4,reaction in aq.solution

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Chapter 4

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Many chemical reactions and virtually all biological processes take place in water.

e.g.: all chemistry that makes life possible occur in aqueous solution, various medical tests involve aqueous reactions like sugar tests..cholesterol .. ect

To understand the chemistry that occur in such diverse places as human body, the atmosphere, the ground water, the oceans, and so on, we must understand how substances dissolved in water react with each other.

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Before we can understand solution reactions, we need to discuss the nature of solutions in which water is the dissolving medium or solvent.

These solutions- (homogeneous mixture) are called aqueous solutions.

aqueous solutions

Soluble- Can be dissolved.

Miscible- liquids dissolve in each other.

SoluteSolvent what gets dissolved

what does the dissolving

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Water, the common solvent� One of the most valuable properties of water

is ability to dissolve many different substances.

� Water is a good solvent for ionic compounds because its molecules are polar. The oxygen atoms have a partial negative charge. The hydrogen atoms have a partial positive charge.

� The angle is 104.5ºC.� It is the polarity that gives water its great

ability to dissolve compounds

+ δ

+ δ

-δ

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� A schematic of an ionic solid dissolving in water is shown in the following figure.

� The “positive ends” of water molecules are attracted to the negatively charged anions and the “negative ends” are attracted to the positively charged cations

� This process is called hydration � The hydration process is breaking the ionic substances

(salts) apart into individual cations and anions and surrounded by water molecules

� When NaCl dissolves in water, the resulting solution contains Na+ and Cl- ions moving around independently.

NaCl(s) Na+(aq) + Cl-

(aq)H2O(ℓ)

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How Ionic solids dissolveHHO

HH

O

H HO

HHO

HH

O

HH

O

H HOH

H OH

HO

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� The solubility is how much of a substance will dissolve in a given amount of water, usually g/100 mL

� The solubility of ionic substances in water varies greatly, e.g: NaCl soluble in water, whereas AgCl insoluble in water

� If the ionic substance do dissolve the ions are separated,

and they can move around.� Water can also dissolve non-ionic compounds if they

have polar bonds like ethanol (C2H5OH).

� “Like dissolve like” is a useful rule for predicting solubility.

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Types of solutions

Solutions are classified three ways� Strong electrolytes- completely dissociate (fall apart into

ions). - Many ions - Conduct electricity well.

e.g.: NaCl give Na+ , Cl- when dissolve in water� Weak electrolytes- Partially fall apart into ions.

– Few ions -Conduct electricity slightly.

e.g.: acetic acid dissociate partially in water.� Non-electrolytes- Don’t fall apart. - No ions- Don’t

conduct electricity.

e.g.: sugar ( C12H22O12) , ethanol (C2H5OH)

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The composition of solutionChemical reactions take place when two solution are mixed.

To perform stoichiometric calculations in such cases, two things must be known:

1.The nature of the reaction

2.The amount of chemicals present in the solution, expressed as concentrations.

Concentration: is how much solute is dissolved in a given amount of solvent.

The concentration of solution can be expressed in many different ways. We will consider only the most commonly used expression of concentration, molarity

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Molarity (M): moles of solute per volume of solution in liters

1 M = 1 mol solute / 1 liter solution

1 M = 1 mol solute dissolved water until the volume of solute and solvent (solution) become 1 liter

The concentration is an intensive property, so its value does not depend on how much of the solution is present.

Example (1):

Calculate the molarity of a solution with 34.6 g of NaCl (molar mass = 58.5g/mol)dissolved in 125 mL of solution.

M = molarity = moles of solutesvolume of solution (L)

mol NaCl = 34.6 g NaCl x 1 mol NaCl58.5 g NaCl

= 0.59 mol NaCl

M = 0.59 mol NaCl125 x 10-3 L = 4.73 mol/L

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Example (2):

How many grams of HCl (molar mass = 36.5 g/mol) would be required to make 50.0 mL of a 2.7 M solution?

