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
Jan 03, 2016
Ch. 4 REACTIONS,SOLUTIONS
Concentration [ ] dilution, molarity (moles/L)
Replacement Rxns activity series, solubility
Electrolytes
Reduction – Oxidation oxidation numbers
RXN:Acid-Base, NeutralizationPPT; REDOX
TERMS
Solute The substance that is dissolved in solution
Solvent The substance that does the dissolving
Solution
A homogeneous mixture of 2 or more pure substances
MOLARITY --CONCENTRATION
Label Mmols liter
moles of soluteliter of solution
Symbol Concentration of ACID
Concentration of BASE
Brackets indicate CONCENTRATION
[H+]
[OH-]
DILUTIONA single solution (homogeneous mixture) of a knownconcentration needs to be diluted to a lower concentration.
This is accomplished by adding a known quantity ofwater (distilled) to the original solution volume.
The terms used indicate:Mo; Vo Initial – Original -- Beginning M1; V1; Mi; Vi concentration & volume
Md; Vd Final – Diluted -- Ending M2; V2; Mf; Vf concentration & volume
Equation: M1 * V1 = M2 * V2
290.0 mL of a 0.560 M Fe(OH)3 solution is requiredto make a 0.250 M solution by dilution. What is thefinal volume.
Step #1: Identify the parts for equation
M1 = 0.560 MV1 = 290.0
M2 = 0.250 MV2 = X ml
Step #2: Set up and solve equation
(0.560 M) (290.0 mL) = (X ml) (0.250 M)
0.250 MNotice: Dilution problems are not volume specific
650 ml
1.25 L of H2O is added to 750 ml of 0.75 M HClsolution. What is the resulting molarity?
Step #1: Identify parts for equation
M1 = 0.75 MV1 = 750 ml
M2 = X V2 = ???
Look at the problem, what does it say!!!1.25 L is added to ….
So, final vol. is amount Start + Added2.0 L or 2000.0 ml
or 0.75 LStep #2: Set up & solve equation
(0.75 M) (0.75 L) = (X) (2.0 L)
2.0 L
0.28 MOR
ml 2000
ml) M)(750 (0.75
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 g3.6 moles
Step 4 : Find molarity of solution 6.3 M M = 3.6 moles 0.575 L
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 1 L
Step 3 : Find moles of NaCl
1L 0.0367 *
L 1mols 0.256 mols
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
mol 1g 58.5 *
1mols10 * 9.4 3
0.55 g NaCl
MOLES
MASS
Mul
tiply
by
Divid
e
by
PRACTICE PROBLEMSMOLARITY1. What is the molarity of a solution containing 2.50 moles of KNO3 dissolved in 5.00 L?
2. How many moles of KCl are present in 100.0 mL of 0.125 M solution?
DILUTION1. What is the molarity of 50.0 mL of a 0.50 M NaOH solution after it has been diluted to 300.0 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?
M 0.5 L 5.00
mols2.50
mol 0.0125 1
L 0.10000 * L
mols0.125
M 0.083 X X
mL 300
M 50.0mL 50.0
M 0.229 X X
mL 700.0
M 400.0mL 400.0
SINGLE REPLACEMENT
FORM: A+B- + E0 ----- E+B- + A0
2 reactants: 1 cmpd. & 1 element forms2 products: 1 new cmpd. & 1 new element
DOUBLE REPLACEMENT
FORM: A+B- + X+Y- ----- A+Y- + X+B-
2 reactants: 2 compounds form2 products: 2 new compds.
SINGLE REPLACEMENT
Reference used for rxn. occurring
ACTIVITIES SERIES of METALSElectrochemical Series
DOUBLE REPLACEMENT
Reference used of rxn. occurring
RULES of SOLUBILITY
ACID + BASE ----- SALT + H2O
Nitric Acid + Potassium Hydroxide --- ??????
HNO3(aq) + KOH (aq) --- KNO3 (aq) + H2O (l)DR
Sulfuric Acid + Barium Hydroxide -----
H2SO4 (aq) + Ba(OH)2 (aq) --- BaSO4 (s) + H2O (l)2
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) ----
Look at relationbetween Zn & Cu
Zn is more active thenCu, therefore, rxn. occurs
ZnSO4 (aq) + Cu (s)
BALANCED???
SOLUTIONSAgNO3 (aq) + NaCl (aq) ---- NaNO3 (aq) + AgCl (s)
Ag+ (aq) + NO3-(aq) + Na+(aq) + Cl-(aq) ----- Na+(aq) + NO3
-(aq) + AgCl(s)
Ag+
Ag+
Ag+
Ag+
Ag+
Ag+
Ag+
Ag+
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
NO3-
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Cl-
Cl-
Cl-
Cl-
Cl-
Cl-
Cl-Now, combine bothsolutions together.
What is the expected effect??????
Silver + Sulfuric Acid -- ?????
Ag (s) + H2SO4 (aq) --
CHECK, “Ag” active metalto replace “H” in an Acid ???
“Ag” is not an activeenough metal toreplace “H” in an Acid
NR
Therefore, this is a “No RXN”
REDOX – Reduction/Oxidation
Lose
Electron
Oxidize
Gain
Electron
Reduce
Oxidize
Is
Lose
Reduce
Is
Gain
OXIDATION - REDUCTION “REDOX”
Reduction: gain electron Oxidation: lose electron charge becomes more neg charge becomes more “+”
H2(g) + O2(g) --------> H2O(g)
Both H2 & O2 are diatomics: charge on each 0In 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)
-212
02
02 OH 2 O H 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
Potassium carbonate reacts w/ strontium nitrate to yield ??????????
Type of rxn; molecular, total, net eqns; oxidized/reduced & agents; spectators
)(1
31
)(2
32
)(1
232
)(2
31
2 NO2K COSr )NO(Sr COK
Molecular
aqsaqaq
)(1
3)(1
)(2
32
)(1
3)(2
)(2
3)(1 2NO 2K COSr 2NO Sr CO 2K
Total
aqaqsaqaqaqaq
)(2
32
)(2
)(2
3 COSr Sr CO
Net
saqaq
Type: double replacement; precipitation
Spectator: K+1 & NO3-1
Reduce, oxidize: none, not a redox rxn
)(2 )(1
22
)(1-1
)(1
22 O2H (CN)Ba CN2H (OH)Ba
Molecular
lsaqaq
)(2 )(1
22
)( )(1
)(2 O2H (CN)Ba 2HCN 2OH Ba
Total
lsaqaqaq
Type: DR; acid-base; neutralization
Spectator: none, since HCN weak acid
Reduce, oxidize: none, not a redox rxn
Barium hydroxide and hydrocyanic acid produces ???????
)(2 )(1
22
)( )(1
)(2 O2H (CN)Ba 2HCN 2OH Ba
Net
lsaqaqaq
)(2 )(12
)(1-1
)(1
22 O2H Cl2Ba Cl2H (OH)Ba
Molecular
lsaqaq
)(2)(1-
)(2
)(1-
)(1
)(1
)(2 O2H Cl2Ba 2Cl2H 2OH Ba
Total
laqaqaqaqaqaq
Type: DR; acid-base; neutralization
Reduce, oxidize: none, not a redox rxn
Same reaction but with a strong acid insteadBarium hydroxide and hydrochloric acid produces ???????
)(2 )(1
)(1 OH OH H
Net
laqaq
Spector: Ba+2 & Cl-1, since HCl strong acid
)(0
)(2
343
2)(2-
42
)(0 3Mn )(SOAl SO3Mn 2Al
Molecular
saqaqs
)s()aq(2
4)aq(3
)aq(2
4 )aq(2
)s(0 3Mn SO32Al 3SO 3Mn 2Al
Total
Type: Single Replacement; Oxidation Reduction
Spectator: SO4-2
Oxidize Reduce Red. Agent Ox. AgentAl (0 ---> +3) Al Mn (+2 ---> 0) Mn
Aluminum metal & manganese II sulfate produce aluminum sulfate & manganese metal
)(0
)(3
)(2
)(0 3Mn 2Al 3Mn 2Al
Net
saqaqs
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 = H =
-2 +1-2 + 1 = -1
Carbon monoxide reacts with diiodine pentaoxide yields iodine & carbon dioxide
5 CO (g) + I2O5 (s) -------> I2 (s) + 5 CO2 (g)
Oxidized:Reduced:
Oxidizing Agent: Reducing Agent:
+2 +5 -2 0 +4-2 -2
C +2 ---> +4; loss 2e-I +5 ---> 0; gain 5 e-
IC
C (s) + CO2 (g) -------> 2 CO (g)
0 -2 -2 +4 +2
loss 2 e- gain 2 e-
Oxidized Reduced Red. Ag Ox. Ag C (s) C in CO2 C (s) CO2
0 ----> +2 +4 ----> +2
Sn (s) + HNO3 (aq) -------> Sn(NO3)2 (aq) + NO2 (g) + H2O (l)
Balance REDOX
0 +1 +1-2 -2 -2 -2+2 +4 +5 +5
loss 2 e- gain 1 e-
Oxidized Reduced Red. Ag Ox. Ag Sn (s) N in HNO3 Sn (s) N0 ----> +2 +5 ----> +4
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. ARE2. 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 H2Ousually dissolves as whole molecule unit
C2H8 (aq) ---------> C2H8 (aq)
STRONG WEAK NONAll ionic cmpds No ionic cmpds Covalent MoleculesStrong Acids Weak Acids -non-acids HCl HBr HI Weak Bases -not contain NH3
HNO3 H2SO4 NH3
HClO3 HClO4
Strong Bases -1A metal hydroxides -2A heavy metal hydroxides: Ca, Ba, Sr
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 forms2 products: 1 new cmpd. & 1 new element
DOUBLE REPLACEMENT
FORM: A+B- + X+Y- ----- A+Y- + X+B- 2 reactants: 2 compounds form2 products: 2 new compds.