N O3 ? 2 S O4 2 X + 3(-2) = -1 N O X + 4(-2) = -2 S O X = +5 X = +6 What is the oxidation states of all atoms in the following ? IF7 F=? 7x-1 + ? = 0 I= NaIO3 Na =? O=? 3x-2 + 1 + ? = 0 I= K2Cr2O7 O=? K=? 7x-2 + 2x+1 + 2x? = 0 Cr =?? 4.10 Balancing oxidation-reduction Equations Two systematic methods for balancing redox equations are available, and are based on the fact that the total electrons gained in reduction equals the total lost in oxidation. The two methods: 1) Use oxidation state changes 2) Use half-reactions (the method to be used her) Balancing redox equations using half-reactions 1: write separate half-reaction equations for oxidation and reduction 2. balance the atoms in the half reactions 3. add enough electrons to one side of each half-reaction to balance the charges 4. multiply each half-reaction by a number to make the electrons equal in both 5. add the balanced half-reactions to show an overall equation and cancel identical species 6. Check that elements and charges are balanced Example 2Mg (s) + O2 (g) 2Mg O2 + 4e- 2MgO (s) 2Mg2+ + 4e- Oxidation half-reaction (lose e-) 2O2- Reduction half-reaction (gain e-) 2Mg + O2 + 4e2Mg + O2 2Mg2+ + 2O2- + 4e2MgO Balancing redox equations in an acidic solution For reactions in acidic solution an 8 step procedure. Write separate half reactions For each half reaction balance all reactants except H and O Balance O using H2O Balance H using H+ Balance charge using e Multiply equations to make electrons equal Add equations and cancel identical species Check that charges and elements are balanced. Example: Balance the following equation I- + OCI- I- Oxidation : I3- + Cl- (acidic solution) 3I- OC1- I3- 3I- Reduction : OC1- I 3- I3- + 2e- C1C1- + H2O OC1- 2H+ + OC1- 1+ C1- + H2O C1- + H2O -1 2H+ + OC1- + 2e- C1- + H2O (2) 3I- I3- + 2e(1) (1) + (2) 2H+ + OC1- + 3I- C1- + I3- + H2O Balancing equations in basic solution • Do everything you would with acid, but add one more step. • Add enough OH- to both sides to neutralize the H+ • For each H atom that is lacking, add one molecule of H2O to the side that requires it. At the same time add one unit of OH- in the opposite side of the half-reaction Example Balance the following equation in basic solution MnO4- + C2O42- MnO2 + CO32First half reaction: MnO4- MnO2 (a) MnO4- MnO2 + 2H2O (b) 4 H2O + MnO4- MnO2 + 2H2O + 4 OH(c) 4 H2O + MnO4- + 3 e- MnO2 + 2H2O + 4 OH(d) 2 H2O + MnO4- + 3 e- MnO2 + 4 OH- (1) Second half reaction: C2O42- CO32(a) C2O42- 2 CO32(b) 2 H2O + C2O42- 2 CO32(c) 4 OH- + 2 H2O + C2O42- 2 CO32- + 4 H2O (d) 4 OH- + 2 H2O + C2O42- 2 CO32- + 4 H2O + 2 e(e) 4 OH- + C2O42- 2 CO32- + 2 H2O + 2 e- 2 H2O + MnO4- + 3 e- MnO2 + 4 OH- (2) (1) We eliminate electrons by multiplying (1) by 2 and (2) by 3. (2) X 3: 12 OH- + 3 C2O42- 6 CO32- + 6 H2O + 6 e- (3) (1) X 2: 4 H2O + 2 MnO4- + 6 e- 2 MnO2 + 8 OH- (4) (3) + (4) to eliminate the electrons and common species. 4 OH- + 3 C2O42- + 2 MnO4- 6 CO32- + 2 H2O + 2 MnO2