Zumdahl Ch 04 Solutions

Zumdahl Chapter 4

Types of chemical reactions and solution chemistry

Vocabulary review

 solution - homogeneous mixture

 solute - gets dissolved

 solvent - does the dissolving

 soluble - can be dissolved

Aqueous solutions



Dissolved in water.



Water is a good solvent because the

molecules are polar.

Hydration

 The process of breaking the ions of salts apart.

Hydration

H

O

H

H

H

O

H

H

O

H

O

H

Like dissolves like.



methanol is a polar molecule and dissolves

Chemical cocktail

 http://www.youtube.com/watch?feature=player_ embedded&v=JE4pwRD8t9Q

Electrolytes

 Electricity is moving charges.

 The ions that are dissolved can move.

 Solutions of ionic compounds can conduct electricity.

Types of solutions



Strong electrolytes - completely

dissociate (separate into ions); conduct well. (Ex: NaCl)



Weak electrolytes - partially separate into ions; conduct slightly. (Ex: HC₂H₃O₂)



_{Non-electrolytes – don’t separate. (No}

Types of solutions



Acids- form H+ ions when dissolved.



Strong acids dissociate completely; mostly

ions



_{H2SO4 HNO3 HCl HBr HI HClO4}



Weak acids - dissociate slightly.



Bases - form OH- ions when dissolved.



Strong bases - mostly ions. Ex: KOH

Ionic compounds

Ionic compounds

“

“

dissociate

dissociate

”

”

NaCl(s)



AgNO

₃

(s)



MgCl

₂

(s)



Na

₂

SO

₄

(s)



AlCl

₃

(s)



Na

+

(aq) + Cl

-

(aq)

Ag

+

(aq) + NO

3

-(aq)

Mg

2+

(aq) + 2 Cl

-

(aq)

2 Na

+

(aq) + SO

4

2-(aq)

Solubility

 How much of a substance will dissolve in a given amount of water.

 Usually in g/100 mL

Measuring molarity

 Concentration- how much is dissolved per unit of volume

 Molarity = moles of solute liters of solution

 abbreviated M

Molarity



A solution is made by dissolving 45.6 g of

Fe₂(SO₄)₃ into 475 mL of solution.



What is the concentration of each ion?

 [Fe3+] = 45.6 g mol 2 mol Fe3+

399.9 g 0.475 L 1 mol Fe₂(SO₄)₃

= 0.480 M Fe3+

Making solutions

 Describe how to make 100.0 mL of a 1.00 M K₂CrO₄ solution.

 (translation: how many grams of the solid are needed to make the solution?)

 1.00 mol K₂CrO₄

L

0.100 L 194.19 g K₂CrO₄

mol K₂CrO₄ = 19.4 g K₂CrO₄

Dilution



_{Adding more solvent to a known solution.}



The moles of solute stay the same.



moles = M x V (volume in liters)



_{M1 V1 = M2 V2}



moles = moles



stock solution (1) - solution of known

Dilution

 How would you prepare 200.0 mL of a 0.100 M HCl solution from a stock solution of 12.0 M HCl?

 M₁ V₁ = M₂ V₂

 V₁ = M₂ V₂ = 0.100 M 200. mL = 1.67 mL

M₁ 12.0 M

 Take 1.67 mL of the stock solution, transfer to

Dilution

End of lesson

4.3 Quiz: Solubility

Fill in the blanks:

1.

Like dissolves ______.

2.

Sodium chloride dissolves in ______.

3.

Ethanol and water are ___________.

4.

Oil and water are __________.

4.4-5 Types of Reactions

precipitation acid-base

reactions reactions reduction reactions

both double replacement

•

single repl.

•

combination

•

combustion

4.5 Precipitation reactions



When aqueous solutions of ionic

compounds are poured together, a solid forms.

