Chapter 4

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.

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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.

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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

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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

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Water Conductivity

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Strong Electrolyte

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Weak Electrolyte

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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)

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Types of Reactions

• synthesis reactions or combination reactions

• decomposition reactions

• precipitation reactions

• neutralization reactions– acid– base

• oxidation-reduction reaction

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Precipitation Reactions

The process of separating a substance from a solution as a solid.

AgNO3 + NaCl ---> AgCl + NaNO3

precipitate

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Neutralization Reactions

• acid

• base

• salt

Household acids and Bases

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• 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

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• 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

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• 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.

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acid + base ---> “salt” + water

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HCl + NaOH ---> NaCl + H2O

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H2SO4 + 2KOH ---> K2SO4 + 2H2O

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Strong vs. Weak Acids and Bases

• strong - completely ionized

• weak - partially ionized

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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

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Synthesis or Combination Reactions

Formation of a compound from simpler compounds or elements.

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Decomposition Reactions

Separation into constituents by chemical reaction.

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Electrolysis

• decomposition caused by an electric current

• anode– electrode where oxidation occurs

• cathode– electrode where reduction occurs

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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:

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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

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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

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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

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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

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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

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Metal Reaction with Acid

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Single Displacement

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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

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Solution

Solutions, in chemistry, homogeneous mixtures of two or more substances.

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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.

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Solvent

The substance present in largest quantity usually is called the solvent. The solvent can be either a liquid or a solid.

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Preparation of a Solution

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Molarity

The number of moles of solute per liter of solution.

molarity => M moles of soluteM = -------------------- liter of solution

units => molar = moles/liter = M

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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

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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 = ----------------------

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(25.0 mL NaOH) (0.400 mol NaOH) #mL H2SO4 = --------------------------------------------- (1 L NaOH)

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(25.0 mL NaOH) (0.400 mol NaOH)(1 L)#mL H2SO4 = -------------------------------------------------------

(1 L NaOH) (1000 mL)

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(25.0) (0.400 mol NaOH)(1) (1 mol H2SO4) #mL H2SO4 = ------------------------------------------------------

(1) (1000) (2 mol NaOH)

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(25.0)(0.400)(1 mol H2SO4)(1000 mL H2SO4)

#mL H2SO4 = -------------------------------------------------------- (1) (1000) (2) (0.200 mol

H2SO4)

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(25.0)(0.400)(1)(1000 mL H2SO4) #mL H2SO4 = ------------------------------------------

(1) (1000) (2) (0.200)

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(25.0)(0.400)(1)(1000 mL H2SO4) #mL H2SO4 = ------------------------------------------

(1) (1000) (2) (0.200)

= 25.0 mL H2SO4

Chapter 4

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