Writing Equations for Aqueous Ionic · PDF fileBa(OH) 2, Group I and Group II ... Sulfuric...

Chapter 4: Reaction Classes

Chapter 4

Major Classes of Chemical Reactions

Slide 4-1 Chapter 4: Reaction Classes

Three Major Classes of Chemical Reactions

4.6 Elements in Redox Reactions

4.1 The Role of Water as a Solvent

4.2 Writing Equations for Aqueous Ionic Reactions

4.3 Precipitation Reactions

4.4 Acid-Base Reactions

4.5 Oxidation-Reduction (Redox) Reactions

Slide 4-2


Electron distribution in molecules of H2 and H2O.

As a result, water is a polar solvent and is able to solvate polar molecules and ionic species


Determining Moles of Ions in Aqueous Ionic Solutions How many moles of each ion are in the following solutions?

(a) 5.0 mol of ammonium sulfate dissolved in water (b) 78.5 g of cesium bromide dissolved in water (c) 7.42 x 1022 formula units of copper(II) nitrate dissolved in water (d) 35 mL of 0.84 M zinc chloride

(a) (NH4)2SO4(s) 2NH4+(aq) + SO4

2-(aq)

5.0 mol (NH4)2SO4 x 2 mol NH4

+

1 mol (NH4)2SO4

= 10. mol NH4+

and 5.0 mol SO42-

7.42 x 1022 formula units x Cu(NO3)2

mol Cu(NO3)2

6.022 x 1023 formula units = 0.123 mol Cu(NO3)2

and 0.123 mol Cu2+

= 0.246 mol NO3-

(c) Cu(NO3)2(s) Cu2+(aq) + 2NO3-(aq)

Slide 4-4


Writing Equations for Aqueous Ionic Reactions

The molecular equation

shows all of the reactants and products as intact, undissociated compounds.

The total ionic equation

shows all of the soluble ionic substances dissociated into ions.

The net ionic equation

omits the spectator ions and shows the actual chemical change taking place.


Writing Equations for Aqueous Ionic Reactions

The molecular equation shows all reactants and products as if they were intact, undissociated compounds.

This gives the least information about the species in solution.

When solutions of silver nitrate and sodium chromate mix, a brick-red precipitate of silver chromate forms.

2AgNO3 (aq) + Na2CrO4 (aq) → Ag2CrO4 (s) + 2NaNO3 (aq)

Slide 4-6


The total ionic equation shows all soluble ionic substances dissociated into ions.

This gives the most accurate information about species in solution.

2Ag+ (aq) + 2NO3- (aq) + 2Na+ (aq) + CrO4

2- (aq) → Ag2CrO4 (s) + 2Na+ (aq) + 2NO3

- (aq)

Spectator ions are ions that are not involved in the actual chemical change. Spectator ions appear unchanged on both sides of the total ionic equation.

2Ag+ (aq) + 2NO3- (aq) + 2Na+ (aq) + CrO4

2- (aq) → Ag2CrO4 (s) + 2Na+ (aq) + 2NO3

- (aq)


The net ionic equation eliminates the spectator ions and shows only the actual chemical change.

2Ag+ (aq) + CrO42- (aq) → Ag2CrO4 (s)

Slide 4-8


An aqueous ionic reaction and the three types of equations.


Figure 4.6 The reaction of Pb(NO3)2 and NaI.

Double-displacement reaction (metathesis)

NaI(aq) + Pb(NO3)2(aq)

PbI2(s) + NaNO3(aq)

2NaI(aq) + Pb(NO3)2(aq)

PbI2(s) + 2NaNO3(aq)

2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq)

PbI2(s) + 2Na+(aq) + 2NO3-(aq)

2NaI(aq) + Pb(NO3)2(aq)

PbI2(s) + 2NaNO3(aq)

Slide 4-10


Predicting Whether a Precipitate Will Form

1. Note the ions present in the reactants.

2. Consider the possible cation-anion combinations (Double Displacement).

3. Decide whether any of the ion combinations is insoluble.

To do this we need to know the solubility rules.


Solubility Rules

Slide 4-12


Predicting Precipitation Reactions Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions.

(a) Potassium fluoride(aq) + strontium nitrate(aq)

(b) Ammonium perchlorate(aq) + sodium bromide(aq)

(a) KF(aq) + Sr(NO3)2 (aq) 2KNO3(aq) + SrF2 (s)

2K+(aq) + 2F-(aq) + Sr2+(aq) + 2NO3-(aq) 2K+(aq) + 2NO3

-(aq) + SrF2 (s)

(b) NH4ClO4(aq) + NaBr (aq) NH4Br (aq) + NaClO4(aq)

All reactants and products are soluble so no reaction occurs.

