properties of aqueous solutions -...

Properties of Aqueous Solutions

Definitions

A solution is a homogeneous mixture of two ormore substances.

The substance present in smaller amount iscalled the solute.

The substance present in larger amount iscalled the solvent.

For now we will discuss only aqueous solutions.

the most import property of water when dealing with aqueous solution is its

polarity

Structure of water

105°δ+ δ+

O—H bonds are covalent but “polar”

Dipole Moments

Dipole Moment

a substance possesses a dipole moment if its centers of positive and negative charge do

not coincide

not polar

+ -polar

μ = e x d Expressed in debye units

examples

H—FO

μ = 1.8 Dμ = 1.7 D

μ = 1.5 D

Solvation

Clustering of molecules of solvent aroundsolute:

Oδ−

hydration is specific termfor solvation when water

is solvent

Water can solvate both cations and anions

Oδ−

δ+ δ+

+– –

Electrolytes vs Nonelectrolytes

An electrolyte is a substance that, when dissolved in water, gives a solution that can

conduct electricity

A nonelectrolyte does not conduct electricity when dissolved in water.

The breaking up of a compound into cations and anions

Dissociation

Na Cl( s )

Cl- Cl-

Cl-Cl-Cl-

Na+Na+ Na+

Cl-+Na+( aq ) ( aq )

Electrolytes vs Nonelectrolytes

Nonelectrolyte

Strong electrolyte

Weak electrolytenot ionized in water

incompletely ionized in water

completely ionized in water

The amount of solute that can be dissolved in a given amount of a saturated solution at a

fixed temperature is the solubility of the solute in the solvent.

Solubility

Some compounds are very soluble : NaCl, KCl, NH4Cl

Solubility

Some are slightly soluble : AgCl

slightly soluble and insoluble can be used interchangeably

Strong electrolytes

Soluble Ionic compoundsStrong acidsStrong bases

The role of the Hydrogen Ion

• HCl (aq) Æ H+(aq) + Cl-(aq)

• What does the neutral H atom consist of?

• What must H+ ion then be?

A PROTON!

Arrhenius definitions of acids

An acid

A base

dissolves in water to yield protons

H—X H+ + X–

dissolves in water to yield hydroxide ions

YOH Y+ + OH–

and bases

Examples of Strong acids

Hydrochloric acid: HCl(aq)

Nitric acid: HNO3(aq)

Sulfuric acid: H2SO4(aq)

HCl H+ + Cl–

HONO2 H+ + NO3–

HOSO2OH H+ + HOSO2O–

Examples of Strong bases

Sodium hydroxide: NaOH(aq) is equivalent to Na+ + OH–(aq)

Likewise: KOH, Ba(OH)2, etc.

Weak electrolytes

Weak acidsWeak bases

A Weak Acid

Acetic acid:

CH3COH

CH3CO– + H+

>99% <1%Reversible reaction

the reaction can occur in both directions

A Weak Base

Ammonia:

H3N: + H2O OH– + NH4+

>99% <1%

Nonelectrolytes

ethanolethylene glycol

sucrose

produce no ions when dissolved in water

Concentration

Molarity ( M )(moles of solute)/(Liters of solution) = n/V

What is the molarity of a solution made up by dissolving 9.52g of NaCl in enough H2O to form 575

mL of solution?

575 mL

103 mL

= 0.284 mol/L

58.4g NaCl1 mol9.52g NaCl x

Important point about concentration

Given: Na2SO4 concentration = 0.683 M

What are [Na+] and [SO42-] ?

[Na+] = 2 x 0.683 M = 1.37 M

= 0.683 M[SO42-]

Na2SO4 Æ 2Na+(aq) + SO42-(aq)

Same total volume but different #s of moles of ions!

Dilution of solutions

more conecntrated solution → less concentrated

= Moles of solutefinalMoles of soluteinitial

At the heart of all dilution problems:

Dilution demo

So... Moles of soluteinitial = Moles of solutefinal

but... Molarity = (moles solute)/(volume solution)

and... Molarity x (volume solution) = (moles solute)

in terms of variables... M x V = n

so... Minitial x Vinitial = Mfinal x Vfinal

The Dilution Equation

orM1V1 = M2V2

Dilution of solutions

How much concentrated HCl (12.0 M HCl) is required in order to prepare 250. mL of a 2.0 M

solution?

