Transcript Aqueous solutions and Net Ionic PPT

Reactions in aqueous solutions
• Aqueous solution:
– Solution in which water is the solvent
(dissolving agent).
• 3 major types of chemical
processes of aqueous solutions:
– Precipitation reactions
– Acid-base reactions
– Redox reactions
• Solution:
–Homogenous mixture of 2 or
more substances.
• Solvent:
–Dissolving medium, usually
present in greater quantity.
• Solute:
–The other substance(s) in the
solution.
• Electrolyte:
– A substance whose aqueous solution
forms ions; conducts electricity.
– Ionic compounds.
• Nonelectrolyte:
– Substance that does not form ions in
an aqueous solution; poor conductor.
– Molecular compounds.
• Ionic compounds in water:
– Dissociate into its component
ions.
dissolving_NaCl_probe.swf
Do not get to wrapped up in the difference between the terms ionization
and dissociation. Consider them to mean the same thing, the separation
of a substances ions.
Equations showing ionization or dissociation


NaCl (aq)  Na (aq)  Cl (aq)
2


CuCl2 (aq)  Cu (aq)  2Cl (aq)


3
Na3 PO4 (aq)  3Na (aq)  PO 4 (aq)
Molecular compounds in H2O
Molecular compounds – nonmetal + nonmetal
Structural integrity of molecule is usually
maintained meaning no ions form (C12H22O11)
Exception:
Some molecular solutes interact with water to
form ions. These would be electrolytes.
Examples: Acids  HCl, H2C3O2
Ammonia  NH3
dissolving_sugar_probe.swf
Strong Electrolytes
• Exists in solution completely or
almost completely as ions.
• All ionic compounds and a few
molecular compounds.(Ex: Strong Acids)
HCl (aq)
NaCl( s)

H

 Na
(aq)
(aq)

 Cl

 Cl
(aq)
(aq)
Weak Electrolytes
• Molecular compounds that produce a
small concentration of ions when
dissolved in H2O.
Ex: Acetic acid (HC2H3O2) only
slightly ionizes when dissolved in
water.
HC2H3O2(aq)
H+(aq)+ C2H3O2-(aq)
Weak acids are better conductors if they are dilute, as you will see in lab. Explain.
• Reactions that result in an
insoluble product.
• Insoluble:
–Substance with solubility less than
0.01 mol/L
–Water molecules cannot overcome
the attraction between the ions.
KI (aq) + Pb(NO3)2 (aq)  PbI2(s) + KNO3 (aq)
You must be able to determine whether a substance is soluble
in water by simple examination of the chemical formula.
To do so, you must memorize how specific
polyatomic ions act in water.
Not as hard as it sounds. We will focus mainly on
10 anions. This will give you the tools to predict the
solubility of many compounds.
Solubility of Ionic Compounds
• All acetates and nitrates are soluble in
water.
• All ionic compounds of alkali metals
and ammonium are soluble.
– (1A goes AWAY)
• Solubility rules are on your reference
sheet.
Soluble
Exceptions
Cl-
Ag+, Hg22+, Pb2+
Acetate ion C2H3O2-
None
Br-
Ag+, Hg22+, Pb2+
I-
Ag+, Hg22+, Pb2+
NO3-
None
SO42-
Sr2+, Ba2+, Hg22+, Pb2+
Insoluble
Exceptions
CO32OH-
PO43S2-
NH4+ and Group 1 metals
Group 1 metals and Ba2+,
Sr2+, Ca2+
NH4+ and Group 1 metals
NH4+ and Group 1 metals
and Ba2+, Sr2+, Ca2+
Equation Types
• Molecular
Pb( NO3 )2  2KI( aq)  PbI 2( s )  2KNO3aq) 
• Complete Ionic
2
( aq)
Pb

 2 NO( aq3 )  2K

( aq)
 2I

( aq)
 PbI 2( s )  2K
• Net ionic equation
2
( aq)
Pb
 2I

( aq)
 PbI 2( s )

