Chapter 4 Chemical Reactions and Solution Stoichiometry DE Chemistry Dr. Walker

Solutions Solute Solvent Homogeneous mixture A solute is the dissolved substance in a solution. Salt in salt water Sugar in soda drinks Carbon dioxide in soda drinks Solvent A solvent is the dissolving medium in a solution. Water in salt water Water in soda

“Like Dissolves Like” Nonpolar solutes dissolve best in nonpolar solvents Fats Benzene Steroids Hexane Waxes Toluene Polar and ionic solutes dissolve best in polar solvents Inorganic Salts Water Sugars Small alcohols Acetic acid

Water Aqueous Solutions Water is a polar molecule Solutions where a solute is dissolved in water Water is a polar molecule Unequal distribution of electrons in molecule (electronegativity difference) Bent shape due to unpaired electrons on oxygen

The Dissolving Process Hydration Process of dissolving in water Ionic compounds dissociate in water Ions interact with the “ends” of the water molecule Same process occurs with polar compounds, no dissociation

Electrolytes Conductivity – Ability to conduct electric current Requires ions to be present in solution Electrolytes - A substance that when dissolved in water produces a solution that can conduct an electric current Strong electrolytes – substances that ionize completely in water Some ionic salts, strong acids/bases Weak electrolytes – substances that ionize partially in water Some ionic salts, weak acids/bases Nonelectrolytes – substance that DO NOT ionize in water Primarily organic compounds

Electrolytes vs. Nonelectrolytes The ammeter measures the flow of electrons (current) through the circuit. If the ammeter measures a current, and the bulb glows, then the solution conducts. If the ammeter fails to measure a current, and the bulb does not glow, the solution is non-conducting.

Molarity Molarity is the ratio of moles of solute to liters of solution -Remember, you may not always get moles and liters -Grams must be converted to moles (divide by molar mass) -Milliliters must be converted to liters (divide by 1000)

Basic Example Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.

Basic Example Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 mL of solution. 1.56 g HCl 1 mol HCl = 0.00428 mol HCl 36.46 g HCl = 1.60 M HCl 26.8 mL 1 L = 0.0268 L HCl 1000 mL

Concentration of Ions in Solution Same calculations overall One more step…multiply the molarity by the number of ions present Give the concentration of each ion present in 0.50 M Co(NO3)2.

Concentration of Ions in Solution Same calculations overall One more step…multiply the molarity by the number of ions present Give the concentration of each ion present in 0.50 M Co(NO3)2. Remember, ionic salts dissociate in water, giving one Co2+ and two NO3- As a result: (0.50 M x 1) Co2+ = 0.50 M Co2+ (0.50 M x 2) NO32- = 1.0 M NO3-

Solutions of Known Concentration Standard solution - a solution whose concentration is accurately known Preparation of Standard solutions How much (volume) x How strong (concentration) x What does it weigh (mass)?

Solutions of Known Concentration Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution?

Solutions of Known Concentration Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution? Step #1: Ask “How Much?” (What volume to prepare?) Step #2: Ask “How Strong?” (What molarity?) Step #3: Ask “What does it weigh?” (Molar mass is?) 1.500 L 0.500 mol 58.44 g = 43.8 g 1 L 1 mol

Stoichiometry Using Molarity Sometimes, you may have to use a solution in a reaction. To do this successfully, you must be able to determine the number of moles that you’re using! After determining moles of reactant, determine moles and grams of product You could be given two solutions, and have to figure out the limiting reagent FIRST!

Stoichiometry Using Molarity - Example What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to excess KOH? Write reaction and balance if necessary __ Al(NO3)3 + __ KOH __ Al(OH)3 + __ KNO3 Al(NO3)3 + 3 KOH Al(OH)3 + 3 KNO3

Stoichiometry Using Molarity - Example What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to excess KOH? Al(NO3)3 + 3 KOH Al(OH)3 + 3 KNO3

Stoichiometry Using Molarity - Example What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to excess KOH? Al(NO3)3 + 3 KOH Al(OH)3 + 3 KNO3 Find moles of Al(NO3)3 Moles of solute = 0.0100 mol

Stoichiometry Using Molarity - Example What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to excess KOH? Al(NO3)3 + 3 KOH Al(OH)3 + 3 KNO3 Find mass of Al(OH)3 0.0100 moles Al(NO3)3 1 mole Al(OH)3 78.01 g Al(OH)3 = 0.780 g Al(OH)3 1 mole Al(NO3)3 1 mole Al(OH)3

Dilutions Dilution of a volume of solution with water does not change the number of moles present Solving dilution problems M1V1 = M2V2

Dilution Example Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution?

Dilution Example Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution? M1V1 = M2V2 (11.6 M)(x Liters) = (3.0 M)(0.250 Liters) x Liters = (3.0 M)(0.250 Liters) 11.6 M = 0.065 L

Reaction Types Precipitation reactions Acid-Base reactions When two solutions are mixed, an insoluble solid forms Acid-Base reactions A soluble hydroxide and a soluble acid react to form water and a salt Oxidation-Reduction reactions (aka “redox”) Reactions in which one or more electrons are transferred

Precipitation Reactions When two solutions are mixed, an insoluble solid forms These are usually double replacement reactions where the ions exchange places in aqueous solution to form new compounds AX + BY  AY + BX One of the compounds formed is usually a precipitate (an insoluble solid), an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

Solubility Rules Soluble Insoluble NO32- NH4+ Alkali metals Halogens Salts of Ag, Pb, Hg, Ba CO32- OH- PO43- http://9f1780.medialib.glogster.com/media/fee651af631818d6a637130235ea2fa77ca9b46eca1a892f36be056b9d76004f/solubility-pyramid.jpg

