role of water chapter 4 role of waterpeople.uwec.edu/bhattas/ppt_ch4_ch3_1.pdf · chemical...

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Chapter 4Chemical Reactions in

Aqueous Solution

Role of WaterRole of H2O(l) as solvent

The polar nature of water moleculeTwo key features:1. The distribution of bonding

electronsO–H covalent bond: the electrons are

sharedB t O t tt t th l t

Chapter 4

H2H–H

But, O atoms attract the electrons strongly

⇒ the shared electrons are closer to O atom

⇒A slightly negative pole (δ–) develops on O the atom

⇒A slightly positive pole (δ+) develops on H atoms

⇒ Polar nature of the (Oδ-—Hδ+) bond

H2OO

Role of WaterChapter 4

H–H H • • H0 0

No net charge separation

Undeformed electron cloud

Deformed

Polar nature of the (Oδ-—Hδ+) bond

δ+ δ–Charge

separationelectron cloud

H–O– ••• •• •

H • •O

2

Role of WaterRole of H2O(l) as solvent

The polar nature of water moleculeTwo key features:

2. Its ‘V’ shape

Chapter 4

Results a net dipole

Two polar (Oδ-—Hδ+) bonds are working at an angle of 105.5°

Ionic Compounds in Water

Role of H2O(l) as a polar solventMany compounds dissolve

=>The attraction between water and ions prevail and provide higher stabilization energy than the solid state

Chapter 4

p g gy

Many don’t=>The attraction between water and ions is insufficient and

the solid state provide higher stabilization energy than the dissolved state


Role of H2O(l) as a polar solventMany compounds dissolve

=>The attraction between water and ions prevail and provide higher stabilization energy than the solid

Chapter 4

p g gystate

3


Role of H2O(l) as a polar solventFor example, AlCl3 (aq)

Chapter 4

Cl– H OAl3+ OH2

Cl H2O

Cl– H2OCl– H2O


Role of H2O(l) as a polar solventBut many don’t

=>The attraction between water and ions is insufficient and the solid state provide higher stabilization

Chapter 4

p genergy than the dissolved state

Ionic Compounds in WaterThey are electrolyte → dissociate to form ions

The solution shows significant conductivity

Chapter 4

NaCl(s) → Na+(aq) + Cl–(aq)water

4

Covalent Compounds in WaterThey are non-electrolyte → do not dissociate to

form ionsThe solution shows no conductivity

e.g. Table sugar, grain alcohol etc.

Chapter 4

They all dissolve in water because…..

They all have polar bond (Oδ-—Hδ+) bond

Solubility RulesCompounds Containing Solubility Exceptions

Group IA (Na+, K+) and NH4+ Soluble

Nitrates Soluble

Acetates (CH3COO−) Soluble

Perchlorates (ClO4−) Soluble

Chapter 4

Chlorides, Bromides, Iodides Soluble Ag+, Pb2+, Hg+

Fluorides Soluble Mg2+, Ca2+, Ag+, Pb2+, Sr2+

Sulfates Soluble Ag+, Ba2+, Sr2+, Pb2+, Hg+

Sulfides Insoluble Group IA and NH4+

Carbonates Insoluble Group IA and NH4+

Phosphates Insoluble Group IA and NH4+

Hydroxides Insoluble Group IA and NH4+

Chemical ReactionsChemical reactions describe processes involving chemical change

The chemical change involves rearranging matterConverting one or more pure substances into new pure substances

Chapter 4

ReactantsSubstances combined in the reaction

ProductsSubstances produced in the reaction

5

Types of ReactionsReaction Types

CombinationDecompositionDouble Replacement

Precipitation

Chapter 4

Acid-base Chemically “a H+ -transfer reaction”Single Replacement

Oxidation-reduction Chemically “an e– -transfer reaction”

Gas Forming

Combination ReactionsCombination reactions have the form

A + B → C

Two or more reactants produce a single product

Chapter 4

2 H2 + O2 2 H2O

= H = O

Combination ReactionsOther combination reactions:

2 Na(s) + S(s) → Na2S(s)

Chapter 4

SO3(g) + H2O(l) → H2SO4(aq)

2Na (s) + Cl2 (g) → 2NaCl (s)

6

Decomposition ReactionsDecomposition reactions have the form

A → B + CSingle reactant breaks down into two or more products

Chapter 4

2 H2O2(g) 2 H2O(l) + O2(g)

= H = O

Decomposition ReactionsFurther examples:

