Chemical Reactions and Solution Stoichiometry Chapter-4

Types of Chemical Reactions and Solution Stoichiometry Main Concepts: Water, the Common Solvent Strong and Weak Electrolytes The Composition of Solutions Types of Chemical Reactions Solution Stoichiometry Oxidation States (numbers) Oxidation-Reduction Reactions Balancing Redox Reactions 2

Aqueous Solutions

Aqueous Solutions Aqueous solutions (aq) are dissolved in water. Water is a good solvent because the molecules are polar and can dissolve nearly all ionic compounds and non-ionic compounds if they have polar bonds. The hydrogen atoms have a partial positive charge and oxygen has a partial negative charge. The molecule is bent with a bond angle of 104.5º.

Solutions Solutions are defined as homogeneous mixtures of two or more pure substances. Solutions can be in the gas phase, the liquid phase, the solid phase, or a combination of these phases. The solvent is usually present in greatest abundance. All other substances are solutes.

Solubility Solubility is normally measured in grams of solute per 100 mL water (g/100 mL or g/100 g). Solubility varies greatly depending upon the temperature (solids, liquids, gases), pressure (gases), and type of solute. Solubility curves are used to determine the mass of solute in 100 g of water at a given temperature. Any point on the curve represents a saturated solution. Above the curve: a supersaturated solution (unstable). Below the curve : unsaturated

Solubility Curves

Dissociation (Dissolution) When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates (hydrates) them. This process is called dissociation. Molecular substances ionize in water.

Dissociation (Dissolution) Dissolution occurs in three steps: Breaking solute-solute attractions (endothermic), ie, lattice energy in salts. Breaking solvent-solvent attractions (endothermic), ie, hydrogen bonding Forming solvent-solute attractions (exothermic), in solvation. The value of the overall enthalpy change is the sum of the individual enthalpy changes of each of these steps and may be exothermic or endothermic.

Dissociation (Dissolution) Solutions with negative enthalpy changes of solution form stronger bonds and have lower vapor pressure. Selected enthalpies of solution (kJ/mol): hydrochloric acid -74.84 ammonium nitrate +25.69 ammonia -30.50 sodium hydroxide -44.51 potassium hydroxide -57.61 cesium hydroxide -71.55 sodium chloride +3.88 potassium chlorate +41.38 acetic acid -1.51

Electrolytes and Nonelectrolytes An electrolyte is a substances that dissociates into ions when dissolved in water and conducts electricity. A nonelectrolyte may dissolve in water, but it does not dissociate into ions and does not conduct electricity.

Types of Solutions Strong electrolytes: completely dissociate (dissolve into ions). Many ions, conduct electricity very well. Weak electrolytes: partially dissolve into ions. Few ions, conduct electricity slightly. Non-electrolytes: may dissolve, but no ions, do not conduct an electric current.

Electrolytes and Nonelectrolytes Strong Electrolytes: Nearly all ionic compounds (soluble salts), strong acids, and strong bases Weak Electrolytes: Weak acids and weak bases Non Electrolyte: All other compounds All Soluble ionic compounds are electrolytes. Molecular (covalent) compounds: tend to be nonelectrolytes, except for acids and bases.

Electrolytes and Nonelectrolytes

Strong Electrolytes (Must Memorize ) Strong acids, Strong bases, and Soluble ionic salts

Strong Electrolytes Strong acids, Strong bases, and Soluble ionic salts

Solublility Two basic rules regarding solubility: Most nitrates (NO3-), sulfates (SO42-), and acetates (CH3COO-) are soluble. So are halides (compounds of halogens) Cl-, Br-, and I-. Most carbonates (CO32-), phosphates (PO43-), chromates (CrO42-), sulfites (SO32-), sulfides (S2-), Ca(OH)2, and AgCl are some of the substances that are only sparingly soluble (less than 0.1 g per 100-mL water). Note: Sparingly soluble, slightly soluble, and insoluble mean the same thing.

Solublility Calcium hydroxide is 100% soluble in 0.01M solutions or lower making it a strong base. In higher concentrations, it is less soluble (as determined by Kb) but remains a strong base because that portion that is soluble dissolves completely. Solubility increases as you go down group-2 The solubility of salt in water depends upon the lattice energy and the hydration energy. if hydration energy > lattice energy, the salt will dissolve. if hydration energy < lattice energy, the salt will not dissolve

Table of Solubility All common compounds made up of alkali metal cations (Li+, Na+, K+, Rb+, Cs+, or Fr+) or NH4+ and negative ions are soluble. 2. All compounds containing C2H3O2- or NO3- and a positive ion are soluble. 3. All compounds containing SO42- are soluble except when combined with ions of Ba2+, Sr2+, Pb2+, Ca+2, or Hg22+.

Table Of Solubility 4. All compounds containing Br-, Cl-, or I- are soluble except when combined with ions of Ag+, Hg22+, or Pb2+. 5. Compounds of HCO3- are soluble with NH4+ and alkali metal/alkaline earth metal ions (Group 1 and 2). Compounds of HCO3- are insoluble with all other positive ions.