Minitial x Vinitial = Mfinal x Vfinal

= ( 2.0 mol / L )( 0.250 L )( 12.0 mole / L ) Vinitial

Vinitial = 0.0417 L = 41.7 ml

Chemical Reactions

Types of Chemical Reactions

PrecipitationAcid-Base

Oxidation-Reduction

Precipitation Reactions

characterized by the formation of an insoluble solid that “falls out of” the solution

AX(aq) + BY(aq) → AY(aq) + BX(s)

usually involve ionic compounds

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

KNO3(aq)

PbI2(s)

Solubility Rules (Table 4.1 in text)

Soluble in water are most compoundscontaining:

• a Group 1A metal ion (usually Na+, K+)

• an ammonium ion (NH4+)

• a nitrate (NO3–), chlorate (ClO3

–), or perchlorate (ClO4

(continued…)

Soluble in water are most compoundscontaining:

• a sulfate ion (SO42–) except when cation is

Ag+, Pb2+, Ca2+, Ba2+, or Hg2+

• a chloride (Cl–), bromide (Br–), or iodide (l–) ion, except when cation is Ag+, Pb2+,

or Hg22+

Insoluble (or slightly soluble) in water are most :

• sulfides (S2–)• carbonates (CO3

2–)• phosphates (PO4

3–)• metal hydroxides

(NaOH , KOH and Ba(OH)2 are soluble,Ca(OH)2 is slightly soluble)

Example

Classify the following ionic compounds assoluble, insoluble or slightly soluble

(a) CuS

(b) Ca(OH)2

(c) Zn(NO3)2

insoluble

slightly soluble

soluble

Molecular EquationsAnd

Ionic Equations

AX(aq) + BY(aq) AY(aq) + BX(s)

AX(aq) = A+(aq) + X–(aq)BY(aq) = B+(aq) + Y–(aq)AY(aq) = A+(aq) + Y–(aq)

Ionic equation showing all of the ions:

A+(aq) + X–(aq) + B+(aq) + Y-(aq) A+(aq) + Y–(aq) + BX(s)

Molecular equation expressed via chemical formulas:

Net ionic equation

an equation that shows all of the ions is notvery useful

identify and discard “spectator” ions

A+(aq) + X–(aq) + B+(aq) + Y–(aq) A+(aq) + Y–(aq) + BX(s)

X–(aq) + B+(aq) BX(s)

2NaI(aq) + Pb(NO3)2(aq) 2NaNO3(aq) + PbI2(s)

2NaI(aq) 2Na+(aq) + 2I–(aq)

Pb(NO3)2(aq) Pb2+(aq) + 2NO3–(aq)

2NaNO3(aq) 2Na+(aq) + 2NO3–(aq)

Equation showing all of the ions: 2Na+(aq) + 2I–(aq) + Pb2+(aq) + 2NO3

–(aq)

2Na+(aq) + 2NO3–(aq) + PbI2(s)

Another example:

Net ionic equation

identify and discard “spectator” ions

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

2Na+(aq) + 2NO3–(aq) + PbI2(s)

2I–(aq) + Pb2+(aq) PbI2(s)

Example

Predict the precipitate formed and write a net ionic equation for this reaction: aluminum nitrate with

sodium hydroxide.

Al(NO3)3(aq) + 3NaOH(aq)

3NaNO3( ) + Al(OH)3( )saq

Example

Al(NO3)3(aq) + 3NaOH(aq)

3NaNO3(aq) + Al(OH)3(s)

Al3+(aq) + 3NO3–(aq) + 3Na+(aq) + 3HO–(aq)

3Na+(aq) + 3NO3–(aq) + Al(OH)3(s)

Rewrite as a complete ionic equation.

Example

Cancel spectator ions.Al3+ + 3NO3

–(aq) + 3Na+(aq) + 3OH–(aq)

3Na+(aq) + 3NO3–(aq) + Al(OH)3(s)

Al3+ + 3HO–(aq) Al(OH)3(s)

derive the net ionic equation

use, from definition of molarity:moles solute = liter volume x molar conc.