2 HgO(s) → 2 Hg(l) + O2(g)This reaction was used by Joseph Priestley in the discovery of oxygen in 1774

Chapter 4

CaCO3(s) → CaO(s) + CO2(g)This reaction is used industrially to produce both lime (CaO) and CO2 from limestone (CaCO3)

Air bags:2 NaN3 (s) 2 Na (s) + 3 N2 (g)

Double ReplacementDouble replacement reactions are also called “metathesis” reactions or “partner swapping” reactions

They have the form AX + BY → BX + AY

Chapter 5

+

HF + NaOH → NaF + H2O

= H = O = Na = F

7

Double ReplacementDouble replacement reactions often take place in water and are of two basic types:

Acid base neutralization reactions:HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Chapter 4

(acid) (base) (salt) (water)H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + H2O(l)

Precipitation reactions (solid forms):Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaNO3(aq)

Here the barium sulfate precipitates out of solution

Mixtures and SolutionsSolute

the component of a solution that is dissolved in another substance

Solventthe medium in which a solute dissolved to form a solution

Chapter 4

Solutiona homogenous mixture in which the components are evenly distributed in each other

AqueousAny solution in which water is the solvent

PrecipitateInsoluble product of a reactant

Mixtures and SolutionsElectrolyte

All ionic compounds that are soluble in water and conduct electricity

Strong electrolyte

Chapter 4

Weak electrolyte

Non-electrolyte

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Mixtures and SolutionsChapter 4

CuCl2 (s) → Cu2+ (aq) + 2Cl- (aq)100 % dissociation (strong electrolyte)

CH3COOH(aq) CH3COO- (aq)+ H+ (aq)<5% ionized (weak electrolyte)

CuCl2 (s) → Cu2+ (aq) + 2Cl- (aq)100 % dissociation (strong electrolyte)

CH3COOH(aq) CH3COO- (aq)+ H+ (aq)<5% ionized (weak electrolyte)

Precipitation ReactionsPrecipitation reactions

Writing EquationsMolecular

Chapter 4

K2SO4 (aq) + Pb(NO3)2 (aq) →

Molecular

Total ionic

Net ionic

http://www.dlt.ncssm.edu/TIGER/Flash/moles/DoubleDisp_Reaction-Precipitation.html

Ionic EquationsWhen an ion appears on both sides of a chemical equation it can be canceled out

“Spectator Ions”

Chapter 4

Net Ionic Equation

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Ionic EquationsMg(ClO4)2(aq) + K2CO3(aq) → MgCO3(s) + KClO4(aq)Total:

Cancel:

Net Ionic:

Chapter 4

Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + NaNO3(aq)

Total:Cancel:

Net ionic:

Ionic EquationsKNO3(aq) + CaCl2(aq) →

Total:

Cancel:

Chapter 4

Net Ionic:

A tip to good problem solving– Break up the problem into parts – Identify the information (that can be used to calculate

and/or derive other unknown quantities)

Whimbey

– Recognize the process to do such derivation

10

Molar MassMolar mass =

Mass in grams of one mole of any element, numerically equal to its atomic weight

Molar mass of molecules can be determined from the chemical formula and molar masses of elements

Each H2O molecule contains 2 H atoms and 1 O

Chapter 3

2atom

Each mole of H2O molecules contains 2 moles of H and 1 mole of OOne mole of O atoms corresponds to 15.9994 gTwo moles of H atoms corresponds to 2 x 1.0079 gSum = molar mass = 18.0152 g H2O per mole

Molar MassCalculate the molar mass of the following

Magnesium nitrate, Mg(NO3)21 Mg = 24.30502 N = 2x 14.0067 = 28.01346 O = 6 x 15.9994 = 95.9964Molar mass of Mg(NO3)2 = 148.3148 g

Chapter 3

Calcium carbonate, CaCO31 Ca = 40.0781 C = 12.0113 O = 3 x 15.9994Molar mass of CaCO3 = 100.087 g

Iron(II) sulfate, FeSO4Molar mass of FeSO4 = 151.909 g

Problem 1: Molar MassThe solid magnesium sulfate has seven water of hydration and is written as formula of magnesium sulfate, 7H2O.A) Calculate the formula weight of magnesium sulfate and provide its molar

Chapter 3

magnesium sulfate and provide its molar mass.

B) Calculate the weight percentages of i) sulfur and ii) water in the salt.

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Problem 2: Molar MassCalculate

A) the molar masses of iron(II)sulphate and iron(III)sulphate

Chapter 3

B) the weight of % iron in both salts

SolutionsSolution: a homogenous mixture in which the components are evenly distributed in each other

Solute: the component of a solution that is dissolved in another substance

Chapter 3

Solvent: the medium in which a solute dissolved to form a solution

Aqueous: any solution in which water is the solvent

The properties and behavior of solutions often depend not only on the type of solute but also on the concentration of the solute.