Table Of Solubility 6. All compounds containing: OH- or S2- are insoluble except with NH4+, Ca2+, Sr2+, Ba2+, and alkali metal cations CO32- are insoluble except with NH4+ and alkali metal cations PO43- are insoluble except with NH4+ and alkali metal cations CrO42- (chromates) are usually insoluble exception Na2CrO4, K2CrO4, (NH4)2CrO4, and MgCrO4.

Solution Concentration Types of Calculations: Molarity Dilution Mixing Neutralization Stoichiometry 22

volume of solution in liters Molarity Two solutions can contain the same substances but be very different because the concentrations of those substances are different. Molarity is one way to measure the concentration of a solution (similar to density for a solid). moles of solute volume of solution in liters Molarity (M) =

Solution Preparation To make a solution of a known molarity, weigh out a known mass (number of moles) of the solute. The solute is added to the solvent (usually water) in a volumetric flask, dissolved, and additional solvent is added to the line on the neck of the flask.

Solution Preparation To make one liter of a 1M CuCl2 solution, weigh out 134.45 g CuCl2 and dissolve it in 200-300 mL water in a 1 L volumetric flask adding water to the line on the neck of the flask. Using a 500 mL flask, dissolve 67.23 g CuCl2 in 100-200 mL water and fill to the line.

Preparation of a Standard Solution

Molarity How many grams of HCl would be required to make 50.0 mL of a 2.7 M solution? 4.93 g HCl 2. What would the concentration be if you used 27g of CaCl2 to make 500.0 mL of solution? 0.49M CaCl2 What is the concentration of each ion in this solution? 0.49M Ca+2, 0.98M Cl- 27

Molarity 3. Calculate the concentration of a solution made by dissolving 45.6 g of Fe2(SO4)3 in 475 mL solution. 0.24M Fe2(SO4)3 What is the concentration of each ion? 0.48M Fe+3, 0.72M SO4-2 4. Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution. 4.73M NaCl 28

Molarity 5. Describe how to make 100.0 mL of a 1.00M K2Cr2O4 solution. 24.6 g (246.2 g/mol) K2Cr2O4, 100 ml mark 6. Describe how to make 250.0 mL of an 2.0M copper(II) sulfate dihydrate solution. 97.75 g (195.5 g/mol) CuSO4∙2H2O, 250 ml mark 29

Dilution 30

Dilution Adding more solvent to a known solution. The moles of solute stay the same. moles = M x V (in liters) M1  V1 = M2  V2 The molarity of the new solution can be determined from this equation where M1 and M2 are the molarity of the concentrated and dilute solutions, respectively, and V1 and V2 are the volumes of the two solutions. A stock solution is a solution of known concentration used to make more dilute solutions (HCl = 12M, H2SO4 = 18M)

Practice Calculation 1) What volume of a 1.7M solution is needed to make 250 mL of a 0.50M solution? 73.5 mL 2. How much solute and solvent are needed to make 375 mL of 0.28M H2SO4 solution starting from stock H2SO4 (18M )? M1V1 = M2V2 V1 = (0.28M)(375 mL) = 5.83 mL H2SO4 18M 369.17 ml H2O

Practice Calculation 3) 18.5 mL of 2.3M HCl is added to 250 mL of water. What is the concentration of the solution? 0.16M 4) How much solute and solvent are needed to make 1200 mL of 1.25M HCl solution starting from stock HCl (12M )? 125 mL HCl and 1075 ml H2O

Dilution You can also prepare very small concentrations of a solution by diluting a more concentrated solution. Serial dilutions are used to accurately create highly diluted solutions. Pipets and volumetric pipets are used to deliver a volume of the solution to a new volumetric flask and then adding solvent to the line on the neck of the new flask.

Serial Dilution Usually, the dilution factor at each step is constant, resulting in a geometric progression of the concentration. Serial dilutions of 10, 100, and 1,000 are common. A ten-fold serial dilution could be 1 M, 0.1 M, 0.01 M, 0.001 M... Serial dilutions allow an accurate way to produce solutions in ppm or ppb.

Serial Dilution Dilution = V of Sample / Total V of (sample + diluent) Dilution Factor = Total V of (sample + diluent) / V of sample ** or we can simply say the reciprocal of Dilution

Serial Dilution

Molarities in Stoichiometric Calculations

Solution Stoichiometry Calculation The following steps are used for calculations involving solutions, limiting reactants, and remaining ions in solution: 2A + 1B  2C + 3D 1) Write the balanced molecular or ionic equation. 2) Calculate the number of moles of each reactant present (moles = MV for solutions). 3) Divide moles of “A” by coefficient of “A” and divide moles of “B” by coefficient of “B” The smallest quantity is your limiting reactant. The other is your excess reactant.

Solution Stoichiometry Calculation Note: The original amount of limiting reactant is then used to calculate the amounts of products formed and the amount of excess reactant remaining. 4) Calculate the amount of product produced. 5) Calculate the amount of excess reactant remaining if required. 6) Determine the starting, ending, and remaining amounts (moles) of ions if necessary. 7) Calculate the ion concentration (molarity) by dividing the moles of ions by the total solution volume.

