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K + (aq) + MnO 4 - (aq). IONIC COMPOUNDS. Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO 4 in water. An Ionic Compound, CuCl 2 , in Water. CCR, page 149. Aqueous Solutions. How do we know ions are present in aqueous solutions?

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ionic compounds

K+(aq) + MnO4-(aq)

IONIC COMPOUNDS

Many reactions involve ionic compounds, especially reactions in water — aqueous solutions.

KMnO4 in water

aqueous solutions
Aqueous Solutions

How do we know ions are present in aqueous solutions?

The solutions conduct electricity!

They are called ELECTROLYTES

HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

aqueous solutions4
Aqueous Solutions

HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

aqueous solutions5
Aqueous Solutions

Acetic acid ionizes only to a small extent, so it is a weak electrolyte.

CH3CO2H(aq) ---> CH3CO2-(aq) + H+(aq)

aqueous solutions6
Aqueous Solutions

Acetic acid ionizes only to a small extent, so it is a weak electrolyte.

CH3CO2H(aq) ---> CH3CO2-(aq) + H+(aq)

aqueous solutions7
Aqueous Solutions

Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes.

Examples include:

sugar

ethanol

ethylene glycol

water solubility of ionic compounds
Water Solubility of Ionic Compounds

If one ion from the “Soluble Compd.” list is present in a compound, the compound is water soluble.

Screen 5.4 & Figure 5.1

Guidelines to predict the solubility of ionic compounds

water solubility of ionic compounds9

Iron pyrite, a sulfide

Orpiment, arsenic sulfide

Azurite, a copper carbonate

Water Solubility of Ionic Compounds

Common minerals are often formed with anions that lead to insolubility:

sulfide fluoride

carbonate oxide

acids

HNO3

ACIDS

An acid -------> H+ in water

Some strongacids are

HCl hydrochloric

H2SO4 sulfuric

HClO4 perchloric

HNO3 nitric

acids11
ACIDS

An acid -------> H+ in water

HCl(aq) ---> H+(aq) + Cl-(aq)

slide12

HCl

-

Cl

H

O

+

H

O

2

3

hydronium

ion

The Nature of Acids

weak acids

Acetic acid

Weak Acids

WEAK ACIDS = weak electrolytes

CH3CO2H

acetic acid

H2CO3 carbonic acid

H3PO4 phosphoric acid

HF

hydrofluoric acid

acids14
ACIDS

Nonmetal oxides can be acids

CO2(aq) + H2O(liq) ---> H2CO3(aq)

SO3(aq) + H2O(liq) ---> H2SO4(aq)

and can come from burning coal and oil.

bases see screen 5 9 and table 5 2
BASESsee Screen 5.9 and Table 5.2

Base ---> OH- in water

NaOH(aq) ---> Na+(aq) + OH-(aq)

NaOH is a strong base

bases
BASES

Metal oxides are bases

CaO(s) + H2O(liq)

--> Ca(OH)2(aq)

CaO in water. Indicator shows solution is basic.

net ionic equations
Net Ionic Equations

Mg(s) + 2 HCl(aq) --> H2(g) + MgCl2(aq)

We really should write

Mg(s) + 2 H+(aq) + 2 Cl-(aq) ---> H2(g) + Mg2+(aq) + 2 Cl-(aq)

The two Cl- ions are SPECTATOR IONS — they do not participate. Could have used NO3-.

net ionic equations20
Net Ionic Equations

Mg(s) + 2 HCl(aq) --> H2(g) + MgCl2(aq)

Mg(s) + 2 H+(aq) + 2 Cl-(aq) ---> H2(g) + Mg2+(aq) + 2 Cl-(aq)

We leave the spectator ions out —

Mg(s) + 2 H+(aq) ---> H2(g) + Mg2+(aq)

to give the NET IONIC EQUATION

chemical reactions in water sections 5 2 5 4 5 6 cd rom ch 5
Chemical Reactions in WaterSections 5.2 & 5.4-5.6—CD-ROM Ch. 5

Pb(NO3) 2(aq) + 2 KI(aq)

----> PbI2(s) + 2 KNO3 (aq)

We will look at EXCHANGE REACTIONS

The anions exchange places between cations.

precipitation reactions
Precipitation Reactions

The “driving force” is the formation of an insoluble compound — a precipitate.