( aq)

3( aq)
 2 NO
Ionic Equations


2  2 NO   2K   2I  
Pb(aq
PbI

2
K

2
NO
3(aq)
(aq)
2( s )
( aq)
3( aq)
)
(aq)
•Those ions that appear on both
sides of a complete ionic equation
are known as Spectator Ions.
•Net ionic equations do not
include spectator ions.
• Exchange Reactions
• Metathesis reactions
• Double displacement
• Double replacement
AX  BY  AY  BX
AgNO3( aq)  KCl( aq)  AgCl( s )  KNO3( aq)
Writing Net Ionic Equations
1) Write a balanced molecular
equation.
2) Rewrite the equation showing
ions of strong electrolytes only.
3) Identify and cancel all
spectator ions.
Acid-Base Reactions
Acids:
• Ionize in H2O, causes
increase in H+ ions.
• H+ ions are bare protons.
• Acids are proton donors.
All this acid
rain is
killing my
complexion!
• Monoprotic Acids: (HCl, HNO3)
–Acids that can only yield one H+ per
molecule upon ionization.
HCl  H+ + Cl-
Diprotic Acids: (H2SO4)
Ionization occurs in 2 steps.
H 2 SO4( aq)  H

4( aq)
HSO

( aq)
H

4( aq)
 HSO

( aq)
2
4( aq)
 SO
Only the first ionization is complete.
Is HF a weak or strong acid?
weak acid
Although it is a weak acid, this acid is
extremely reactive because of the F- ion.
Must be kept in special polypropylene
container because it eats through glass.
Used to etch glass.
Has caused major accidents in lab.
• Substances that increase the OH- when
added to water. (NaOH)
• NH3 is a base. In water it accepts an
H+ ion from HOH, leaving an OH- in
solution.
– NH3 is a weak electrolyte
– About 1% ionizes to form NH4+/OH-
Strong acids and bases
• Acids and bases that ionize
completely in solution are strong
acids and bases.
• Those that only ionize partially are
weak acids and bases.
• You must memorize these.
Strong Acids
Hydrochloric Acid –
HCl
Hydrobromic Acid –
HBr
Hydroiodic Acid –
HI
Nitric Acid –
HNO3
Sulfuric Acid –
H2SO4
Chloric Acid –
HClO3
Perchloric Acid –
HClO4
Strong Bases
All group 1 Metal Hydroxides
(LiOH, NaOH, KOH, RbOH, CsOH)
Heavy Group 2 Metal Hydroxides
Ca(OH)2, Sr(OH)2, Ba(OH)2
Once you memorize the strong acids and bases, you
will have enough information to determine if a
substance is a strong or weak electrolyte.
Example
problems:
KF
Na3PO4
NH3
CH3CH2OH
HCl
NO2
HC2H3O2
CH4
NH4Cl
CH3Cl
strong electrolyte
strong electrolyte
weak electrolyte
nonelectrolyte
strong electrolyte
nonelectrolyte
weak electrolyte
nonelectrolyte
strong electrolyte
nonelectrolyte
Acid + Base
Neutralization
• Products of a neutralization reaction have none
of the properties of an acid or a base.
• An acid reacts with a metal hydroxide to form a
salt plus water.
Neutralization Reactions
• Acid + Base (Metal Hydroxide)  Salt + Water
• HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
• H+ + Cl- + Na+ + OH-  Na+ + Cl- + H2O(l)
H+ + OH-  H2O(l)
Write the net ionic equation for the following
reaction. It might help to first write the molecular
equation, and then the complete ionic equation,
followed by the net ionic equation.
Potassium Hydroxide + Sulfuric Acid
• Ionic equation:
2K(aq)  2OH (aq)  2H (aq)  SO4(2aq)  2 HOH (l )  2K(aq)  SO4(2aq)
• Net Ionic equation:
H