Precipitation reactions

 _{3NaOH(aq) + FeCl3(aq)  3NaCl(aq) + Fe(OH)3(s)}  is really

 3 Na+(aq) + 3 OH-(aq) + Fe3+(aq) + 3 Cl-(aq) _

3 Na+(aq) + 3 Cl-(aq) + Fe(OH)

3(s)  … so all that really happens is:

 _{3 OH}-(aq) + Fe+3 _ Fe(OH)

Precipitation reaction



We can predict the products



_{Can only be certain by experimenting}

Precipitation reactions



Only happen if one of the products is insoluble



_{Otherwise all the ions stay in solution - nothing}

happens.

Solubility rules



All nitrates are soluble



Compounds containing alkali metal ions

and NH₄+ ions are soluble



Halides are soluble except Ag+, Pb2+, and

Hg₂2+



Most sulfates are soluble, except Pb2+,

Ba2+, Hg

Solubility Rules



Most hydroxides are slightly soluble

(insoluble) except NaOH and KOH



Most sulfides, carbonates, chromates,

and phosphates are insoluble



Lower number rules supersede, so Na₂S is

Are these ions soluble? Or do

they form a ppt?

a.NaCl f. MgI2

b.K₃PO₄ g. Ca(OH)₂

c. LiBr h. Ag₂CrO₄

d.AgCli. PbCO₃

Does a precipitate form?

 Ca(NO₃)₂ + Na₂C₂O₄  ppt??

 Ca2+ NO

3- Na+ C2O4

2- Ca(NO₃)₂? CaC₂O₄? NaNO₃? Na₂C₂O₄?

What ppt will form, if any?

a.

NaOH + Cu(NO₃)₂ 

b.

FeSO₄ + BaCl₂ 

c.

KCl + NaNO₃ 

d.

CuCl₂ + AgNO₃ 

Cu(OH)₂

BaSO₄

no ppt

Write balanced equations;

identify ppts as (s).



AgNO

_3(aq)

+ KCl

_(aq)



4.6 Net ionic equations



Objective: Students are able to write

Three Types of Equations

 Balanced formula equation- written as whole formulas, not separated into ions.



_{Ex: K2CrO4(aq) + Ba(NO3)2(aq)  ?}

2KNO_3(aq) + BaCrO_4(s)

• Complete ionic equation shows dissolved electrolytes as ions.



_{Ex: 2K}+ + CrO

42- + Ba2+ + 2NO3- 

BaCrO₄(s) + 2K+ + 2NO 3

 Net ionic equations show only those ions that react, omitting the spectator ions



Ex: Ba

2+

+ CrO

How do you know what to write in

ionic versus molecular form for the

complete ionic equation?

 Use ionic form for

 soluble ionic compounds  strong acids and bases

 Use molecular form for

 insoluble ionic compounds  weak acids and bases

Practice with net ionic

equations



Write the three types of equations for

the reactions when these solutions are

mixed:

a)

iron (II) sulfate and potassium

sulfide

b)

lead (II) nitrate and sulfuric acid

a) iron (II) sulfate + potassium sulfide

BFE (balanced formula equation):



FeSO_4(aq) + K₂S_(aq)  FeS_(s) + K₂SO_4(aq)

CIE (complete ionic equation):



Fe2+ + SO

42- + 2K+ + S2- 

2K+ + SO

42- + FeS(s)

NIE (net ionic equation):



Fe2+ + S2-  FeS

b) lead (II) nitrate and sulfuric acid

BFE:



Pb(NO₃)₂ + H₂SO_4(aq)  PbSO_4(s) + 2HNO_3(aq)

CIE:



Pb2+ + 2NO

3- + 2H+ + SO42-  PbSO4(s)

+ 2H+ + 2NO 3

-NIE:



Pb2+ + SO

BFE:



HCl + KOH



KCl + H

₂

O

_(l)

CIE:



H

+

+ Cl

-

+ K

+

+ OH

-



K

+

+ Cl

-

+ H

2

O

(l)

NIE:



H

+

+ OH

-



H

2

O

(l)

4.7 Stoichiometry of

precipitation reactions



Objective: Students will be able to

Stoichiometry of precipitation



Similar to other stoichiometry problems,

Sample problem 1:



What volume of 0.204 M HCl is needed to

precipitate the silver from 50.0 mL of 0.0500 M silver nitrate solution ?



