PRINCIPLES OF CHEMISTRY I

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PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 4
DR. AUGUSTINE OFORI AGYEMAN Assistant professor
of chemistry Department of natural
sciences Clayton state university
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CHAPTER 4 CHEMICAL REACTIONS IN SOLUTION
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SOLUTION
- A homogeneous mixture of two or more
substances Solvent - The substance present in
the greatest quantity Solute - The other
substance(s) dissolved in the solvent
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SOLUTION
- Solutions can exist in any of the physical
states Solid Solution - dental fillings, metal
alloys (steel), polymers Liquid Solution - sugar
in water, salt in water, wine, vinegar Gas
Solution - air (O2, Ar, etc. in N2), - NOx, SO2,
CO2 in the atmosphere
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AQUEOUS SOLUTION
- A solution in which water (H2O) is the
solvent NaCl solution solvent is H2O and solute
is NaCl Hydrophilic - Substances that dissolve
in water - Water loving (NaCl) Hydrophobic -
Substances that do not dissolve well in water -
Water fearing (hydrocarbons)
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AQUEOUS SOLUTION
- Ions makes aqueous solutions good conductors of
electricity - Solution conductivity indicates
the presence of ions
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AQUEOUS SOLUTION
Ionic Compounds - Form ions in aqueous solution
(dissociate into component ions) Example - NaCl
solution contains Na and Cl- ions NaCl(aq) ?
Na(aq) Cl-(aq) - Each ion is surrounded by
water molecules - Good conductor of electricity
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AQUEOUS SOLUTION
Solvation Process - Ions in aqueous solution are
surrounded by the H2O molecules - The O atom in
each H2O molecule has partial negative charge
(d-) - Attract positive ions - The H atoms have
partial positive charge (d) - Attract negative
ions - Cations and anions are prevented from
recombining - Ions disperse uniformly
throughout the solution (homogeneous)
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AQUEOUS SOLUTION
Molecular Compounds - Most molecular compounds
do not form ions in aqueous solution - The
molecules disperse throughout the solution -
Molecules are surrounded by H2O
molecules Example - Sucrose solution contains
neutral sucrose molecules - Each molecule is
surrounded by water molecules - Poor conductor
of electricity - A few molecular compounds form
ions in aqueous solution - HCl dissociates into
H(aq) and Cl-(aq) - HNO3 dissociates into H(aq)
and NO3-(aq)
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NONAQUEOUS SOLUTION
- A solution in which another substance other
than water is the solvent Examples Alcohol p
etroleum ether Pentane Carbon tetrachloride
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RATE OF DISSOLUTION
The rate at which solutes dissolve can be
increased by - Grinding or crushing solute
particles (size reduction) - Heating - Stirring
or agitation
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ELECTROLYTES
- Substances whose aqueous solutions contain ions
NaCl(aq) ? Na(aq) Cl-(aq) - Two
categories strong and weak electrolytes Strong
Electrolytes - Solutes that completely or nearly
completely ionize when dissolved in
water Salts NaCl, NH4Cl, KBr, NaNO3 Strong
acids HCl, HNO3, H2SO4 Strong Bases NaOH, KOH,
Ca(OH)2
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ELECTROLYTES
- Substances whose aqueous solutions contain
ions NaCl(aq) ? Na(aq) Cl-(aq) - Two
categories strong and weak electrolytes Weak
Electrolytes - Only a small fraction of solutes
ionize when dissolved in water (exhibit a small
degree of ionization) Weak acids acetic acid
(HC2H3O2), citric acid (C6H8O7) Weak bases
ammonia (NH3) methylamine, cocaine, morphine
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ELECTROLYTES
- Single arrow is used to represent ionization of
strong electrolytes H2SO4(aq) ? H(aq)
HSO4-(aq) - Ions have no tendency of recombining
to form H2SO4 - Double arrow is used to
represent ionization of weak electrolytes HC2H3O2(
aq) ? H(aq) C2H3O2-(aq) - This implies
reaction occurs in both directions - Chemical