M = moles of HCl

volume of solution (L)

2.7 (mol/L) = moles of HCl50.0 x10-3 L

moles of HCl = 2.7 (mol/L) x 50.0 x10-3 L = 0.135 mol HCl

Mass of HCl = 0.135 mol HCl x 1 mol HCl36.5 g HCl = 4.93 g HCl

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Example(3):

Typical blood serum is about 0.14 M. what volume of blood contain 1.0 mg NaCl (molar mass = 58.5 g/mol).

0.14 mol/L =

mol of NaCl =1.0 x 10-3g NaCl x 1 mol NaCl58.5 g NaCl= 1.71 x 10-5 mol NaCl

1.71 x 10-5 mol NaClvolume of blood serum

volume of blood serum (L) = 1.71 x 10-5 mol NaCl0.14 mol/L

= 1.2 x 10-4 L solution = 0.12 ml solution

Thus: 0.12 ml of blood contain 1.71x10-5 mol NaCl or 1.0mg NaCl

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Solution concentration is always given in term of the form of the solute before it dissolves and does not take into account any subsequent processes, such as the dissociation of salts.

e.g.: 1.0 M NaCl, this mean that 1.0 mol (58.5g) solid NaCl dissolved in enough water to make 1.0 L solution; it does not mean that the solution contain 1.0 mol of NaCl units.

Actually, the solution contains 1.0 mol Na+ ions and 1.0 mol Cl- ions.

Example (4):

What is the concentration of each ions in 0.5M CaCl2 solution: CaCl2(s) Ca+

(aq) + 2 Cl-(aq)

1 mol CaCl2 that dissolved, the solution contain 1 mol Ca+2 and 2 mol Cl- ions. Thus 0.5 M CaCl2 contain 0.5 M Ca+2 and 2 x 0.5 M Cl- or 1 M Cl-

H2O

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Example (5):

A 27 g of CaCl2 (molar mass = 111 g/mol) dissolved in 500 ml solution. Calculate: a) concentration of CaCl2

b) concentration of each ion

mol of CaCl2 = 27 g CaCl2 x 111 g CaCl2

1 mol CaCl2 = 0.24 mol CaCl2a)

b) 1 mol CaCl2

1 mol Ca+2

2 mol Cl-

0.48 mol CaCl2

0.48 mol Ca+2

2 x 0.48 mol Cl- = 0.96 mol Cl-

M (CaCl2) = 0.24 mol CaCl2

500 x 10-3 L = 0.48 M CaCl2

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Example (6):

A 45.6 g of Fe2(SO4)3 (molar mass = 400 g/mol) dissolved in 475 ml solution. Calculate: a) concentration of Fe2(SO4)3

b) concentration of each ion M (Fe2(SO4)3) =

(mass/molar mass) Fe2(SO4)3

V (L) solution

M (Fe2(SO4)3) =45.6 g/(400g/mol)

475 x 10-3 L= 0.24 M or mol/L

0.24 (mol/L) Fe2(SO4)3

2 x 0.24 M Fe+3 = 0.48 M Fe+3

3 x 0.24 M SO4-2 = 0.72 M SO4

-2

a)

b)

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The number of moles of solute present in a given volume of solution of known molarity can be calculated as follows:

Molarity (mol/L)= mol of solutevolume of solution (L)

mol of solute = molarity (mol/L) x volume of solution (L)

Example (7):Calculate the number of mol of Cl- ions in 1.75L of 1x10-3 M ZnCl2

1x10-3 M ZnCl2

1x10-3 M Zn+2

2 x(1x10-3)M Cl- = 2 x 10-3 M Cl-

mol of Cl- = 2 x 10-3 mol Cl-

Lx 1.75 L = 3.5 x10-3 mol Cl-

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Preparing a Solution of Known Concentration

A standard solution: is a solution whose concentration is accurately known. It can be prepared as follows. First, the solute is accurately weighed and transferred to volumetric flask as shown in the following figure. Next water is added to the flask, to dissove the solid. After all the solid has dissolved, morewater is added to bring the level of solution exactly to the volume mark.

