CHM 108 SUROVIEC SPRING 2014 Chapter 4 Part 2. I. Solution Stoichiometry According to the following reaction, how many moles of Fe(OH)2 can form from.

CHM 108Suroviec

Spring 2014

Chapter 4Part 2

I. Solution Stoichiometry

According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that there is excess FeCl2.

FeCl2(aq) + 2 LiOH(aq) → Fe(OH)2(s) + 2 LiCl(aq)

Determine the number of grams H2 formed when 250.0 mL of 0.743 M HCl solution reacts with 3.41 x 1023 atoms of Fe according to the following reaction.

2 HCl(aq) + Fe(s) → H2(g) + FeCl2(aq)

II. Aqueous Solution and Solubility

Consider salt dissolving in water and sugar dissolving in water.

A. Electrolyte

The way ionic compounds vs. molecular compounds dissolve in water shows the difference between types of solution.

A. Electrolyte

Electrolytes – ions that act at charge carriersSolutes that completely dissociate into ions

are called strong electrolytes

B. Solubility of Ionic Compounds

Most ionic compounds when dissolved in water the solute breaks into ions.

Not true for all ionic compounds

Determine the insoluble compounds

AgClNaNO3

PbCl2

Ba(OH)2

III. Precipitation Reactions

Precipitate: insoluble solid that separates from solution where no solid existed before reaction

Hard water contains Ca2+ and Mg2+

Laundry detergent contains Na2CO3

Examples

1. silver nitrate and potassium chloride2. lead (II) nitrate and potassium chromate3. potassium chromate and silver nitrate4. sodium carbonate and copper (II) chloride5. nickel (II) chloride and potassium hydroxide

IV. Molecular and Ionic Equations

A. Molecular EquationsConsider the following equation:CaCl2 (aq) + Na2SO4 (aq) CaSO4 (s) + 2NaCl

(aq)

B. Ionic Equations

In these equations, see that some of the ions are present on both sides of the arrow

Example

Given:2AgNO3(aq) + MgCl2(aq) 2AgCl (s) + Mg(NO3)2 (aq)

What is the ionic equation? Net Ionic?

Write NET ionic equations

1. AlCl3 (aq) + Na3PO4 (aq)

2. lead (II) nitrate and potassium chloride

IV. Acid and Base Reactions

Bronstead definition of acid: proton donorBronstead definition of base: proton acceptor

A.AcidsHCl (aq) + H2O (l) → H3O+ (aq) + Cl- (aq)

H2SO4 (aq) H+ (aq) + HSO4

- (aq)HSO4

- (aq) H+ (aq) + SO42- (aq)

B. Bases

Proton acceptors

Strong bases ionize completely to OH-

NaOH (s) → Ca(OH)2 →

Weak bases ionize only partially NH3 (aq) + H2O ⇌

C. Reactions of Acids and Bases

1. Neutralization

acid + base salt + water

HCl (aq) + NaOH (aq)

C. Reactions of Acids and Bases

2. Weak Acid/Base reactions

CH3COOH + NaOH

V. pH

Concentration scale for acids and bases

Vinegar: [H+] = 1.610-3MPure Water: [H+] = 1.010-7M Ammonia: [H+] = 1.010-11M

pH = -log[H+]

Determine the pH of the above. What is the trend of acids and bases?

VI. Acid-Base Titrations

Commonly used to determine the concentration of a dissolved species or its percentage in a mixture

Titration Measuring the volume

of a standard solution (known concentration) needed to react with a measured quantity of a sample

Titrant (in the buret)

Analyte (in the Erlenmeyer flask)

VI. Acid-Base Titrations

Equivalence point is where the number of moles of acid equals the number of moles of base

The endpoint is indicated by a color change in the acid-base indicator

Example

1. What volume (in mL) of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution?

2. What volume (in mL) of 0.955 M HCl is required to titrate 2.152g of Na2CO3 to the equivalence point?

VI. Redox Reactions

A. Oxidation Numbers Needed when we are looking at reactions

between 2 nonmetals. The oxidation number of an atom in a

compound is the “charge” that it would have it all shared electrons were assigned to the atom with higher electronegativity.

III. Oxidation-Reduction Reactions

Short name: Redox reactionsElectron exchange

Oxidation is a loss of electrons Reduction is a gain of electrons

III. Redox Reactions

Fe (s) Fe? (aq) + 2e- 2H? (aq)+ 2e- H2 (g)

Examples

A. Fe3+ (aq) + H2 (g) Fe2+ (aq) + H+ (aq)

B. Au (s) + F2 (aq) F- (aq) + Au3+(aq)

1. Break into 1/2 reactions2. Mass balance 1/2 reactions3. Combine and check for neutrality and check again for mass

balance