equilibrium is when there is a balance in both
directions
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NONELECTROLYTES
- Substances whose aqueous solutions do not
contain ions Examples Many molecular
compounds Sucrose (C12H22O11) ethanol (C2H5OH)
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SOLUBILITY
- A measure of how much of a solute can be
dissolved in a solvent at a given temperature -
Units grams/100 mL Example Solubility of sugar
in water at 20 oC is 204 g/100 mL H2O Three
factors that affect solubility - Temperature -
Pressure - Polarity
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SOLUBILITY
Unsaturated Solution - More solute can still be
dissolved at a given temperature Saturated
Solution - No more solute can be dissolved at a
given temperature Supersaturated Solution - Too
much solute has temporarily been dissolved (more
than solute solubility) Precipitate - Solute
(solid) that falls out of solution
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SOLUBILITY RULES
The best way to determine the solubility of a
substance is by experiment - Most nitrates
(NO3-) are soluble - Most salts of alkali metals
(Group 1A), ammonium (NH4), acetates (C2H3O2-),
and perchlorates (ClO4-) are soluble - Most
salts containing Cl-, Br-, and I- are soluble
Exceptions salts of Ag, Hg22, Pb2
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SOLUBILITY RULES
The best way to determine the solubility of a
substance is by experiment - Most sulfates
(SO42-) are soluble Exceptions BaSO4, PbSO4,
Hg2SO4, SrSO4 - Most hydroxides (OH-) are
slightly soluble Hydroxides of Ba2, Sr2, and
Ca2 are marginally soluble - Most salts
containing S2-, CO32-, PO43-, CrO42- are
insoluble Exceptions salts of alkali metals
and NH4
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PRECIPITATION REACTIONS
- Reactions that result in the formation of an
insoluble product - The insoluble product
(solid) is known as the precipitate - These
products have very low solubility in water -
Attraction between the oppositely charged ions is
so strong that water molecules cannot separate
them - A solute is insoluble if less than 0.01
mol of the solute dissolves in 1 L of solvent
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PRECIPITATION REACTIONS
To predict solubility - Examine the reactants -
Identify the ions present - Predict the
products - Identify which are soluble and which
are insoluble
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PRECIPITATION REACTIONS
Example AgNO3(aq) KCl(aq) ? white
precipitate - Ions present Ag, NO3-, K,
Cl- - Possible combinations AgNO3, AgCl, KCl,
KNO3 - Predict products AgCl and KNO3 - KNO3
is soluble and AgCl is not AgNO3(aq) KCl(aq) ?
AgCl(s) KNO3(aq)
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IONIC EQUATIONS
- When all soluble strong electrolytes are shown
as ions - Chemical equation is balanced -
Soluble compounds (aq) are separated into ions
(only strong electrolytes) - Insoluble
compounds (s), liquids (l), and gases (g) are
NOT separated into ions
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IONIC EQUATIONS
Complete ionic equation - When all ions in both
reactants and products are shown AgNO3(aq)
KCl(aq) ? AgCl(s) KNO3(aq) Ag(aq)
NO3-(aq) K(aq) Cl-(aq) ? AgCl(s)
K(aq) NO3-(aq)
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IONIC EQUATIONS
Net Ionic Equation - When spectator ions are
cancelled from the complete ionic equation - Net
charge on reactant side must equal net charge on
product side Ag(aq) NO3-(aq) K(aq)
Cl-(aq) ? AgCl(s) K(aq) NO3-(aq) Ag(aq)
Cl-(aq) ? AgCl(s) - Some ions appear on both
reactant and product sides - These ions play no
direct role in the reaction - These ions are
called spectator ions
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IONIC EQUATIONS
Neutralization Reaction HCl(aq) NaOH(aq) ?
NaCl(aq) H2O(l) Complete 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)
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CONCENTRATION OF SOLUTIONS
- The amount of solute dissolved in a given
quantity of solution MOLARITY (M) - The number
of moles of solute per liter of solution
- A solution of 1.00 M (read as 1.00 molar)
contains 1.00 mol of solute per liter of solution
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CONCENTRATION OF SOLUTIONS

