Download g moles molarity

Reactions in Aqueous Solutions
“A” students work
(without solutions manual)
~ 10 problems/night.
Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Water – the stuff of life and of the earth.
Reactions – carried out in water, either as a bucket or glass of water
or as adsorbed water onto surfaces.
Types of common reactions
precipitation
acid-base
oxidation-reduction reactions
Office Hours W – F 2-3 pm
All of above reactions involve IONS in
solution and are affected by
1. Charge of ions
2. Size of ions
3. Number of ions/volume
particles
mass → particles →
volume
g → moles → molarity
molarity ( M ) ≡
[ A] =
moles of solute
liters of solution
moles A
liters of solution
Example:
Solve either stepwise or chainwise.
Stepwise
If a bottle is labeled as concentrated hydrochloric acid (12.0 M) how many
Moles are there in 50.0 mL?
Concentrated
[HCl] = 12.0M
Moles in 50.0 mL
[ HCl] =
[ A] =
moles A
liters of solution
12.0molesHCl
L
⎡ 12.0molesHCl ⎤
x= ⎢
⎥⎦ 0.0500 L
L
⎣
L ⎤
= 0.0500 L
⎣ 10 mL ⎥⎦
x = 0.600molesHCl
( 50mL) ⎡⎢
3
12.0molesHCl xmolesHCl
=
L
0.0500 L
Sig Figs
Concentrated – relatively large (>?>6M)
Dilute – relatively small (<?<0.1 M)
1
Example:
Chainwise
If a bottle is labeled as concentrated hydrochloric acid (12.0 M) how many
Moles are there in 50.0 mL?
[HCl] = 12.0M= known
Moles in 50.0 mL= unknown
1L 12.0molesHCl
( 50.0mL) ⎡⎢ 3 ⎤⎥
= x
⎣ 10 mL ⎦
Suppose you need to react 0.25 mol of HCl with a second substance.
What volume of 12.0 M HCl would you use?
Need 0.25 mol HCl
Stepwise solution
⎛ 12.0molesHCl ⎞ 0.25molHCl
⎜
⎟ =
⎝
⎠
1L
xL
Chainwise solution
xL =
0.25molHCl
3
3
1L
⎡
1LL⎤ ⎤ ⎡⎤ 10 mL ⎤
⎡⎤⎡⎡ 10 1mL
.8333mL⎥
= 20
⎛ 12.0molesHCl ⎞ 0.25molHCl0⎢0.025
molHCl
.25
25
molHCl
.12
molHCl
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
L ⎦ ⎦ ⎣⎦ L ⎦
⎣ moleHCl⎣⎦12
⎜
⎟
moleHCl
⎣⎣12moleHCl
⎝
⎠
1L
xL =
0.25
L
12.0
L
x = 0.600molesHCl
Second substance
Vol 12.0 M HCl unknown
20.8333mL = 21mL
xL = 2.083x10−2 L
⎡ 10 3 mL ⎤
x mL = 2.083x10 − 2 L ⎢
⎥
⎣ L ⎦
Sig figs
x mL = 20.83mL = 21mL
Example. If you have 3.0 M PbCl2 what concentration of Cl- ions do you have?
Most ionic compounds dissolve completely in water to the individual ions.
NaCl( s) → NaCl( aq ) ⎯completely
⎯⎯⎯
⎯→ Na(+aq ) + Cl(−aq )
NaCl( s) ⎯completely
⎯⎯⎯
⎯→ Na(+aq ) + Cl(−aq )
+
−
PbCl2( s) → PbCl2( aq ) ⎯completely
⎯⎯⎯
⎯→ Pb(2aq
) + 2Cl( aq )
+
−
PbCl2( s) ⎯completely
⎯⎯⎯
⎯→ Pb(2aq
) + 2Cl( aq )
OJO!!!! – VERY IMPORTANT – Must conserve mass
- Equation must balance
+
−
PbCl2( s) ⎯completely
⎯⎯⎯
⎯→ Pb(2aq
) + 2Cl( aq )
Treat as conversion factors
−
−
⎡ 3.0molesPbCl2 ⎤ ⎡ 2molesCl ⎤ ⎡ 6.0molesCl ⎤
⎥
⎢
⎥ ⎢ 1molePbCl ⎥ = ⎢
1L
1L
⎣
⎦⎣
⎣
⎦
2 ⎦
[Cl ] = 2[ PbCl ]
−
2
The concentration of Cl- ion is double that of PbCl2
[ Cl ] = 2( 3.0) = 6.0 M
−
2
Another example:
Write the chemical equation for the solution of ammonium phosphate
NH 4 + PO43−
( NH 4 ) 3 PO4 ( s )
We memorized these ions
Molecular charge nuetrality requires
These are the most common charges of
Transition metal ions and aluminum
completely
((NH
⎯
NH44))33PO
PO
⎯⎯⎯⎯
⎯⎯⎯⎯
⎯→
→ 3 NH4+( aq ) + PO43(−aq )
4 (4s()s )⎯
completely
1:
stoichiometry
3:
1
+
+
⎡ molesPO4 ( s ) ⎤ ⎡ 3molesNH 4 ( aq ) ⎤ ⎡ molesNH 4 ( aq ) ⎤
=
⎢
⎥
⎢
⎥
⎢
⎥ 1molePO
L
L
⎥⎦
⎣
⎦ ⎢⎣
4( s) ⎥
⎦ ⎣⎢
3+
3+
2+,
4+
3+ 2+
2+
2+ 2+ 2+ 2+
[ PO ]3 = [ NH ]
3−
4
1+ 2+
+
4
2+
What is the concentration of ammonium in solution compared to phosphate?
