UNIT 2: CHEMICAL REACTIONS. 4.1, 4.2 Aqueous Solutions _ = _ +...

UNIT 2:

CHEMICAL REACTIONS

UNIT 2: CHEMICAL REACTIONS 4.1, 4.2 Aqueous Solutions _____________ = _____________ + _______________

Water is polar!!!

solution solute solvent

O - 3.5H - 2.1 1.4 EN Diff so VERY Polar bond

+d

-dH2O strong dipole

must be soluble!!

AqueousReactions

© 2012 Pearson Education, Inc.

Solutions

• Solutions are defined as homogeneous mixtures of two or more pure substances.

• The solvent is present in greatest abundance.

• All other substances are solutes.

Solubility of a substance in water is determined by

the AFFINITY of the water molecules for the solute particles

For INSOLUBLE cmpds: the attraction of the (+) and (-) ions for each other is stronger than their attraction for POLAR water molecules

An ELECTROLYTE is a substance

whose aqueous solution conducts electric current. The current is carried by IONS (+) and (-)

Key: The more ions in solution, the more current is carried.

Thus if NO ions in a solution (just neutral molecules) then NO current can be conducted.

STRONG ELECTROLYTES: are excellent “conductors” because ___ solute particle are in the form of ______all ions

1) soluble ionics NaCl NH4NO3 KISee Solubility Table!

2) strong acids neutral molecules that “ionize”(react with) H2O

HI, HBr, HCl, HClO3, HClO4, HNO3, H2SO4

“Is Bright and Clear, No Snow, 3,4”

3) strong bases soluble hydroxides

LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2

backwards “J” “Can, Sara, Bat”

7

8

all end in “ic”

AqueousReactions


Acids

There are only seven strong acids:• Hydrochloric (HCl)• Hydrobromic (HBr)• Hydroiodic (HI)• Nitric (HNO3)

• Sulfuric (H2SO4)

• Chloric (HClO3)

• Perchloric (HClO4)

AqueousReactions


Bases

The strong bases are the soluble metal salts of hydroxide ion:• Alkali metals• Calcium• Strontium• Barium

AqueousReactions


Dissociation

• An electrolyte is a substances that dissociates into ions when dissolved in water.

AqueousReactions


Solutions

• An electrolyte is a substance that dissociates into ions when dissolved in water.

• A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

AqueousReactions


Electrolytes and Nonelectrolytes

Soluble ionic compounds tend to be electrolytes.

WEAK ELECTROLYTES: are poor “conductors” because ________ solute particles are in the form of _____.very few ions

(Almost all solute remains as neutral molecules)

1) weak acids

HF HC2H3O2 HNO2

2) weak (molecular) bases NH3 CH3NH2

NaCl(s) --------> Na+(aq) + Cl-

(aq)

“hydrated” or (“solvated”) ions

H2O

IowaState visual

strong electrolyte

weak electrolyte

non-electrolyte

ALL ions VERY FEW ions NO ions

+

+

+---

+-

-

-+-

-+

+

Str Acids Str BasesSoluble Ionics

Weak AcidsWeak Bases

POLAR molecularcompounds

AqueousReactions



Soluble ionic compounds tend to be electrolytes.

AqueousReactions



Molecular compounds tend to be nonelectrolytes, except for acids and bases.

AqueousReactions


Electrolytes

• A strong electrolyte dissociates completely when dissolved in water.

• A weak electrolyte only dissociates partially when dissolved in water.

AqueousReactions


Strong Electrolytes Are…

• Strong acids• Strong bases• Soluble ionic salts

BaCl2(s)

dissolves

BaCl2(s) ------> Ba2+(aq) + 2Cl-

(aq)H2O

“DISSOCIATION” of an ionic compound in water:

ionic solid solvated ions

SOLUBLE IONIC

breaking bonds between ions in the solid then making bonds between each ion and polar H2O molecules that surround them

AqueousReactions


Dissociation

• When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them.

• This process is called dissociation.

HCl solution

Str ACID ionizes 100% in H2O

HCl(g) + H2O(l) ---> H3O+(aq) + Cl-

(aq)

“IONIZATION” of an ACID in H2O.

ALL solute is ions.

STRONG ACID

NaOH solution

Str BASE dissociates 100% in H2O

NaOH(s) -------> Na+(aq) + OH-

(aq)

H2O

(ionic)

ALL solute is ions

STRONG BASE

acetic acid solution

mostly molecules & very few ions

HC2H3O2(l) + H2O(l) <---> H3O+(aq) + C2H3O2

-(aq)

Weak ACID ionizes only slightly in H2O

WEAKACID

Methods of Expressing Solution Concentration

1) Mass % = mass of solute x 100 (mass of solute + mass of solvent)

Mass %A = mass of component A ratio x 100 total mass of solution “10% MgSO4(aq)” means 10.g MgSO4 + 90.g H2O = 100.g solution

ratio: 10.g solute 100.g solution

100% - 6.80% = 93.20%

Problem: How many grams H2O are needed to prepare 525g of 6.80%NaCl(aq) ?

525g x 6.80% = 35.7g NaCl solute so 525g solution - 35.7g NaCl = 489.3g H2O

solute solvent

OR you could find %H2O first, then do a proportion:

93.20g = x g 100g 525g

x = 489.3g H2O

2) Volume% = volume of solute x 100 (volume of solute + volume of solvent)

Volume%A = volume of component A x 100 total volume of solution

again it is a ratio mL of A 100mL solution

Problem: How many liters of water are needed to prepare 1.00 liter of 23% by volume CH3OH(aq)?

1.00L x 23% = 0.23L CH3OH so 1.00L solution - 0.23L CH3OH = 0.77L H2O

OR 100% - 23% = 77%77L = x L

100L 1.00L“Proof” is a volume %. Proof = 2 x volume % alcohol C2H5OH) also called “EtOH” so “180 proof” whiskey means 90% ethanol by volume!!

solute solvent

3) Parts per million or parts per billion (ppm) (ppb) used for very small solute concentrations (for mass or volume)

ppm(mass) = mass of solute x 106 ppb….x 109 mass of solution

OR ppm = mg solute ppb = µg solute kg solution kg solution

“0.1ppm by mass of Pb2+ ions” in drinking water means 0.1mg Pb2+ in 1kg water

Problem: Find ppm of 0.5g of Ca2+ ions in 2500g tapwater

ppm = 0.5g x 106

2500g

= 200ppm Ca2+

4) MOLARITY (M) = moles of solute Liter of solution

mol

M L

AqueousReactions


Molarity

• Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different.

• Molarity is one way to measure the concentration of a solution:

moles of solute

volume of solution in litersMolarity (M) =

AqueousReactions


Mixing a Solution

• To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute.

• The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

Problem: Find Molarity when 53.3grams FeCl3 are dissolved in 755mL of solution.

Now find Molarity of chloride ions in this solution:

FeCl3 Fe3+ + 3Cl-

= 53.3g x 1 mol162.20g0.755L

Fe – 55.85Cl – 3(35.45) 162.20 g/mol to convert

to moles

0.43524…….

= 0.435M (means mol/L FeCl3)

_ _ _ 3

M

1 to 3 ratio3(0.435M) = 1.305M Cl-

1 to 1 ratio

Problem: How many moles of K+ ions are present in 25.0mL of 2.50M K2O?

moles = Molarity x VolumeLiters

= (2.50M) (0.0250L)

= 0.625mol K2O

2 mol K+ ions 1 mol K2O

0.625mol K2O = 1.250mol K+ ionsx

mg 52.00

10-3

1

1. Calculate Molarity of 2.0ppm Cr6+ ions in well water. (find mol/L)

(in this very dilute solution, 1mL has a mass of ~1g)

2.0ppm means _______________ ______ = ______

______ = ______

Convert: numerator ______----> ________

denominator _____----> ________

2.0mg Cr6+ ions

1kg solution1kg H2O 1L H2O1g H2O 1mL H2O

mg mol

kg L

2.0mg x g x 1 mol

g

= mol3.846 x 10-5

3.8 x 10-5mol1 L

= 3.8 x 10-5M1kg = 1L

2. How much LiF is needed to prepare 450.mL of a 1.20M sol’n. (M is ratio of mol solute to L solution so find mol…then mol-->g)

_____________ = _______ or mol = M x L

1.20mol LiF1 L

x mol 0.450L

0.54X = 0.540mol

0.540mol

Li – 6.94F – 19.00 25.94 g/mol

x 25.94 g 1 mol

= 14.0076

14.0g LiF

3. Find Molarity of a 5.00% Pb(NO3)2 solution by mass. (find mol/L)

D = 1.05g/mL

5.00% Pb(NO3)2 means ______________

Convert: numerator _____----> ______

denominator _____----> ______

5.00g Pb(NO3)2

100g solution

g

g

mol

LPb – 207.20N – 2(14.00)O - 6(16.00) 331.20g/mol 5.00gx 1 mol =

331.20g

0.0151mol Pb(NO3)2

100g x mL1.05g

x 1 L = 1000mL

0.0952L sol’n

Pb(NO3)2(s) Pb2+(aq) + 2NO3

-

(aq)

NONE of this ALL of this in solution in solution [Pb2+] =

_____________

[NO3-] =

_____________

[total ion] = _____________

M = 0.0151mol 0.0952L

0.15861= 0.159M

0.159M

+ 0.318M (x 2)

0.477M (x 3)

H2O

1) Find total moles of iodide ions in 250.mL of a 0.750M BaI2 sol’n.

____________ = _________

0.750mol

1 Lx mol 0.250L

BaI2(s) ------> Ba2+(aq) + 2I-

(aq)

H2O

0.1875x = 0.188mol BaI2

0.188mol 0.188mol 2(0.188)mol

0.376mol of I- ions

2. Find [Cl-] when 9.82g CuCl2 are dissolved in 600.mL of sol’n.

M = ________ X __________

9.82g0.600L

Cu – 63.55Cl – 2(35.45) 134.45g/mol

134.45g1 mol

0.12173= 0.122M CuCl2

x __________ 0.122M CuCl22 mol Cl-

1 mol CuCl2

= 0.244mol Cl-

3. Which solution of strong electrolytes contains the largest # of chloride ions?

A or B

50.0mL of 0.60M MgCl2 200.mL of 0.40M NaCl

mol = M x L mol = M x L

= (0.60M)(0.0500L)

= 0.030mol MgCl2

MgCl2(s)---> Mg2+(aq) + 2Cl-

(aq)

0.030mol 2(0.030)mol

0.060mol Cl-

= (0.40M)(0.200L)

= 0.080mol NaCl

NaCl(s)---> Na+(aq) + Cl-

(aq)

0.080mol

0.080mol Cl-

0.080mol 1 to 2 1 to 1

4. How would you prepare: 1.00L of a 2.5M HNO3 solution from conc. (16M) nitric acid?

dilution formula:

Mconc x Vconc = Mdil x Vdil

(16M) (2.5M) (1000.00mL)(x mL) =

x = 160mL)

1000mL

2) add H2O to line3) cap & mix

1) add 160mL conc acid

AqueousReactions


Dilution• One can also dilute a more concentrated

solution by– Using a pipet to deliver a volume of the solution to a

new volumetric flask, and– Adding solvent to the line on the neck of the new flask.

AqueousReactions


DilutionThe molarity of the new solution can be determined from the equation

Mc Vc = Md Vd,

where Mc and Md are the molarity of the concentrated and dilute solutions, respectively, and Vc and Vd are the volumes of the two solutions.

How would you prepare: 1.00L of a 0.250M NaNO3 solution from the pure solid?

Na – 22.99N – 14.00O - 3(16.00) 85.00g/mol

0.250M is 0.250mol NaNO3 in 1Liter sol’n

0.250mol x 85.00g 1mol

=

21.25

21.3g

5. Find [Li+] when 20.0mL of 1.0M LiCl is added to 80.0mL of 2.0M LiBr

1) find moles of each cmpd (solute) 2) find total moles of solute ions

20.0mL of 1.0M LiCl (0.0200L)(1.0M) = 0.020mol LiCl

80.0mL of 2.0M LiBr

(0.0800L)(2.0M) = 0.16mol LiBr

0.020mol Li+

0.16mol Li+

0.18mol Li+

total

100.0mLtotal volume

[Li+] = mol L

= 0.18mol 0.100L

= 1.8M

TYPES OF REACTIONS IN AQUEOUS SOLUTION

1) precipitation (formation of insoluble solid product)2) acid-base (formation of water, a neutralization reaction)3) oxidation-reduction reaction (changing oxidation #s) METATHESIS reaction = DOUBLE REPLACEMENT

AB + CD -----> _____( ) + _____( )

make it go!

AD CB

(s) ppt(l) H2O or other molec cmpd (WA)(g) GAS

“spectator ions” are (aq) product

AqueousReactions


Metathesis (Exchange) Reactions

• Metathesis comes from a Greek word that means “to transpose.”

AgNO3(aq) + KCl(aq) AgCl(s) + KNO3(aq)

AqueousReactions


Metathesis (Exchange) Reactions

• Metathesis comes from a Greek word that means “to transpose.”

• It appears as though the ions in the reactant compounds exchange, or transpose, ions:


AqueousReactions


Precipitation Reactions

When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

1) NaCl(aq) + Pb(NO3)2(aq)---->

Writing equations:1) The “MOLECULAR” equation is a balanced equation with

all formulas & state of matter symbols

Ex:________( ) + _______ ( ) PbCl2 saqNaNO322

Pb2+ Cl-Na+ NO3-

AqueousReactions


Molecular Equation

The molecular equation lists the reactants and products in their molecular form:


2) The “COMPLETE IONIC” equation shows all reactants and products separated into individual ions as they exist in solution: strong acids, strong bases and soluble ionics (remember SA,SB,SI)

+ PbCl2(s)

Note: Never separate the ppt, H2O or gas

2Na+(aq) + 2Cl-(aq) + Pb2+

(aq) + 2NO3-(aq) ----->

2Na+(aq) + 2NO3

-(aq)

AqueousReactions


Ionic Equation

• In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions.

• This more accurately reflects the species that are found in the reaction mixture:

Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq)

AgCl(s) + K+(aq) + NO3−(aq)

3) The “NET IONIC EQUATION” is missing the spectator ions, (cancelled)

showing only the particles involved in the reaction that made PRODUCTS. (made it go!!)

Net ionic equation: 3) Pb2+

(aq) + 2Cl-(aq) ----> PbCl2(s)

AqueousReactions


Net Ionic Equation

• To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right:

Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq)


AqueousReactions


Net Ionic Equation

• To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.

• The only things left in the equation are those things that change (i.e., react) during the course of the reaction:

Ag+(aq) + Cl−(aq) AgCl(s)

AqueousReactions


Net Ionic Equation

• To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.

• The only things left in the equation are those things that change (i.e., react) during the course of the reaction.

• Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions:Ag+(aq) + NO3

−(aq) + K+(aq) + Cl−(aq)


AqueousReactions


Writing Net Ionic Equations

1. Write a balanced molecular equation.

2. Dissociate all strong electrolytes.

3. Cross out anything that remains unchanged from the left side to the right side of the equation.

4. Write the net ionic equation with the species that remain.

SHORTCUT TO NET IONIC EQUATIONS:

1) BaCl2(aq) + K2SO4(aq) Net ionic equation:

______________________________________

insoluble

soluble spect ions

BaSO4(s)Ba2+(aq) + SO4

2-(aq)

BaSO4(s)

2) FeCl2 (aq) + Na3PO4(aq) Net ionic equation:

______________________________________

Fe3(PO4)2

insoluble

solublespect ions

3Fe2+(aq) + 2PO4

3-(aq) Fe3(PO4)2

1) HCl(aq) + KOH(aq) Net ionic equation:

______________________________________

water

soluble

H+(aq) + OH-

(aq) H2O(l)

H2O(l)

spect ions

SOLUBILITY RULES

Any COMPOUNDS containing these ions are SOLUBLE: “CLAAAN” !

Cl A A A NChlorates Acetates Ammonium Alkali metals Nitrates

ClO3- C2H3O2

- NH4+ Li+ NO3

-

or K+

ClO4- CH3COO- Na+

Rb+

Cs+

Also all chlorides, bromides and iodides are SOLUBLE except those of Pb2+, Ag+, Hg2

2+ “PAH” !

All 8 strong bases (hydroxides) are SOLUBLE.

Precipitation Reactions in Aqueous Solution

Steps for Determining the Mass of Product (ppt) Formed:

1) Identify the substances present in the mixture of solutions

and determine what reaction occurs.

2) Write the balanced molecular equation with state of matter symbols.

3) Calculate moles of each reactant given.(moles = M x VL )

4) Solve to find the “Limiting Reactant”.

5) Convert moles of product to grams of product.

Problem: What mass of iron(III) hydroxide is produced when 35.0mL of 0.250M Fe(NO3)3 is mixed with

55.0mL of 0.180M KOH?BAL. EQ:

__________ + ________ ----> _________ +________

Fe(NO3)3 KOH Fe(OH)3 KNO33 3(0.250M)(0.0350L) (0.180M)(0.0550L)

0.00875mol 0.00990mol

(aq)(s)(aq) (aq)

0.00875mol Fe(NO3)3

1 mol Fe(NO3)3

3 mol KOH = 0.2625 mol KOH

Need this much KOH, but don’t have so Fe(NO3)3 in excessAnd KOH is LR

KOH Fe(OH)3

ppt

= 0.00330mol Fe(OH)3

x 106.88g = 1 mol

Fe – 55.85O – 3(16.00)H - 3(1.01) 106.88g/mol

0.353g Fe(OH)3

0.00990mol mol KOH 3 mol KOH

1 mol Fe(OH)3

0.00330mol

Now, calculate the Molarity of all ions left in solution:

Fe(NO3)3 3KOH Fe(OH)3 3KNO3

ppt!0.00875mol 0.00990mol 0.00330mol

subscripts coefficientsBEFORE AFTERFe3+ NO3

- K+ OH- Fe(OH)3 K+ NO3-

0.00875 mol

0.02625 mol

1 to 3

0.00990 mol

disregard!!

0.00990 mol

spectator ions No CHANGE

0.02625 mol

0.00990 mol

0.00330 mol

1 to 1

Since KOH is LR, then only OH- ions got used up as they formed Fe(OH)3 ppt. Some Fe3+ ions will be left over.

Now to make solving clearer, write “net ionic equation”: & work backwards!!

Fe3+ + 3OH- ---------> Fe(OH)3

0.00330mol ppt formed [Note: 1:1 ratio between

Fe3+mol used & Fe(OH)3 made]

Fe3+ given: Total Vol. Sol’n: mLFe3+ used: mLFe3+ left: mL =

______L

0.00330molused up to make ppt

3(0.00330)mol 0.00990molALL USED UP

0.00875mol- 0.00330mol

0.00545mol

35.055.0

90.0

0.0900

Ions remaining in Sol’n:

K+ _________mol [K+] = _______

NO3- ________mol [NO3

-] = _______

Fe3+ _________mol [Fe3+] = ________

0.00990

0.0900L

0.0900L

0.0900L

0.02625

0.00545

0.11 0.110M

0.29166 0.292M

0.06055 0.0606M

__________ + ________ ----> _________ +________

When 200.0mL of 1.0M NaOH is mixed with 300.0mL of 0.50M AlBr3, find mass of ppt & concentration of all ions left in solution.

NaOH AlBr3NaBr Al(OH)33 3

(1.0M)(0.2000L) (0.50M)(0.3000L)

0.20mol 0.15mol

(s)(aq)(aq) (aq)

NaOH ---> ppt AlBr3 ---> ppt

3 mol = 1 mol0.20mol x mol


x = 0.067mol ppt x = 0.15mol ppt

LR = NaOH

Mass of ppt:

0.067mol Al(OH)3

Al – 26.98O – 3(16.00)H - 3(1.01) 78.01g/mol

x 78.01g = 1 mol

5.22667

5.2g Al(OH)3

Al3+ given: Total Vol. Sol’n: mLAl3+ used: mLAl3+ left: mL =

______L

BEFORE AFTERNa+ OH- Al3+ Br- Al(OH)3 Na+ Br-

0.20 mol

0.20 mol

0.15 mol

0.45 mol

spectator ions

0.45 mol

0.20 mol

0.067 mol

Al3+ + 3OH- ------> Al(OH)3

LR

0.067 mol

0.067 mol

3(0.067) 0.20mol all used up

0.15mol- 0.067mol0.08mol

200.0

500.0 0.5000

0.083300.0

Ions Conc left::

Na+ = _____mol [Na+] = _______

Br- = _____mol [Br-] = _______

Al3+ = ____mol [Al3+] = ________

0.200.5000L

0.5000L

0.5000L

0.45

0.08

0.4 0.40M

0.9 0.90M

0.16 0.2M

ACID-BASE REACTIONS

are ______________________reactions.

The “driving force” behind neutralization reactions is the

___________________ or a__________________. !!!!

ACID + BASE -----> SALT + H2O _____ + _____ -----> _____ + _____

neutralization

formation of water weak electrolyte

HA BOH BA H2O

Definitions:

Arrhenius acid _____________________________ Acid increases _________________________________

Ex: recall: HCl(g) + H2O(l) -----> H3O+(aq) + Cl-(aq)

simplified:

Arrhenius base _____________________________ Ex: recall: Ca(OH)2(s) ------> Ca2+

(aq) + 2OH- (aq)

produces H3O+ ions in water

[H+] (molarity of H+ ions) in solution

HCl(aq) -----> H+(aq) + Cl-(aq)

releases OH- ions in water

H2O

Bronsted-Lowry acid ______________________________

Bronsted-Lowry base _____________________________

Ex: recall:

NH3(g) + H2O(l) -----> NH4

+(aq) + OH-

(aq) _____ ______ ______ _______

is a proton donor (H+)

is a proton acceptor (H+)

WEAK BASE

BLB BLA Conj BaseConj Acid

AqueousReactions


Acids

• The Swedish physicist and chemist S. A. Arrhenius defined acids as substances that increase the concentration of H+ when dissolved in water.

• Both the Danish chemist J. N. Brønsted and the British chemist T. M. Lowry defined them as proton donors.

AqueousReactions


Bases

• Arrhenius defined bases as substances that increase the concentration of OH− when dissolved in water.

• Brønsted and Lowry defined them as proton acceptors.

+ Br-(aq)

STRONG ACID-STRONG BASE

Molec. EQ:

Complete Ionic EQ:

Net Ionic EQ:

HBr(aq) + KOH(aq)----> ________ + _____ (aq)(l)H2O KBr

H2O(l) H+(aq) + Br-

(aq) + K+(aq) + OH-

(aq) ---> + K+(aq)

(spect will cancel)

SA,SB,SI get pulled apart

SAME for every SA/SB reaction.