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Example (8): Describe how to make 100.0 mL of a 1.0 M K2Cr2O4 (molar mass =246 g/mol) solution.

mol of K2Cr2O4 =1.0 mol K2Cr2O4

Lx 100 x 10-3 L = 0.1 mol K2Cr2O4

mass of K2Cr2O4 = 0.1 mol K2Cr2O4 x1 mol K2Cr2O4 246g K2Cr2O4 =24.6 g K2Cr2O4

Thus, to prepare 100 ml of 1.0 M K2Cr2O4, weigh out 24.6g K2Cr2O4 , add distilled water until the volume of solution = 100ml Example (9): Describe how to make 250. mL of an 2.0 M copper (II) sulfate dihydrate (CuSO4.2H2O, molar mass = 200g/mol) solution.mol of CuSO4 = 2.0 mol CuSO4 x 250 x 10-3 L = 0.5 mol CuSO4

Lmass of CuSO4 = 0.5 mol CuSO4

200 g CuSO4 1 mol CuSO4

= 100 g CuSO4

Thus, to prepare 250.0 ml of 2.0 M CuSO4.2H2O, weigh out 100 g CuSO4.2H2O , add distilled water until the volume of solution = 250 ml

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Dilution� In the lab. a concentrated solutions are present, these solutions are called Stock solution.� Water is added to these solutions to achieve the molarity desired for a particular solution. � This process is called dilution: the procedure for preparing a less concentrated solution from more concentrated one� The moles of solute stay the same.

DilutionAdd Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)

=MiVi MfVf=

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Example:

You have a 4.0 M stock solution. Describe how to make 1.0L of a 0.75 M solution from the stock solution.

(M V)conc. = (M V)dil.

0.4 M x V = 0.75 M x 1.0 L

Vconc. = 0.75 M x 1.0 L/ 4 M = 0.1875 L = 187.5 ml

Take 187.5 ml of the stock solution, add water until the volume of the solution (solute and solute) become 1.0 L

Example:

18.5 mL of 2.3 M HCl is diluted to 250 mL with water. What is the concentration of the solution?

M1 V1 = M2 V2 2.3 M x 18.5 ml = M2 x 250 ml

M2 = 2.3 M x 18.5 ml / 250 ml

M2 = 0.17 M

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Example:

A solution is prepared by dissolving 10.8 g of (NH4)2SO4 (molar mass = 132 g/mol) in enough water to make 100 ml stock solution. A 10.0 ml of this stock solution is added to water to make 50.0 ml solution. Calculated the concentration of NH4

+ and SO4-2 ions in the final solution.

moles of (NH4)2SO4 = 10.8 g132 g /mol = 0.0818 moles of (NH4)2SO4

molarity of (NH4)2SO4 = 0.818 moles 100 x 10-3 L

= 0.818 M (NH4)2SO4

(M V) conc. = (M V) dil.0.818 M x 10.0 ml = M x 50 mlM dil. = 0.16 M of (NH4)2SO4

0.16 M of (NH4)2SO4 0.16M SO4

-2

2 x 0.16M NH4+

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Classes of Reactions

Chemical reactionsChemical reactions

Precipitation reactions

Precipitation reactions

Acid-BaseReactionsAcid-BaseReactions

Oxidation-ReductionReactions

Oxidation-ReductionReactions

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1) Precipitation reactions:� Occur in aqueous solutions of ionic compounds which results in

the formation of an insoluble product, or a precipitate.� A precipitate is an insoluble solid that separates from the solution.� For example: Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + KNO3(aq)

� This reaction is an example of a metathesis reaction (also called double-displacement reaction): a reaction that involves the exchange of parts between the two compounds.

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Pb2+NO3

1–

Na1+ I1–

Ions in Aqueous Solution

Pb(NO3)2(s) Pb(NO3)2(aq)

Pb2+(aq) + 2 NO31–(aq)

addwater

NaI(s)

+ H2O(ℓ)

dissociation:

+ H2O(ℓ)

Na1+(aq) + I1–(aq)

Mix them and get…

Balance to get overall ionic equation…

Cancel spectator ions to get net ionic equation…

NaI(aq)

NO31–

Pb2+

NO31–

NO31–

in solution

Na1+ I1–

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Mix them and get…

Pb2+(aq) + 2 NO31–(aq) + 2 Na1+(aq) + 2 I1–(aq) PbI2(s) + 2 NO3

1–(aq) + 2 Na1+(aq)