Example Give concentration in molarity of each ion in 0.080 M zinc phosphate
Zn3 ( PO4 ) 2 ( s ) → 3Zn(2aq+ ) + 2 PO43(−aq )
Stoichiometry
1:
3:
Water – the stuff of life and of the earth.
Reactions – carried out in water, either as a bucket or glass of water
or as adsorbed water onto surfaces.
21
Types of common reactions
⎡ 0.080molesZn3 ( PO4 ) ⎤ ⎡ 3molesZn 2 + ⎤
2
⎢
⎥⎢
⎥ = 0.24molesZn 2 +
L
⎢⎣
⎥⎦ ⎢⎣ 1moleZn3 ( PO4 ) 2 ⎥⎦
⎡ 0.080molesZn3 ( PO4 ) ⎤ ⎡ 2molesPO 2 − ⎤ 016
. molesPO42 −
2
4
⎢
⎥⎢
⎥=
. MPO43−
= 016
L
L
⎢⎣
⎥⎦ ⎢⎣ 1moleZn3 ( PO4 ) 2 ⎥⎦
precipitation
acid-base
oxidation-reduction reactions
Which Rules Apply?
3
General
Chemistry
Low Charge Density
Intermediate Charge Density
High Charge Density
Who precipitates and who stays soluble?
Depends upon who reacts with whom?
FITCH Rules
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
C1. It’s all about charge
C2. Everybody wants to “be like Mike”
⎛ qq ⎞
C3. Size Matters
⎟
E = k⎜
⎝r +r ⎠
C4. Still Waters Run Deep
C5. Alpha Dogs eat first
1 2
No
NO3Group 1 cations (1+)
Clean
Cl-
SO42-
Oh
Card me
OH-
CO32-
PleeeeaSe!!
S2-
PO43-
Weak Electrostatic
Interaction = Soluble
BaSO4 Mg(OH)2
Group 2 cations (2+)
Transition metal cations
(usually 2+)
Socks
AgCl
Strong Electrostatic Interaction
results in precipitation
el
1
2
Example
What happens when a solution of CuSO4 is mixed with a solution of NaNO3?
What happens when a solution of Na2CO3 is mixed with a solution of CaCl2?
First assume everbody goes into solution or already in solution
Identify who is charge dense and who is not charge dense
High/high always precipitates
Low/low never precipitates
Low/High generally does not precipitate
No Clean Socks
2+
2−
+
−
aq
4 aq
aq
3aq
Group 1
Transition
No Clean Socks
Metal ion
CuSO4 ( aq ) + NaNO3( aq ) → Cu
+ SO
+ Na + NO
Extremes
1. Low -Low charge density ion interactions - weak electrostatic energy Stay soluble
2. High -High charge density ion interactions - strong electrostatic energy Precipitate
3. High- Intermediate charge density ion interactions – generally strong electrostatic energy – precipitate
In Between
1. Low – High charge density ion interactions – generally weak electrostatic energy: Soluble with exceptions
2. Low – Intermediate charge density ion interactions – generally weak electrostatic energy: Soluble with
exceptions
3. Intermediate – Intermediate – charge density ion interactions – generally weak electrostatic energy: soluble
with exceptions
Example.
What happens when a solution of CuSO4 is mixed with a solution of NaNO3?
What happens when a solution of Na2CO3 is mixed with a solution of CaCl2?
Rx
Oh Card me PleaSe
+
2+
Na 2 CO3( aq ) + CaCl2 ( aq ) → 2 Na aq
+ CO32aq− + Ca aq
+ 2Claq−
Group 1
Group 2
Group 17
+
2+
+
2 Na aq
+ 2Claq− + CO 2 − 3( aq ) + Ca aq
→ CaCO32s− + 2 Na aq
+ 2Claq−
no reaction
4
Chemistry
General
FITCH Rules
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
C1. It’s all about charge
C2. Everybody wants to “be like Mike”
⎛ qq ⎞
C3. Size Matters
⎟
E = k⎜
⎝r +r ⎠
C4. Still Waters Run Deep
C5. Alpha Dogs eat first
Example.
What happens when a solution of CuSO4 is mixed with a solution of NaNO3?
What happens when a solution of Na2CO3 is mixed with a solution of CaCl2?