( aq)
 OH

( aq)
 H 2O(l )
Neutralization Reaction of Weak Acid
*Remember, only strong electrolytes are written as ions.*
Acetic Acid + Sodium Hydroxide 
HC2H3O2(aq) + NaOH(aq)  NaC2H3O2(aq) + H2O(l)
Weak acid
strong base soluble salt
water
HC2H3O2 + Na+ + OH-  Na+ + C2H3O2- + H2O(l)
HC2H3O2(aq) + OH (aq)  C2H3O2 (aq)+ H2O(l)
Acid/Base Rx’s with gas formation
• Other bases besides OH- react
with H+ to form molecular
compounds.Two common
-2
-2
bases are CO3 and S .
• Carbonates and bicarbonates
react with acid to form CO2.
Hydrochloric acid + Sodium Sulfide 
2HCl (aq) + Na2S(aq)  H2S(g) + 2NaCl(aq)
2H+ (aq) + S2-(aq)  H2S(g)
Hydrochloric acid + Sodium Hydrogen Carbonate 
HCl (aq) + NaHCO3(aq)  NaCl(aq) + H2CO3(aq)
H2CO3(aq)  H2O(l) + CO2(g)
HCl (aq) + NaHCO3(aq)  NaCl(aq) + H2O(l) + CO2(g)
H+ (aq) + HCO3-(aq)  H2O(l) + CO2(g)
Reactions in which electrons are
transferred between substances
Use of Oxidation numbers in
determining redox reactions is
basically a bookkeeping method
for keeping track of electrons
You must be able to identify an oxidation-reduction
reaction. But first, we must learn the rules for
assigning oxidation #’s to different species.
Rules for oxidation numbers
1) Atoms in elemental form are 0.
2) Monatomic ion; charge of the ion is its oxidation
number.
3) Nonmetals; usually negative numbers.
a.) oxygen = -2 unless a peroxide = -1
b.) Hydrogen +1 with nonmetals, -1 with metals
c.) Halogens (-1) unless bonded to oxygen (+)
in a polyatomic ion (Ex: ClO3-; Cl = +5)
4) Sum of oxidation numbers must = 0
5) Most electronegative (furthest to right and up)
element gets a negative charge.
See pages 128 – 129 for more on this.
1) Atoms in elemental form are 0.
Examples
Ag
Pb
Cl2
O2
Oxidation # = 0 for 7 diatomic elements and for all
other elements when by themselves.
2) Monatomic ion-charge of the ion is its oxidation number.
Examples
AgCl
Ag = +1
PbI2
Pb = +2
Fe2O3
Fe = +3
Cl = -1
I = -1
O = -2
3) Nonmetals; usually negative numbers.
a.) oxygen = -2 unless a peroxide = -1
b.) Hydrogen +1 with nonmetals, -1 with metals
c.) Halogens (-1) unless bonded to oxygen (+)
in a polyatomic ion (Ex: ClO3-; Cl = +5)
PbO
H2S
KI
Examples
oxygen = -2
Na2O2
hydrogen = +1
NaH
iodine = -1
KIO2
oxygen = -1
hydrogen = -1
iodine = + 3
Determine Oxidation # of element red
element in each of the following:
MnO2
+4
Br2
0
KMnO4
HClO4
+7
+7
BrO2-
H2SO4
+3
+6
BrO3-
PO33-
+5
+3
CaH2
-1
SO42-
+6
Na2S
-2
Mg(NO3)2
+5
Again, oxidation reduction reactions occur
when there is a transfer of electrons from
one species to another in a reaction.
• If one reactant gains electrons
another must lose electrons.
• Reduction is always
accompanied by oxidation.
Oxidation-Reduction Reactions
• An atom that becomes more
positively charged is oxidized.
–This is due to loss of e-.
• The gain of electrons by an
atom is called reduction.
Two mnemonics for
remembering which substance
is undergoing oxidation and
which is undergoing reduction?
OIL -- RIG
Oxidation Involves Loss -- Reduction Involves Gain
“Leo the lion says Ger”
Loss of electrons oxidation -- Gain of electrons reduction
Many metals react with O2 in
the air to form metal oxides.
Metals lose electrons to oxygen.
2 Fe + O2  2 FeO
As Fe is oxidized (loses e-),
oxygen is reduced (gains e-).
Reduction is gain
2 Fe
+
oxidation
O2