C t ti th t f l t di l d i i

SolutionsChapter 3

Concentration: the amount of solute dissolved in a given quantity of solvent or solution– many different concentration units

• (%, ppm, g/L, etc)

– often expressed as Molarity

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Solution ConcentrationsMolarity = moles of solute per liter of solution

Designated by a capital M (mol/L)

6.0 M HCl 6 0 moles of HCl per liter of solution

Chapter 3

6.0 moles of HCl per liter of solution.

9.0 M HCl 9.0 moles of HCl per liter of solution.

Molarity can be used as a conversion factor.

The definition of molarity contains 3 quantities:

Solution ConcentrationsChapter 3

Molarity = moles of solutevolume of solution in liters

If you know two of these quantities, you can find the third.

Determine the molarity of each solution2.50 L of solution containing 1.25 mol of solute


Solution ConcentrationsChapter 3

225 mL of solution containing 0.486 mole of solute

100. mL of solution containing 2.60 g of NaCl

Strategy:g → mol → molarity

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Problem 3: Solution ConcentrationsChapter 3

Calculate the molarity of

A)225 mL of solution containing 0.486 mole of aluminum chloridealuminum chloride.

B)100. mL of solution containing i) 2.6 moles of iron(III) chloride and ii) 5.83 moles of iron(II) chloride.

Problem 4: Solution ConcentrationsChapter 3

A)Calculate how many gram of HCl is needed to make a liter of 1.5 molar solution.

B) Calculate the molarity of 117g of sodium chloride dissolved in 5 L solution

.


Gi en 2 5 L of soln

Solution PreparationChapter 3

Example: How many moles of HCl are present in 2.5 L of 0.10 M HCl?

Given: 2.5 L of soln0.10M HCl

Find: mol HClUse molarity as a conversion factor

Mol HCl = 2.5 L soln x 0.10 mol HCl1 L of soln

= 0.25 mol HCl

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Example: What volume of a 0.10 M NaOH solution is needed to provide 0.50 mol of NaOH?

Given: 0.50 mol NaOH0 10 M N OH

Use M as a


0.10 M NaOHFind: vol soln

conversion factor

Vol soln = 0.50 mol NaOH x 1 L soln0.10 mol NaOH

= 5.0 L solution

Solution PreparationSolutions of exact concentrations are prepared by dissolving the proper amount of solute in the correct amount of solvent – to give the desired final volume

Determine the proper amount of solute

Chapter 3

How is the final volume measured accurately?

Example: How many grams of CuSO4 are needed to prepare 250.0 mL of 1.00 M CuSO4?

Given: 250.0 mL solution 1.00 M CuSO4

Conversion factors:Molarity molar mass


Find: g CuSO4Molarity, molar mass

Strategy:mL L mol grams

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g CuSO4 = 250 0 mL soln x 1 L x 1 00 mol


Given: 250.0 mL solution, 1.00 M CuSO4Find: g CuSO4

Strategy:mL L mol grams

g CuSO4 250.0 mL soln x 1 L x 1.00 mol1000 mL 1 L soln

x 159.6 g CuSO41 mol

= 39.9 g CuSO4

Describe how to prepare the following:a) 500. mL of 1.00 M potassium dichromate

(K2Cr2O7)

Chapter 3

Solution Concentrations

b) 100. mL of 3.00 M glucose(C6H12O6).

Steps involved in preparing solutions from pure solids


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Steps involved in preparing solutions from pure solids

– Calculate the amount of solid required– Weigh out the solid

Place in an appropriate volumetric flask


– Place in an appropriate volumetric flask– Fill flask about half full with water and mix.– Fill to the mark with water and invert to mix.

Solution PreparationSolutions of exact concentrations can also be prepared by diluting a more concentrated solution of the solute to the desired concentration

Chapter 3

Many laboratory chemicals such as acids are purchased as concentrated solutions (stock solutions).

12 M HCl


12 M H2SO4

More dilute solutions are prepared by taking a certain quantity of the stock solution and diluting it with water.

17

A given volume of a stock solution contains a specific number of moles of solute.– 25 mL of 6.0 M HCl contains 0.15 mol HCl

(How do you know this???)


If 25 mL of 6.0 M HCl is diluted with 25 mL of water, the number of moles of HCl present does not change.– Still contains 0.15 mol HCl

Moles solute moles solutebefore dilution after dilution

Although the number of moles of solute does not change the volume of solution


=

does not change, the volume of solution does change.