Practice Calculations Calculate: a) The mass of lead (II) iodide formed from mixing 50.0 mL 0.500 M lead (II) nitrate with 75 mL of 0.750 M potassium iodide? Pb(NO3)2 + 2KI → PbI2 + 2KNO3 0.025 mol Pb(NO3)2 vs 0.05625 mol KI present 0.025 mol Pb(NO3)2 vs 0.028125 mol KI PbI2 formed: 0.025 mol Pb(NO3)2 x 1 x (461 g PbI2) = 11.52 g 1 mol

Practice Calculations Calculate: b) The concentrations of K+ and NO3- remaining in solution? Pb(NO3)2 + 2KI → PbI2 + 2KNO3 0.025 mol Pb(NO3)2 and 0.05625 mol KI present Neither of these ions (K+ and NO3- ) were converted to product. K+ = 0.05625 = 0.45 M NO3- = 0.050 = 0.40 M 0.125 L 0.125L

Stoichiometry of Precipitation 2) What mass of solid is formed when a 380 mL solution of 0.82M NaOH is mixed with 400 mL of 0.73M ZnSO4 and what is the concentration of the remaining ions in solution? 2NaOH + ZnSO4 → Na2SO4 + Zn(OH)2 0.312 mol 0.292 mol 0.156 mol NaOH limiting and 15.5 g Zn(OH)2 (0.156 mol) is formed Ions Start Used End Conc. Na+ 0.312 -0- 0.312 0.40M OH- 0.312 0.312 -0- -0- Zn+2 0.292 0.156 0.136 0.174M SO4-2 0.292 -0- 0.292 0.374M

Solution Stoichiometry Calculation 3) 25 mL 0.67M of H2SO4 is added to 35 mL of 0.40M CaCl2. What mass CaSO4 is formed and what is the concentration of the remaining ions in solution? H2SO4 + CaCl2 → CaSO4(s) + 2HCl 0.01675 mol 0.014 mol 0.014 mol CaCl2 is limiting: CaSO4 0.014 mol = 1.9 g formed Ions Start Used End Conc. H+ 0.0335 none 0.0335 0.56M SO4-2 0.01675 0.014 0.00275 0.046M Ca+2 0.014 0.014 -0- -0- Cl- 0.028 none 0.028 0.47M

Stoichiometry of Precipitation 4) What mass of precipitate will be formed and what is the concentration of the remaining ions in solution when 125 mL 0.75M of Pb(NO3)2 is added to 300 mL of 0.50M KI. ? Pb(NO3)2 + 2KI → PbI2(s) + 2KNO3 0.0938 mol 0.15 mol → 0.075 KI is limiting: 0.075 mol PbI2 formed (34.5 g) Ions Start Used End Conc. Pb+2 0.0938 0.075 0.0188 0.044M 2NO3- 0.1876 -0- 0.1876 0.44M K+ 0.15 -0- 0.15 0.35M I- 0.15 0.15 -0- -0-

Practice Calculation 5) Calculate the volume of steam produced by the complete combustion of 1.48 x 104 L of 6.25 M ethanol (C2H5OH). C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) Vg = (1.48 x 104 L )(6.25 M)(3/1)(22.4 L/mol) Vg = 6.22 x 106 L steam 9.25 x 104 mol C2H5OH produces 2.775 x 105 mol H2O = 6.22 x 106 L steam.

Mixing M = M1V1 + M2V2 +… V1 + V2 +… Calculate the final concentration of a solution made by mixing 125 mL of 2M HCl with 45 mL of 0.30M HCl. 1.55M 2. Calculate the final concentration of a solution made by mixing 350 mL of 2M HCl with 100 mL of 1.2M HCl and adding an additional 500 mL of water. 0.86M

Neutralization MaVa = MbVb naMaVa = nbMbVb where n = # H’s for the acid and # OH’s for the base. Calculate the amount of 2.0M HCl required to neutralize 650 mL of 0.45M NaOH. 146 mL 2. Calculate the amount of 1.8M Ba(OH)2 required to neutralize 450 mL of 3.5M HNO3. 437.5 mL

Major Types of Chemical Reactions

Major Types of Chemical Reactions Combustion Reactions: Combination Reactions (synthesis/composition): Decomposition Reactions (separation): Single Replacement Reactions Double Replacement Reactions (Metathesis) Ionic Reactions/Reactions in Aqueous Solutions Oxidation–Reduction (Redox Reactions): Nuclear Reactions: reactions that involve changes in the nucleus.

Metathesis Reactions Double Replacement Metathesis (double replacement) comes from a Greek word that means “to transpose.” These reactions include all double replacement reactions in which the cations (or anions) exchange partners ( cation replaces cation or anion replaces anion). AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)

Metathesis Reactions Products of metathesis reactions may include a precipitate, an insoluble gas, water, a weak electrolyte, or a covalent compound while the other compound is usually soluble and remains dissolved in solution as determined by the table of solubility. AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)

Metathesis Reactions The chemistry takes place in several steps: When potassium chloride and silver nitrate dissolve, they become hydrated ions: KCl(s) + 12H2O ® K(H2O)6+ + Cl(H2O)6- AgNO3(s) + 12H2O ® Ag(H2O)6+ + NO3(H2O)6- When the silver ions and chloride ions meet in solution, they combine and form a solid, which appears as a white precipitate: Ag+(H2O)6 + Cl(H2O)6- ® AgCl(s) + 12H2O

Metathesis Reactions 1. Precipitation Reaction: When ions are mixed and form compounds that are insoluble, a precipitate is formed. NaCl(aq) + AgNO3(aq) ® AgCl(s) + NaNO3(aq)

Types of Metathesis Reactions 2. Neutralization Reactions (Acid-Base Reactions) acid + base ® salt + water HCl + NaOH ® NaCl + H2O Traditional acids follow the Arrhenius definition: any substance that dissolves in water increasing the H+ concentration. Traditional acids start with “H” (HCl, HF, HNO3..). Organic acids are organic compounds with acidic properties. The most common organic acids are carboxylic acids having the carboxyl functional group (–COOH: formic acid HCOOH, acetic acid CH3COOH, lactic acid C3H6O3…).

