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
H H
O
105°δ+ 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
HH
μ = 1.8 Dμ = 1.7 D
μ = 1.5 D
Solvation
Clustering of molecules of solvent aroundsolute:
+
H H
Oδ−
H H
Oδ−
hydration is specific termfor solvation when water
is solvent
Water can solvate both cations and anions
+
H H
Oδ−
H H
Oδ−
H H
O
δ+ δ+
H H
O
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 )
Na+
Cl-
Na+
Na+
Na+
Na+
Na+
Cl-
Cl- Cl-
Cl-
Cl-
Cl-Cl-Cl-
Na+
Na+Na+ Na+
H HO
HH
O
H HO
HH
O Cl-
H HO
HH
O
H HO
HH
O
Cl-+Na+( aq ) ( aq )
H2O
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:
O
CH3COH
O
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
1x
103 mL
1 Lx
= 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
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
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…)
Solubility Rules (Table 4.1 in text)
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+
Solubility Rules (Table 4.1 in text)
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–
(aq)
Arrhenius definitions of acids
dissolves in water to yield hydroxideions
YOH Y+ (aq) + OH–
(aq)
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
O
H
H
: H+
Monoprotic acids
A few examples:
HF, HCl, HBr, HI
HNO3
CH3COOH (or HC2H3O2)
Sulfuric acid is a diprotic acid
H2SO4 ( aq ) H+ HSO4-+( aq ) ( aq )
HSO4-
( aq ) H+ SO42-+( aq ) ( aq )
strong - 100% dissociation of 1st proton
weak - partial dissociation of 2nd proton
Sulfuric acid is a diprotic acid
Sulfuric acid is a diprotic acid
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
Phosphoric acid is a triprotic acid
Phosphoric acid is a triprotic acid
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
OH:
:H:N
H
H
H
O H
::
:
HN
H
H
H+
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-
OH- OH-
OH-
H+
H+
H+
H+
H+
H+
buret
H+
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)
H+
equivalence point
buret
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
Lx
L1000mlx
1mol H2SO4
2mol NaOHx
16.1ml=0.610mol NaOH
1Lx
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
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, . . . . .
Reference points for assigning oxidation numbers
(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
Reference points for assigning oxidation numbers
(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.
Reference points for assigning oxidation numbers
(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.
Reference points for assigning oxidation numbers
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.
Reference points for assigning oxidation numbers
(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.
Reference points for assigning oxidation numbers
(6) The sum of oxidation numbers in a neutral molecule must equal 0 .
K2CO3Example:
2(+1) = +2 3(-2) = -6
+4
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
Oxidation-Reduction Reactions
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
Oxidation-Reduction Reactions
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
Cl2 +
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)
1. Write the unbalanced equation in ionic form.
MnO4- + Fe2+ Mn2+ + Fe3+
2. Separate the equation into two half-reactions
Fe2+ Fe3+
MnO4- Mn2+
3. Balance each half reaction (except for O and H).
4. In acid solution, balance O by adding H2O and H by adding H+
Fe2+ Fe3+
MnO4- Mn2+8H++ 4H2O+
5. Balance the charges by adding electrons.
Fe2+ Fe3+ + 1e-
MnO4- Mn2+8H++ 4H2O++ 5e-
5 x
1 x
You need the same number of electrons on both sides of the equation.
(
( )
)
5. Balance the charges by adding electrons.
5Fe2+ 5Fe3+ + 5e-
MnO4- Mn2+8H++ 4H2O++ 5e-
5Fe2+ 5Fe3+ + 5e-
MnO4- Mn2+8H++ 4H2O++ 5e-
6. Add the half reactions
MnO4-
Mn2+
8H++
4H2O ++
+ 5Fe2+
5Fe3+
This is the balanced equation
Example
Balance the following equation for the reaction in acid solution. *
HNO3(aq) + H3PO3(aq) NO(g ) + H3PO4 (aq)
1. Write the unbalanced equation in ionic form.
2. Separate the equation into two half-reactions
NO3 - NO
H3PO3 H3PO4
H+ (aq) +NO3- (aq) + H3PO3(aq) NO(g ) + H3PO4aq)
3. Balance each half reaction (except for O and H).
4. In acid solution, balance O by adding H2O and H by adding H+
NO3 - NO
H3PO3 H3PO4
+ 2H2O4H+ +
H2O + + 2H+
5. Balance the charges by adding electrons.
NO3 - NO +
H3PO3 H3PO4
2H2O4H+ +
H2O + + 2H+
3e- +
-+ 2e-
5. Balance the charges by adding electrons.
NO3 - NO +
H3PO3 H3PO4
2H2O4H+ +
H2O + + 2H+
3e- +
+ 2e-
2 x
3 x
(
( )
)
5. Balance the charges by adding electrons.
2NO3 - 2NO +
3H3PO3 3H3PO4
4H2O8H+ +
3H2O + + 6H+
6e- +
+ 6e-
6. Add the half reactions
2NO3 - 2NO +
3H3PO3 3H3PO4
4H2O8H+ +
3H2O + + 6H+
6e- +
+ 6e-
6. Add the half reactions
2NO3 - 2NO +
3H3PO3 3H3PO4
4H2O2H+ +
3H2O + + 6H+
6e- +
+ 6e-
6. Add the half reactions
2NO3 - 2NO +
3H3PO3 3H3PO4
H2O2H+ +
3H2O + + 6H+
6e- +
+ 6e-
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
+ 2e-
2e- +
HS- + NO3- S + NO2
-
NO3- NO2
-
+5-2 0 +3
Example
Balance the following equation for the reaction in basic solution. *
HS- S
NO3- NO2
-H+ + + H2O
HS- + NO3- S + NO2
-
NO3-H+ +HS- + NO2- + H2O+ SOH- + + OH-
Example
Balance the following equation for the reaction in basic solution. *
HS- S
NO3- NO2
-H+ + + H2O
HS- + NO3- S + NO2
-
NO3-H+ +HS- + NO2- + H2O+ SOH- + + OH-
Example
Balance the following equation for the reaction in basic solution. *
HS- S
NO3- NO2
-H+ + + H2O
HS- + NO3- S + NO2
-
NO3-HS- + NO2- + H2O+ S + OH-H2O +
Example
Balance the following equation for the reaction in basic solution. *
HS- + NO3- S + NO2
-
NO3-HS- + NO2
- + S + OH-
Example
Balance the following equation for the reaction in acid solution. *
Fe(s) + HCl(aq) HFeCl4(aq) + H2 (aq )
H+ H2
Fe HFeCl44Cl- +H+ +
22e- +
+ 3e-( )2
( )3
Example
Balance the following equation for the reaction in acid solution. *
Fe(s) + HCl(aq) HFeCl4(aq) + H2 (aq )
H+ 3H2
2Fe 2HFeCl48Cl- +2H+ +
66e- +
+ 6e-
Example
H+ 3H2
2Fe 2HFeCl48Cl- +2H+ +
66e- +
+ 6e-
+ 3H22Fe 2HFeCl48Cl- +8H+ +
+ 3H22Fe 2HFeCl48HCl +
or