Stoichiometry of Precipitation Reactions

Example

Calculate the mass of solid NaCl that must be added to 1.50L of a 0.100 M AgNO3 solution to precipitate all the

Ag+ ions in the form of AgCl.

+Cl- (aq) + Ag+ (aq) NO3- (aq)

AgCl(s) + NO3- (aq)Na+ (aq) +

Na+ (aq) +

NaCl(aq) + AgNO3 (aq)? g 1.50 L, 0.100 M

Example

Cl-(aq) + Ag+ (aq) AgCl(s)

1.50LL

x0.100 mol Ag+

1 mol Cl-

1 mol NaClx

1 mol Ag+

1 mol Cl-x

58.4g NaCl

1 mol NaClx = 8.76g NaCl

Roadmap: liters AgCl x concentration AgCl → mol AgCl → mol Ag+ → mol Cl- → mol NaCl → g NaCl

Acid and Base Reactions

Properties of Acids

• sour taste

• change color of litmus from blue to red

• give carbon dioxide on reaction with carbonates and bicarbonates

• electrolytes ( some strong, some weak )

• give hydrogen on reaction with certain metals

Properties of bases

• bitter taste

• slippery to the touch

• change litmus from red to blue

• electrolytes ( some strong, some weak )

Definitions of acids and bases

Svante Arrhenius (Sweden) 1859-1927

Johannes Bronsted (Denmark) 1879-1947

G. N. Lewis (U.S.) 1875-1946

dissolves in water to yield protons

H—X H+ (aq) + X–

Arrhenius definitions of acids

dissolves in water to yield hydroxideions

YOH Y+ (aq) + OH–

A base

An acid

and bases

Bronsted Definition

An acid is a proton donor(Right on, Arrhenius!)

A base is a proton acceptor(Much more general def.)

Hydronium Ion ( H3O+)

for convenience we write H+(aq) to represent the hyrated proton

But H3O+ is closer to reality

Monoprotic acids

A few examples:

HF, HCl, HBr, HI

CH3COOH (or HC2H3O2)

Sulfuric acid is a diprotic acid

H2SO4 ( aq ) H+ HSO4-+( aq ) ( aq )

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

strong - 100% dissociation of 1st proton

weak - partial dissociation of 2nd proton

Sulfuric acid is a diprotic acid

Phosphoric acid is a triprotic acid

H3PO4 ( aq ) H+ H2PO4-+( aq ) ( aq )

H2PO4-

( aq ) H+ HPO4 2-+( aq ) ( aq )

HPO4 2-

( aq ) H+ PO4 3-+( aq ) ( aq )

Phosphoric acid is a triprotic acid

A base is a proton acceptor

an example is NaOH

a source of hydroxide ions ( OH- )

Ammonia is a Bronsted base

A Weak Base

Acid-Base Neutralization

Complete ionic equation

HCl(aq) + NaOH(aq)

Na+(aq) + Cl–(aq) + H2O(l)

Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq)

Cancel Spectator ions

Acid + Base Salt + Water

OH–(aq) + H+(aq) H2O(l )

Net ionic equation

Acid-Base NeutralizationNaCl(aq) + H2O (l )

How is Acid + Base neutralization similar to a precipitation reaction?

Acid-Base Titrations

a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete(the equivalence

point).

Indicators are substances that have distinctly different colors in a basic or acidic environment

used to indicate the equivalence point.

OH- OH-

titrant of known concentration

(could be an acid or a base - here

it’s an acid)

Unknown conc. of base, plus addition of an Indicator

(unknown is opposite of titrant -

here it’s a base)

equivalence point

titrant

H+H+H+

salt + water

known volume of titrant added

Acid-Base Titrations

if we know*the volumes of the standard and the unknown

*the concentration of the standard solution

then we can calculate the concentration of the unknown

Neutralization

Neutralization happens at the equivalence point when moles of H+ added = moles of base

OH–(aq) + H+(aq) H2O(l )

ExampleIn a titration experiment a student finds that 0.5468g

KHP(a monoprotic acid MW 204.2g) is needed to completely neutralize 23.48ml of a NaOH solution.