Start with balanced equation:

 Carry out a stoichiometric calculation.

 = 0.0123 L = 12.3 mL

0.0500 mol AgNO₃

L

0.0500 L 1 mol HCl 1 mol AgNO₃

L

Sample problem 2:



(a) What mass of lead (II) carbonate

can be produced when 20.8 mL of

0.500 M lead (II) nitrate reacts with

30.7 mL of 0.400 M potassium

carbonate?



(b) Calculate the concentrations of

all ions remaining in solution after

precipitation is complete.

 _{Carry out two stoichiometric calculations for (a).}

0.500 mol Pb(NO₃)₂ 0.0208 L 1 mol PbCO₃ 267.2 g PbCO₃

L 1 mol Pb(NO₃) mol PbCO₃

= 2.78 g PbCO₃

 0.400 mol K₂CO₃ 0.0307 L 1 mol PbCO₃ 267.2 g PbCO₃

L 1 mol K₂CO₃ mol K₂CO₃

= 3.28 g PbCO₃

 _{The smaller mass is the theoretical yield.}

 Pb(NO₃)₂ is limiting.

 (b) Calculate the concentrations of all ions remaining in solution after precipitation is complete.

 [Pb2+] = 0 M because it is all used up (limiting

reactant)

 [NO₃-] = 0.500 M 2 0.0208 L = 0.404

M

 _[CO32-] = mol initial – mol used

L total

= (0.400 M x 0.0307 L) - (0.500 M x 0.0208 L) (0.0307 + 0.0208) L

= 0.0123 - 0.0104 mol

4.8 Stoichiometry of

acid-base reactions



Objectives: Students will be able to

write complete ionic equations and

net ionic equations for acid-base

reactions

Acid - base reactions



An acid is a proton donor.



A base is a proton acceptor, usually OH

-

What is the net ionic equation for the

reaction of HCl(aq) and KOH(aq)?



Strong acid + strong base



salt + water



NIE: H

+

+ OH

-



H



Often called a neutralization reaction

because the acid neutralizes the base.



We use titration to determine

concentrations.



A solution of known concentration

(titrant) …



is added to the unknown (analyte),



until the equivalence point is reached,

Titration



endpoint: where the indicator changes

color.



not always at the equivalence point.



A 50.00 mL sample of aqueous Ca(OH)

₂

requires 34.66 mL of 0.0980 M nitric acid

for neutralization. What is [Ca(OH)

₂

]?



# of H+ x M

A 50.00 mL sample of aqueous Ca(OH)

₂

requires 34.66 mL of 0.0980 M nitric acid

for neutralization. What is [Ca(OH)

₂

]?



# of H+ x M

A x VA = # of OH- x MB x VB



In this case: M_A V_A = 2 M_B V_B



M_B = M_A V_A

2 V_B

= 0.0980 M 34.66 mL

Review of net ionic equations for

acid-base reactions



Write the BFE, CIE, and NIE for the reaction of

solid lead (II) hydroxide and sulfuric acid.



BFE:



_{Pb(OH)2(s) + H2SO4(aq)  PbSO4(s) + 2H2O(l)}



CIE:



_{Pb(OH)2(s) + 2H}+ + SO

42-(aq)  PbSO4(s) + 2H2O(l)

Left-over concentration problem



When 50.0 mL of 0.200 M NaOH reacts

with 150.0 mL of 0.300 M H

₂

SO

₄

,

calculate the concentration of H

+

or OH

-ion left over.



2NaOH + H

₂

SO

₄



Na

₂

SO

₄

+ 2H

₂

O



2OH

-

+ 2H

+



2H

OH

-

+ H

+



H

2

O

0.0500L x 0.200 M 0.150 L x 0.300 M x 2

= 0.0100 mol = 0.0900 mol

- 0.0100 mol - 0.0100 mol

0 0.0800 mol

M = 0.0800 mol = 0.400 M H+ in excess



75.0 mL of 0.250 M HCl is mixed with 225 mL

of 0.055 M Ba(OH)₂ . What is the concentration

of the excess H+ or OH- ?

