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.