AqueousReactions


Acid-Base Reactions

In an acid–base reaction, the acid donates a proton (H+) to the base.

AqueousReactions


Neutralization Reactions

Generally, when solutions of an acid and a base are combined, the products are a salt and water:

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

________ + _____

WEAK ACID-STRONG BASE

Molec. EQ:

Complete Ionic EQ:

Net Ionic EQ:

HF(aq) + NaOH(aq)----> (aq)(l)H2O NaF

+ F-(aq)

H2O(l) HF(aq) + Na+(aq) + OH-

(aq) ---> + Na+(aq)

WA

+ F-(aq)

H2O(l) HF(aq) + OH-

(aq) --->

STRONG ACID-WEAK BASE

Molec. EQ:

Complete Ionic EQ:

Net Ionic EQ:

________ + _________ HNO3(aq) + Mg(OH)2(aq)----> (aq)(l)H2O Mg(NO3)22 2

2H2O(l) 2H+(aq) + 2NO3

-(aq)

+ Mg(OH)2(aq) ---> + Mg2+(aq)

+ 2NO3-(aq)

2H2O(l)+ Mg(OH)2(aq) ---> + Mg2+(aq) 2H+

(aq)

FINDING MOLARITY OF IONS

5.25g of Ba metal is placed in enough water to make 45.0mL. Find [OH-]

_________ + _______ ------> ______ + _________

Find moles Ba reacted:

Find moles OH-:

Find Molarity of OH-: M = mol L

Ba(s) H2O(l) H2(g) Ba(OH)2(aq)21 mol ----------------------------------------------------------> 1 mol Ba2+ ions

2 mol OH- ions

5.25g Ba x 1 mol = 137.33g 0.0382291

0.0382 mol Ba

0.0382 mol Ba x 2 mol OH- = 1 mol Ba

0.0764 mol OH-

0.0764 0.0450

[OH- ] = 1.70M

1.697777

TITRATION: is an analytical technique whereby a solution of known concentration and volume (titrant) is reacted with a solution of known volume (analyte) to determine its concentration.

AqueousReactions


Titration

Titration is an analytical technique in which one can calculate the concentration of a solute in a solution.

MH+ x Vacid = MOH- x Vbase

moles of H+ = moles of OH-

H+(aq) + OH-

(aq) ------> H2O(l)

Neutralization equation:

MOLARITYacid ion x VOLUMEacid = MOLARITYbase ion x VOLUMEbase

1. Find molarity of HCl if 36.7mL of acid is needed to titrate 43.2mL of 0.236M Ca(OH)2

Note: HCl ---> H+ + Cl- Ca(OH)2 ----> Ca2+ + 2OH-


[H+] = [HCl] [OH-] = 2[Ca(OH)2]

[OH-] = 2(0.236M) = 0.472M

(0.472M) (43.2mL)(36.7mL) =(x)

x = (0.472M)(43.2mL) (36.7mL)

0.555596

x = 0.556M = [H+] = [HCl]

2. What volume of 0.80M Ba(OH)2 is needed to neutralize 50.0mL of a 0.20M H3PO4 solution? Note: H3PO4 ----> 3H+ + PO4

3- Ba(OH)2 ----> Ba2+ + 2OH-


[H+] = 3[H3PO4]

[H+] = 3(0.20M) = 0.60M

[OH-] = 2[Ba(OH)2]

[OH-] = 2(0.80M) = 1.60M

(x)(1.60M)(0.60M)(50.0mL) =

x = 19mL Ba(OH)2

18.75

FINDING MOLARITY OF ALL IONS LEFT IN SOLUTIONWhen 12.5g KOH is added to 450.0mL of 0.250M H2SO4, find concentration of all ions left in solution following this neutralization reaction. Is solution acidic? basic/ or neutral?1) Initially, find moles of base added & moles of acid present:

H2SO4(aq) + KOH(s) -----> H2O(l) + K2SO4(aq)

( )( ) moles moles

20.250M 0.4500L

0.113 0.223

12.5g KOH K – 39.10O – 16.00H - 1.01 56.11g/mol

x 1 mol = 56.11g

0.223mol

2spect

2) Find LR and moles H2O(l) formed:

H2SO4 H2O KOH H2O



x = 0.226mol H2O x = 0.223mol H2O

LR = KOH

3) Before & After moles of all ions:

H+ SO42-

K+ OH- -----> H2O K+ SO42-

4) Net Ionic EQ:

2(0.113) 0.226 mol

0.113 mol

0.223 mol

0.223 mol

0.223 mol

0.223 mol

0.113 mol

H+(aq) + OH-

(aq) H2O(l)0.223 mol

0.223 mol

0.223 mol ALLused up LR

H+ given: Total Vol. Sol’n: mL =H+ used: H+ left:

0.226 mol0.223 mol0.003 mol

450.0

0.4500L

_1

5) now find Molarity of all ions left is solution spectators: K+ _____mol = [K+] = L SO4

2- _____mol = [SO42-] =

L excess ion when forming water: H+ ______mol = [H+] = L Since acid ion is left over , the solution is therefore ______ Formula: pH = - log [H+] so pH = ____

Acidic solutions: pH < 7Basic solutions: pH > 7Neutral solutions: pH = 7

0.223

0.113

0.4500

0.4500

0.0030.4500

0.49555 0.496M

0.25111 0.251M

0.006666 0.007M

acidic

= - log 0.007M

2.1549

2.2_1

AqueousReactions


Oxidation-Reduction Reactions

• An oxidation occurs when an atom or ion loses electrons.• A reduction occurs when an atom or ion gains electrons.• One cannot occur without the other.

AqueousReactions


Oxidation Numbers

To determine if an oxidation–reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.

OXIDATION-REDUCTION REACTIONS “REDOX”

OIL-RIG

OXIDATION is

REDUCTION is

A “redox” reaction always involves a

RULES for ASSIGNING OXIDATION NUMBERS:

1) Any free element is __

2) Monatomic ions is

loss of electrons & an increase in oxidation #.

gain of electrons & a decrease in oxidation #.

change in oxidation #s.

0 H2 Fe C P4

o o o o

the charge on the ion

Na+ Pb4+ N3- Br-+1 +4 -3 -1

3) H in a molecular cmpd is ___ H in a metal hydride is ___

4) O in a compound is ___ except peroxides is ___

5) F in a compound is always ____

In ionic compounds, it is the ion’s charge anyway, but in a molecular compound, F atom gets a -1 because it has the highest EN of all atoms (ie greatest attraction for electrons in a shared pair)

+1 -1 +3 -1 NaH FeH3

+1 -1

-2 -1

Na2O2 H2O2+1 -1 +1 -1

-1

NF3+3 -1

BrCl+1 -1

IBr+1 -1

highest EN halogen gets the -1

6) sum of oxidation #s in a compound is ___ 7) sum of oxidation #s in a polyatomic ion is its _________

0

charge

CrO4 2-

= -2

-2

-8

Cr = +6

NO2 -

= -1

-2

-4

N= +3

REACTION TYPES. Yes, ONLY IF oxid #s change after the reaction!

1) Synthesis:

E + E ---> C E + C ---> C C + C ---> C __________ ___________ ___________

2) Decomposition:

C ---> E + E C ---> E + C C ---> C + C __________ __________ ____________

3) Single Replacement: E + C ----> E + C

_______________

0 0 + - 0 +- + -

Yes Yes Usually Not

Yes Yes Usually Not

+ - 0 0 + - 0 + -

0 +- 0 + -

Yes always

4) Double Replacement (metathesis):

C + C ----> C + C ____________

5) Combustion: CxHy + O2 ----> CO2 + H2O

__________________

(2 ions switching places)

NO

Yes always

-?+1 0 +4-2 +1-2

A REDUCING AGENT causes __________ in another reactant,

but itself undergoes __________.

Metal elements love to _____ electrons (________) and

form (+) ions, so they are good reducing agents.

2Na(s) + O2(g) -----> 2Na2O(s)

oxidation

reduction

lose oxidize

0 0 +1 -2

oxid

red

Oxid. Agent:

Red. Agent: Na

O2

An OXIDIZING AGENT causes __________ in another reactant,

but itself undergoes __________. Non-metal elements love to _____ electrons (_________)

and form (-) ions, so they are good oxidizing agents.

oxidation

reduction

reducegain

Note: Atoms in POLYATOMIC IONS with HIGH oxidation #s are

______________________________________________

good reducers therefore strong oxidizing agents

KMnO4 = 0

-2

-8

Mn = +7+1

+1

It is possible to get a “non-integer” oxidation #:

Means

Fe3O4

+8

4

-8

= 0

-2

-8+8

+8/3

Broader Definition: OXIDATION is the ______________

REDUCTION is the _______________

Ex: PbO(s) + CO(g) ---> Pb(s) + CO2(g)

PbO _______________ and _______________

CO _______________ and _______________

Common method for reducing metal oxides to get the metal element isolated.