Types of Metathesis Reactions 3) Hydrolysis Reactions: Hydrolysis is the reverse of neutralization and results when a salt plus water yields an acid plus a base. Salts are a product of neutralization, but salts that undergo hydrolysis usually are not neutral. Strong acid base reactions yield neutral salts and the pH is 7. salt + water ® acid + base NH4Cl + HOH ® HCl + NH4OH

Types of Metathesis Reactions 4. Gas Forming Reactions: Metal carbonates and metal hydrogen carbonates (bicarbonate) react with acids to produce carbon dioxide gas and water. CaCO3(s) + HCl(aq)  H2CO3(aq) + CaCl2(aq) however carbonic acid, H2CO3, is unstable and decomposes to form CO2(g) and H2O(l): therefore, CaCO3(s) + HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l) NaHCO3(aq) + HBr(aq) NaBr(aq) + CO2(g) + H2O(l) Remember this! H2CO3, is unstable and decomposes to form CO2(g) and H2O(l)

Gas-Forming Reactions Metal sulfites (SO3-2) and metal hydrogen sulfites react with acids to produce sulfur dioxide gas and water. SrSO3(s) + 2HI(aq) ® SrI2(aq) + SO2(g) + H2O(l) 2KHSO3(s) + H2SO4(aq) ® 2SO2(g) + 2H2O(l) + K2SO4(aq)

Gas-Forming Reactions Metal sulfides react with acids to produce hydrogen sulfide gas. CaS(s) + 2HNO3(aq) ® H2S(g) + Ca(NO3)2(aq) Net Ionic Equation: CaS(s) + 2H+(aq) ® H2S(g) + Ca2+(aq)

Three Types of Equations 1. Molecular (formula) Equations: written as whole formulas with no ions. 2. Ionic Equations: only strong electrolytes dissolved in water are written in ionic form. 3. Net Ionic Equations: no spectator ions are shown.

Writing Net Ionic Equations Write a balanced molecular equation. Rewrite the equation to show the ions that form in solution when each soluble strong electrolyte dissociates into its component ions. Only strong electrolytes dissolved in aqueous solution are written in ionic form (strong acids, strong bases, and soluble salts) no solids or liquids. Cross out anything that remains unchanged from the left side to the right side of the equation (spectator ions). Solubility table. Write the net ionic equation with the species that remain.

Three Types of Equations 1) Write the three types of equations when solutions of potassium chromate and barium nitrate are mixed: Molecular equation: written as whole formulas with no ions: K2CrO4(aq) + Ba(NO3)2(aq) ® BaCrO4(s) + 2KNO3 Ionic equations: show only dissolved strong electrolytes as ions. 2K+(aq) + CrO4-2(aq) + Ba+2(aq) + 2 NO3-(aq) ® BaCrO4(s) + 2K+(aq) + 2 NO3-(aq)

Three Types of Equations Net ionic equation: show only the product and the ions that form the product; no spectator ions are shown. 2K+(aq) + CrO4-2(aq) + Ba+2(aq) + 2NO3-(aq) ® BaCrO4(s) + 2K+(aq) + 2NO3-(aq) Spectator ions CrO4-2(aq) + Ba+2(aq) ® BaCrO4(s)

Net Ionic Equation Practice 2) Write the three types of equations for the reaction when hydrochloric acid reacts with solid ferrous sulfide. molecular equation: 2HCl(aq) + FeS(s)  FeCl2(aq) + H2S(g) ionic equation: 2H+ + 2Cl- + FeS  H2S + Fe2+ + 2Cl- net ionic equation: 2H+ + FeS  H2S + Fe2+

Ionic Equations Practice 3) Write the net ionic equation when lead(II) nitrate and sodium iodide are mixed. Molecular equation: Pb(NO3)2 + 2NaI  PbI2 + 2NaNO3 Ionic equation: Pb+2 + 2NO3- + 2Na+ + 2I-  PbI2(s) + 2Na+ + 2NO3- Net ionic equation: Pb+2 + 2I-  PbI2(s)

Ionic Equations Practice 4) Write the net ionic equation when solutions of acetic acid (CH3COOH) and sodium sulfite are mixed. Molecular equation: 2CH3COOH(aq) + Na2SO3(aq)  2NaCH3COO(aq) + H2O(l) + SO2(g) Ionic equation: 2CH3COOH + 2Na+ + SO32-  2Na+ + 2CH3COO- + HOH + SO2 Net ionic equation: 2CH3COOH + SO32-  CH3COO- + HOH + SO2

Writing Net Ionic Equations Note: The first proton in sulfuric acid (H2SO4) is ionized completely; the second proton is only partially ionized. Sulfuric acid is the only polyprotic acid that exhibits this property. All other polyprotic acids are weak and are written in their molecular form. For sulfuric acid ionization (dissolved in water): H2SO4  H+ + HSO4- Aqueous solutions of sulfuric acid contain a mixture of H+, HSO4- and SO42.