 75.0 mL of 0.250 M HCl is mixed with 225 mL of 0.055 M Ba(OH)₂ . What is the concentration of the excess H+ or OH- ?

H

+

+ OH

-



H

2

O

0.0750 L x 0.250 M 0.225 L x 0.055 M x 2

= 0.0188 mol H+ = 0.0248 mol

- 0.0188 mol - 0.0188 mol

0 0.0060 mol

OH-M = 0.0060 mol = 0.020 OH-M OH- in excess

Pre-lab questions for

juice titration lab

1.

Write the complete chemical equation

for the reaction of a solution of NaOH with HCl.

2.

How many mL of 0.10 M HCl are required to react completely with 5.00 mL of 0.10

M NaOH?

3.

If equal molar amounts of NaOH and HCl are mixed, when the reaction is complete what will be the chemical species in the resulting solution?

4.

What will be the pH of the mixture in question 3, acidic, neutral, or basic? Explain.

5.

Write the chemical equation for the

reaction of a 0.10 M solution of acetic acid (CH₃COOH) with a 0.10 M solution of

Indicator pH range Color Change methyl orange

 

3.1–4.4 orange to yellow

methyl red  

4.2–6.2 red to yellow

bromthymol blue

 

6.0–7.6 yellow to green-blue

phenolphthalei n

 

8.3–10.0 colorless to pink

thymol blue  

1.2–2.8  

red to yellow  

6. How many mL of 0.10 M NaOH solution are required to react completely with 5.00 ml of 0.10 M acetic acid (CH₃COOH)

solution? Explain.

7. At the equivalence point, what will be the pH (acidic, neutral, basic) of the solution in question 6? Explain.

8. How is it possible to determine when an acid-base reaction is complete?

Hydrates



_{Not part of the AP test…}



_{Hydrates are compounds that have water}

chemically incorporated into the crystal. There are usually whole numbers of water molecules per one crystal formula unit.



_{hydrated copper (II) sulfate = CuSO4•5H2O.}



For every mole of copper (II) sulfate, there are

5 moles of water attached chemically. The

 Ex. 1 A student takes 1.082g of hydrated nickel (II) sulfate and heats it in a crucible. After heating, the anhydrous material (no water) has a mass of 0.596g. What is the

complete formula for this hydrated salt? What is its name?

 Answer: Subtract to get mass of H₂O.

 _{1.082 g (hydrate) – 0.596 g (anhydrous) =}

 0.486 g H₂O

 Divide grams by molar mass.

 0.596 g mol NiSO₄ = 0.00385 mol NiSO₄

 Repeat for the water.

 0.486 g mol H₂O = 0.0270 mol H₂O

18.02 g H₂O

 Divide moles of water by moles of anhydrous compound to get a whole number ratio.

 0.0270 mol H₂O = 7.01 : 1

0.00385 mol NiSO₄

Formula = NiSO₄ •7 H₂O



Ex. 2 A hydrated crystal is found to be 89.2% barium bromide and 10.8% water by weight. What is the formula for this hydrate? What is its name?

 Answer: assume a 100 g sample.

 89.2 g anhydrous; 10.8 g water

 Divide grams by molar mass.

 89.2 g mol BaBr₂ = 0.300 mol BaBr₂

 Repeat for the water.

 10.8 g mol H₂O = 0.599 mol H₂O

18.02 g H₂O

 Divide moles of water by moles of anhydrous compound to get a whole number ratio.

 0.599 mol H₂O = 2.00 : 1

0.300 mol BaBr₂

Formula = BaBr₂ •2 H₂O

4.9 Red-ox reactions

 Objective: Students will be able to determine the oxidation number of any element in the formula of a compound.