Pb2+(aq) + 2 I1–(aq) PbI2(s)

Pb(NO3)2(aq) + 2 NaI(aq) PbI2(s) + 2 NO31–(aq) + 2 Na1+(aq)

Balance to get overall ionic equation…

Cancel spectator ions to get net ionic equation…

solid

in solution

Pb2+NO3

1–

Na1+ I1–

NO31–

Na1+ I1–

Pb2+

NO31–

Na1+

I1–

NO31–

Na1+

I1–

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Ba2+

OH1–

OH1–

NO31–

NO31–

Mix together Zn(NO3)2(aq) and Ba(OH)2(aq):

Zn2+(aq) + 2 NO3

1–(aq) Ba2+

(aq) + 2 OH1–(aq)

Ba(OH)2(aq)Zn(NO3)2(aq)

Balance to get overall ionic equation…

Zn2+

Zn(NO3)2(aq) + Ba(OH)2(aq) Zn(OH)2(s) + 2 NO31–

(aq)+ Ba2+

(aq)

Zn2+(aq) + 2 NO3

1–(aq) + Ba2+

(aq) + 2OH1–(aq) Zn(OH)2(s) + 2 NO3

1–(aq) + Ba2+

(aq)

Mix them and get…

Zn2+(aq) + 2 OH1–

(aq) Zn(OH)2(s)

Cancel spectator ions to get net ionic equation…

Ba(NO3)2 and Zn(OH)2(aq) (ppt)

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Three Types of Equations� Molecular Equation- written as whole formulas, not the ions.

K2CrO4(aq) + Ba(NO3)2(aq) → BaCrO4(s) + 2KNO3(aq)

� Complete Ionic equation show dissolved electrolytes as the ions.2K+

(aq) + CrO4-2

(aq) + Ba+2(aq) + 2 NO3

-(aq) →

BaCrO4(s) + 2K+(aq) + 2 NO3

-(aq)

� Spectator ions are those that don’t react.

� Net Ionic equations show only those ions that react, not the spectator ions

Ba+2(aq)

+ CrO4-2

(aq) → BaCrO4(s)

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� How can we predict whether a precipitate will form or not when two aqueous solutions are mixed? It depends on the solubility of the solute.

� The solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at specific temperature.

� Substances can be classified as soluble, slightly soluble, or insoluble

� All ionic compounds are strong electrolytes, but they are not equally soluble

� the solubility rules for common ionic compounds in water are as follows:

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Solubility Rules

1. Most nitrates (NO3-) are soluble.

2. Most salts containing Group (I) ion and ammonium ion, NH4

+, are soluble.

3. Most chloride (Cl-), bromide (Br-), and iodide (I-) salts are soluble, except Ag+, Pb2+ and Hg2

2+.

4. Most sulfate (SO4-2) salts are soluble, except BaSO4, PbSO4,

Hg2SO4, Ag2SO4 and CaSO4.5. Most hydroxides (OH-) except Group (1) and Ba(OH)2.

6. Most sulfides (S-2), carbonates (CO3-2), chromates (CrO4

-2), and phosphates (PO4

-3)are only slightly soluble.

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Examples:

Predict if a precipitate is formed when the following solutions are mixed. Write the three types of equations for the reactions that lead to the formation of a precipitate.

� iron (III) sulfate and potassium sulfide � Lead (II) nitrate and sulfuric acid.

� NH4Cl(aq) + NaNO3(aq) →

� AgNO3(aq) + KCl(aq) →

� Zn(NO3)2(aq) + BaCr2O7(aq) →

� CdCl2(aq) + Na2S(aq) →

� NaOH(aq) + FeCl3(aq) →

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Stoichiometry of Precipitation ReactionsThe procedure for doing stoichiometric calculations for solutions reactions are very similar to those for other types of reactions.The following steps are used:� Write the balanced net ionic equation for the reaction� Calculate the moles of the reactants.� Determine which reactant is limiting.� Calculate the moles of product or products, as required.� Convert to grams, as required

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Example (1)

What mass of Na2CrO4 (molar mass = 162 g/mol) is required to precipitate all Ag+ ions from 75.0ml of 0.1 M solution AgNO3.