Oh Card me PleaSe
+
aq
2+
Na 2 CO3( aq ) + CaCl2 ( aq ) → 2 Na + CO32aq− + Ca aq
+ 2Claq−
Group 1
1 2
Do not write redundant (on both sides) ions
Write NET ionic equation
2
2+
CO 2 − 3( aq ) + Ca aq
→ CaCO32s−
Stoichiometry is important in these reactions
Stoichiometry is important in these reactions
Example : Consider the net ionic equation for the reaction that
occurs when solutions of Cu(NO3)2 and NaOH are mixed. What
volume of 0.106 M Cu(NO3)2 solution is required to form 6.52 g
of solid Cu(OH)2?
1. Check for charge dense ions that can precipitate
2. Write a net ionic reaction which excludes spectators (low charge dense ions)
3. Count total moles of all precipitating reactants
Cu(2+aq )
OH(−aq )
NO3(− aq )
Na
Charge Dense Transition metal
Cu(2+aq ) + OH (−aq ) + → CuOH x ( s)
Oh, Card me Please – charge dense
2+
( aq )
No Clean Socks – not charge dense
+
( aq ) Group – not charge dense
Cu
+ 2OH(−aq ) + → Cu( OH ) 2( s)
Group 17
+
2+
+
2 Na aq
+ 2Claq− + CO 2 − 3( aq ) + Ca aq
→ CaCO32s− + 2 Na aq
+ 2Claq−
el
1
Group 2
Example : Consider the net ionic equation for the reaction that
occurs when solutions of Cu(NO3)2 and NaOH are mixed. What
volume of 0.106 M Cu(NO3)2 solution is required to form 6.52 g
of solid Cu(OH)2?
−
2+
Cu( aq ) + 2OH( aq ) + → Cu( OH ) 2( s)
1:
2:
1
⎡ 1moleCu(OH ) 2 ⎤
−2
⎥ = 6.6823x10 moleCu(OH ) 2
2⎦
⎣
[6.52 gCu(OH ) ]⎢ 97.57 gCu(OH )
2
⎡ moleCu ⎤ ⎡ moleCu( NO3 ) 2 ⎤
= 6.6823x10 − 2 moleCu( NO3 ) 2
6.6823x10 − 2 moleCu(OH ) 2 ⎢
⎥⎢
moleCu ⎥⎦
⎣ moleCu(OH ) 2 ⎦ ⎣
⎡ 1LCu( NO3 ) 2
⎤
−1
6.6823x10 − 2 moleCu( NO3 ) 2 ⎢
⎥ = 6.304133849 x10 L
0106
.
moleCu
(
NO
)
⎣
3 2⎦
? = 0.630L
5
Stoichiometry is important in these reactions
Stoichiometry is important in these reactions
Example : Consider the net ionic equation for the reaction that
occurs when solutions of Cu(NO3)2 and NaOH are mixed. What
volume of 0.106 M Cu(NO3)2 solution is required to form 6.52 g
of solid Cu(OH)2?
Cu 2+ + 2OH − + → Cu( OH )
( aq )
2( s)
( aq )
Chainwise solution
52 gCu(OH ) ]⎡ 1moleCu(OH )
[[66..52
gCu(OH ) ]⎢
97.57 gCu(OH )
2
2
⎤ ⎡ moleCu ⎤ ⎡ moleCu( NO3 ) 2 ⎤ ⎡ 1LCu( NO3 ) 2
⎤
⎢
⎥=?
⎥⎢
⎥⎢
moleCu ⎥⎦ ⎣ 0106
. moleCu( NO3 ) 2 ⎦
2 ⎦ ⎣ moleCu( OH ) 2 ⎦ ⎣
2
⎣
? = 6.304133849 x101 L
? = 0.630L
Example: When aqueous solutions of sodium hydroxide and iron(III)
nitrate are mixed, a red gelatinous precipitate forms. Calculate the
mass of precipitate formed when 50.00 mL of 0.200 M NaOH and
30.00 mL of 0.125 M Fe(NO3)3 are mixed
1. Check for charge dense ions that can precipitate
2. Write a net ionic reaction which excludes spectators (low charge dense ions)
3. Count total moles of all precipitating reactants
Fe(III) Fe(3+aq ) + OH(−aq ) + → Fe(OH ) x ( s) Fe 3+ + 3OH − + → Fe(OH )
3( s )
( aq )
( aq )
OHCharge balance both sides of equation
⎡ 1L ⎤ ⎡ 0.200moleNaOH ⎤
3
⎥⎦ = 0.0100moleNaOH
LNaOH
⎣ 10 mL ⎥⎦ ⎢⎣
[50mLNaOH ] ⎢
⎡ 0125
. moleFe( NO3 ) 3 ⎤
⎥ = 0.00375moleFe( NO3 ) 3
LFe( NO3 ) 3
⎣
⎦
[30.00mLFe( NO ) ]⎡⎢⎣ 101LmL ⎤⎥⎦ ⎢
3 3
3
Stoichiometry is important in these reactions
Water – the stuff of life and of the earth.
