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Chapter 4. Chemical Reactions. 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. - PowerPoint PPT Presentation
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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, NH4
+, are soluble.
3. All halides are soluble except: AgX, Hg2X2 and PbX2, where X may be Cl, Br, or I.
4. All sulfates are soluble except: PbSO4, BaSO4, and SrSO4.
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(NO3)2 is added to a solution of Na2SO4 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 EquationsMolecular Equation
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
Complete or Total Ionic Equation
Ag+(aq) + NO3
-(aq) + Na+
(aq) + Cl-(aq)
AgCl(s) + Na+
(aq) + NO3-(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.
AgNO3 + NaCl ---> AgCl + NaNO3
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 + H2O
Dr. S. M. Condren
Neutralization Reactions
acid + base ---> “salt” + water
H2SO4 + 2KOH ---> K2SO4 + 2H2O
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 Al3+
(aq) + 3 H2(g)
Al, H+, Al3+, H2
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 -1
5. O is -2 except when combined with F (then +1 or +2) or in peroxides, -1.
6. sum of oxidation states equals charge on ion or molecule
Dr. S. M. Condren
Oxidation State
What is the oxidation state of S in H2SO4?
• 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 HClO4?
• 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) + O2 (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 soluteM = -------------------- 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 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?
2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O
Dr. S. M. Condren
EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O
(25.0 mL NaOH) #mL H2SO4 = ----------------------
Dr. S. M. Condren
EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O
(25.0 mL NaOH) (0.400 mol NaOH) #mL H2SO4 = --------------------------------------------- (1 L NaOH)
Dr. S. M. Condren
EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O
(25.0 mL NaOH) (0.400 mol NaOH)(1 L)#mL H2SO4 = -------------------------------------------------------
(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 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O
(25.0) (0.400 mol NaOH)(1) (1 mol H2SO4) #mL H2SO4 = ------------------------------------------------------
(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 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O
(25.0)(0.400)(1 mol H2SO4)(1000 mL H2SO4)
#mL H2SO4 = -------------------------------------------------------- (1) (1000) (2) (0.200 mol
H2SO4)
Dr. S. M. Condren
EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O
(25.0)(0.400)(1)(1000 mL H2SO4) #mL H2SO4 = ------------------------------------------
(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 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O
(25.0)(0.400)(1)(1000 mL H2SO4) #mL H2SO4 = ------------------------------------------
(1) (1000) (2) (0.200)
= 25.0 mL H2SO4