Pb(NO3)2(aq) + 2 KI(aq) ----->

2 KNO3(aq) + PbI2(s)

Net ionic equation

Pb2+(aq) + 2 I-(aq) ---> PbI2(s)

acid base reactions
Acid-Base Reactions
  • The “driving force” is the formation of water.

NaOH(aq) + HCl(aq) --->

NaCl(aq) + H2O(liq)

  • Net ionic equation

OH-(aq) + H+(aq) ---> H2O(liq)

  • This applies to ALL reactions of STRONG acids and bases.
slide24

Acid-Base Reactions

CCR, page 162

acid base reactions25
Acid-Base Reactions
  • A-B reactions are sometimes called NEUTRALIZATIONSbecause the solution is neither acidic nor basic at the end.
  • The other product of the A-B reaction is a SALT, MX.

HX + MOH ---> MX + H2O

Mn+ comes from base &Xn- comes from acid

This is one way to make ionic compounds!

gas forming reactions
Gas-Forming Reactions

This is primarily the chemistry of metal carbonates.

CO2 and water ---> H2CO3

H2CO3(aq) + Ca2+ --->

2 H+(aq) + CaCO3(s) (limestone)

Adding acid reverses this reaction.

MCO3 + acid ---> CO2 + salt

gas forming reactions27
Gas-Forming Reactions

CaCO3(s) + H2SO4(aq) --->

2 CaSO4(s) + H2CO3(aq)

Carbonic acid is unstable and forms CO2 & H2O

H2CO3(aq) ---> CO2 (g) + water

(Antacid tablet has citric acid + NaHCO3)

slide28

See also: Gas Forming Reactions in Biological Systems

Three of the pioneers in working out the roles of NO forming reactions

shared a Nobel Prize in 1988 for their discoveries.

terminology
Terminology

In solution we need to define the -

  • SOLVENT

the component whose physical state is preserved when solution forms

  • SOLUTE

the other solution component

concentration of solute

moles solute

(

M

)

=

Molarity

liters of solution

Concentration of Solute

The amount of solute in a solution is given by its concentration.

Concentration (M) = [ …]

slide33
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate molarity.

Step 1: Calculate moles of NiCl2•6H2O

Step 2: Calculate molarity

[NiCl2•6 H2O] = 0.0841 M

the nature of a cucl 2 solution ion concentrations
The Nature of a CuCl2 SolutionIon Concentrations

CuCl2(aq) -->

Cu2+(aq) + 2 Cl-(aq)

If [CuCl2] = 0.30 M, then

[Cu2+] = 0.30 M

[Cl-] = 2 x 0.30 M

using molarity

moles = M•V

USING MOLARITY

What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution?

Because Conc (M) = moles/volume = mol/V

this means that

using molarity36
USING MOLARITY

What mass of oxalic acid, H2C2O4, is

required to make 250. mL of a 0.0500 M

solution?

Step 1: Calculate moles of acid required.

(0.0500 mol/L)(0.250 L) = 0.0125 mol

Step 2: Calculate mass of acid required.

(0.0125 mol )(90.00 g/mol) = 1.13 g

moles = M•V

preparing solutions
Preparing Solutions
  • Weigh out a solid solute and dissolve in a given quantity of solvent.
  • Dilute a concentrated solution to give one that is less concentrated.
problem you have 50 0 ml of 3 0 m naoh and you want 0 50 m naoh what do you do
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?

Add water to the 3.0 M solution to lower its concentration to 0.50 M

Dilute the solution!

problem you have 50 0 ml of 3 0 m naoh and you want 0 50 m naoh what do you do39
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?

But how much water

do we add?

problem you have 50 0 ml of 3 0 m naoh and you want 0 50 m naoh what do you do40

moles of NaOH in ORIGINAL solution =

moles of NaOH in FINAL solution

PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?

How much water is added?

The important point is that --->

problem you have 50 0 ml of 3 0 m naoh and you want 0 50 m naoh what do you do41
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?

Amount of NaOH in original solution =

M • V =

(3.0 mol/L)(0.050 L) = 0.15 mol NaOH

Amount of NaOH in final solution must also = 0.15 mol NaOH

0.15/Volume of final solution = 0.5 M/ 1 L

Volume of final solution =

(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L

or 300 mL

problem you have 50 0 ml of 3 0 m naoh and you want 0 50 m naoh what do you do42
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?

