CHEM 200/202Professor Jing GuOffice: EIS-210

All emails are to be sent to:chem200@mail.sdsu.edu

My office hours will be held in GMCS-212 on Monday from 9 am to 11 am or by

appointment.

ANNOUNCEMENTS• The volumetric report will be due 9/16-9/17 and

the prelab for qual analysis is due 9/16-9/17.

• Quiz 1 due 9/19/19.

• Homework 3 and 4 due 9/20/19.

• Exam 1 is on 9/21/19 at 2-4 pm. •

EXAM 1

• Things to bring:

• bring pencils

• erasers

• Non-programmable and non-

• graphing calculators

• RedID

Fall 2019 ROOM ASSIGNMENTS EXAMS 1—3

ENS-280 AL-201Lab TA: Section: Lab TA: Section:

Sun

Chem 200-03

Schroeder

Chem 200-01

Chem 200-08 Chem 200-04

Chem 200-21 Chem 200-11

Tranvo

Chem 200-07 Chem 200-18

Chem 200-10 Chem 200-20

Chem 200-22 Chem 202-08

Mendez

Chem 200-09Pearce

Chem 202-01

Chem 200-12 Chem 202-06

Chem 200-19Breyer

Chem 200-05

Rashid

Chem 202-02 Chem 200-14

Chem 202-03 Bowles Chem 200-13

Chem 202-04Williams

Chem 200-16

LiChem 200-02 Chem 200-23

Chem 200-06

ACIDS AND BASES

• Acids - produce H3O+(aq) when dissolved in water (often expressed as H+(aq))

• Bases - produce OH–(aq) when dissolved in water

• Strong acid/base - completely dissociates in water

• Weak acid/base - incompletely dissociates in water

ACID BASE THEORY• H+ (proton) forms H3O+ in water (hydrogen bonding)

• H+ is electron deficient, wants electrons, electron pair acceptor

• OH– is electron rich, can donate a pair of electrons

Water self-ionizes: 2H2O(l) → H3O+(aq) + OH–(aq)

Acid-Base definitions:Arrhenius:

•Acid increases conc. of H3O+ when added to water•Base increases conc. of OH- when added to water

Brønsted-Lowry:•Acid = proton donor•Base = proton acceptor

Lewis:•Acid: electron pair acceptor•Base: electron pair donor

SELECTED ACIDS & BASESStrong AcidsHydrochloric acid, HClHydrobromic acid, HBrHydroiodic acid, HINitric acid, HNO3

Sulfuric acid, H2SO4

Perchloric acid, HClO4

Strong BasesSodium hydroxide, NaOHPotassium hydroxide, KOHCalcium hydroxide, Ca(OH)2

Strontium hydroxide, Sr(OH)2

Barium hydroxide, Ba(OH)2

Weak AcidsHydrofluoric acid, HFPhosphoric acid, H3PO4

Acetic acid, CH3COOH (or HC2H3O2)

Weak BasesAmmonia, NH3

ACIDS• Monoprotic: one ionizable hydrogen

• HCl + H2O → H3O+ + Cl–

• Diprotic: two ionizable hydrogens

• H2SO4 + H2O → H3O+ + HSO4–

• HSO4– + H2O → H3O+ + SO42–

• Triprotic: three ionizable hydrogens

• H3PO4 + H2O → H3O+ + H2PO4–

• H2PO4– + H2O → H3O+ + HPO42–

• HPO42– + H2O → H3O+ + PO43–

Polyprotic, generic term meaning that

there is more than one ionizable hydrogen on

the molecule.

BASES• Monobasic: yields one OH- ion

• KOH → K+ + OH–

• NH3 +H2O → NH4+ + OH–

• Dibasic: yields two OH– ions

• Ba(OH)2 → Ba2+ + 2OH–

• Ca(OH)2 → Ca2+ + 2OH–

ACID-BASE NEUTRALIZATIONAcid + Base → Water + Salt

Molecular equation

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

Total ionic equation

H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → H2O(l) + Na+(aq) + Cl–(aq)

Net ionic equationH+(aq) + OH–(aq) → H2O(l)

Hint: Balance the H+ with OH- the rest will work itself out.