Slide 4-13 Chapter 4: Reaction Classes Slide 4-14

Acids and Bases: Definitions-1

All definitions are operationally equivalent; however the Lewis definition is by far the most general

14


Acids and Bases: Definitions-2

“H+”, as a discrete species, does NOT exist in water; rather, it combines with water to give hydronium ion:

It is the magnitude of the hydronium ion concentration ��([H3O+]; in mol/L) that determines how acidic a solution is!

Since H3O+ is generated from H+, and since H+ comes directly from the acid under consideration, measurement of the [H3O+] will have a direct correlation

to the strength of the acid!


The H+ ion as a solvated hydronium ion.

H+ interacts strongly with H2O, forming H3O+ in aqueous solution (NOT H+).

Slide 4-16


Acids and Bases: Definitions-3 PROTON: an archaic term still used to describe a hydrogen ion (“H+”); in acid/base chemistry it is still used to denote an acidic hydrogen and should NOT be confused with the more traditional definition of a proton (a positively charged nuclear particle).

PROTON TRANSFER: in acid/base chemistry this is used to denote the transfer of a hydrogen ion from an acid (H-donor) to an acceptor (H-acceptor).

WEAK ACID or BASE VS. STRONG ACID or BASE: The strength of an acid or a base is a direct function of the degree of ionization of the acid or base (or, how well it does proton transfer reactions). A relative term.


Acid-Base Reactions In the most general sense, an acid reacts with a base by transferring an ionizable hydrogen (an “acidic H”) to it:

The term “conjugate” is used exclusively to describe the materials on the product side of this reaction. Conjugates are easy

to write:

If the base is a hydroxide base, a neutralization reaction occurs:

Slide 4-18


Strong Acids: HBr, HCl, HI, HNO3, H2SO4, HClO4

Strong Bases: NaOH, KOH, Ca(OH)2, Sr(OH)2, ��Ba(OH)2, Group I and Group II oxides


Acid Strength: Determining the Molarity of H+ Ions in an Aqueous Solution of a Strong Acid

Sulfuric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4 M sulfuric acid?

Slide 4-20


Writing Ionic Equations for Acid-Base Reactions Write balanced molecular, total ionic, and net ionic equations for each of the following acid-base reactions and identify the spectator ions.

(a) Strontium hydroxide(aq) + perchloric acid(aq)

(b) Barium hydroxide(aq) + sulfuric acid(aq)


An acid-base titration.

Start of titration Excess of acid

Point of neutralization

Slight excess of base

Slide 4-22


Finding the Concentration of Acid from an Acid-Base Titration

You perform an acid-base titration to standardize an HCl solution by placing 50.00 mL of HCl in a flask with a few drops of indicator solution. You put 0.1524 M NaOH into the buret, and the initial reading is 0.55 mL. At the end point, the buret reading is 33.87 mL. What is the concentration of the HCl solution?

Strategy:


SOLUTION:

volume of base = 33.87 mL – 0.55 mL = 33.32 mL

= 5.078x10-3 mol NaOH

Since 1 mol of HCl reacts with 1 mol NaOH, the amount of HCl = 5.078x10-3 mol.

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

1L 103 mL

33.32 mL soln x 0.1524 mol NaOH 1 L soln

x

5.078x10-3 mol HCl

50.00 mL x 103 mL

1 L = 0.1016 M HCl

Slide 4-24


Oxidation-Reduction Reactions Oxidation: the LOSS (via electron transfer) of one or more electrons:

Reduction: the GAIN (via electron transfer) of one or more electrons:


The redox process in compound formation.

Slide 4-26

Chapter 4: Reaction Classes Slide 4-27 Chapter 4: Reaction Classes

Oxidation Number (ON) Determination 1.  The ON of an atom as an element is zero: Na(s), C(gr), H2(g),

I2(s); 2.  For a monatomic ion, its charge is its ON; 3.  F is always –1; 4.  Group I elements are always +1; Group II elements are always

+2; 5.  Oxygen is usually assigned an ON of –2; the only exception is

for peroxides (species where you have O–O bonds) where oxygen has an ON of –1;

6.  Hydrogen is +1 when combined with non-metals and –1 when combined with metals (hydrides);

7.  The sum of the oxidation states of all atoms in a compound must equal the charge on the compound or ion of interest;

8.  Non-integer oxidation states DO exist (Fe3O4) Slide 4-28


Oxidation Number (ON) Determination •  A compound (ionic or covalent) is neutral and has a charge of

zero. •  An ion WILL have a TOTAL charge; •  The sum of all the oxidation states in a compound must equal the

overall charge on the ion or molecule. •  Set up and solve for the unknown algebraically; For Fe2O3:

For Cr2O7 –2

:

# irons( ) ! charge of iron( )[ ] + # oxygens( ) ! charge of oxygen( )[ ] = charge of Fe2O3

2x( )[ ] + 3 "2( )( )[ ] = 0; solve to get x = +3

# chromiums( ) ! charge of chromium( )[ ] + # oxygens( ) ! charge of oxygen( )[ ] = charge of Cr2O7–2

2x( )[ ] + 7 "2( )( )[ ] = –2; solve to get x = +6


Balancing Redox Reactions: Half Reaction Method 1. Write separate equations for both the oxidation and reduction

half reactions; 2. Balance under acidic conditions FIRST by doing the following

for EACH half reaction separately: a.  Balance all elements EXCEPT H and O; b.  Balance O using H2O; c.  Balance H using H+; d.  Balance charge using electrons (e–)

3. Multiply each half reaction by an integer to ensure the number of electrons lost = number of electrons gained**;

4. Add half reactions and cancel species that occur on both the reactant AND product side;

5. Check to ensure balance!

Slide 4-30


Balancing Redox Reactions: Basic Conditions 1.  Balance under acidic conditions FIRST ; get overall balanced

equation; 2.  Add an equal number of OH–’s as you have H+ present to

BOTH sides of the equation; 3.  When H+ and OH– are present on the same side, make H2O; 4.  Eliminate equal numbers of waters from both sides of the

equation until only one side has waters left; 5.  Check to ensure balance!


Redox Balance: An Example Balance the following:

The half-reactions are:

For the carbon species: C balanced

O balanced

H balanced

charge balanced

C7H8 + MnO4– !"! C7H6O2 + MnO2

C7H8 !"! C7H6O2 and MnO4– !"! MnO2

C7H8 !"! C7H6O2

2 H2O + C7H8 !"! C7H6O2

2 H2O + C7H8 !"! C7H6O2 + 6 H+

2 H2O + C7H8 !"! C7H6O2 + 6 H+ + 6 e–

Slide 4-32


Redox Balance: An Example For the manganese species:

Mn balanced

O balanced

H balanced

charge balanced

Multiply the Mn equation by 2 to get correct number of electrons:

MnO4– !"! MnO2

MnO4– !"! MnO2 + 2 H2O

4 H+ + MnO4– !"! MnO2 + 2 H2O

3 e– + 4 H+ + MnO4– !"! MnO2 + 2 H2O

2 H2O + C7H8 !"! C7H6O2 + 6 H+ + 6 e–

6 e– + 8 H+ + 2 MnO4– !"! 2 MnO2 + 4 H2O


Redox Balance: An Example Combine reactions and eliminate common species:

Final balanced equation:

2 H2O + C7H8 + 6 e– + 8 2 H+ + 2 MnO4– !"!

C7H6O2 + 6 H+ + 6 e– + 2 MnO2 + 4 2 H2O

C7H8 + 2 H+ + 2 MnO4

– !"! C7H6O2 + 2 MnO2 + 2 H2O

in base : C7H

8 + 2 H+ + 2 OH– + 2 MnO

4– !"!

C7H

6O

2+ 2 MnO

2 + 2 H

2O+ 2 OH–

overall :C

7H

8 + 2 MnO

4– !"! C

7H

6O

2+ 2 MnO

2 + 2 OH–

Slide 4-34


Highest and lowest oxidation numbers of reactive main-group elements.


A summary of terminology for oxidation-reduction (redox) reactions.

Slide 4-36


Recognizing Oxidizing and Reducing Agents Identify the oxidizing agent and reducing agent in each of the following:

(a) 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)

(b) PbO(s) + CO(g) Pb(s) + CO2(g)

(c) 2H2(g) + O2(g) 2H2O(g)

Strategy: Assign an O.N. for each atom and see which atom gained and which atom lost electrons in going from reactants to products.

An increase in O.N. means the species was oxidized (and is the reducing agent) and a decrease in O.N. means the species was reduced (is the oxidizing agent).

(a) 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)

0 0 +6 +1 -2 +3 +6 -2

The O.N. of Al increases; Al is oxidized; it is the reducing agent. The O.N. of H decreases; H is reduced; H2SO4 is the oxidizing agent.


Figure 4.12 An active metal displacing hydrogen from water.

Slide 4-38


Displacing one metal by another.


The activity series

of the metals.

Slide 4-40


End of Chapter 4

Slide 4-41

Writing Equations for Aqueous Ionic · PDF fileBa(OH) 2, Group I and Group II ... Sulfuric...

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