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

Net ionic equation: 2Ag+ (aq)+ Na2CrO4(s)→Ag2CrO4(s) + 2Na+

(aq)

moles of Ag+ = M x V = 0.1 (mol/L) x 75.0 x 10-3 L = 7.5 x 10-3 molmol of Na2CrO4 = 7.5 x 10-3 mol Ag+ x 1 mol of Na2CrO4

2 mol Ag+

= 3.8 x 10-3 mol Na2CrO4

mass of Na2CrO4 = 3.8 x 10-3 mol Na2CrO4 x 162 g/mol

= 0.616 g Na2CrO4

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Example (2):Calculate the mass of Ba(OH)2 (molar mass = 171g/mol ) is formed when 100.00 mL of 0.100 M BaCl2 is mixed with 100.00 mL of 0.100 M NaOH. BaCl2(aq) + 2 NaOH(aq) → Ba(OH)2(s) + 2 NaCl(aq)

mol of of BaCl2 = M x V = 0.1(mol/L) x 100 x 10-3 L

= 0.01 mol

mol of NaOH = 0.1(mol/L) x 100 x 10-3 L = 0.01 mol NaOH

BaCl2 : 2 NaOH

0.01 0.01/2

NaOH is the limiting reactant

mol of Ba(OH)2 = 0.01 mol NaOH x 1 mol Ba(OH)2

2 mol NaOH= 0.005 mol Ba(OH)2

mass Ba(OH)2 = 0.005 mol Ba(OH)2 x 171 g Ba(OH)2

1 mol Ba(OH)2

= 0.855 g Ba(OH)2

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Example (3):What volume of 0.204 M HCl is needed to precipitate all the silver ions from 50.ml of 0.0500 M silver nitrate solution.

AgNO3(aq) + HCl(aq) → AgCl(s) +

mol of AgNO3 = M x V = 0.05 M x 50 x 10-3 L = 2.5 x 10-3 mol

mol of HCl = 2.5 x 10-3 mol AgNO3 x 1 mol AgNO3

1 mol HCl = 2.5 x 10-3 mol

Molarity of HCl = mol of HCl Volume (L)

Volume (L) = mol of HCl molarity of HCl

volume of HCl = 2.5 x 10-3 mol 0.204 mol/L = 0.01225 L = 12.23 ml

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2) Acid-Base reactions Arrhenius’s concept of acids and bases:Acid: is a substance that produce H+ ions when dissolved in

water.Base: is a substance that produce OH- ions.A more general definition of a base (which include

substances that do not contain OH-) was provided by Brǿnsted-Lowry whose defined acids and bases as follows:

an acid is a proton donor.a base is a proton acceptor.

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When aqueous solution of HCl (strong electrolyte) is mixed with aqueous solution of NaOH(strong electrolyte), the reaction can be represented as follows:Molecular equation: HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(ℓ)

Ionic equation : H+

(aq) + Cl-(aq) + Na+

(aq) + OH-(aq) → Na+

(aq) + Cl-(aq)+ H2O(ℓ)

Net ionic equation: H+(aq) + OH-

(aq) → H2O(ℓ)

So, all acid-base reaction can be represented as :

Acid + Base → salt + water

Acid - Base Reaction is often called a neutralization reaction. When enough acid is added to react exactly with the base, we say that the acid has been neutralized.

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When aqueous solution of weak acid such as HCN is mixed with aqueous solution of NaOH, the reaction can be represented as follows:Molecular equation: HCN(aq) + NaOH(aq)→ NaCN(aq) + H2O(ℓ)

Ionic equation : HCN

(aq) + Na+(aq) + OH-

(aq) → Na+(aq) + CN-

(aq)+ H2O(ℓ)

Net ionic equation: HCN (aq) + OH-

(aq) → CN-(aq) + H2O(ℓ)

Note that only Na+ is a spectator ion.