QUESTIONWhat is the concentration of a sulfuric acid solution if it requires 26.05 mL of a 2.045 M

sodium hydroxide solution to titrate 12.05 mL of the sulfuric acid solution?

GAS FORMATION WITH AN ACID-BASE REACTION

Molecular equation

NaHCO3(aq) + CH3CO2H(aq) → H2O(l) + CH3CO2Na(aq) + CO2(g)

Total ionic equation

Na+(aq) + HCO3-(aq) + H+(aq) + CH3CO2-(aq)→ H2O(l) + Na+(aq) + CH3CO2-(aq) + CO2(g)

Net ionic equationHCO3-(aq) + CH3CO2H(aq) → CH3CO2- + H2O(l) + CO2(g)

• An aqueous solution of H2SO4 is added to an aqueous solution of Ba(OH)2. The reaction is monitored using a conductivity meter. Predict the correct statement(s):

1. Both H2SO4 & Ba(OH)2 are strong electrolytes

2. This is a neutralization reaction

3. This is a precipitation reaction

4. The light bulb will glow at the neutralization point

•Statement 2•Statements 1 & 2•Statements 1, 2 & 3•All of the statements are true•All of the statements are lies

AnswersABCDE

SOLUBILITY RULES1. All common compounds of Group 1A(1) ions (Li+, Na+, K+...)

and ammonium ions (NH4+)

2. All common nitrates (NO3-), acetates (CH3CO2-) and most perchlorates (ClO4-)

3. All common chlorides (Cl-), bromides (Br-) and iodides (I-); except those of Ag+, Pb2+, Cu+ and Hg22+. All common fluorides (F-) are soluble; except for Pb2+ & Group2A(2)

4. All common sulfates (SO42-); except Ca2+, Sr2+, Ba2+, Ag+ & Pb2+

Soluble

LECTURE OBJECTIVES

• Chapter 4.2-4.4

• Determine the oxidation states of elements in compounds.

• Identify the oxidizing and reducing agents in redox reactions.

• Perform stoichiometric calculations involving mass, moles, and solution molarity.

• Calculate theoretical, and percent yields for chemical reactions.

REDOX REACTIONS IN COMPOUND FORMATION

Electrons are transferred in the formation of ionic

compounds.

Electrons are shifted in the formation of

covalent compounds.

OXIDATION NUMBER RULESGeneral Rules1. For an atom in its elemental form (e.g. Na, O2, Cl2,...) the O.N. = 0.2. For a monoatomic ion (e.g. Br-, Cu2+,...) the O.N. = ion charge.3. The sum of the O.N. values for atoms in a compound equals zero. For polyatomic ions the sum equals the charge of the ion.Specific Rules

1. For Group 1(A)1 - O.N. is +1 in all compounds2. For Group 2(A)2 - O.N. is +2 in all compounds3. For hydrogen - O.N. is +1 when bound to nonmetals4. For fluorine - O.N. is -1 when bound to metals & boron5. For oxygen - O.N. is -1 when in peroxides (e.g. H2O2)

- O.N. is -2 for all others (except with fluorine)6. For Group 7(A)17 - O.N. is -1 when with metals, nonmetals

(except O) & for other halogens lower in group

OXIDATION NUMBERSThe main group elements can

have different oxidation numbers depending on the molecule they are part of.

Compound O.N. of nitrogenNH3 -3N2H4 -2

NH2OH -1N2 0

N2O +1NO +2NO2- +3NO2 +4NO3- +5

REDOX TERMINOLOGY

• Mg loses electrons

• Mg is oxidized

• Mg is the reducing agent

• The oxidation number of Mg is increased

2Mg(s) + O2(g) → 2MgO(s)

2Mg → 2Mg2+ + 4e- O2 + 4e- → 2O2-

• O gains electrons

• O is reduced

• O is the oxidizing agent

• The oxidation number of O is decreased

O.N.: 0 +2 O.N.: 0 -2

OXIDATION REDUCTIONOIL RIG

Oxidationisloss of electrons

Reductionisgain of electrons

LEO GER

Loseelectrons isoxidation

Gainelectrons isreduction

QUESTIONWhat is the oxidation number of carbon in Na2C2O4?