Example: When aqueous solutions of sodium hydroxide and iron(III)
nitrate are mixed, a red gelatinous precipitate forms. Calculate the
mass of precipitate formed when 50.00 mL of 0.200 M NaOH and
30.00 mL of 0.125 M Fe(NO3)3 are mixed
Moles present
Fe 3+
OH-
0.00375
0.0100
1:
Fe
3:
3+
( aq )
+ 3OH
1
−
( aq )
0.00375
0.0100/3 = 0.00333
Smallest #
is answer
1moleFe(OH ) 3 ⎤ ⎡ 10687
. gFe(OH ) 3 ⎤
3+
⎥ ⎢ 1moleFe(OH ) ⎥ = 0.356 gFe(OH ) 3
1
moleFe
⎣
⎦⎣
3 ⎦
[ 0.00333moleFe ]⎡⎢
3+
could make
moles product
Reactions – carried out in water, either as a bucket or glass of water
or as adsorbed water onto surfaces.
Types of common reactions
precipitation
acid-base
oxidation-reduction reactions
Predicting
who gets the proton (acid-base)
depends on most charge dense of
competing reactions
+ → Fe(OH ) 3( s)
6
Acid-Base Reactions
Acid is a sour taste
Foliate papillae are situated on the edge of the tongue slightly anterior of the
circumvallate line. They are predominantly sensitive to sour tastes. Innervated by
the glossopharyngeal (lXth cranial) nerve. On average 5.4 foliate papillae per side
of the tongue, 117 taste buds per foliate papillae, total = 1280 foliate taste buds
per tongue.
Fatty acids
and oils fill
your skin.
Carboxylate group – low charge
density
Bases dissolve
Can be “bumped” by charge
Dense OH-
These
compounds
making
soaps (feel
slippery)
+ 3NaOH
In 1849 a Harvard chemistry professor, John Webster argued with a colleague, Dr.
Parkman, about money he owed the colleague for some dinosaur bones. He killed his
fellow Professor and attempted to hide the crime by dissolving the body in a vat
Of “lime” (Ca(OH)2) and then burning the remaining bones.
(Or KOH)
saponification
+ 3 CH3(CH2)14CO2Na
glycerol
A crude soap
7
Low Charge Density
Intermediate Charge Density
High Charge Density
Who gives up and who holds onto a proton?
Chemical Definition of Acid/Base
No
Svante Arrhenius
H+
Socks
SO42-
Low Charge Density
Anions Give up protons
X- is an anion who is
partnered with the
cation, H+
Oh
Card me
OH-
CO32-
H2O
High Charge Density
Anions hold onto protons
PleeeeaSe!!
PO43-
S2-
WEAK acids
STRONG acids
HX aq → Haq+ + X aq−
Acid: HX
Strong
acid
Cl-
NO3-
An acid is a species that produces proton ions in solution
Solution
Clean
Weak to
Intermediate
HX aq → Haq+ + X aq−
Weak
acid
Completely falls apart
→
HX aq ← Haq+ + X aq−
X- tries to hold onto some protons
Double arrow
In a STRONG acid the anion can not provide enough electrostatic attraction to
hold onto the H+ (proton)
In a WEAK acid the anion is very charge dense and holds onto the H+
Another very common type of Weak Acid - ….COOH
Some Weak Acids with COO- functional group (Carboxylate)
Condensed Structural Formula gives a clue as to the
Important Functional Group
Blue represents negative charge
causes potential field
HCOOH
What do you see?
HCOOH
Is the negative charge fully localized?
O
O
Formic Acid
(Ants)
O
Acetic Acid
Vinegar
HOAc (OAc = acetate)
OH
H3C
Butanoic Acid
O
O
H
C
H
Dissociate temporarily
H
H
Proton actually resides
Somewhere between
Two highly negatively
Charged oxygen
OH
H3C
OH
HCOOH
CH3COOH
Will Be Used Over and
Over In Example Problems
CH3CH2CH2COOH
Different modeler
Gave opposite color scheme (Red = large neg charge)
http://academic.reed.edu/chemistry/alan/ED/JCE/figlist.html
8
Low Charge Density
Intermediate Charge Density
High Charge Density
Who gives up and who holds onto a hydroxide?
Chemical Definition of Acid/Base
No
Svante Arrhenius
MOH → OH + M
Solution
Strong
acid
H+
+
aq
Weak
acid
Strong
base
Socks
SO42-
Oh
Card me
OH-
CO32-
PleeeeaSe!!
PO43-
S2-
WEAK acids
STRONG acids
Group 1 cations (1+)
Strong
Bases
Base:
Acid: HX
Cl-
NO3-
An acid is a species that produces proton ions in solution
A base is a species that produces hydroxide ions in solution
−
aq
Clean
Group 2 cations (2+)
Weak
base
Weak bases are produced by an alternative manner
Where M+ is some cation
http://core.ecu.edu/phys/flurchickk/AtomicMolecularSystems/molecularStructures/molecularStructures.html
net neutral Molecule ammonia, NH3
Is the charge evenly distributed?
Is there a region of high charge
Density?