Writing Net Ionic Equations 4) However, the net ionic equation for the neutralization reaction between sulfuric acid and sodium hydroxide goes to completion: molecular equation: H2SO4(aq) + 2NaOH(aq)  2HOH(l) + Na2SO4(aq) ionic equation: 2H+(aq) + SO4-2(aq) + 2Na+(aq) + 2OH-(aq)  2HOH(l) + 2Na+(aq) + SO42- net ionic equation: 2H+(aq) + 2OH-(aq)  2H2O(l)

Writing Net Ionic Equations 5) Write the molecular, ionic, and net ionic equation for the reaction between: solid calcium fluoride and sulfuric acid CaF2(s) + H2SO4  CaSO4 + 2HF CaF2(s) + H+ + HSO4- → CaSO4(s) + 2HF Net: CaF2(s) + H+ + HSO4- → CaSO4 + 2HF or CaF2(s) + 2H+ + SO4-2 → CaSO4 + 2HF

Writing Net Ionic Equations 6) Write the molecular, ionic, and net ionic equation for the reaction between aluminum and hydrobromic acid. 2Al(s) + 6HBr(aq)  2AlBr3(aq) + 3H2(g) 2Al(s) + 6H+ + 6Br-  2Al3+ + 6Br- + 3H2(g) Net: 2Al(s) + 6H+(aq)  2Al3+(aq) + 3H2(g)

Writing Net Ionic Equations 5) Write the net ionic equation when solutions of Ferric bromide and ammonium acetate are mixed: NR. All are soluble FeBr2 + NH4CH3COO 

Acid-Base Reactions

Acids and Bases Types of Acids and Bases Arrhenius (traditional) Acids and Bases Bronsted-Lowry Acids and Bases Lewis Acids and Bases 73 73

Arrhenius Acids and Bases Arrhenius defined traditional acids as substances that increase the concentration of H+ when dissolved in water. Arrhenius defined traditional bases as substances that increase the concentration of OH- when dissolved in water.

Arrhenius Acid/Base Reaction Acid + Base = Salt + Water 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)

Brønsted - Lowry Acids and Bases Brønsted - Lowry defined acids as anything that donates a H+ (proton donor). Brønsted-Lowry defined bases as anything that accepts a H+ (proton acceptor). A Brønsted - Lowry acid-base reaction produces a conjugate acid-base pair Reaction: acid + base  conj. acid + conj. base HNO2(aq) + H2O(aq)  NO2-(aq) + H3O+(aq) The difference between any acid and its conjugate base is one H+. The stronger the acid, the weaker the conjugate base….

Brønsted - Lowry Acids and Bases In a Brønsted - Lowry acid-base reaction, the acid (proton donor) donates a proton (H+) to the base (proton acceptor). NH3 + H2O  NH4+ + OH-

Lewis Acids and Bases Lewis: acid: accepts an electron pair base: donates an electron pair. The Lewis base must have an unshared pair of electrons on the central atom (same as a Brønsted base). The advantage of this theory is that many more reactions can be considered acid-base reactions because they do not have to occur in solution. Reaction: CO + BH3  COBH3

Bases Bases also include substances that do not contain hydroxide ions (OH-) but react with H+ ions in solution forming hydroxide ions. Ammonia (NH3) is a common weak base. When added to water, it accepts H+ ions from water molecules producing OH- ions. NH3(aq) + HOH(l) ↔ NH4+(aq) + OH-(aq)

Bases In nearly all reactions in aqueous solutions, NH3 ↔ NH4+ or NH4+ ↔ NH3 Solid ammonium chloride + NaOH (aq) NH4Cl(s) + OH-(aq) → NH3(aq) + HOH(l) + Cl-(aq) Solutions of ammonia and hydrofluoric acid are mixed NH3 + HF → NH4+ + F-

Bases Exceptions are reactions of ammonia with solutions containing transition metals especially Cu, Ni, Ag, Zn, Sn, and Au to form complex ions called coordination compounds. Concentrated ammonia plus cupric nitrate NH3 + Cu2+ → [Cu(NH3)4]2+ ammonia plus zinc hydroxide NH3 + Zn(OH)2 → [Zn(NH3)4]2+ + OH-

Acid/Base Reactions When equal amounts of a strong acid reacts with a strong base, the pH = 7.0 and the solution is neutral (neutralization). The net ionic equation is: H+(aq) + OH-(aq)  H2O(l) The conjugate base of a strong acid is a weak base that has no effect on the solution pH.

Acid-Base Reactions 2) When a strong acid reacts with a weak base, the solution is acidic. NH3(aq) + HCl(l)  NH4+(aq) + Cl-(aq) The conjugate of a weak base is a weak acid that lowers the pH of the final solution.

Acid-Base Reactions 3) When a weak acid reacts with a strong base, the solution is basic. CH3COOH(aq) + NaOH(aq)  NaCH3COO(aq) + H2O(l) CH3COOH(aq) + OH-(aq)  CH3COO-(aq) + H2O(l) The conjugate of a weak acid is a weak base that raises the pH of the final solution.