The concentration of the solution will change since Molarity = mol solute

volume solution

When a solution is diluted, the concentration of the new solution can be found using:

Mc• Vc = moles = Md• Vd

Where, Mc = concentration of concentrated solution (mol/L)V l f t t d l ti


Vc = volume of concentrated solutionMd = concentration of diluted solution (mol/L)Vd = volume of diluted solution

18

Example: What is the concentration of a solution prepared by diluting 25.0 mL of 6.00 M HCl to a total volume of 50.0 mL?

Given: V = 25 0 mL


Given: Vc = 25.0 mLMc = 6.00 MVd = 50.0 mL

Find: Md

Mc x Vc = Md x Vd

Mc • Vc = Md • Vd

6.00 M x 25.0 mL = Md x 50.0 mL


Md = 6.00 M x 25.0 mL = 3.00 M50.0 mL

Note: Vc and Vd do not have to be in liters, but they must be in the same units.

Solution PreparationDescribe how to prepare 500. mL of 0.250 M NaOH solution using a 6.00 M NaOH solution.Given: Mc = 6.00 M

Md = 0.250 MVd = 500.0 mL

Find: V

Chapter 3

Find: Vc

Mc• Vc = Md• Vd

What volume of 2.30 M NaCl should be diluted to give 250. mL of a 0.90 M solution?

A) 152.3 mLB) 97.8 mLC) 639 mLD) 252.3 mL

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Solution Preparation ReviewIf you dissolve 9.68 g of potassium chloride in 1.50 L, what is the final molar concentration?

Chapter 3

A) 0.087 B) 0.129 C) 0.194 D) 0.968

How many grams of sodium sulfate are contained in (dissolved in) 45.0 mL of 3.00 M solution?

)

A) 0.124 B) 12.9 C) 42.6D) 19.2

Solution Preparation ReviewWhat volume of 2.06 M potassium permanganate contains 322 g of the solute?

Chapter 3

Conversion Factors

Number of particles Moles Mass

Chapter 3

Avogadro’s number6.022 x 1023

Molarmass

20

Conversion Factors

Number of particles

M l

Mass

Molar mass

Chapter 3

Avogadro’s number6 022 x 1023 Moles

Solution of a certain concentration

6.022 x 1023

Volume of solution in L

WhimbeyProblem Solving

– Problem solver: The person will solve the assigned problems using pencil, paper, calculator, and any other required items. During the entire process of solving the solver will articulate what she/he is doing out loudout loud.

– Listener: The person listens to the solver and prompts solver to keep talking. Prompts such as “What next?”, “Does that sound right?”, and “Talk to me.” are appropriate. NEVER give the right answer, only prompt the solver.

Source: Beyond Problem Solving and Comprehension by Whimbey and Lochhead, 1984

Molar ConversionsDetermine the following:

The moles of FeCl3 in a 50.0 g sample

Chapter 3

The mass of MgCl2 in a 2.75 mole sample

21


How many FeCl3 units are there in a 50.0 g sample?

Chapter 3

How many MgCl2 units are there in a 2.75 mole sample?


Molarity of 50.0 g FeCl3 solution in 500 mL?

Chapter 3

How many grams of MgCl2 are there in a 250 mL of 2.75 M mole sample?

Acids and BasesAcida substance that, when dissolved in water, increases the concentration of hydrogen ions (H+) in solution

Baseb t th t h di l d i t i th t ti f

Chapter 4

a substance that, when dissolved in water, increases the concentration of hydroxide ions (OH–) in solution

Acid base neutralization reactions:

HCl (aq) + NaOH(aq) → NaCl(aq) + H2O(l)(acid) (base) (salt) (water)H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + H2O(l)

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Acids and Bases as Electrolytes

• Strong acids and bases are strong electrolytes

• Weak acids and bases are weak electrolytes

Chapter 4

Acids and BasesChapter 4

Acid-Base ReactionsHNO3(aq) + NaOH(aq) →

Total:

Cancel:

Chapter 4

Net Ionic:

23

Acid-Base ReactionsH2SO4(aq) + KOH(aq) →

Total:

Cancel:

Chapter 4

Net Ionic:

Acid-Base Titrations• Standardized solution

• Acid-base indicator

• Equivalence point

Chapter 4

A solution whose concentration is known

A chemical which changes color when the pH of the solution changes

q p

• End point

- when an exactequivalent of acid is added to a base or vice-a-versa

- when the indicator indicates the complete neutralization of an acid by a base or vice-a-versa

Problem on Acid-Base Titration

• Sample problem 4.6

Chapter 4

24

HCl (aq) → H+(aq) + Cl–(aq)

Acid-Base Reactions Are Proton (H+) Transfer Reaction

Chapter 4

H+ OH Cl H OH+ OH2 Cl– H2O

How is the HCl molecule in gas phase, where there is no water?