Conclusion:

add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

preparing solutions by dilution
Preparing Solutions by Dilution

A shortcut

Cinitial • Vinitial = Cfinal • Vfinal

ph a concentration scale
pH, a Concentration Scale

pH: a way to express acidity -- the concentration of H+ in solution.

Low pH: high [H+]

High pH: low [H+]

Acidic solution pH < 7

Neutral pH = 7

Basic solution pH > 7

the ph scale
The pH Scale

pH = log (1/ [H+])

= - log [H+]

Remember : log a = b if 10b=a

In a neutral solution, [H+] = [OH-] = 1.00 x 10-7 M at 25 oC

pH = - log [H+] = -log (1.00 x 10-7) = - (-7)

= 7

See CD Screen 5.17 for a tutorial

h and ph
[H+] and pH

If the [H+] of soda is 1.6 x 10-3 M, the pH is ____?

Because pH = - log [H+]

then

pH= - log (1.6 x 10-3)

pH = - (-2.80)

pH = 2.80

What’s the origin of the name of the soda 7up ?

ph and h
pH and [H+]

If the pH of Coke is 3.12, it is ____________.

Because pH = - log [H+] then

log [H+] = - pH

Take antilog and get

[H+] = 10-pH

[H+] = 10-3.12 = 7.6 x 10-4 M

slide48

SOLUTION STOICHIOMETRYSection 5.10

  • Zinc reacts with acids to produce H2 gas.
  • Have 10.0 g of Zn
  • What volume of 2.50 M HCl is needed to convert the Zn completely?
general plan for stoichiometry calculations

Mass

HCl

Mass

zinc

Stoichiometric

factor

Moles

zinc

Moles

HCl

Volume

HCl

GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS
slide50
Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

Step 1: Write the balanced equation

Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g)

Step 2: Calculate amount of Zn

Step 3: Use the stoichiometric factor

slide51
Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

Step 3: Use the stoichiometric factor

Step 4: Calculate volume of HCl req’d

acid base reactions titrations

Oxalic acid,

H2C2O4

ACID-BASE REACTIONSTitrations

H2C2O4(aq) + 2 NaOH(aq) --->

acidbase

Na2C2O4(aq) + 2 H2O(liq)

Carry out this reaction using a TITRATION.

titration
Titration

1. Add solution from the buret.

2. Reagent (base) reacts with compound (acid) in solution in the flask.

3. Indicator shows when exact stoichiometric reaction has occurred.

4. Net ionic equation

H+ + OH- --> H2O

5. At equivalence point

moles H+ = moles OH-

lab problem 1 standardize a solution of naoh i e accurately determine its concentration
LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentra-tion of the NaOH?

slide56

1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

Step 1: Calculate amount of H2C2O4

Step 2: Calculate amount of NaOH req’d

slide57

1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

Step 1: Calculate amountof H2C2O4

= 0.0118 mol acid

Step 2: Calculate amount of NaOH req’d

= 0.0236 mol NaOH

Step 3: Calculate concentration of NaOH

[NaOH] = 0.663 M

lab problem 2 use standardized naoh to determine the amount of an acid in an unknown
LAB PROBLEM #2: Use standardized NaOH to determine the amount of an acid in an unknown.

Apples contain malic acid, C4H6O5.

C4H6O5(aq) + 2 NaOH(aq) --->

Na2C4H4O5(aq) + 2 H2O(liq)

76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?

76 80 g of apple requires 34 56 ml of 0 663 m naoh for titration what is weight of malic acid
76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?

Step 1: Calculate amount of NaOH used.

C • V = (0.663 M)(0.03456 L)

= 0.0229 mol NaOH

Step 2: Calculate amount of acid titrated.

= 0.0115 mol acid

76 80 g of apple requires 34 56 ml of 0 663 m naoh for titration what is weight of malic acid60
76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?

Step 1: Calculate amount of NaOH used.

= 0.0229 mol NaOH

Step 2: Calculate amount of acid titrated

= 0.0115 mol acid

Step 3: Calculate mass of acid titrated.

76 80 g of apple requires 34 56 ml of 0 663 m naoh for titration what is weight of malic acid61
76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?

Step 1: Calculate amount of NaOH used.

= 0.0229 mol NaOH

Step 2: Calculate amount of acid titrated

= 0.0115 mol acid

Step 3: Calculate mass of acid titrated.