In this model blue is negative
charge causing a potential field
and red is Positive charge
opposite potential field
R + H2 O → RHaq+ + OH aq−
R + H2 O → RHaq+ + OH aq−
Write reaction to show the unbonded electrons on R (where R is N in ammonia)
H 2 O+ : R → OHaq− + RHaq+
Write the reaction to show that electrons on R are attracted to proton on water
HO ⋅ ⋅ H + : R → OH aq− + RH aq+
NH 3 + H2 O →← NH 4+ + OH −
Nitrogen typically contains a set of unbonded electrons
which give Large charge density near the N atom
Compounds in which H on NH3 has been replaced by –C groups react
Similarly
NH3
CH3NH2
R
C2H5NH2
(C2H5)2NH
C6H5NH2
9
Equations for ACID/BASE Reactions
NH3
NH3 + H 2 O →← NH 4+ + OH −
In order to KNOW How MUCH acid or base is present we
often react acids and bases together
CH3NH2
CH 3 NH 2 + H2 O →← CH 3 NH 3+ + OH −
(CH ) NH + H O (CH )
3
2
→
←
3 2
NH 2+ + OH −
Three types of reactions:
Three types of reactions:
Strong Acid + Strong Base
Strong Acid + Weak Base
Weak Acid + Strong Base
SA/SB
SA/WB
WA/SB
Chemists Are Lazy Notation
(CH3)2NH
Weak Acid Strong Base Reaction
Strong Acid Strong Base Reaction
WA/SB
SA/SB
SA
SB
HNO3( aq ) → H
+
( aq )
−
3( aq )
+ NO
NaOH( aq ) → Na (+aq ) + OHaq− )
H(+aq ) + OH (−aq ) → H2 O
→
HX aq ← Haq+ + X aq−
WA
Do not react
(Low Charge Density)
Spectators
Reactants are H+ and OH-
Net ionic equation SA/SB
Remember: double
arrow means
X- is charge dense
and tries to hold
H+
SB
NaOH( aq ) → Na (+aq ) + OHaq− )
Add
These 2 rx
Spectator
H aq+ + OHaq− → H 2 O
Why is there a single arrow?
We are indicating that the protons and hydroxides we added as HNO3 and NaOH
Are not going back to NO3- and Na+ . Proton and hydroxide must find each other and
react
HX aq + OHaq− → H 2 O + X aq−
Net Ionic equation SB/WA
HFaq + OH aq− → H2 O + Faq−
example
10
Strong Acid Weak Base Reaction
WA/SB
Remember: double
arrow means an
incomplete
reaction
→
Raq + H2 O ← OHaq− + RHaq+
WB
SA
HX → H aq+ + X aq−
add
Ous = least number of oxygen
NaOH; Na+ is group 1
HNO3 is nitric acid so
nitrous acid would be HNO2
Low charge density cation
Na+ can not hold onto OH-
NO3-
Strong base
Low charge density anion
Higher charge density anion
Probability: wants to hold onto H+
Probability: weak acid
HX ( aq ) + OH
−
( aq )
→ X
Net Ionic Eq
Strong Acid
Strong Base
SA/SB
H+, OH-
H (+aq ) + OH(−aq ) → H2 O
Weak Acid
Strong Base
WA/SB
HX, OH- HX ( aq )
+ OH (−aq ) → X (−aq ) + H2 O
Strong Acid
Weak Base
SA/WB
H +, R
R( aq ) + H(+aq ) → HR(+aq )
example
Example : Write a net ionic equation for each of the
following reactions in dilute water solution
Nitrous acid with sodium hydroxide
vs
Reacting Species
Net Ionic equation SB/WA
NH3aq + Haq+ → NH4+aq
NO2-
Reactants
Spectator
H aq+ + OHaq− → H 2 O
Raq + Haq+ → RHaq+
Table
Reaction is WA/SB
−
( aq )
+ H2 O
HNO2 ( aq ) + OH(−aq ) → NO2−( aq ) + H2 O
Example : Write a net ionic equation for each of the
following reactions in dilute water solution
Ethylamine (CH3CH2NH2) with perchloric acid HClO4
CH3CH2NH2 is an amine
No Clean Socks
Cl- and ClO4-
N on amine has local
negative charge
perchlorate ion is 5 atom with single
charge, electrons spread out over oxygen
implies it is low charge dense, can’t hold
N reacts with water to
proton.
steal a proton and leave
Strong acid
a hydroxide
Reaction is SA/WB
Weak Base
R( aq ) + H (+aq ) → HR(+aq )
CH3CH 2 NH 2 + H(+aq ) → CH3 CH 2 NH 3+( aq )
11
Example : Write a net ionic equation for each of the
following reactions in dilute water solution
Hydrobromic acid (HBr) with potassium hydroxide
Br is group 17
K is group 1
Group 17 are singly charged
Group 1 are singly charged
Br is one charge/1 atom
K is one charge/1 atom
HBr = strong acid
KOH= strong base
Knowing HOW MUCH stuff is present
How acidic is vinegar?
Is tasting the way to go?