Example: write the ionic, and the net ionic equation for the following acid-base reaction:

Na2CO3(aq) + 2 HCl → 2 NaCl(aq) + H2CO3(aq)

H2CO3(aq) decompose as follows:H2CO3(aq) → CO2(g) + H2O(ℓ)

Ionic equation:

2 Na+(aq) + CO-3

(aq) + H+(aq) + Cl-

(aq) → CO2(g) + H2O(ℓ) +2 Na+(aq) + 2 Cl-

(aq)

Net ionic equation: CO-3(aq) + H

+(aq) → CO2(g) + H2O(ℓ)

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Acid-Base Titration:

Quantitative studies of acid-base neutralization reactions are carried out using a technique known as titration.

In titration, a solution of accurately known concentration (standard solution) is added gradually by using the buret to another solution of unknown concentration, until the chemical reaction between the two solutions is complete. The reaction is said to be complete when we reach the Equivalence point.

Equivalence point – the point at which the reaction is complete

The equivalence point is usually

signaled by a sharp change in the

Color of an indicator

Indicator – substance that changes

color at (or near) the equivalence point

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Example (1):

75 mL of 0.25M HCl is mixed with 225 mL of 0.055 M Ba(OH)2 . calculate : a) the amount of H2O produce.

b) the concentration of the excess H+ or OH-

Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2 H2O(ℓ)

moles of HCl = 0.25 M x 75x10-3L = 0.01875 mol

mol of Ba(OH)2 = 0.055 M x 225x10-3L = 0.012375 mol

Ba(OH)2 : 2 HCl

0.012375 : 0.01875/2 limiting reactant is HCl

Mass of H2O = 0.01875 mol HCl x 2 mol HCl2 mol H2O x 18 g H2O

1 mol H2O = 0.3375 g H2O

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b) moles of Ba(OH)2 reacted = 0.01875 mol HCl x 2 mol HCl

1 mol Ba(OH)2

= 0.009375 mol Ba(OH)2

Total mol of Ba(OH)2 = mol Ba(OH)2reacted + mol Ba(OH)2 unreacted 0.012375 mol = 0.009375 mol + mol Ba(OH)2unreacted

mol Ba(OH)2 unreacted = 0.003 mol

mol of OH- = 2 x 0.003 mol = 0.006 mol

Molarity of OH- = 0.006 mol OH- total volume (L)

= 0.006 mol300 x 10-3 L

= 0.02 M OH-

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Example (2):

A 50.00 mL sample of aqueous Ca(OH)2 requires 34.66 mL of 0.0980 M HNO3 for neutralization. What is the concentration of Ca(OH)2.

Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(ℓ)

mol of HNO3 = 0.098 (mol/L) x 34.66x10-3 L = 3.4 x10-3 mol

mol of Ca(OH)2 = 3.4 x 10-3 mol HNO3 x 2 mol HNO3

1 mol Ca(OH)2

= 1.7 x 10-3 mol Ca(OH)2

molarity of Ca(OH)2= 1.7 x 10-3 mol 50 x 10-3 L

= 0.034 M

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3) Oxidation-Reduction (called Redox) Reactions:

For the following reaction:

2 Na(s) + Cl2(g) → 2 NaCl(s)

Na is a neutral atom react with Cl2 which is also a neutral compound to form NaCl which contain Na+ and Cl- ions.

Reactions like this one, which involves the transfer of electrons, are called oxidation-reduction (or redox) reactions.

Many important redox reactions occur in our daily life, e.g.:

Photosynthesis, combustion of sugar in our body,..etc.

To explain how electrons are transfer, the concept of oxidation states must be introduced.

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Oxidation StatesThe concept of oxidation states (or oxidation numbers) provides a way of keeping track of the electrons in oxidation reduction reactions.� need the rules for assigning oxidation states (memorize).� The sum of the oxidation states must be zero in compounds or equal the charge of the ion.� The following table summarized the rules for assigning the oxidation states

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The oxidation state of… summary examples

• elements in their standard states is zero

Element: 0 Na(s), O2(g) , Hg(ℓ)

•monoatomic ions are the same as their charge.

