Do Now: Please begin completing the “Error Analysis” section of your Unknown Hydrate laboratory report utilizing the correct empirical formula for the unknown hydrate which is: AlK(SO4)2 x 12H2O Objective: Define and discuss the nature of aqueous solutions: Strong and Weak electrolytes. Discuss Types of chemical reactions. HW: Do problems pg. 171 #10, 22, 24 (b and d). Pg. 172 #36, 44 and 46

Types of Chemical Reactions and Solution Stoichiometry

Classification of Matter Solutions are homogeneous mixtures

Solute Solvent 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

Saturation of Solutions A solution that contains the maximum amount of solute that may be dissolved under existing conditions is saturated. A solution that contains less solute than a saturated solution under existing conditions is unsaturated. A solution that contains more dissolved solute than a saturated solution under the same conditions is supersaturated. One useful property for characterizing a solution is its electrical conductivity

Definition of Electrolytes and Nonelectrolytes An electrolyte is: A substance whose aqueous solution conducts an electric current. A nonelectrolyte is: A substance whose aqueous solution does not conduct an electric current. Try to classify the following substances as electrolytes or nonelectrolytes…

Electrolytes? Pure water Tap water Sugar solution Sodium chloride solution Hydrochloric acid solution Lactic acid solution Ethyl alcohol solution Pure, solid sodium chloride

Answers… ELECTROLYTES: NONELECTROLYTES: Tap water (weak) NaCl solution HCl solution Lactate solution (weak) Pure water Sugar solution Ethanol solution Pure, solid NaCl But why do some compounds conduct electricity in solution while others do not…?

Ionic CompoundsDissociate NaCl(s)  Na+(aq) + Cl-(aq) AgNO3(s)  Ag+(aq) + NO3-(aq) MgCl2(s)  Mg2+(aq) + 2 Cl-(aq) Na2SO4(s)  2 Na+(aq) + SO42-(aq) AlCl3(s)  Al3+(aq) + 3 Cl-(aq)

Ions tend to stay in solution where they can conduct a current rather than re-forming a solid. The reason for this is the polar nature of the water molecule… Positive ions associate with the negative end of the water dipole (oxygen). Negative ions associate with the positive end of the water dipole (hydrogen).

Some covalent compounds IONIZE in solution Covalent acids form ions in solution, with the help of the water molecules. For instance, hydrogen chloride molecules, which are polar, give up their hydrogens to water, forming chloride ions (Cl-) and hydronium ions (H3O+).

Strong acids such as HCl are completely ionized in solution. Other examples of strong acids include: Sulfuric acid, H2SO4 Nitric acid, HNO3 Hydriodic acid, HI Perchloric acid, HClO4

Weak acids such as lactic acid usually ionize less than 5% of the time. Many of these weaker acids are “organic” acids that contain a “carboxyl” group. The carboxyl group does not easily give up its hydrogen.

Because of the carboxyl group, organic acids are sometimes called “carboxylic acids”. Other organic acids and their sources include: Citric acid – citrus fruit Malic acid – apples Butyric acid – rancid butter Amino acids – protein Nucleic acids – DNA and RNA Ascorbic acid – Vitamin C This is an enormous group of compounds; these are only a few examples.

However, most covalent compounds do not ionize at all in solution. Sugar (sucrose – C12H22O11), and ethanol (ethyl alcohol – C2H5OH) do not ionize - That is why they are nonelectrolytes!

Molarity The concentration of a solution measured in moles of solute per liter of solution. mol = M L

Preparation of Molar Solutions 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 is its mass?” (Molar mass is?) 1.500 L 0.500 mol 58.44 g = 43.8 g 1 L 1 mol

Preparation of Molar Solutions Problem: 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. Try this at your workstations!