Reaction is SA/SB
H (+aq ) + OH(−aq ) → H2 O
Types of problems:
At Equivalence Point
Solve for volume
Solve for Molarity
Solve for Mass
Solve for moles
[ A] =
molarity
nA
VA
[ A] =
nA
VA
m A = n A MM A
m A = n A MM A
Moles of acid
mass
Stoichiometry
n H ,eq pt = nOH ,eq pt
Moles protons
Volume
Na+ - weak charge density = Strong base
Acetic Acid CH3COOH = weak acid
HX ( aq ) + OH(−aq ) → X (−aq ) + H 2 O
[
]
aq
Moles of Base
moles
⎛ 1 ⎞
mB ⎜
⎟ = nB
⎝ MM B ⎠
0.8333mol
= 0.833 M
L
+ OH(−aq ) → CH 3COO(−aq ) + H 2 O
1:
nB
[ B]
n H ,eq pt = nOH ,eq pt
25.0 mL 0.500 M NaOH
Titrate
Vinegar = Acetic Acid
15.0 mL Acetic Acid
Molarity?
H CH3 COO
Moles OH
VB =
Example : In a titration it is found that 25.0 mL of 0.500 M NaOH is
required to react with a 15.0 mL sample of vinegar. What is the
molarity of acetic acid in the sample?
1;
[[
]]
1:
1
CH 3COO
COOequivalencepo
equivalencepoint
mole⎞⎞
intmole
moleNaOH ⎞⎛⎛ H CH
⎛ 1L ⎞ ⎛ 0.500moleNaOH
⎞ ⎟
HCH3 COOequivalencepo
int mole
3
25mLNaOH
mLNaOH )⎛⎛⎜ 11LL3 ⎞⎞⎟⎟⎛⎛⎜⎜00..500
500moleNaOH ⎞⎞⎟⎟⎛⎜⎜⎜H
−3
(( (25
3
.152
( 25mLNaOH
LNaOH ⎟⎠⎠⎜⎜⎝ NaOHequivalencepo
NaOHequivalencepoint
mole ⎟ ⎟⎟⎟⎠ = 152
intmole
1033 mL
mL⎟⎠⎠⎜⎝⎝ 11LNaOH
25
mLNaOH )) ⎜⎜⎝)⎝10
x10 −molH
molHCH
CH3COO
. x10
3 COO
1LNaOH ⎠ ⎝ NaOHequivalencepo int mole ⎠ ⎠ =
10 mL ⎠ ⎝
0.015 L
0.015 LH CH 3COO [
[
]
[[
]]
12
Example : The principal ingredient of certain commercial antacids is
calcium carbonate, CaCO3. A student titrates an antacid tablet
weighting 0.542 g with hydrochloric acid; the reaction is
Example : The principal ingredient of certain commercial antacids is
calcium carbonate, CaCO3. A student titrates an antacid tablet
weighting 0.542 g with hydrochloric acid; the reaction is
CaCO3( s) + 2 H(+aq ) → Ca (2aq+ ) + CO2 ( g ) + H2 O
CaCO3( s) + 2 H(+aq ) → Ca (2aq+ ) + CO2 ( g ) + H2 O
If 38.5 mL of 0.200 M HCl is required for complete reaction, what is
the percentage of CaCO3 in the antacid table?
If 38.5 mL of 0.200 M HCl is required for complete reaction, what is
the percentage of CaCO3 in the antacid table?
⎡ 0.3853gCaCO3( s) ⎤
⎤
.
stoichiometry
⎢ 0.542 gtablet
⎥ 100 = 71097%
100
⎢ gtablet ⎥
⎣
⎦
⎣
⎦
711%
.
Want: ⎡ gCaCO
Before we start
1. Problem is slightly unusual
because of gas production
CO32aq− + H(+aq ) → HCO3−aq1
H2 O
H+
OHaq− + CO2 g
http://academic.reed.edu/chemistry/alan/ED/JCE/figlist.html
Notice the “pinch point”
Where there is little electron density =
Weak bond
What do you predict happens?
CO32-
Aqueous reactions with
In presence of protons result in gas
3( s )
Need g CaCO3
+
⎡ 1L ⎤ ⎡ 0.200molesHCl ⎤ ⎡ 1moleH ⎤ ⎡ 1moleCaCO ⎤ ⎡ 100.09 gCaCO ⎤
⎡ 1L ⎤ ⎡ 0.200molesHCl ⎤
3
3
[ 38
38..55mLHCl
mLHCl ] ⎢ 3 mL ⎥ ⎢
⎥⎦ ⎢ moleHCl ⎥ ⎢⎣ 2moleH + ⎥⎦ ⎢ moleCaCO ⎥ = gCaCO3
⎣ 10
1LHCl
⎦⎣
⎣
⎦
3 ⎦
⎣
FITCH Rules
Chapter 4: Reactions in Aqueous Solutions
General
Water – the stuff of life and of the earth.
Reactions – carried out in water, either as a bucket or glass of water
or as adsorbed water onto surfaces.