2 FeO
reduction
Oxidation of metals by acids
and salts
• Reaction of a metal with
either an acid or metal salt
follows general form of:
A + BX
AX +B
• Single displacement reaction
+2+6-2
0
+2+6-2
0
CuSO4(aq) + Zn(s)  ZnSO4(aq) + Cu(s)
Reduced
Oxidized
What are the products?
What are the charges on each species?
What is oxidized and what is reduced?
For the following reactants:
1) Write the reaction that occurs.
2) Identify what is being oxidized and reduced.
Magnesium + Hydrochloric Acid
Aluminum + Cobalt(II) Nitrate
Mg(s) + HCl (aq)  MgCl2 (aq) + H2(g)
oxidation
reduction
Al(s) + Co(NO3)2 (aq)  Al(NO3)3(aq) + Co(s)
oxidation
reduction
Types of Redox Reactions
• Combination (synthesis)
• Decomposition
• Displacement
– hydrogen, metal, halogen
• Disproportionation (When an element is
simultaneously oxidized and reduced).
• Ex: H2O2  H2O + O2
Activity Series
• List of metals in order of decreasing
ease of oxidation.
• Alkali and alkaline earth metals are at
the top. (active metals)
• Gold, Silver, Platinum, and palladium
are considered to be (noble metals)
because they resist oxidation.
Using activity series to
predict reactions
• Activity series can be used to
predict reactions between metals
and metal salts or acids.
• Any metal listed on the series
can be oxidized by the ions of
elements below it on the list.
Using activity series of metals, which metals
from the list below can be oxidized by H+?
Ni
Al
Cu
Pb
Ag
Mg
Au
This should be a review before handing out All reaction types worksheet. Must edit all reaction type
worksheet. It has to many weak acids reacting with strong bases, which is confusing for students.
Make sure most have reactions (see note in folder on handout).
Chemical Reaction Types
• Decomposition
• Synthesis
• Single Replacement
• Precipitation
• Neutralization
• Combustion
Types of
Redox Reactions
Types of Double
Replacement Reactions
What are the 7 Diatomic Elements
H2 – hydrogen
N2 – nitrogen
O2 – oxygen
F2 - fluorine
Cl2 – chlorine
Br2 – bromine
I2 – Iodine
Synthesis
A + B  AB
Examples
H2 (g) + O2 (g)  H2O (g)
Mg (s) + O2 (g)  MgO (s)
Na (s) + Cl2 (g)  NaCl (s)
Decomposition
AB  A + B
Examples
NaCl (s)  Na (s) + Cl2 (g)
elec
KClO3 (s)  KCl (s) + O2 (g)
Single Replacement
A + BC  AC + B
Examples
Na (s) + HOH (l)  NaOH (aq) + H2 (g)
sodium replaces hydrogen in water
Cl2 (g) + NaBr (aq)  Br2 (l) + NaCl (aq)
chlorine replaces bromine in sodium bromide
Double Replacement (Metathesis)
AB + CD  AD + BC
Examples
AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
NH4Cl (aq) + NaOH (aq)  NH3 (g) + H2O (l) + NaCl (aq)
Blue color for the products represents the driving force
which allows the chemical reaction to occur.
Combustion
hydrocarbon + oxygen  carbon dioxide + water
Examples
CH4 (g) + O2 (g)  CO2 (g) + H2O (l)
C3H8 (g) + O2 (g)  CO2 (g) + H2O (l)
CH3OH (g) + O2 (g)  CO2 (g) + H2O (l)
CO2 + H2O
Neutralization Reactions
Strong Acid + Strong Base  Salt + Water
Example
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
H+
+
+
+ Cl + Na + OH  Na + Cl + H2O(l)
H+ + OH-  H2O
Concentration (Molarity)
• Concentration — the amount of solute per unit of solution.
• Molarity (M) — expresses concentration:
Moles of solute
(M ) 
Volume of solution in Liters
Calculate the molarity of a solution
that contains 20.0g copper(I) chloride
and has a total volume of 300.0 mL.
mol
M
L
Must have the
correct units.
Given: 20.0 grams CuCl; 300.0 ml solution
Need: Moles CuCl; L of solution
20.0 g CuCl
1 mol CuCl
99.0 g CuCl
300.0 mL
1 L
1000 mL
= 0.202 mol CuCl
= 0.3000 L
mol
M
L
0.202 mol CuCl
M 
0.3000 L
0.673 M CuCl
How many grams of NaCl are needed to
make 2.5 L of 0.20 molar solution?
Given: 0.20 M solution; 2.5 L solution
Need: grams of NaCl
mol (L)
(L) M 
L
Must calculate # of moles, and then convert it into grams.
mol = ML
Mol = ML
Mol = 0.20 mol NaCl x 2.5 L = 0.50 mol NaCl
L
0.50 mol NaCl
55.84 g NaCl
1
=
mol NaCl
28 g NaCl
Dilution of Stock Solutions
• Chemicals are purchased in concentrated form.
They need to be diluted for most lab use.
• Formula for dilution:
Mi Vi = Mf Vf
How much stock (12 M) HCl (aq) is
required to make 200.0 mL of 3M HCl (aq)?
Mi Vi = Mf Vf
Mi = 12 M
Vf = 200.0 mL
Mf = 3 M
Vi = ? mL
Must rearrange equation above to
solve for initial volume
.
Vi 
M fVf
Mi
(3.0M )( 200 .0ml )
Vi 
 50. mL
(12 M )
Measure 150 mL of water in a beaker. Slowly add 50.0
mL of 12 M HCl for a final volume of 200.0 mL.
Two things to note:
1) Always add concentrated acid to water, and not
the reverse to avoid unwanted splashing due to
the heat generated.
2) When diluting a solution, the amount of solute
doesn’t change, only the final volume.
If diluting a solution other than acids,
start with initial volume of concentrated
solution, and then dilute with distilled
water until you have the desired volume.
You try one!
We want to prepare 500. mL of 1.00 M acetic
acid from a 17.5 M stock solution of acetic acid.
What volume of the stock solution is required?
Mi Vi = Mf Vf
Mi = 17.5 M
Vf = 500. mL
Mf = 1.00 M
Vi = ? mL
M fVf
Vi 
Mi
(1.00 M )(500 .0ml )
Vi 
 28.6 mL
(17 .5M )
Pour 471.4 mL of distilled water into a beaker. Slowly
pour the 28.6 ml of acid into the water and swirl. Fill
the container with distilled water to 500. mL.
There may be times when you must consider the
concentration of ions in a solution.
(You must consider the subscripts for this)
MgCl2  Mg2+ + 2Cl-
In a solution of 0.25 M MgCl2 you have:
M of Mg2+ = 0.25 M
M of Cl- = 2 x 0.25 M = 0.50 M
What is the concentration of each ion in the following?
0.15 Na3P
M of Na+ = 0.45 M
M of P3- = 0.15 M
Titrations
• Determining the concentration of an unknown
solution.
• Use a 2nd solution of known concentration
(standard solution) that undergoes a reaction with
the unknown solution.
• Use the ratios in the balanced equation along with
the M = mol/L equation to determine molarity of
unknown.
• The point at which the two
solutions are stoichiometrically
equal is known as the
equivalence point.
–The reaction is complete and no
excess reactant is present.
–How do we know when this
occurs during the reaction?
• In acid base reactions dyes known
as indicators are used.
– Phenolphthalein is colorless in acid
solution, and pink in basic solution.
– End point is reached when a drop of
the base remains pink. There is no
acid for this drop to react with and
the solution is now basic.