What is the molarity of the NaOH solution?

204.2g KHP0.5468g KHP x

1mol KHP

= 0.1141 mol/L OH–

1mol KHP

1mol H+x

1mol H+

1mol OH–x

1x0.02348L

g KHP Æ mol KHP Æ mol H+ added (same as moles of KHP b/c monoprotic acid) = mol OH- neutralized (at equivalence pt.) Æ moles NaOH (same as moles of OH- b/c NaOH is “monobasic”)

Roadmap:

Example

H2SO4(aq) + 2NaOH(aq) Na2SO4 (aq) + 2H2O (l )

How many milliliters of a 0.610M NaOH solution are needed to completely neutralize 20.0ml of a 0.245M

H2SO4 solution.

.020L0.245mol H2SO4

L1000mlx

1mol H2SO4

2mol NaOHx

16.1ml=0.610mol NaOH

volume & conc. of H2SO4 Æ mol H2SO4 Æ mol H+ added (factor of 2 involved b/c diprotic acid) = mol OH- neutralized (at equivalence pt.) Æ moles NaOH (same as moles of OH- b/c “monobasic”) Æ L of NaOH (using conc.)

Roadmap:

Oxidation-Reduction Reactions

Acid-base reactions can be characterized as proton-transfer

oxidation-reduction reactions or redox, reactions are considered electron transfer reactions

Oxidation number

signifies the number of charges the atom would have in a molecule ( or an ionic compound ) if

electrons were transfered completely.

useful tool for understanding electron transfer in oxidation-reduction reactions

also called oxidation state

Reference points for assigning oxidation numbers

(1) The oxidation number of an element is 0.

Na, Fe, Cl2, F2, P4, . . . . .

(2) The oxidation numbers in compounds of the metals in group 1A are always +1, those in group 2A

are always +2, those in group 3A are +3.

Sodium is +1 NaCl

Calcium is +2 CaOAluminum is +3 Al2O3

(3) Oxygen has an oxidation number of -2 in most of its compounds.

Since the oxidation number of hydrogen of +1, each oxygen in H2O2 must have an

oxidation number of -1.

(4) The oxidation number of hydrogen is usually +1.

Since the oxidation state of florine is -1, in all of its compounds, hydrogen must be +1 in

HF. Likewise, the oxidation number of hydrogen is +1 in HCl, HBr, and HI.

A notable exception is that when hydrogen is bonded to a metal, the oxidation number of hydrogen is -1.

The compounds NaH and CaH2 contain hydrogen in the -1 oxidation state.

(5) The halogens (fluorine, chlorine, bromine, and iodine ) have an oxidation number of -1 in most of their compounds Fluorine always has an oxidation

number of -1 in its compounds.

The oxidation number of bromine is -1 in NaBr, CaBr2, AlBr3, and NiBr2.

(6) The sum of oxidation numbers in a neutral molecule must equal 0 .

K2CO3Example:

2(+1) = +2 3(-2) = -6

Cr2(SO4)3Example:

2(+3) = +6 12(-2) = -24

3x = +18

x = +6Or do just SO42-

K2Cr2O7Example:

2(+1) = +2 7(-2) = -14

2x = +12

x = +6

reduction

oxidation

gain of electrons by atoms or ions

loss of electrons from an atom or ion

is an increase in oxidation state

is a decrease in oxidation state

reducing agent

oxidizing agent gains electrons,

got reduced

gives up electrons, got oxidized

Types of Redox Reactions

combinationdecompositiondisplacement

of hydrogenof a metalof halogens

A + B C

Combination reactions

redox if A or B is an element

S + O2 SO2

0 0 +4 -2

got oxidized &is reducing agent

got reduced &is oxidizing agent

A + BC

Decomposition reactions

2KCl + 3O22KClO3

+5 -2 -1 0

got oxidized & is reducing agent

got reduced & is oxidizing agent

spectator

Disproportionation reactions

ClO- + Cl- + H2O2HO-0 +1 -1

Balancing Oxidation-Reduction Equations

Half-Reaction Method in Acid

1. Write the unbalanced equation in ionic form.

2. Separate the equation into two half-reactions.

3. Balance each half reaction (except for O and H).

5. Balance the charges by adding electrons.

4. In acid solution, balance O by adding H2O and H by adding H+

6. Add the half reactions

7. Check to make sure atoms and charges are balanced

Example

Balance the following equation for the reaction in acid solution. *

MnO4- + Fe2+ Mn2+ + Fe3+

*All species are (aq)