Why you must spend taxes
to maintain not to build
Chemistry
Types of common reactions
precipitation
acid-base
oxidation-reduction reactions
Moles HCl
equivalence pt
Need Moles CaCO3
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
C1. It’s all about charge
C2. Everybody wants to “be like Mike”
⎛ qq ⎞
C3. Size Matters
⎟
E = k⎜
⎝r +r ⎠
C4. Still Waters Run Deep
C5. Alpha Dogs eat first
1 2
el
1
2
13
Net ionic reaction (excludes spectators)
zinc reduces
the protons
= a reducing agent
Zn( s) + 2 H(+aq ) → Zn(2aq+ ) + H2 ( g )
protons
oxidize the zinc
= oxidizing agent
For convenience
split
Zn( s) → Zn(2aq+ ) + 2e
2 H (+aq ) + 2e → H2 ( g )
Oxidation (loss of electrons)
increase in positive chargec
Reduction (gain of electrons)
(reduction in the positive
charge on the ion)
VERY IMP #1. - can’t really have 1/2 reaction
because the electrons need to go somewhere!!!!
OIL - Oxidation is Losing Electrons
RIG
Reduction is Gaining Electrons
They are TOO
Charge dense
To exist in solution
VERY IMP #2 - no net change in number of electrons
(corollary of VERY IMP #1 - if we “lose” 2 need to “gain” 2
Writing the Net Ionic Reaction of
an Oxidation-Reduction Reaction
Track the oxidation numbers of the various species
Writing the Net Ionic Reaction of
an Oxidation-Reduction Reaction
Track the oxidation numbers of the various species
Oxidation Numbers
Oxidation Numbers
1. Ox # of an element in an elementary substance is zero
O2 Ox# = 0
Cl2
Ox# Cl =0;
2. Ox # of an element in a monatomic ion = charge of ion
Cl- Na+
3. Group 1 always +1 = always Ox # +1
Group 2 always +2 = always Ox# +2
1.
F = -1
(always) Other halogens = -1 (usually) unless oxyanions
ClO4- Ox# of Cl = +7
O= -2, with exceptions
H = +1, with exceptions (when bonded to a nonmetal)
2.
3.
Ox # of an element in an elementary substance is zero
Cl2
Ox# Cl =0; O2 Ox# = 0
Ox # of an element in a monatomic ion = charge of ion
ClNa+
Group 1 always +1 = always Ox # +1
Group 2 always +2 = always Ox# +2
F = -1
O= -2, with exceptions
H = +1, with exceptions
4. Sum of Ox #=charge
charge on molecule
or ion
14
Writing the Net Ionic Reaction of an Oxidation-Reduction Reaction
Oxidation numbers in
Split into Reactions to be balanced
PbO
PbO2
Pb3O4
litharge, lead oxide
platternite, lead dioxide
minium, trilead tetraoxide
Oxygen (with some exceptions) is -2
Pb3O4
PbO
PbO2
? + -2= 0
? = +2
3(?) +4(-2) = 0
? + 2(-2) = 0
3(?) + -8 = 0
? = +4
This one is an oddity!!!!
Need to know that lead only
adopts +2 or +4 oxidation numbers
Write oxidation half reaction
Write reduction half reaction
A. Balance atoms of element
oxidized
b. Balance Rx sides by adding e
c. Balance charge by adding H+
or OHd. Balance hydrogen by adding
H2O
e. Balance oxygen
A. Balance atoms of element
reduced
b. Balance Rx sides by adding e
c. Balance charge by adding H+
or OHd. Balance hydrogen by adding
H2O
e. Balance oxygen
2(+2) + (+4) = -8
Writing the Net Ionic Reaction of an Oxidation-Reduction Reaction
Combine and balance electrons
Example: Balance the following redox equation in an acidic solution
Fe(2aq+ ) + MnO4−( aq ) → Fe(3aq+ ) + Mn(2aq+ )
Reaction to be balanced
Write oxidation half reaction
Ox #balance
2+
3+
Write reduction half reaction
A. Balance atoms of element
A. Balance atoms of element
reduced
oxidized
b. Balance Ox number with e
b. Balance Ox number with e
c. Balance charge by adding H+
c. Balance charge by adding H+
or OHor OHProtons
are
added
d. Balance hydrogen by adding
d. Balance hydrogen by adding
in an acidic solution
H2O
H2O
e. Balance oxygenand hydroxyls are e. Balance oxygen
added in a basic
solution.
Combine and balance electrons
When in doubt add protons
Fe
2+
( aq )
→ Fe
+7
−
4 ( aq )
MnO
3+
( aq )
+2
→ Mn(2aq+ )
Mass balance
Mass balance
Fe
2+
( aq )
3+
( aq )
→ Fe
ox# balance
+e
-1 = 4(-2) +?
8-1 = ?
7=?
