Ch. 4 REACTIONS,SOLUTIONS
Concentration [ ]
dilution, molarity (moles/L)
Replacement Rxns
activity series, solubility
Electrolytes
Reduction – Oxidation
oxidation numbers
RXN:
Acid-Base, Neutralization
PPT; REDOX
TERMS
Solution
A homogeneous mixture of 2
or more pure substances
Solute
The substance that is dissolved in solution
Solvent The substance that does the dissolving
MOLARITY --CONCENTRATION
Label
mols
liter
M
moles of solute
liter of solution
Symbol
[H+]
Concentration of ACID
[OH-] Concentration of BASE
Brackets indicate CONCENTRATION
DILUTION
A single solution (homogeneous mixture) of a known
concentration needs to be diluted to a lower concentration.
This is accomplished by adding a known quantity of
water (distilled) to the original solution volume.
The terms used indicate:
Mo; Vo
Initial – Original -- Beginning
M1; V1; Mi; Vi
concentration & volume
Md; Vd
M2; V2; Mf; Vf
Final – Diluted -- Ending
concentration & volume
Equation:
M1 * V1 = M2 * V2
290.0 mL of a 0.560 M Fe(OH)3 solution is required
to make a 0.250 M solution by dilution. What is the
final volume.
Step #1: Identify the parts for equation
M1 = 0.560 M
V1 = 290.0
M2 = 0.250 M
V2 = X ml
Step #2: Set up and solve equation
(0.560 M) (290.0 mL) = (X ml) (0.250 M)
650 ml
0.250 M
Notice: Dilution problems are not volume specific
1.25 L of H2O is added to 750 ml of 0.75 M HCl
solution. What is the resulting molarity?
Step #1: Identify parts for equation
M1 = 0.75 M
So, final vol. is amount
M2 = X
V1 = 750 ml
Start +
2.0 L or 2000.0
mlAdded
V2 = ???
or 0.75 L
Look at the problem, what does it say!!!
Step #2: Set up & solve equation
1.25 L is added to ….
(0.75 M) (0.75 L) = (X) (2.0 L)
2.0 L
OR
0.28 M
(0.75 M)(750 ml) 
2000 ml
Given mass & volume, find molarity
Determine the M of 752.0 g Barium
Chloride in 575 ml of solution.
Step 1: Need to change 575 ml to 0.575 L
Step 2: Find formula wt. of CMPD. BaCl2 = 208.3 g/mol
Step 3 : Find moles of CMPD.
752.0 g * 1 mol
1
208.3 g
.
3.6 moles
Step 4 : Find molarity of solution M = 3.6 moles
0.575 L
6.3 M
Given molarity & volume, find moles & mass
How many moles of NaCl are in
36.7 ml of a 0.256 M solution?
Step 1: Need to change 36.7 ml to 0.0367 L
Step 2: Convert M label to mols/L 0.256 mols
1L
Step 3 : Find moles of NaCl
mols  0.256 mols * 0.0367 L
1 L
1
Using given concentration
& volume, find moles
0.0094 or 9.4*10-3 moles NaCl
Continuing, next we need to find
the MASS of NaCl in0 .0094 moles
Step 1: find formula wt. of NaCl
58.5 g /mol
Step 2 : Find mass
9.4 *103 mols * 58.5 g
1
1 mol
MOLES
MASS
0.55 g NaCl
PRACTICE PROBLEMS
MOLARITY
1. What is the molarity of a solution containing 2.50 moles of KNO3
2.50 mols
dissolved in 5.00 L?
 0.5 M
5.00 L
2. How many moles of KCl are present in 100.0 mL of 0.125 M solution?
DILUTION
0.125 mols 0.10000 L
*
 0.0125 mol
L
1
1. What is the molarity of 50.0 mL of a 0.50 M NaOH solution after it has
been diluted to 300.0 mL?
50.0 mL 0.50 M 
 X X  0.083 M
300 mL
2. If 300.0 mL of water is added to 400.0 mL of a 0.400 M Na2CrO4
solution, what is the molarity of the resulting solution?
400.0 mL 0.400 M 
 X X  0.229 M
700.0 mL
SINGLE REPLACEMENT
FORM: A+B- + E0 ----- E+B- + A0
2 reactants: 1 cmpd. & 1 element
forms
2 products: 1 new cmpd. & 1 new element
DOUBLE REPLACEMENT
FORM:
A+B- + X+Y- ----- A+Y- + X+B2 reactants: 2 compounds
form
2 products: 2 new compds.
SINGLE REPLACEMENT
Reference used for rxn. occurring
ACTIVITIES SERIES of METALS
Electrochemical Series
DOUBLE REPLACEMENT
Reference used of rxn. occurring
RULES of SOLUBILITY
ACID
+ BASE ----- SALT + H2O
Nitric Acid + Potassium Hydroxide --- ??????
DR
HNO3(aq) + KOH (aq) --- KNO3 (aq) + H2O (l)
Sulfuric Acid + Barium Hydroxide -----
H2SO4 (aq) + Ba(OH)2 (aq) --- BaSO4 (s) + 2 H2O (l)
Zinc + Copper II Sulfate -----
?
1st is a reaction going to occur or not????
What type of RXN??? Which players will trade places????
We have an element plus a compound.
Single Replacement
Check the METAL REACTIVITY list
Zn (s) + CuSO4 (aq) ----ZnSO4 (aq) + Cu (s)
BALANCED???
Look at relation
between Zn & Cu
Zn is more active then
Cu, therefore, rxn. occurs
SOLUTIONS
AgNO3 (aq) + NaCl (aq) ---- NaNO3 (aq) + AgCl (s)
Ag+ (aq) + NO3-(aq) + Na+(aq) + Cl-(aq) ----- Na+(aq) + NO3-(aq) + AgCl(s)
NO3-
Ag+
Ag+
-
-
NO3
Ag+
NO3
Ag+
NO3-
Ag+
Ag+
Now, combine both
solutions together.
ClNa+
What is the expected
effect??????
Na+
Cl-
Na+
Na+
Cl- Na+
Na+
-
NO3
- Ag+
NO3- NO3
Ag+
Cl-
Na+
Cl-
Cl+
Cl- Na
Silver + Sulfuric Acid -- ?????
Ag (s) + H2SO4 (aq) --
NR
CHECK, “Ag” active metal
to replace “H” in an Acid ???
Therefore, this is a
“No RXN”
“Ag” is not an active
enough metal to
replace “H” in an Acid
REDOX – Reduction/Oxidation
Lose
Electron
Oxidize
Oxidize
Is
Lose
Gain
Electron
Reduce
Reduce
Is
Gain
OXIDATION - REDUCTION
“REDOX”
Reduction: gain electron
charge becomes more neg
Oxidation: lose electron
charge becomes more “+”
H2(g) + O2(g) --------> H2O(g)
Both H2 & O2 are diatomics: charge on each 0
In the cmpd. H is +1 & O is -2
H: from 0 to +1 charge, loses e-, oxidized (reducing agent)
O: from 0 to -2 charge, gain e-, reduced (oxidizing agent)
1
2
2H O 2H O
0
2
0
2
-2
Ionic Reaction -- Ionic Equations
3 diff types of equations for same reaction
molecuar; total ionic; net ionic
Spector ions
- not involved in reaction
- no D in charge & state
STATES:
(aq): aqueous; soln
(s): solid; precipitate
(g): gas
(l): liquid: H2O
Type of rxn; molecular, total, net eqns; oxidized/reduced & agents; spectators
Potassium carbonate reacts w/ strontium nitrate to yield ??????????
Type: double replacement; precipitation
Molecular
K 21CO3
2
( aq )
 Sr  2 ( NO3 ) 21 ( aq)  Sr  2CO3
2
(s)
 2K 1NO3
1
( aq )
Total
2K 1 ( aq)  CO3
2
( aq )
 Sr  2 ( aq)  2NO3
1
( aq )
 Sr  2CO3
2
(s)
 2K 1 ( aq)  2NO3
Net
CO3
2
( aq )
 Sr  2 ( aq)  Sr  2CO3
2
(s)
Spectator: K+1 & NO3-1
Reduce, oxidize: none, not a redox rxn
1
( aq )
Barium hydroxide and hydrocyanic acid produces ???????
Type: DR; acid-base; neutralization
Molecular
Ba  2 (OH) 2
1
( aq )
 2H 1CN-1 ( aq)  Ba  2 (CN)2
1
(s)
 2H 2O(l )
Total
Ba  2 ( aq)  2OH1 ( aq)  2HCN ( aq)  Ba  2 (CN)2
1
(s)
 2H 2O(l )
Net
Ba  2 ( aq)  2OH1 ( aq)  2HCN ( aq)  Ba  2 (CN)2
1
(s)
 2H 2O(l )
Spectator: none, since HCN weak acid
Reduce, oxidize: none, not a redox rxn
Same reaction but with a strong acid instead
Barium hydroxide and hydrochloric acid produces ???????
Type: DR; acid-base; neutralization
Molecular
Ba  2 (OH) 2
1
( aq )
 2H 1  Cl-1 ( aq)  Ba  2  2Cl1 ( s )  2H 2O(l )
Total
Ba  2 ( aq)  2OH1 ( aq)  2H 1 ( aq)  2Cl-1 ( aq)  Ba  2 ( aq)  2Cl-1 ( aq)  2H 2O(l )
Net
H 1 ( aq)  OH1 ( aq)  H 2O(l )
Spector: Ba+2 & Cl-1, since HCl strong acid
Reduce, oxidize: none, not a redox rxn
Aluminum metal & manganese II sulfate produce aluminum sulfate & manganese metal
Type: Single Replacement; Oxidation Reduction
Molecular
2Al 0 ( s )  3Mn  2SO-42 ( aq)  Al 2 3 (SO 4 )3
2
( aq )
 3Mn 0 ( s )
Total
2Al0 (s )  3Mn 2 ( aq )  3SO4
2
( aq )
 2Al 3 ( aq )  3SO4
2
( aq )
 3Mn (s )
Net
2Al 0 ( s )  3Mn  2 ( aq)  2Al 3 ( aq)  3Mn 0 ( s )
Spectator: SO4-2
Oxidize
Al (0 ---> +3)
Reduce
Mn (+2 ---> 0)
Red. Agent
Al
Ox. Agent
Mn
OXIDATION NUMBERS
IA:+1
IIA: +2
IIIA: +3
PO4-3: sum “P” charge + “O” charge = -3
P + 4(-2) = -3
P = +5
OH-1: O = -2
H = +1
-2 + 1 = -1
Carbon monoxide reacts with diiodine pentaoxide yields iodine & carbon dioxide
+2 -2
+5 -2
0
+4 -2
5 CO (g) + I2O5 (s) -------> I2 (s) + 5 CO2 (g)
Oxidized:C +2 ---> +4; loss 2eReduced: I +5 ---> 0; gain 5 e-
Oxidizing Agent: I
Reducing Agent: C
+2
0
+4 -2
-2
C (s) + CO2 (g) -------> 2 CO (g)
loss 2 e-
Oxidized
C (s)
0 ----> +2
gain 2 e-
Reduced
C in CO2
+4 ----> +2
Red. Ag
C (s)
Ox. Ag
CO2
Balance REDOX
+5
0
+1 -2
+5
+2
+4 -2
-2
+1 -2
Sn (s) + HNO3 (aq) -------> Sn(NO3)2 (aq) + NO2 (g) + H2O (l)
gain 1 e-
loss 2 e-
Oxidized
Sn (s)
0 ----> +2
Reduced
N in HNO3
+5 ----> +4
Red. Ag
Sn (s)
Ox. Ag
N
PRACTICE PROBLEMS
1. Calcium Acetate (aq) + Aluminum Sulfate (aq) ------
write and balance the molecular, total ionic, & net ionic equations
reduced? oxidized?????
3 Ca(C2H3O2)2(aq) + Al2(SO4)3 (aq) ----> 3 CaSO4 (s) + 2 Al(C2H3O2)3 (aq)
3 Ca+2 + 6 C2H3O2-1 + 2 Al+3 + 3 SO4-2 ----> 3 CaSO4 (s) + 2 Al+3 + 6 C2H3O2-1
Ca+2(aq) + SO4-2 (aq) ----> CaSO4 (s)
NOT a redox rxn
ELECTROLYTES:
Subst. that produces ions when in H2O;
conducts electrical current
1. All ionic subst. ARE
2. Most covalent ARE NOT;
acids ARE
DISSOCIATE: Completely separates into ions
ionic cmpds
STRONG ELECTROLYTE: Ionize completely in H2O
NaCl (aq) ---------> Na+1 (aq) + Cl-1 (aq)
H2SO4 (aq) ---------> 2 H+1 (aq) + SO4-2 (aq)
WEAK ELECTROLYTE: Partially ionize in H2O
HC2H3O2 (aq) <---------> HC2H3O2 (aq) + H+1 (aq) + C2H3O2-1 (aq)
NONELECTROYLTE: Not produce ions in H2O
usually dissolves as whole molecule unit
C2H8 (aq) ---------> C2H8 (aq)
STRONG
WEAK
All ionic cmpds
No ionic cmpds
Strong Acids
Weak Acids
HCl HBr HI
Weak Bases
HNO3 H2SO4
NH3
HClO3 HClO4
Strong Bases
-1A metal hydroxides
-2A heavy metal hydroxides: Ca, Ba, Sr
NON
Covalent Molecules
-non-acids
-not contain NH3
COMPOSITION REACTIONS
FORM: A0 + B0 ----- A+B-
2 reactants form 1 product
DECOMPOSITION REACTIONS
FORM:
A+B- ----- A0 + B0
1 reactants breaks apart into 2 or more product
SINGLE REPLACEMENT
FORM:
A+B- + E0 ----- E+B- + A0
2 reactants: 1 cmpd. & 1 element
forms
2 products: 1 new cmpd. & 1 new element
DOUBLE REPLACEMENT
FORM:
A+B- + X+Y- ----- A+Y- + X+B-
2 reactants: 2 compounds
form
2 products: 2 new compds.