OH2H+ H+ Cl– Cl– H2O HCl (g)

HCl (g) + H2O (l) → H3O+(aq) + Cl–(aq)


Chapter 4

Proton (H+) Transfer

HH

H+ OH2

OH2H 2

O

H2 O

H2 O

H+ OH2

OH2H 2

O

H2 O

H2 O

Solvated hydronium cation

When NaOH(aq) is added

H O+( ) Cl ( ) [N +( ) OH ( )]

Chapter 4

Proton (H+) Transfer


H3O+(aq) + Cl–(aq) + [Na+(aq) + OH– (aq)] →

H2O(l) + Cl–(aq) + Na+(aq) + HOH (l)]

Getting rid of spectator ions

H3O+(aq) + OH– (aq) → H2O(l) + HOH (l) or,2H2O (l) Or, H+(aq) + OH–(aq) → H2O(l)

25

How many moles of H+ ions are present in 2.5 moles of H2SO4 ?

Given: 2.5 moles H2SO4 Conversion factor:

Molar ConversionsChapter 3

Given: 2.5 moles H2SO4Find: moles H+

Conversion factor:molar ratio

Mol H+ = 2.5 mol H2SO4 x 2 mol H+

1 mol H2SO4Mol H+ = 5.0 mol H+

Concentrations of AcidsThe pH scale

Chapter 3

pH = -log [H+]

pH of vinegar = -log (1.6 x 10-3 M) = - (-2.80) = 2.80

pH of pure water = -log (1.0 x 10-7 M) = - (-7.00) = 7.00

pH of blood = -log (4.0 x 10-8 M) = - (-7.40) = 7.40

pH of ammonia = -log (1.0 x 10-11M) = - (-11.00) = 11.00

acidic substance, pH< 7Basic substance, pH >7neutral, pH = 7

Lower the concentration of H+, higher the pH

On serial dilution of acid solution, the pH increases because the concentration of H+ ions decreases with dilution

Concentration pH0 1M HCl 1

Concentrations of AcidsChapter 3

The H+ concentration of a solution of known pH can be calculated using the following equation:

0.1M HCl 10.01 M HCl 20.001 M HCl 3

[H+] = 10-pH

26

Concentrations of AcidsChapter 3

Calculate pH of 0.0065 M HCl solution.

Calculate the concentration of H+ ion in a solution of pH 7.5.

Solution Preparation ReviewWhat volume of 8.00 M sulfuric acid should be diluted to produce 0.500 L of 0.250 M solution?

Chapter 3

What’s the pH of the final solution?

Redox ReactionsRedox reactions are also called Single replacementor substitution reactions

They have the form A + BX → B + AX, where A and B are elements and BX and AX are compounds

Chapter 4

= H = O

+

H2 + CuO → H2O + Cu = Cu

27

Redox ReactionsOther redox reactions

3 C(s) + 2 Fe2O3(s) → 4 Fe(s) + 3 CO2(g)

Cu(NO3)2(aq) + Zn(s) → Zn(NO3)2(aq) + Cu(s)

Chapter 4

Single replacement reactions are also oxidation-reduction (REDOX) reactions.

Oxidation

Reduction

OXIDATION NUMBER

Redox Reactions Are Electron (e–) Transfer Reactions

Oxidation of Mg (0) to Mg (II):2M ( ) + O ( ) 2M O ( )

Chapter 4

The oxidation number of an atom is the charge that atom would have if the compound was composed of ions.

2Mg (s) + O2 (g) → 2MgO (s)

Reduction of Fe (III) to Fe(0):Fe2O3 (s) + 3CO (g) →2Fe (s) + 3 CO2 (g)

Reduction of Ag (I) to Ag and oxidation of Cu (0) to Cu (II):2Ag+ (aq) + Cu(s) → 2 Ag (s) + Cu2+ (aq)

Chapter 4

Oxidation Number

28

Oxidation Number

• SF6 and FeS

Chapter 4

Oxidation Number Some Reactions Produce GasReactions leading to the formation of an insoluble gas

Examples

Chapter 4

Na2CO3 (aq) + 2HCl (aq) → 2NaCl (aq) + CO2 (g) + H2O (l)

Some Reactions Produce GasChapter 4

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