Acid-Base Reactions 4) When a weak acid reacts with a weak base, the solution can be acidic or basic depending upon the degree of ionization (Ksp) of the acid and base. CH3COOH(aq) + NH3(aq) → NH4+(aq) + CH3COO-(aq)

Neutral Aqueous Ions Neutral Aqueous Anions Neutral Aqueous Cations (anion from strong acids) (cation from strong bases) chloride Cl- lithium ion Li+ bromide Br- sodium ion Na+ iodide I- potassium ion K+ nitride NO3- rubidium ion Rb+ sulfate SO4-2 cesium ion Cs+ clorate ClO3- calcium ion Ca+2 perchlorate ClO­4- strontium ion Sr+2 barium ion Ba+2

Acid-Base Reactions Will the solution formed from the following reagents have a pH > 7, pH < 7, or pH = 7? HNO3 + CsOH Neutral, pH = 7 strong acid/strong base 2) HF + Sr(OH)2 pH > 7, basic, weak acid/strong base 3) HClO + Ni(OH)2 undetermined, weak acid/weak base 4) HClO4 + Mg(OH)2 pH < 7, acidic, strong acid/weak base

Titration

Titration In an acid-base titration, a known concentration of acid (standard solution) is used to measure an unknown concentration of base and vice-versa.

Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point.

Titration Endpoint: the point in the titration where the indicator changes color. Equivalence point (stoichiometric point): the point in the titration where the amount of titrant and analyte exactly neutralize each other. The endpoint is not always at the equivalence point.

Titration Titration is an analytical technique used to calculate the concentration of a solute in a solution.

Titration Often called a neutralization reaction because the acid neutralizes the base and vice-versa. A solution of known concentration (titrant) is added to the unknown (analyte) containing an indicator until the end point is reached where enough titrant has been added to neutralize the analyte. An aliquot is a portion of a total amount of a solution.

Indicators: Measurement of pH:

Acid-Base Reaction naMaVa = nbMbVb A 50.00 mL sample of aqueous Ca(OH)2 requires 34.66 mL of 0.26M nitric acid for neutralization. Calculate the concentration of the Ca(OH)2. 0.09M Ca(OH)2

Acid-Base Reaction 2) Calculate the pH of a solution when a 150.00 mL sample of 0.4 M LiOH is mixed with 250 mL of 0.26 M HCl. Moles of OH- = MV = (0.4 M)(0.150 L) = 0.06 mol Moles of H+ = MV = (0.26 M)(0.250 L) = 0.065 mol Excess H+ = 0.005 mol [H+] = 0.005 mol = 0.0125 M 0.4 L pH = [H+] = -log 0.0125 M = 1.9

Acid-Base Reaction 3) 75 mL of 0.25M HCl is mixed with 225 mL of 0.055 M Ba(OH)2 . What is the concentration of the excess ion (H+ or OH-) and the pH of the solution? H+ = 0.01875 mol, OH- = 2(0.012375) = 0.02475 mol OH- in excess (0.02475 - 0.01875) = 0.006 mol [OH- ] = 0.006 mol = 0.02M 0.3 L pOH = -log [OH- ] = -log [0.02M ] = 1.7 and pH = 12.3

Stoichiometry of Precipitation 4) What mass of solid is formed when a 540 mL solution of 0.83M NaOH is mixed with 350 mL of 0.54M ZnSO4 and what is the concentration of the remaining ions in solution? 2NaOH + ZnSO4 → Na2SO4 + Zn(OH)2 0.4482 mol 0.189 mol 0.189 mol ZnSO4 is limiting (0.2241 NaOH/0.189 mol ZnSO4) and 18.78 g Zn(OH)2 (0.189 mol) is formed Ions Start Used End Conc. Na+ 0.448 -0- 0.448 0.50M OH- 0.448 0.378 0.07 0.079M Zn+2 0.189 0.189 -0- -0- SO4-2 0.189 -0- 0.189 0.21M

Redox Reactions Oxidation-Reduction Reactions

Oxidation-Reduction Reactions Oxidation occurs when an atom or ion loses electrons. Reduction occurs when an atom or ion gains electrons. One cannot occur without the other.

Oxidation Numbers To determine if an oxidation-reduction reaction has occurred, oxidation numbers are assigned to each element in a neutral compound or charged entity.

Oxidation Number Rules: Made Simple Oxidation numbers are assigned in the following order: 1, 2, 13, H, F, O, 17, 16 Group-1, group-2, group-13, hydrogen, fluorine, oxygen, group-17, and finally group-16.

Oxidation States Assign the oxidation states to each element in the following. CO2 NO3- H2SO4 NaH Fe2O3 Fe3O4 Rb2O

Balancing with Half Reactions Cu(s) + AgCl(aq)  CuCl2(aq) + Ag(s) The two half reactions are: Oxidation: Cu(s)  Cu2+(aq) + 2e- Reduction: 2[Ag+(aq) + e-  Ag(s)] Adding: 2Ag+(aq) + Cu(s) ® 2Ag(s) + Cu2+(aq) Cu is being oxidized (reducing agent) and Ag is being reduced (oxidizing agent).