= 1.54 g acid

Step 4: Calculate % malic acid.

slide62

EXCHANGE: Precipitation Reactions

EXCHANGE

Gas-Forming

Reactions

EXCHANGE

Acid-Base

Reactions

REACTIONS

REDOX REACTIONS

slide63

REDOX REACTIONS

Redox reactions are characterized by ELECTRON TRANSFER between an electron donor and electron acceptor.

Transfer leads to—

1. increase in oxidation number of some element = OXIDATION

2. decrease in oxidation number of some element = REDUCTION

slide64

REDOX REACTIONS

Cu(s) + 2 Ag+(aq)

---> Cu2+(aq) + 2 Ag(s)

In all reactions if something has been oxidized then something has also been reduced

why study redox reactions
Why Study Redox Reactions

Batteries

Corrosion

Manufacturing metals

Fuels

oxidation numbers
OXIDATION NUMBERS

The electric charge an element APPEARS to have when electrons are counted by some arbitrary rules:

1. Each atom in free element has ox. no. = 0.

Zn O2 I2 S8

2. In simple ions, ox. no. = charge on ion.

-1 for Cl-

+2 for Mg2+

oxidation numbers67
OXIDATION NUMBERS

3. F always has an oxidation number of -1 when forming compounds with other elements.

4. Cl, Br and I have oxidation numbers of -1 when forming compounds with other elements, except when combined with oxygen and fluorine.

5a. O has ox. no. = -2

(except in peroxides: in H2O2, O = -1)

oxidation numbers68
OXIDATION NUMBERS

5b. Ox. no. of H = +1

(except when H is associated with a metal as in NaH where it is -1)

6. Algebraic sum of oxidation numbers

= 0 for a compound

= overall charge for an ion

oxidation numbers69

Oxidation number of F in HF?

OXIDATION NUMBERS

NH3 N =

ClO- Cl =

H3PO4 P =

MnO4- Mn =

Cr2O72- Cr =

C3H8 C =

recognizing a redox reaction
Recognizing a Redox Reaction

Corrosion of aluminum

2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s)

Al(s) --> Al3+(aq) + 3 e-

  • Ox. no. of Al increases as e- are donated by the metal.
  • Therefore, Al is OXIDIZED
  • Al is the REDUCING AGENT in this balanced half-reaction.
recognizing a redox reaction71
Recognizing a Redox Reaction

Corrosion of aluminum

2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s)

Cu2+(aq) + 2 e- --> Cu(s)

  • Ox. no. of Cu decreases as e- are accepted by the ion.
  • Therefore, Cu is REDUCED
  • Cu is the OXIDIZING AGENT in this balanced half-reaction.
recognizing a redox reaction72
Recognizing a Redox Reaction

Notice that the 2 half-reactions add up to give the overall reaction —if we use 2 mol of Al and 3 mol of Cu2+.

2 Al(s) --> 2 Al3+(aq) + 6 e-

3 Cu2+(aq) + 6 e- --> 3 Cu(s)

-----------------------------------------------------------

2 Al(s) + 3 Cu2+(aq) ---> 2 Al3+(aq) + 3 Cu(s)

Final eqn. is balanced for mass and charge.

common oxidizing and reducing agents see table 5 4

Metals (Cu) are reducing agents

Metals (Na, K, Mg, Fe) are reducing agents

HNO3 is an oxidizing agent

Common Oxidizing and Reducing AgentsSee Table 5.4

Cu + HNO3 --> Cu2+ + NO2

2 K + 2 H2O --> 2 KOH + H2

recognizing a redox reaction see table 5 4
Recognizing a Redox ReactionSee Table 5.4

In terms of oxygen gain loss

In terms of halogen gain loss

In terms of electrons loss gain

Reaction Type

Oxidation

Reduction

examples of redox reactions
Examples of Redox Reactions

Metal + halogen

2 Al + 3 Br2 ---> Al2Br6

examples of redox reactions76
Examples of Redox Reactions

Nonmetal (P) + Oxygen

Metal (Mg) + Oxygen

examples of redox reactions77
Examples of Redox Reactions

Metal + acid

Mg + HCl

Mg = reducing agent

H+ = oxidizing agent

Metal + acid

Cu + HNO3

Cu = reducing agent

HNO3 = oxidizing agent