Oxidation number:0

+1+2+3+4

QUESTIONIdentify the oxidizing agent and reducing agent

in the following reaction:Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g)

Oxidizing agent Reducing agent AnswerH+ Sn AH+ Sn2+ BSn H+ CSn H2 D

Sn2+ H2 E

TYPES OF REDOX REACTIONS

• The different types of redox reactions are classified by the components of the reaction and what happens to those components.

• There are four types of redox reactions which involve elements - combination, decomposition, displacement and combustion.

• In these reactions, elements may be reagents, products or transferred during the reaction.

COMBINATION REACTION2K(s) + Cl2(g) → 2KCl(s)

2NO(g) + O2(g) → 2NO2(g)

DECOMPOSITION REACTION2HgO(s) → 2Hg(l) + O2(g)

∆ ∆ = heat

2H2O(l)electricity 2H2(g) + O2(g)

DISPLACEMENT REACTIONAn active metal displacing

hydrogen from water2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)

COMBUSTION REACTIONS•Combustion reactions always involve elemental oxygen.•The reactions reduce oxygen and release energy, frequently as heat and light.

2CO(g) + 2O2(g) → 2CO2(g)

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

C6H12O6(g) + 6O2(g) → 6CO2(g) + 6H2O(g)

REACTION YIELDS• The reaction yield is a measure of the completeness of a

reaction; quantifying how much of the possible product was formed.

• Determining the theoretical yield for a reaction requires a balanced chemical reaction, and the identification of the limiting reagent.

• The limiting reagent is the reagent that will be entirely consumed first, stoping the reaction (limiting the amount of product formed).

LIMITING REAGENT

• The Haber-Bosch process produces ammonia from nitrogen and hydrogen gas (unbalanced reaction below).

• _N2(g) + _H2(g) → _NH3(g)

• Hydrogen limiting reagent: How many grams of ammonia would be produced if you 3.00 mol of H2 and an infinite amount of N2?

• Nitrogen limiting reagent:How many grams of ammonia would be produced if you 3.00 mol of N2 and an infinite amount of H2?

PROBLEM• Considering this reaction: _N2(g) + _H2(g) → _NH3(g)

• If 1.83 mol of N2 is reacted with 5.92 mol of H2, how many grams of NH3 will be produced?

• How many moles of the non-limiting (excess) reagent, are left over after the reaction?

• What mass of each gas will you have once the reaction is completed?

REACTION YIELDS

• Not every reaction proceeds perfectly to produce 100% of the maximum product.

• Reactions that are imperfect has reaction yields of less than 100%.

• Considering the reaction: _N2(g) + _H2(g) → _NH3(g)

• The reaction is performed with 1.84 mol of N2 and 2.84 mol of H2. At the end of the reaction you collect 0.932 mol of NH3. What is the yield of this reaction?

PROBLEMS

• 8.09 g of H2 react with 4.82 g of N2. The reaction has a 72.5% yield. What mass of NH3 is produced?( Do it together)

• 45.92 g of N2 is reacted with an excess of H2. The reaction produces 36.05 g of NH3. What is the yield of this reaction? How much mass of N2 was left unreacted? (Do it by yourself)

QUANTITATIVE ANALYSES• Knowledge of the stoichiometry of a reaction, along

with the composition and amount of one of the reagents, we can quantify the other component in the reactions.

• This is the basis of titrations, where carefully measured amounts of a reagent (e.g. a base) is added to a solution of unknown composition to quantify the amount of a reagent (e.g. an acid).

PROBLEM

• You have 4.912 g of a solid mixture. The mixture is known to contain both NaCl, and NaOH. You dissolve all the solid in 50.00 mL of water. (Do it together)

• You titrate the solution with HCl, to neutralize the NaOH, while not reacting with the NaCl.

• You need 32.67 mL of 0.5638 M HCl to neutralize the solution.

• What percentage of the solid was NaOH?

PROBLEM

• A 25.00 mL solution contains an unknown amount of Pb(NO3)2. An excess of aqueous KI is added to the solution, precipitating all the lead as lead(II) iodide. The amount of lead(II) iodide is determined to be 1.042 g. What was the concentration of Pb(NO3)2 in the original 25 mL solution?( Do it by yourself)