Serial Dilution It is not practical to keep solutions of many different concentrations on hand, so chemists prepare more dilute solutions from a more concentrated “stock” solution. Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution? MstockVstock = MdiluteVdilute (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

Single Replacement Reactions A + BX  AX + B BX + Y  BY + X Replacement of: Metals by another metal Hydrogen in water by a metal Hydrogen in an acid by a metal Halogens by more active halogens

The Activity Series of the Metals Lithium Potassium Calcium Sodium Magnesium Aluminum Zinc Chromium Iron Nickel Lead Hydrogen Bismuth Copper Mercury Silver Platinum Gold Metals can replace other metals provided that they are above the metal that they are trying to replace. Metals above hydrogen can replace hydrogen in acids. Metals from sodium upward can replace hydrogen in water

The Activity Series of the Halogens Fluorine Chlorine Bromine Iodine Halogens can replace other halogens in compounds, provided that they are above the halogen that they are trying to replace. 2NaCl(s) + F2(g)  2NaF(s) + Cl2(g) ??? MgCl2(s) + Br2(g)  No Reaction ???

Double Replacement Reactions The ions of two compounds exchange places in an aqueous solution to form two 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.

Double replacement forming a precipitate… 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) Net ionic equation eliminates the spectator ions Pb2+(aq) + 2 I-(aq)  PbI2(s)

Solubility Rules – Mostly Soluble Ion Solubility Exceptions NO3- Soluble None ClO4- Na+ K+ NH4+ Cl-, I- Pb2+, Ag+, Hg22+ SO42- Ca2+, Ba2+, Sr2+, Pb2+, Ag+, Hg2+

Solubility Rules – Mostly Insoluble Ion Solubility Exceptions CO32- Insoluble Group IA and NH4+ PO43- OH- Group IA and Ca2+, Ba2+, Sr2+ S2- Groups IA, IIA, and NH4+

Oxidation and Reduction (Redox) Electrons are transferred Spontaneous redox rxns can transfer energy Electrons (electricity) Heat Non-spontaneous redox rxns can be made to happen with electricity

Oxidation and Reduction An old memory device for oxidation and reduction goes like this… LEO says GER Lose Electrons = Oxidation Gain Electrons = Reduction

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

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

Rules for Assigning Oxidation Numbers Rules 1 & 2 The oxidation number of any uncombined element is zero 2. The oxidation number of a monatomic ion equals its charge

Rules for Assigning Oxidation Numbers Rules 3 & 4 3. The oxidation number of oxygen in compounds is -2 4. The oxidation number of hydrogen in compounds is +1

Rules for Assigning Oxidation Number Rule 5 5. The sum of the oxidation numbers in the formula of a compound is 0 2(+1) + (-2) = 0 H O (+2) + 2(-2) + 2(+1) = 0 Ca O H

Rules for Assigning Oxidation Numbers Rule 6 6. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its charge X + 4(-2) = -2 S O X + 3(-2) = -1 N O  X = +5  X = +6

Reducing Agents and Oxidizing Agents The substance reduced is the oxidizing agent The substance oxidized is the reducing agent Sodium is oxidized – it is the reducing agent Chlorine is reduced – it is the oxidizing agent

Trends in Oxidation and Reduction Active metals: Lose electrons easily Are easily oxidized Are strong reducing agents Active nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents

Not All Reactions are Redox Reactions Reactions in which there has been no change in oxidation number are not redox rxns. Examples:

Balancing Oxidation-Reduction Equations Half-Reaction Method separate the full equation into 2 half-reactions: oxidation and reduction balance each half-reaction separately add them together to get full balanced equation

Steps for acidic solution Write two separate equations for oxidation and reduction include any compound containing the element involved does not need to include everything

Steps for acidic solution For each half-reaction: balance all elements but H and O balance O using H2O balance H using H+ balance charge using electrons (e-)

Steps for acidic solution If needed, multiply one or both half reactions by an integer so that number of electrons is equal in both half reactions

Steps for acidic solution Add the half-reactions together and simplify Check to be sure all is balanced.

Steps for basic solution Balance using acidic method Add OH- ions equal to #H+ ions to both sides Form water on the side of equation that contains both ions Simplify # water molecules Check balancing

Example

Example