MnO4- + Fe2+ Mn2+ + Fe3+

2. Separate the equation into two half-reactions

Fe2+ Fe3+

MnO4- Mn2+

Fe2+ Fe3+

MnO4- Mn2+8H++ 4H2O+

Fe2+ Fe3+ + 1e-

MnO4- Mn2+8H++ 4H2O++ 5e-

You need the same number of electrons on both sides of the equation.

5Fe2+ 5Fe3+ + 5e-

MnO4- Mn2+8H++ 4H2O++ 5e-

5Fe2+ 5Fe3+ + 5e-

MnO4- Mn2+8H++ 4H2O++ 5e-

4H2O ++

+ 5Fe2+

This is the balanced equation

Example

HNO3(aq) + H3PO3(aq) NO(g ) + H3PO4 (aq)

2. Separate the equation into two half-reactions

NO3 - NO

H3PO3 H3PO4

H+ (aq) +NO3- (aq) + H3PO3(aq) NO(g ) + H3PO4aq)

NO3 - NO

H3PO3 H3PO4

+ 2H2O4H+ +

H2O + + 2H+

NO3 - NO +

H3PO3 H3PO4

2H2O4H+ +

H2O + + 2H+

-+ 2e-

NO3 - NO +

H3PO3 H3PO4

2H2O4H+ +

H2O + + 2H+

2NO3 - 2NO +

3H3PO3 3H3PO4

4H2O8H+ +

3H2O + + 6H+

2NO3 - 2NO +

3H3PO3 3H3PO4

4H2O8H+ +

3H2O + + 6H+

2NO3 - 2NO +

3H3PO3 3H3PO4

4H2O2H+ +

3H2O + + 6H+

2NO3 - 2NO +

3H3PO3 3H3PO4

H2O2H+ +

3H2O + + 6H+

2NO3 + -2H+ + 3H3PO3 2NO +H2O + 3H3PO4

This is the balanced equation

Half-Reaction Method in Base

1. Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H

+ ions were present.

2. Add the number of OH- ions to BOTH sides of the equation to turn the remaining H+ ions to H2O

3. Eliminate waters that appear on both sides of the equation.

Example

Balance the following equation for the reaction in basic solution. *

*All species are (aq)

HS- S + H+

2H+ + + H2O

HS- + NO3- S + NO2

NO3- NO2

+5-2 0 +3

Example

NO3- NO2

-H+ + + H2O

HS- + NO3- S + NO2

NO3-H+ +HS- + NO2- + H2O+ SOH- + + OH-

Example

NO3- NO2

-H+ + + H2O

HS- + NO3- S + NO2

NO3-H+ +HS- + NO2- + H2O+ SOH- + + OH-

Example

NO3- NO2

-H+ + + H2O

HS- + NO3- S + NO2

NO3-HS- + NO2- + H2O+ S + OH-H2O +

Example

HS- + NO3- S + NO2

NO3-HS- + NO2

- + S + OH-

Example

Fe(s) + HCl(aq) HFeCl4(aq) + H2 (aq )

Fe HFeCl44Cl- +H+ +

22e- +

+ 3e-( )2

Example

Fe(s) + HCl(aq) HFeCl4(aq) + H2 (aq )

H+ 3H2

2Fe 2HFeCl48Cl- +2H+ +

66e- +

Example

H+ 3H2

2Fe 2HFeCl48Cl- +2H+ +

66e- +

+ 3H22Fe 2HFeCl48Cl- +8H+ +

+ 3H22Fe 2HFeCl48HCl +

properties of aqueous solutions -...

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