+2
MnO4−( aq ) + 5e → Mn(2aq+ )
15
Example: Balance the following redox equation in an acidic solution
Fe(2aq+ ) + MnO4−( aq ) → Fe(3aq+ ) + Mn(2aq+ )
2+
3+
+7
−
4 ( aq )
MnO
Fe(2aq+ ) → Fe(3aq+ )
Example: Balance the following redox equation in an acidic solution
Fe(2aq+ ) + MnO4−( aq ) → Fe(3aq+ ) + Mn(2aq+ )
+2
→ Mn(2aq+ )
Fe(2aq+ ) → Fe(3aq+ ) + e
Balance hydrogen by adding water
Mass balanced
Mass balanced
Fe(2aq+ ) → Fe(3aq+ ) + e
−
Ox numbers 2 +
4 ( aq )
( aq )
MnO
+ 5e → Mn
MnO4−( aq ) + 5e + 8 H(+aq ) → Mn(2aq+ ) + 4 H2 O
Check oxygen balance
4O
+2
-1 + -5 = -6
Charge balance
MnO4−( aq ) + 5e + 8 Haq + → Mn(2aq+ )
Example: Balance the following redox equation in an acidic solution
Fe(2aq+ ) + MnO4−( aq ) → Fe(3aq+ ) + Mn(2aq+ )
Fe(2aq+ ) → Fe(3aq+ ) + e
MnO4−( aq ) + 5e + 8 H(+aq ) → Mn(2aq+ ) + 4 H2 O
Recombine while balancing electrons
5Fe(2aq+ ) → 5Fe(3aq+ ) + 5e
MnO4−( aq ) + 5e + 8 H(+aq ) → Mn(2aq+ )
=+2
=4O
Check total charge balance: (-1)+5(-1)+8(+1)=+2
Example: Balance the following redox equation in an acidic solution
Fe(2aq+ ) + MnO4−( aq ) → Fe(3aq+ ) + Mn(2aq+ )
Fe(2aq+ ) → Fe(3aq+ ) + e
MnO4−( aq ) + 5e + 8 H(+aq ) → Mn(2aq+ ) + 4 H2 O
5Fe(2aq+ ) → 5Fe(3aq+ ) + 5e
MnO4−( aq ) + 5Fe(2aq+ ) + 8 H(+aq ) → 5Fe(3aq+ ) + Mn(2aq+ ) + 4 H2 O
Final equation does not show any electrons
because electrons do not exist in solution
16
MnO4−( aq ) + 5Fe(2aq+ ) + 8 H(+aq ) → 5Fe(3aq+ ) + Mn(2aq+ ) + 4 H2 O
0.250moleFe( NO3 ) 2 ⎡
⎤
moleFe
⎡⎡ ⎡⎡ 1111LLLL ⎤⎤ ⎡⎡⎤⎤ 0⎡⎡ .00250
250moleFe
moleFe
NO)3 )) 2⎤⎤ ⎡⎤⎤ ⎡⎡ 11moleFe
moleMnO4⎤ ⎤⎤ ⎡⎡ 11moleKMnO
moleKMnO4 ⎤⎤ ⎡ LKMnO4
1moleFe ⎤ ⎡⎤⎤ ⎡⎡ 11moleMnO
(( NO
moleFe( NO
1L ⎤⎥ ⎢ ..250
mLFe
NO
[[27
⎢ 1moleFe ⎥⎥ ⎢⎥ 1⎢moleMnO
27
((( NO
( NO
)))3 )]]2)⎡⎢⎣⎢]10
[27[...27505050.50.mLFe
⎥ ⎥⎢
50mLFe
mLFe
( NO
⎥⎢
moleFe(( NO
NO )) ) ⎦ 5moleFe
mL
27
mLFe
NO
⎥⎦⎥ ⎢⎣⎥ 11⎢moleFe
3 mL⎥⎥⎦ ⎢⎣⎥ ⎢
⎢⎣⎣]10
1111LLLL
mL
⎣⎣⎢ 10
⎦ ⎣⎦⎦ ⎣ 11moleFe
moleFe(( NO
NO ) ⎦ ⎣⎦ ⎣ 55moleFe
10mL
mL⎦⎦ ⎣⎦⎦ ⎣⎣
moleFe 2+ ⎦ ⎦ ⎣ 11moleMnO
moleMnO − ⎦ 0.684moleKMnO
10
3 2
333 3 2223 2 2
333 3
3
2 +2 +
2 +2 +
33 3 22 2
⎣
3 2
3 332 22
⎦⎣
− −−
44
2 +2 +
4
−
⎦⎣
44
⎦⎣
4
⎤
−3
⎥ = 2.01x10 LKMnO4
⎦
= 2.01x10 −3 LKMnO4
Vol KMnO4?
Moles
MnO4-
Moles Fe2+
General
What volume of 0.684 M KMnO4 is required to react completely
with 27.50 mL of 0.250 M Fe(NO3)2
Chemistry
Having balanced the equation we can know predict or measure
quantities that should react.
FITCH Rules
G1: Suzuki is Success
G2. Slow me down
G3. Scientific Knowledge is Referential
G4. Watch out for Red Herrings
G5. Chemists are Lazy
C1. It’s all about charge
C2. Everybody wants to “be like Mike”
⎛ qq ⎞
C3. Size Matters
⎟
E = k⎜
⎝r +r ⎠
C4. Still Waters Run Deep
C5. Alpha Dogs eat first
1 2
el
1
2
stoichiometry
“A” students work
(without solutions manual)
~ 10 problems/night.
Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours W– F 2-3 pm
17