Precipitation Reactions Ionic compounds dissolve in water and the ions separate and move independently Notice the four distinct ions in solution after dissociation

Precipitation Reactions After dissociation, we can predict the products of this reaction Reactants do not reform or there is no reaction Like charges do not bond Overall reaction becomes: AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Whether the products are aqueous or insoluble depends on solubility rules – I’m not making you memorize them

Describing Reactions There are three ways to describe reactions: Lead(II) nitrate + potassium iodide  lead(II) iodide + potassium nitrate Double replacement (ionic) equation Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq) Complete ionic equation shows compounds as aqueous ions Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) +2 I-(aq)  PbI2(s) + 2K+(aq) + 2 NO3-(aq) - notice the precipitate cannot dissociate!! Net ionic equation eliminates the spectator ions Pb2+(aq) + 2 I-(aq)  PbI2(s)

Another Example Give the complete ionic equation and net ionic equation for the following reaction:

Another Example Give the complete ionic equation and net ionic equation for the following reaction: Complete ionic equation Net ionic equation

Neutralization (Acid-Base) Reactions

Properties of Acids Acids are proton (hydrogen ion, H+) donors Acids have a pH < 7 Acids taste sour Acids react with active metals, producing H2 Acids react with carbonates, neutralize bases

Acids are Proton (H+) Donors Strong acids are assumed to be 100% ionized in solution (good H+ donors). HCl H2SO4 HNO3 Weak acids are usually less than 5% ionized in solution (poor H+ donors). H3PO4 HC2H3O2 Organic acids

Types of Acids Bronsted-Lowry: Acids are proton donors, bases are proton acceptors Arrhenius: Acids produce H+ ions in water, bases produce OH- ions in solution

Acids React with Active Metals Acids react with active metals to form salts and hydrogen gas. Mg + 2HCl  MgCl2 + H2(g) Zn + 2HCl  ZnCl2 + H2(g) Mg + H2SO4  MgSO4 + H2(g)

Acids Neutralize Bases Neutralization reactions ALWAYS produce a salt and water. HCl + NaOH  NaCl + H2O H2SO4 + 2NaOH  Na2SO4 + 2H2O 2HNO3 + Mg(OH)2  Mg(NO3)2 + 2H2O

Properties Of Bases Bases are proton (hydrogen ion, H+) acceptors Bases have a pH > 7 Bases taste bitter Solutions of bases feel slippery Bases neutralize acids

Bases are Proton (H+ ion) Acceptors Sodium hydroxide (lye), NaOH Potassium hydroxide, KOH Magnesium hydroxide, Mg(OH)2 Calcium hydroxide (lime), Ca(OH)2 OH- (hydroxide) in base combines with H+ in acids to form water H+ + OH-  H2O

Bases Neutralize Acids Milk of Magnesia contains magnesium hydroxide, Mg(OH)2, which neutralizes stomach acid, HCl. 2 HCl + Mg(OH)2 MgCl2 + 2 H2O

Volumetric Analysis Technique for determining the amount of a certain substance by doing a titration. A titration involves delivery (from a buret) of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte).

Titrations Vocabulary Titrant - Solution of known concentration Analyte - Solution of unknown concentration Equivalence point - Point at which the amount of titrant added to analyte results in perfect neutralization Indicator - a substance added at the beginning of the titration that changes color at the equivalence point Endpoint - the point at which the indicator changes color Ideally, the endpoint and equivalence point occur at the same time!!!

Successful Titrations The following three requirements must be met for a titration to be successful: The exact reaction between titrant and analyte must be known (and rapid). The stoichiometric (equivalence) point must be marked accurately. The volume of titrant required to reach the stoichiometric point must be known accurately. Titration data is usually determined using dilution calculations

More To Come…. … on acid/base chemistry in Chapter 14

“Redox” reactions Electrons are transferred (LEO says GER) Gain electrons = reduction Lose electrons = oxidation Spontaneous redox rxns can transfer energy Electrons (electricity) Heat Non-spontaneous redox rxns can be made to happen with electricity

Examples of redox rxns Photosynthesis Combustion of fuels Oxidation of sugars, fats, proteins for energy

Oxidation Reduction Reactions (Redox) Each sodium atom loses one electron: Each chlorine atom gains one electron:

Lose Electrons = Oxidation LEO says GER : Lose Electrons = Oxidation Sodium is oxidized Gain Electrons = Reduction Chlorine is reduced

Oxidation States Keeps track of electrons in redox reactions Similar to concept of charge in ionic compounds, but assigns numbers to elements in organic molecules as well

Oxidation State Examples H2SO4 H = S = O =

Oxidation State Examples H2SO4 H = + 1 S = + 6 O = -2 (+ 1) + (4 x -2) + (+6) = 0 The sum of all compounds is zero!!

More Practice

More Practice A) K = +1, Mn = +7, O = -2 F) O = -2, Fe = +8/3 B) Ni = +4, O = -2 G) F = -1, O = -2, Xe = +6 C) Na = +1, Fe = +2, O = -2, H = +1 H) F = -1, S = +4 D) H = +1, O = -2, N = -3, P = +5 I) O = -2, C = +2 E) O = -2, P = +3 J) H = +1, O = -2, C = 0

Balancing Redox Reactions Balanced by “half-reaction” method Split into oxidation and reduction reactions and balanced separately

Steps to Balance Redox Reactions (Acidic)

Redox Practice “Chalk talk”

Problem Set Chemical Reactions and Solution Stoichiometry (4)