Monoatomic ions:Charge of ion

Na+ , Cl-, Ca+2

•Oxygen is assigned an oxidation state of -2 in its covalent compounds except as a peroxide (O2

-2, oxygen is -1, e.g. H2O2)

Oxygen : -2H2O , CO2 , CO

•hydrogen is assigned the oxidation state +1, except when it bonded to metals; LiH, NaH …, its oxidation state is -1

Hydrogen: +1 HCN, HCl, NH3

•fluorine is always –1 in its compounds.

Flourine: -1 HF, PF3

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The following figure shows oxidation numbers of familiar elements, arranged according to their position in the periodic table.

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Example:

Assign the oxidation states to each element in the following compounds.

� CO2 : O is -2, so -2 x 2 + C = 0 C = +4

� NO3- : O is -2, so (3 x -2)+ N = -1 N = +5

� H2SO4 : H is +1, O is -2, so [2x(+1)]+[(4x(-2)]+S =0

S = +6

� Fe2O3 : O is -2, so [3x(-2)]+ 2xFe = 0 Fe = +3

� Cr2O7 -2 : 2xCr +[7x(-2)]= -2 Cr = +6

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Oxidation-reduction reactions are characterized by the transfer electrons, so the oxidation states change.For the following reactions:1) 2 Na + Cl2 → 2NaCl

Oxidation state 0 0 +1 -1

2) CH4 + 2O2 → CO2 + 2H2O

Oxidation state

-4 +1 (each H) 0

+4 -2(each O)

+1each H

-2

Carbon undergoes a change in oxidation state from -4 in CH4 to +4 in CO2, 8 electrons (symbol e- stands for electrons) are lost

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CH4 → CO2 + 8 e-

Each oxygen change from an oxidation state 0 in O2 to -2 in H2O and CO2 , each atom gain 2 electrons. Since four oxygen atoms are involved , this is a gain of 8 electrons.

O2 + 8 e- → CO2 + 2 H2O

No change occurs in the oxidation state of hydrogen .

-4 +4

0 4 x (-2)= - 8

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We can now define some important terms:� Oxidation is the loss of electrons or an increase in

oxidation state - lose electrons. .� Reduction is the gain of electrons or a decrease in

oxidation state - gain electrons.� The substance that is oxidized is called the reducing agent.

Reducing agent gets oxidized. Loses electrons.

More positive oxidation state.� The substance that is reduced is called the oxidizing agent.

Oxidizing agent gets reduced. Gains electrons. More negative oxidation state.

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Example:For the following reactions, identify� Substance oxidized� Substance reduced� Oxidizing agent� Reducing agent1) Fe (s) + O2(g) → Fe2O3(s)

0 0 + 3 -2The oxidation state for Fe increase from 0 to +3. thus Fe is oxidized. The oxidation state for O decrease from 0 to -2. thus O is reduced. The oxidizing agent is O2.The reducing agent is Fe.

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2) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)

3) SO32- + H+ + MnO4

- → SO42- + H2O + Mn2+

The oxidation state for C increase from +2 to +4. thus C is oxidized. The oxidation state for Fe decrease from +3 to 0. thus O is reduced. The oxidizing agent is Fe.The reducing agent is CO.

The oxidation state for S increase from +4 to +6. thus S is oxidized. The oxidation state for Mn decrease from +7 to +2. thus Mn is reduced. The oxidizing agent is MnO4

-.The reducing agent is SO3

2-.

+3 -2 +2 -2 0 +4 -2

+4 -2 +1 +7 -2 +6 -2 +1 -2 +2

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Balancing oxidation-reduction equationsIt is difficult to balance redox equations by simple inspection, the technique used for balancing redox reactions called the half-reaction method. � All redox reactions can be thought of as happening in two halves.� One produces electrons - Oxidation half.� The other requires electrons - Reduction half.� Write the half reactions for the following.1) Na + Cl2 → Na+ + Cl-

2) Ce+4 + Sn+2 → Ce+3 + Sn+4

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1) Na + Cl2 → Na+ + Cl-

Na → Na+ oxidation

Cl2 → Cl- reduction

2) Ce+4 + Sn+2 → Ce+3 + Sn+4

Ce+4 → Ce+3 reduction

Sn+2 → Sn+4 oxidation

The method for balancing redox equations differ slightly depending on whether the reaction takes place in acidic or basic solution.