• Calculate the molarity of a solution made by
  dissolving 2.56 g of
• NaCl in enough water to make 2.00 L of solution
• - Calculate moles of NaCl using grams and molar
  mass
• Convert volume of solution to liters
• - Calculate molarity using moles and liters

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CONCENTRATION OF SOLUTIONS
After dissolving 1.56 g of NaOH in a certain
volume of water, the resulting solution had a
concentration of 1.60 M. Calculate the volume of
the resulting NaOH solution - Convert grams NaOH
to moles using molar mass - Calculate volume (L)
using moles and molarity
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CONCENTRATION OF IONS
Consider 1.00 M NaCl 1.00 M Na and 1.00 M
Cl- 1.00 M ZnCl2 1.00 M Zn2 and 2.00 M
Cl- 1.00 M Na2SO4 2.00 Na and 1.00 M SO42-
Square brackets are commonly used to represent
concentration The concentrations of Na and Cl-
above may be represented as Na 1.00 M and
Cl- 1.00 M
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CONCENTRATION OF IONS
Calculate the number of moles of Na and SO42-
ions in 1.50 L of 0.0150 M Na2SO4
solution 0.0150 M Na2SO4 solution contains 2 x
0.0150 M Na ions and 0.0150 M SO42- ions moles
Na 2 x 0.0150 M x 1.50 L 0.0450 mol
Na moles SO42- 0.0150 M x 1.50 L 0.0225 mol
SO42-
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DILUTION
Consider a stock solution of concentration M1 and
volume V1 If water is added to dilute to a new
concentration M2 and volume V2 moles before
dilution moles after dilution M1V1
M2V2 Calculate the volume of 3.50 M HCl needed
to prepare 500.0 mL of 0.100 M HCl (3.50 M)(V1)
(0.100 M)(500.0 mL) V1 14.3 mL
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CHEMICAL ANALYSIS (TITRATIONS)
Volumetric Analysis - Analysis by volume -
Acid-base titrations Gravimetric Analysis -
Analysis by mass - Determination of halides by
addition of silver nitrate Cl- AgNO3 ?
AgCl (white ppt) NO3- - Determination of
sulfates by addition of barium chloride BaCl2
SO42- ? BaSO4 (white solid) 2Cl-
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CHEMICAL ANALYSIS (TITRATIONS)

• Calculate the concentration of NaOH solution if
  24.50 mL of this
• base is needed to neutralize 12.00 mL of 0.225 M
  HCl solution
• - Write balanced equation and determine mole
  ratio
• - Calculate moles of HCl (convert mL to L)
• - Determine moles of NaOH
• Calculate molarity of NaOH

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CHEMICAL ANALYSIS (TITRATIONS)
NaOH HCl ? NaCl H2O 1 mol NaOH 1
mol HCl Volume HCl 12.00 mL 0.01200 L mol
HCl 0.225 M x 0.01200 mL 0.00270 mol mol
NaOH
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CHEMICAL ANALYSIS (TITRATIONS)

• How many grams of KOH are needed to neutralize
  25.00 mL of
• 0.250 M H2SO4 solution
• - Write balanced equation and determine mole
  ratio
• - Calculate moles of H2SO4
• - Determine moles of KOH
• - Calculate grams of KOH using molar mass

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CHEMICAL ANALYSIS (TITRATIONS)
2KOH H2SO4 ? K2SO4 2H2O 2 mol KOH 1 mol
H2SO4 mol H2SO4 0.250 M x 0.02500 L 0.00625
mol mol KOH 2 x 0.00625 mol 0.0125 mol