Activity Series

Oxidation-Reduction 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 Practice For the following reactions identify: Oxidizing agent, reducing agent, substance oxidized, and substance reduced. Fe(s) + O2(g) ® Fe2O3(s) Fe2O3(s) + 3CO(g) ® 2 Fe(l) + 3CO2(g) SO32- + H+ + MnO4- ® SO42- + H2O + Mn2+

Balancing Redox Reactions

Half-Reactions All redox reactions can be thought of as happening in two halves: One produces electrons - Oxidation half. The other requires electrons - Reduction half.

Balancing Redox Reactions: Molecular 1) I2 + HNO3  HIO3 + NO2 + H2O oxidation I2  I+5 + 5e I2  2I+5 + 10e reduction N+5 + e-  N+4 Adding: 10[N+5 + e-  N+4] I2  2I+5 + 10e- I2 + 10HNO3  2HIO3 + 10NO2 + 4H2O

Balancing Redox Reactions: Molecular 2) HNO3 + KI  KNO3 + I2 + NO + H2O oxidation 2I-  I2 + 2e 3[2I-  I2 + 2e] reduction N+5 + 3e N+2 2[N+5 + 3e N+2] 6I- + 2N+5  3I2 + 2N+2 2HNO3 + 6KI  KNO3 + 3I2 + 2NO + H2O 2HNO3 + 6KI  6KNO3 + 3I2 + 2NO + H2O must adjust coefficient for HNO3 on left to account for the “N” on right that is not being oxidized or reduced. 8HNO3 + 6KI  6KNO3 + 3I2 + 2NO + 4H2O

Balancing Redox Reactions: Ionic SO32- + H+ + MnO4- ® SO42- + H2O + Mn+2 For balancing ionic reactions in acidic or basic solution, use the following steps:

Balancing Redox Reactions: Ionic For reactions in acidic solution, 8 steps: 1. Write separate half reactions (use entire species) 2. For each half reaction balance all reactants and products for mass except H and O. 3. Balance O by adding H2O. 4. Balance H by adding H+. 5. Balance charge using e- 6. Multiply equations to make electrons equal 7. Add equations and cancel identical species 8. Check that the charges and elements are balanced.

Basic Solutions Do everything you would with acid solution, but add two more steps. 9. Add enough OH- to both sides to neutralize the H+: this makes water 10. Cancel the water

Balancing Redox Reactions 3) Balance the following reaction that occurs in acidic solution: SO32- + H+ + MnO4- ® SO42- + H2O + Mn+2 Half reactions: Oxid: SO32- ® SO42- Red: MnO-4 ® Mn+2 Balanced for mass: Oxid: H2O + SO32- ® SO42- + 2H+ Red: 8H+ + MnO-4 ® Mn+2 + 4H2O Balanced for charge: Oxid: H2O + SO32- ® SO42- + 2H+ + 2e- Red: 8H+ + MnO-4 + 5e- ® Mn+2 + 4H2O

Balancing Redox Reactions Equalize Charges: multiply oxidation by 5 and reduction by 2 Oxid: 5[H2O + SO32- ® SO42- + 2H+ + 2e-] Red: 2[8H+ + MnO-4 + 5e- ® Mn+2 + 4H2O] 5H2O + 5SO32- ® 5SO42- + 10H+ + 10e- 16H+ + 2MnO-4 + 10e- ® 2Mn+2 + 8H2O Combine: 5H2O + 5SO32- + 16H+ + 2MnO-4 ® 5SO42- + 10H+ + 2Mn+2 + 8H2O Eliminate: 5SO32- + 6H+ + 2MnO-4 ® 5SO42- + 2Mn+2 + 3H2O Check: -10 + 6 -2 ® -10 + 4 -6 ® -6

Redox Practice 4) Balance: Pb + PbO2 + SO4-2 ® PbSO4 Oxid: Pb ® Pb+2 + 2e- Red: PbO2 + SO4-2 ® PbSO4 Pb ® Pb+2 + 2e- 4H+ + PbO2 + SO4-2 ® PbSO4 + 2H2O Pb ® Pb+2 + 2e- 2e- + 4H+ + PbO2 + SO4-2 ® PbSO4 + 2H2O 4H+ + PbO2 + SO4-2 + Pb ® Pb+2 + PbSO4 + 2H2O

Redox Practice 5) Balance: Mn+2 + NaBiO3 ® Bi+3 + MnO4- + Na+ Oxid: Mn+2 ® MnO4- Red: NaBiO3 ® Bi+3 + Na+ 4H2O + Mn+2 ® MnO4- + 8H+ 6H+ + NaBiO3 ® Bi+3 + Na+ + 3H2O 2[4H2O + Mn+2 ® MnO4- + 8H+ + 5e-] 5[6H+ + NaBiO3 + 2e- ® Bi+3 + Na+ + 3H2O]

Redox Practice 8H2O + 2Mn+2 ® 2MnO4- + 16H+ + 10e- 30H+ + 5NaBiO3 + 10e- ® 5Bi+3 + 5Na+ + 15H2O 14H+ + 5NaBiO3 + 2Mn+2 ® 2MnO4- + 5Bi+3 + 5Na+ + 7H2O