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Balancing Redox Equations in acidic solution� In aqueous solutions the key is the number of electrons

produced must be the same as those required.� For reactions in acidic solution an 8 step procedure.1. Write separate equations for the oxidation-reduction half reactions2. For each half reaction balance all reactants except H and O3. Balance O using H2O

4. Balance H using H+

5. Balance charge using e-

6. Multiply equations to make electrons equal

7. Add the half-reactions, and cancel identical species

8 Check that charges and elements are balanced.

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Examples:

Balance the following oxidation-reduction reactions occur in acidic aqueous solution.

1) MnO4- + Fe+2 → Mn+2 + Fe+3

1) Write the half reactions

MnO4- → Mn+2 Fe+2 → Fe+3

2) Balance each half-reaction:

MnO4- → Mn+2

-Mn is balanced

-balance oxygen by adding 4H2O to the write side of the equation.

- Balance hydrogen by adding 8H+ to the left side of equation. 8H+ + MnO4

- → Mn+2 + 4H2O

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-Balance the charge using electrons

8H+ + MnO4- → Mn+2 + 4H2O

+8 -1 → +2 0

+7 +2

equalize the charges by adding five electrons to the left side.

5e- + 8H+ + MnO4- → Mn+2 + 4H2O reduction

+2 +2

both the elements and charges are now balanced.

For the oxidation reaction: Fe+2 → Fe+3 + e-

+2 +3-1

+2 +2

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Equalize the electron transfer in the two half-reactions

5e- + 8H+ + MnO4- → Mn+2 + 4H2O

5 Fe+2 → 5 Fe+3 + 5e-

8H+ + MnO4- + 5 Fe+2 → Mn+2 + 4H2O + 5 Fe+3

check that elements and charges are balanced.

elements balance: 5Fe, 1Mn, 4O, 8H→ 5Fe, 1Mn, 4O, 8H

charge balance: 8(+1)+(-1)+5(+2)→(+2)+0+5(+3)

+17 → +17

the equation is balanced.

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2) Cu + NO3- → Cu+2 + NO(g)

3 / Cu → Cu+2 + 2e-

2 / 3e- + 4H+ + NO3- → NO + 2H2O

3Cu + 8H+ + 2NO3- → 3Cu+2 + 2NO + 4H2O

3) Mn+2 + BiO3- → Bi+3 + MnO4

-

2 / 4 H2O + Mn+2 → MnO4- + 8H+ +5e-

+2 +7

5/ 2e- + 6H+ +BiO3- → Bi+3 + 3H2O

+5 +3

2Mn+2 + 5BiO3- + 14H+ → 2MnO4

- + 5Bi+3 + 7H2O

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Balancing Redox Equations in basic solution

� Do everything you would with acid, but add one more step.

� Add enough OH- to both sides to neutralize the H+ � Makes water

Examples: Balance the following equation in basic solutions:

Cr(OH)3 + OCl- + OH- → CrO42- + Cl- + H2O

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Cr(OH)3 + OCl- + OH- → CrO42- + Cl- + H2O

2/ Cr(OH)3 + H2O → CrO42- + 5H++3e-

0 +3

3/ 2e- + OCl- + 2H+ → Cl- + H2O

+1 -1

2Cr(OH)3 + 3OCl- → 2CrO42- + 3Cl- +H2O +4H+ (acidic solution)

4OH- +2Cr(OH)3 + 3OCl- → 2CrO42- + 3Cl- +H2O + 4H+ + 4OH-

4OH- +2Cr(OH)3 + 3OCl- → 2CrO42- + 3Cl- +5H2O (basic solution)

6161

0.1327 mol KMnO4

1 Lx

5 mol Fe2+

1 mol KMnO4

x1

0.02500 L Fe2+x0.01642 L

= 0.4358 M Fe2+

5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O

Example: 16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of an acidic FeSO4 solution. What is the molarity of the iron solution: