john a. schreifels chemistry 211-notes 1 chapter 4 chemical reactions

John A. SchreifelsChemistry 211-notes

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CHAPTER 4

Chemical Reactions


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Overview

• Ions in Aqueous Solution– Ion theory in solutions; precipitation reactions– Molecular and ionic equation

• Typical Reactions– Precipitation– Acid-Base– Oxidation-Reduction (Balancing)

• Working with solutions• Quantitative analysis


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Ionic Thory of Solutions• Pure water is a very poor conductor of electricity.• Solutions from dissolving NaCl or KCl in water are very

conductive.• Solutions from dissolving substances such as sugar (sucrose)

C12H22O11 are non-conductive. • Electrolyte: substance that produces ions when dissolved in

water.– Strong- good electrical conductor when dissolved in water (completely

ionized). E.g. NaCl, KNO3, Mg(NO3)2, etc.

– Weak-poor conductor when dissolved in water (partial ionization):

• Non-electrolyte: substance that does not produce ions when dissolved in water.

)aq(Cl)aq(Na)s(NaCl OH 2

OH)aq(NH)l(OH)aq(NH 423


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Aqueous Reactions and Net Ionic Equations

Three forms for writing chemical reaction:• Molecular: :

– AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq).• Ionic:

– Spectator ions are not directly involved in the reaction

• Net ionic: exclude spectator ions:– Ag+(aq) + Cl(aq) AgCl(s).

AgCl(s)(aq)3NO(aq)Na(aq)Cl(aq)Na(aq)3NO(aq)Ag


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Precipitation Reactions• Metathesis reaction: reaction in which two

substances react through exchange of their components. Driving force is often a precipitation.

AX + BY AY + BX E.g. Predict if precipitation occurs for the mixture:

AgNO3(aq) + KI(aq)

NaClO4(aq) + Pb(NO3)2(aq)

Na2SO4(aq) + BaCl2(aq),

Ni(NO3)2(aq) + (NH4)2S(aq) Hint: Use the solubility rules to determine if either

product is insoluble.


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Solubility RulesRule ExceptionSoluble

Group 1 elements, NH4+

NO3 , ClO3

, ClO4

Chlorides, bromides, iodides Ag+, Pb2+, Hg22+

Acetates Ag+, Hg22+

Sulfates Sr2+, Ba2+, Pb2+, Ca2+

Insoluble Compounds

Carbonates, phosphates, oxalates, chromates, sulfides

Group 1, NH4+

Hydroxides, oxides Group 1, Ba2+


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Metathesis Reaction (cont.) Driving force is• sometimes formation of weak or non – electrolyte.

E.g. acid – base reactions

CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l)– CuO is normally insoluble in water, but readily dissolves in

aqueous nitric acid.

• sometimes formation of gas; – most common is CO2 from carbonates or H2S from sulfides

E.g.

CuCO3(s) + 2HNO3(aq) CO2(g) + Cu(NO3)2(aq) + H2O(l)

CuS(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2S(g)


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Acids, Bases, and SaltsArrhenius definition most often used: • Acid = a hydrogen containing compound that

releases hydrogen ions (H+) in solution. – HA(aq) + H2O(l) H3O+(aq) + A(aq) where

• HA = HCl, HNO3,etc. and • H3O+ = hydronium ion often written as H+.

• Base = compound that releases hydroxide ions (OH) in solution. The general reaction for a base is:– MOH(s) OH(aq) + M+(aq) where

• M = some metal such as Na, K, etc.

• Acids and bases can be strong or weak electrolytes. • A base/acid that is a strong electrolyte is a strong

base/acid.


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Polyprotic acids and weak bases

• Some acids have more than one acidic proton. – Sulfuric:

– Phosphoric:

• Most weak bases produce hydroxide ions by reaction with water. – Ammonia

(aq)O3H(aq)4SO O(l)2H (aq)4HSO

(aq)O3H(aq)4HSO O(l)2H (l)4SO2H

2

(aq)O3H(aq)4PO O(l)2H (aq)4HPO

(aq)O3H(aq)4HPO O(l)2H (aq)4PO2H

(aq)O3H(aq)4PO2H O(l)2H (l)4PO3H

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2

)aq(OH)aq(NHOH)aq(NH 423


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Strong and Weak Acids and Bases

• Organic acids are weak (usually have –COOH).• Amines (containing nitrogen) are weak.• In water they are completely dissociated:

– HCl(aq) + H2O(l) H3O+(aq) + Cl(aq)

Strong Acids Strong Bases

Chloric, HClO3 Grp 1A hydroxides (LiOH, NaOH, KOH, RbOH, CsOH)Hydrobromic, HBr

Hydroiodic, HI Grp 2A metal hydroxides (Ca(OH)2, Sr(OH)2, Ba(OH)2Perchloric, HClO4

Sulfuric, H2SO4

Nitric, HNO3


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Neutralization Reaction• Acids react with bases to form a salt and

possibly water (called the neutralization reaction):– HA(aq)+MOH(aq)M+(aq)+A(aq)+H2O(l).– If either the acid or base is a strong electrolyte,

exclude spectator ions in the ionic form. E.g. HCN = weak acid; NaOH = strong base;

neutralization reaction is– HCN(aq) + OH(aq) CN(aq) + H2O(l)

Eg. 2 HCl neutralized by NaOH; net ionic equation:– H+(aq) + OH(aq) H2O(l)


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Oxidation – Reduction• Oxidation = loss of at least one electron during a

reaction..– Ni(s) + H+(aq) Ni2+(aq) + H2(g)

• Reduction = gain of at least one electron during a reaction.– In above example, H+ gains an electron to become reduced.

• Every reaction must have an oxidation and reduction.• Metals react with acids to form salts and hydrogen

gas.– Cu(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2(g)

• Metals also oxidized with salts– Fe(s) + Cu(NO3)2(aq) Fe(NO3)2(aq) + Cu(s)


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Oxidation Number• Oxidation number (state): the charge on an

atom in a substance or monatomic ion.• Rules:

– Elemental form: 0– Monatomic ions: charge of ion– Oxygen: 2, except in H2O2 and other

peroxides.– Hydrogen: +1, except with metal

hydrides when it is 1.– Halogens: 1 (except when bound to

oxygen or a halide above it)– Alkali and alkaline earth metal ions have

a charge of +1 and +2, respectively.– Compounds and ions: sum of the

charges on the atoms in a compound add up to 0 and to the ion charge in the ion.

Ca in CaO +2

Ca2+(aq) +2

Cl(aq) 1

Cr in CrO3 +6

Fe in Fe2O3 +3

Cr in K2Cr2O7 +6


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Displacement Reactions: Activity series of the elements

• A relative reactivity scale allows us to predict if reaction will occur when two substances are mixed together.

E.g. Copper ions in solution are reduced to the metal when an iron nail is placed in the solution.

– Cu2+(aq) + Fe(s) Fe2+(aq) + Cu(s) Iron displaces copper.

– Fe2+(aq) + Cu(s) NR copper will not displace iron.

• Iron more reactive than copper.E.g. Predict which reaction will occur when:

– Li is mixed with K+ and – Li+ is mixed with K.

E.g. In which of the following mixtures will reaction occur:

– Li+ + Mg– Al + Mn2+ – Fe + Cd2+ – Cr + Zn2+

Li

Reacts vigorously with acids to give H2

Reacts with H2O to give H2

K

Ba

Ca

Na

Mg Reacts with acids to give H2

Reacts slowly with H2O to give H2

; more vigorous with steam

Al

Zn

Cr

Fe

Cd

Etc.


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Balancing: Oxidation-Number Method

• Determine oxidation # for each atom- both sides of equation.• Determine change in oxidation state for each atom.• Left side: make loss of electrons = gain. • Balance other side.• Insert coefficients for atoms that don't change oxidation state. E.g. Balance

– FeS(s)+CaC2(s) + CaO(s) Fe(s)+ CO(g)+ CaS(s)• In acidic or basic solution balance as above, then balance

charge with H+ or OH on one side and water on other side.E.g. Balance:

– Acidic solution: 2

42 CrOClCrOClO


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Balancing: Half-Reaction Method

Write unbalanced half reactions for the oxidation and the reduction• Balance the number of elements except O and H for each. • Balance O's with H2O to the deficient side. • Balance H's with H+ to the hydrogen deficient side

– Acidic: add H+

– Basic: add H2O to the deficient side and OH to the other side.• Balance charge by adding e to the side that needs it. • Multiply each half-reaction by integers to make electrons

cancel.• Add the two half-reactions and simplify.E.g. Balance:

– Acidic: Zn(s) + VO2+(aq) Zn2+(aq) + V3+(aq).– Basic: Ag(s) + HS(aq) + CrO4

2(aq) Ag2S(s) + Cr(OH)3(s).


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Solution Composition, Molarity• Most reactions performed in solution (homogeneous mixture) since

reactants mobile.• Solute dissolved substance.• Solvent substance in which solute is dissolved.• Concentration: amount of solute dissolved in a given amount of

solvent.• Concentrated solution: large amount of solute in solvent.• Dilute solution: very little solute in solvent. Often obtained by dilution.• Molar concentration ( Molarity, M ): moles of solute dissolved in a liter

of solution. E.g. An aqueous solution of 0.25 M NaCl can be prepared by dissolving:

– 0.25 mol NaCl in a 1-Liter flask– 0.50 mol NaCl in a 2-Liter flask– 0.125 mol in 1/2 liter flask (500 mL).

E.g. 1 Determine mass needed to prepare exactly 2 liters of 0.150M NaCl. E.g. 2 Determine the concentration when 12.5 g NaCl is dissolved and

diluted to 500.0 mL.


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Ion Concentrations in Solutions

• Concentrations of ions after dissolution depends on formula

• Determine concentration of each ion in the following solutions: 0.100 M CaSO4, 0.100 M Cu(NO3)2, 0.100 M FeCl3


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Mass % To MolarityOften manufacturers provide us with mass % of a

compound in solution, but it is more convenient to use molarity.

E.g. Determine the molarity of NH3(aq) if the mass% = 28.0 % NH3 and the density = 0.898 g/mL. – Assume 100g of solution – From mass of solute (28 g NH3) in 100 g determine .– From the mass of solution (100 g) and the density,

determine V = the volume of the solution.– From above steps determine the molar concentration:

[NH3] = n(NH3)/V = 1.65 mol NH3/ 0.111L = 14.96 M NH3.


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DILUTIONS• Dilute solutions are prepared from more concentrated

ones by adding solvent to the concentrated one.• The concentration of the dilute solution can be

determined if we know:– The volume of the concentrated solution, Vi.– The concentration of the concentrated solution, Mi.– The volume of the dilute solution, Vf.

• The relationship between the molarities and volumes is:

• E.g. Determine the volume needed to prepare 500.0 mL of a 0.100 M HCl solution from a 12.40 M stock solution.

fVfM iViM solute of moles


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REACTIONS IN SOLUTION• Reactions usually carried out in solution. • Amounts of reactants and products (m or n) must be determined from the

volume and molarity of the solution. • Start with the stoichiometric relationship for any reaction:

– aA + bB cC

• Depending upon what is given in the problem substitute for mol A, B or C. E.g., if we are dealing with a solution we substitute MAVA for the mol A.

E.g. Calculate the volume of 0.200 M KI required to react with 50.0 mL of 0.300 M Pb(NO3)2.

• Strategy:• Balance reaction: Pb(NO3)2 + 2KI PbI2 + 2KNO3.• From stoichiometry:

• Substitute for mol: • Solve

caC mol Amol

baB mol Amol

12Pb molK mol

12

PbVPbMKVKM


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TITRATIONS• Titration: a procedure for determining the amount of one

substance A by adding a carefully measured amount of a solution B until A is just consumed.

• Calculations are the same as in the last overhead.• E.g. What is the molarity of HCl if 25.00 mL of it was titrated to

the equivalence point with 33.33 mL of 0.1000 M Ba(OH)2?

• The stoichiometric relationship is:• Substitute the given quantities and solve for the [HCl].• The ratio of stoichiometric coefficients tells how much of one

compound will react if we know the amount of the other:aA + bB cC

– Solutions: n = MV; solids: n = m/FM

12

HClV2Ba(OH)V

2Ba(OH)MHClM

ba

Bn ?

john a. schreifels chemistry 211-notes 1 chapter 4 chemical reactions

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