Redox Practice 6) Balance: MnO4- + Fe+2 ® Mn+2 + Fe+3 8H+ + MnO4- + 5Fe+2 ® Mn+2 + 5Fe+3 + 4H2O 7) Balance: Cu + NO3- ® Cu+2 + NO 8H+ + 3Cu + 2NO3- ® 3Cu+2 + 2NO + 4H2O

Basic Solutions Do everything you would with acid solution, but add two more steps. Add enough OH- to both sides to neutralize the H+: this makes water Cancel water

Redox Practice: Basic Solution Balance: CN-(aq) + MnO4-(aq) ® CNO-(aq) + MnO2(s) Oxid: CN- ® CNO- Red: MnO4- ® MnO2 3[H2O + CN- ® CNO- + 2H+ + 2e-] 2[4H+ + MnO4- + 3e- ® MnO2 + 2H2O] 3H2O + 3CN- ® 3CNO- + 6H+ + 6e- 8H+ + 2MnO4- + 6e- ® 2MnO2 + 4H2O

Redox Practice: Basic Solution 8H+ + 2MnO4- + 3H2O + 3CN- ® 3CNO- + 6H+ + 2MnO2 + 4H2O 2H+ + 2MnO4- + 3CN- ® 3CNO- + 2MnO2 + H2O 2OH- + 2H+ + 2MnO4- + 3CN- ® 3CNO- + 2MnO2 + H2O + 2OH- H2O + 2MnO4- + 3CN- ® 3CNO- + 2MnO2 + 2OH-

Redox Practice: Basic Solution 2) Balance: Ag + CN- + O2 → Ag(CN)2- + 2H2O(l) Oxid: Ag + CN- → Ag(CN)2- Red: O2 → 2H2O Ag + 2CN- → Ag(CN)2- + e- 4H+ + O2 + 4e- → 2H2O 4[Ag + 2CN- → Ag(CN)2- + e-] 4H+ + O2 + 4e- → 2H2O

Redox Practice: Basic Solution 4[Ag + 2CN- → Ag(CN)2- + e-] 4H+ + O2 + 4e- → 2H2O 4Ag + 8CN- → 4Ag(CN)2- + 4e- 4H+ + O2 + 4e- → 2H2O 4Ag + 8CN- + 4H+ + O2 → 4Ag(CN)2- + 2H2O 4Ag + 8CN- + 4H+ + 4OH- + O2 → 4Ag(CN)2- + 2H2O + + 4OH- 4Ag + 8CN- + 2H2O + O2 → 4Ag(CN)2- + 4OH-

Redox Practice: Basic Solution 3) Balance: MnO4-(aq) + S2-(aq) ® MnS2(s) + S(s) Oxid: S2-(aq) ® S(s) Red: MnO4-(aq) + 2S2-(aq) ® MnS2(s) 3[S2-(aq) ® S(s) + 2e-] 2[8H+ + MnO4- + 2S2- + 3e- ® MnS2 + 4H2O] 3S2- ® 3S(s) + 6e- 16H+ + 2MnO4- + 4S2- + 6e- ® 2MnS2 + 8H2O

Redox Practice: Basic Solution 16H+ + 2MnO4- + 7S2- ® 2MnS2 + 3S(s) + 8H2O 16OH- + 16H+ + 2MnO4- + 7S2- ® 2MnS2 + 3S + 8H2O + 16OH- 8H2O + 2MnO4- + 7S2- ® 2MnS2 + 3S + 16OH-

Redox Practice: Basic Solution 4) Balance: MnO4-(aq) + C2O42-(aq) ® MnO2(s)+ CO32-(aq) Oxid: MnO4- ® MnO2 Red: C2O42- ® 2CO32- 2(3e- + 4H+ + MnO4- ® MnO2 + 2H2O) 3(2H2O + C2O42- ® 2CO32- + 4H+ + 2e-) 2MnO4- + 2H2O + 3C2O42- ® 2MnO2 + 6CO32- + 4H+ adding OH- and combining: 2MnO4- + 3C2O42- + 4OH- ® 2MnO2 + 6CO32- + 2H2O

Stoichiometry Calculation What mass of solid is formed when 100.00 mL of 0.100M iron chloride is mixed with 100.00 mL of 0.100M sodium hydroxide and what is the concentration of the remaining ions in solution? FeCl2 + 2NaOH ® Fe(OH)2 + 2NaCl NaOH limiting and 0.449 g Fe(OH)2 formed. Ion Start Used End Conc. Fe+2 0.01 0.005 0.005 0.025M 2Cl- 0.02 -0- 0.02 0.10M Na+ 0.01 -0- 0.01 0.05M OH- 0.01 0.01 -0- -0-

Stoichiometry Calculation 2) What volume of 0.204 M HCl is needed to precipitate the silver from 50.0 ml of 0.0500 M silver nitrate solution? HCl(aq) + AgNO3(aq) ® AgCl (s) + HNO3(aq) 12.3 mL

Practice Calculation 3) The average concentration of bromide ion in seawater is 65 mg of bromide ion per kg seawater. Calculate the molarity of the bromide ion if the density of seawater is 1.025 g/mL. density of seawater = 1.025 g/mL 1 kg water/1.025 g/mL = 975.6 mL concentration of bromide ion: 65 x 10-3 g/kg = 8.14 x 10-4 mol Br/975.6 mL = 8.14 x 10-4 mol Br /0.976 L = 8.3 x 10-4 M