gain of oxygen

loss of oxygen

+2 -2 +2-2 0 +4 -2

reduces from +2 to 0

oxidizes from +2 to +4

loses oxygen

gains oxygen

6FeCl2(aq) + 6HCl(aq) + NaClO3(aq) --> 6FeCl3(aq) + NaCl(aq) +3H2O(l)

6Fe2+(aq) + 12Cl-(aq) + 6H+

(aq) + 6Cl-(aq) + Na+(aq) + ClO3

-(aq) ----->

6Fe3+

(aq) + 18Cl-(aq) + Na+(aq) + Cl-(aq) + 3H2O(l)

Net ionic EQ:

6Fe2+(aq) + 6H+

(aq) + ClO3-(aq) ---> 6Fe3+

(aq) + Cl-(aq) + 3H2O(l)

Oxidation:

Reduction:

6Fe2+ ---> 6Fe3+ + 6e-

ClO3- + 6e- --> Cl-

Everything balances: elements, e-s & charges

(+12) + (+6) + (-1) = (+18) + (-1) + (0)

(+17) = (+17)

We can consider a balanced net ionic equation as the sum of

an oxidation ½ reaction + a reduction ½ reaction

today

tomorrow

BALANCE by the Half-Reaction Method (see steps!) MnO4

- + C2O42- ----> Mn2+ + CO2

oxidation reduction

+7 -2 +3 -2 +2 +4-2

MnO4- -->Mn2+

C2O42- --> CO2

+ 4H2O 8H+ +

(-2) = (0) + 2e-

(+8) + (-1) = (+2) + (0)5e- +2

(+7) = (+2)

25

(in acidic solution)

5C2O42- --> 10CO2 + 10e-

10e- + 16H+ + 2MnO4

- --> 2Mn2+ + 8H2O

5C2O42- + 16H+ + 2MnO4

- --> 10CO2 + 2Mn2+ + 8H2O

Redox Balancing in BASIC solution

Al + MnO4- ----> MnO2 + Al(OH)4

-0 +7 -2 +4 -2 +3 -2+1

Al-->Al(OH)4- MnO4

- -->MnO2 4H2O+ + 2H2O4H+++4H+

+4OH-

4OH-++4OH- +4OH-

4H2O 4H2O(-4) + (0) + (0) = (-1) + (0)

+ 3e- (0) + (-1) = (0) +(0) + (-4)

+ 3e-

4OH-+ 4H2O+ Al + 4H2O+ MnO4- --> Al(OH)4

- +4H2O+ MnO2

+ 2H2O +4OH-+ 2H2O

Al + 2H2O + MnO4- ----> Al(OH)4

- + MnO2

oxidation reduction

1)

Cl2 ----> Cl- + OCl-0 -1 -2+1

oxidation reduction

2)

Cl2---> OCl- Cl2 ----> Cl-

2 2 2H2O+ +4H++4OH-

4OH-+

4H2O(-4) + (0) + (0) = (-2) + (0) (0) = (-2)

+ 2e-

2e- +

4OH-+ 2H2O + + Cl2 ----> 2OCl- + 4H2O + 2Cl-

2H2O

4OH- + 2Cl2 ----> 2OCl- + 2H2O + 2Cl-

Cl2

hypochlorite ion

SINGLE-REPLACEMENT REACTIONS all are redox!!!

E + C ----> E + CWhen Element is a METAL:

spect. ion_____

____ + _____ ----> ____ + _____

The metal element _________________________________

When Element is a NON-METAL:

spect. ion_____ ____ + _____ ----> ____ + _____

The non-metal element ______________________________ _

M AB A MB0 + - 0 + - B-

M oxidizes ↑ while metal ion A+ reduces↓

A+

N AB ANB0 + - 0 + -

N reduces ↓ while nonmetal ion B-↑ oxidizes

Problem: Zn(s) + AgNO3(aq) ---> _____ + _________

Eooxid = _____

Eored = _____

Eocell = _____

Complete ionic EQ:

____________________________________________________

Net Ionic EQ:

Ag(s) Zn(NO3)2(aq)2 20 + 0 2+

0.76V0.80V1.56V

yes

Zn(s) + 2Ag+(aq) + 2NO3

-(aq) -----> 2Ag(s) + Zn2+

(aq) + 2NO3-(aq)

Zn(s) + 2Ag+(aq) -----> 2Ag(s) + Zn2+

(aq)

Spect ion: _____NO3-

Date post:	01-Apr-2015
Category:	Documents
Upload:	lila-tolly
View:	237 times
Download:	4 times

UNIT 2: CHEMICAL REACTIONS. 4.1, 4.2 Aqueous Solutions _____________ = _____________ +...

Documents

UNIT 2: CHEMICAL REACTIONS. 4.1, 4.2 Aqueous Solutions _ = _ +...