Dr. S. M. Condren Chapter 4 Chemical Reactions Dr. S. M. Condren Solubility Rules 1. All nitrates are soluble. 2. All compounds of Group IA metals and.

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Presentation transcript:

Dr. S. M. Condren Chapter 4 Chemical Reactions

Dr. S. M. Condren Solubility Rules 1. All nitrates are soluble. 2. All compounds of Group IA metals and the ammonium ion, NH 4 +, are soluble. 3. All halides are soluble except: AgX, Hg 2 X 2 and PbX 2, where X may be Cl, Br, or I. 4. All sulfates are soluble except: PbSO 4, BaSO 4, and SrSO 4.

Dr. S. M. Condren Solubility Rules 5. All hydroxides and sulfides are insoluble except those of the Group IA metals and the ammonium ion. 6. All carbonates and phosphates are insoluble except those of the Group IA metals and the ammonium ion.

Dr. S. M. Condren A solution of Ba(NO 3 ) 2 is added to a solution of Na 2 SO 4 to make a precipitate. From a table of solubility rules, the product is barium sulfate, sodium nitrate

Dr. S. M. Condren Electrolytes Conduct electricity in solution due to the presence of ions Strong electrolyte – completely ionized in solution Weak electrolyte – partially ionized in solution Non-electrolyte – nonionic solution

Dr. S. M. Condren Water Conductivity

Dr. S. M. Condren Strong Electrolyte

Dr. S. M. Condren Weak Electrolyte

Dr. S. M. Condren Chemical Equations Molecular Equation AgNO 3(aq) + NaCl (aq)  AgCl (s) + NaNO 3(aq) Complete or Total Ionic Equation Ag + (aq) + NO 3 - (aq) + Na + (aq) + Cl - (aq)  AgCl (s) + Na + (aq) + NO 3 - (aq) Net Ionic Equation Ag + (aq) + Cl - (aq)  AgCl (s)

Dr. S. M. Condren Types of Reactions synthesis reactions or combination reactions decomposition reactions precipitation reactions neutralization reactions –acid –base oxidation-reduction reaction

Dr. S. M. Condren Precipitation Reactions The process of separating a substance from a solution as a solid. AgNO 3 + NaCl ---> AgCl + NaNO 3 precipitate

Dr. S. M. Condren Neutralization Reactions acid base salt Household acids and Bases

Dr. S. M. Condren Neutralization Reactions acid –Any of a large class of sour-tasting substances whose aqueous solutions are capable of turning blue litmus indicators red, of reacting with and dissolving certain metals to form salts, and of reacting with bases or alkalis to form salts. –Substance that donates H + ions to solution

Dr. S. M. Condren Neutralization Reactions base –Any of a large class of compounds, including the hydroxides and oxides of metals, having a bitter taste, a slippery solution, the ability to turn litmus blue, and the ability to react with acids to form salts. –Substance that donates a OH -1 ion to solution

Dr. S. M. Condren Neutralization Reactions salt –The term salt is also applied to substances produced by the reaction of an acid with a base, known as a neutralization reaction. –Salts are characterized by ionic bonds, relatively high melting points, electrical conductivity when melted or when in solution, and a crystalline structure when in the solid state.

Dr. S. M. Condren Neutralization Reactions acid + base ---> “salt” + water

Dr. S. M. Condren Neutralization Reactions acid + base ---> “salt” + water HCl + NaOH ---> NaCl + H 2 O

Dr. S. M. Condren Neutralization Reactions acid + base ---> “salt” + water H 2 SO 4 + 2KOH ---> K 2 SO 4 + 2H 2 O

Dr. S. M. Condren Strong vs. Weak Acids and Bases strong - completely ionized weak - partially ionized

Dr. S. M. Condren Oxidation-Reduction Reaction Oxidation - loss of electrons Reduction - gain of electrons Redox reaction oxidizing agent - substance that causes oxidation reducing agent - substance that cause reduction

Dr. S. M. Condren Synthesis or Combination Reactions Formation of a compound from simpler compounds or elements.

Dr. S. M. Condren Decomposition Reactions Separation into constituents by chemical reaction.

Dr. S. M. Condren Electrolysis decomposition caused by an electric current anode –electrode where oxidation occurs cathode –electrode where reduction occurs

Dr. S. M. Condren Identify the oxidizing agent in the reaction: 2Al (s) + 6 H + ==> 2 Al 3+ (aq) + 3 H 2(g) Al, H +, Al 3+, H 2 Identify the oxidizing agent in the reaction:

Dr. S. M. Condren Oxidation States Rules for Assigning Oxidation States 1. zero for uncombined element 2. charge on monatomic ion 3. F is always -1; other halogens -1 except when combined with more electronegative halogen or oxygen

Dr. S. M. Condren Oxidation States Rules for Assigning Oxidation States 4. H is +1 except in metal hydrides, where H is O is -2 except when combined with F (then +1 or +2) or in peroxides, sum of oxidation states equals charge on ion or molecule

Dr. S. M. Condren Oxidation State What is the oxidation state of S in H 2 SO 4 ? H => +1 O => -2 neutral compound, thus sum equals zero 4O => 4*-2 = -8 2H => 2*+1 = +2 0 = +2 + (x) + (-8) x = +6

Dr. S. M. Condren Oxidation State What is the oxidation state of Cl in HClO 4 ? H => +1 O => -2 neutral compound, thus sum equals zero 4O => 4*-2 = -8 H => 1*+1 = +1 0 = +1 + (y) + (-8) y = +7

Dr. S. M. Condren Activity Series of Metals Highest metal in series is the most reactive A reactive metal will react with ions of less reactive metal to produce ions of reactive metal and atoms of the less reactive metal

Dr. S. M. Condren Metal Reaction with Acid

Dr. S. M. Condren Single Displacement

Dr. S. M. Condren 2Cu (s) + O 2 (g) ---> 2CuO (s) The oxidation number of copper in the product is Cu(0), Cu(I), Cu(II) In the reaction, copper metal is reduced, oxidized, unchanged in oxidation state

Dr. S. M. Condren Solution Solutions, in chemistry, homogeneous mixtures of two or more substances.

Dr. S. M. Condren Solute The substance that is present in smallest quantity is said to be dissolved and is called the solute. The solute can be either a gas, a liquid, or a solid.

Dr. S. M. Condren Solvent The substance present in largest quantity usually is called the solvent. The solvent can be either a liquid or a solid.

Dr. S. M. Condren Preparation of a Solution

Dr. S. M. Condren Molarity The number of moles of solute per liter of solution. molarity => M moles of solute M = liter of solution units => molar = moles/liter = M

Dr. S. M. Condren EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of M H 2 SO 4 are needed to react completely with 25.0 mL of M NaOH? 2 NaOH (aq) + H 2 SO 4(aq) -----> Na 2 SO 4(aq) + 2 H 2 O

Dr. S. M. Condren EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of M H 2 SO 4 are needed to react completely with 25.0 mL of M NaOH? 2 NaOH (aq) + H 2 SO 4(aq) -----> Na 2 SO 4(aq) + 2 H 2 O (25.0 mL NaOH) #mL H 2 SO 4 =

Dr. S. M. Condren EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of M H 2 SO 4 are needed to react completely with 25.0 mL of M NaOH? 2 NaOH (aq) + H 2 SO 4(aq) -----> Na 2 SO 4(aq) + 2 H 2 O (25.0 mL NaOH) (0.400 mol NaOH) #mL H 2 SO 4 = (1 L NaOH)

Dr. S. M. Condren EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of M H 2 SO 4 are needed to react completely with 25.0 mL of M NaOH? 2 NaOH (aq) + H 2 SO 4(aq) -----> Na 2 SO 4(aq) + 2 H 2 O (25.0 mL NaOH) (0.400 mol NaOH)(1 L) #mL H 2 SO 4 = (1 L NaOH) (1000 mL)

Dr. S. M. Condren EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of M H 2 SO 4 are needed to react completely with 25.0 mL of M NaOH? 2 NaOH (aq) + H 2 SO 4(aq) -----> Na 2 SO 4(aq) + 2 H 2 O (25.0) (0.400 mol NaOH)(1) (1 mol H 2 SO 4 ) #mL H 2 SO 4 = (1) (1000) (2 mol NaOH)

Dr. S. M. Condren EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of M H 2 SO 4 are needed to react completely with 25.0 mL of M NaOH? 2 NaOH (aq) + H 2 SO 4(aq) -----> Na 2 SO 4(aq) + 2 H 2 O (25.0)(0.400)(1 mol H 2 SO 4 )(1000 mL H 2 SO 4 ) #mL H 2 SO 4 = (1) (1000) (2) (0.200 mol H 2 SO 4 )

Dr. S. M. Condren EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of M H 2 SO 4 are needed to react completely with 25.0 mL of M NaOH? 2 NaOH (aq) + H 2 SO 4(aq) -----> Na 2 SO 4(aq) + 2 H 2 O (25.0)(0.400)(1)(1000 mL H 2 SO 4 ) #mL H 2 SO 4 = (1) (1000) (2) (0.200)

Dr. S. M. Condren EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of M H 2 SO 4 are needed to react completely with 25.0 mL of M NaOH? 2 NaOH (aq) + H 2 SO 4(aq) -----> Na 2 SO 4(aq) + 2 H 2 O (25.0)(0.400)(1)(1000 mL H 2 SO 4 ) #mL H 2 SO 4 = (1) (1000) (2) (0.200) = 25.0 mL H 2 SO 4