 An oxidation-reduction (red-ox) reaction involves the transfer of electrons.

Rules for assigning

oxidation states

 A system of keeping track of the electrons.

 Not absolutely accurate for where the electrons are, but it works.

 Written as +1, +2, etc.

 Memorize these rules:

 Elements in their standard states have an oxidation state of zero.

 Oxygen is assigned an oxidation state of -2 in its compounds, except peroxides (-1).

 Hydrogen +1 in compounds, except hydrides (-1).

 Fluorine always –1 in compounds.



Assign oxidation states to each element

in the following:



CO

₂



PF

₅



CO

₃

2-

H

₂

SO

₄



Fe

₂

O

₃



Fe

₃

O

₄

+4, -2

+5, -1

+4, -2

+1, +6, -2

+3, -2

+8/3, -2



Red-ox reactions are reactions in

which one element is oxidized and

one is reduced.



2Na + Cl

₂



2NaCl



CH

₄

+ 2O

₂



CO

₂

+ 2H

₂

O



Oxidation is the loss of

electrons/increase in ox state.



Reduction is the gain of



OIL RIG

or



LEO the lion

says “GER!”

(Oxidation is loss; reduction is gain!)



The oxidizing agent contains the element being reduced.



The reducing agent contains the element

being oxidized.



Identify whether the reaction is

red-ox, which element is oxidized, which

is reduced, the oxidizing agent, and

the reducing agent:



(1) Zn + HCl



ZnCl

₂

+ H

₂

0 +1 -1 +2 -1 0

OXIDATION

OXIDATION

REDUCTION

Reducing



Identify whether the reaction is

red-ox, which element is oxidized, which

is reduced, the oxidizing agent, and

the reducing agent:



(2) HCl + NaOH



H

₂

O + NaCl

+1 -1 +1 -2 +1 +1 -2 +1 -1



Identify whether the reaction is

red-ox, which element is oxidized, which

is reduced, the oxidizing agent, and

the reducing agent:



(3) CH

₄

+ 2O

₂



CO

₂

+ 2H

₂

O

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

OXIDATION OXIDATION

REDUCTION

Reducing



(4) (more challenging!)

Cr₂O₇2- + 3Sn2+ + 14H+  2Cr3+ + 3Sn4+ +7H 2O

+6 -2 +2 +1 +3 +4 +1 -2

OXIDATION

OXIDATION

REDUCTION

Reducing agent Oxidizing



(5) (super challenging, almost impossible!)

FeS + 3NO₃- + 4H+  3NO + SO

42- + Fe3+

+2 -2 +5 -2 +1 +2 -2 +6 -2 +3

REDUCTION

Reducing agent

Oxidizing agent OXIDATION

4.10 Balancing red-ox

equations

Half-Reactions



All red-ox reactions can be thought of as

happening in two halves.



Oxidation produces electrons



Reduction requires electrons



Write half reactions for the following

(unbalanced) equations.



_{Na + Cl2  Na}+ + Cl

-

_SO32- + H+ + MnO

Balancing red-ox equations in

acidic solution



The key is the number of e-s given off in

oxidation must be the same as those required for reduction.



In acidic solution, use an 8-step procedure:



Write separate half reactions



For each half reaction balance all elements

except H and O



Balance H using H+



Balance charge using e-s



Multiply equations to make # of electrons equal



Add equations and cancel identical species



Check that charges and elements are

Practice

 Balance the following reactions occurring in acidic solution.  _MnO4- + Fe2+ _ Mn2+ + Fe3+

 Answer:

 _{Red: MnO4}-  Mn2+

 _{Ox: Fe}2+  Fe3+

 _MnO4- + 5Fe2+ + 8H+ + 5e-

 Mn2+ + 5Fe3+ + 4H

2O +

5+

e-+ 4H₂O

+ 8H+ +

Practice

2 Cr(OH)₃ + 3 OCl-  2 CrO

42- + 3 Cl- + 4H+ + H2O

+ 5 H+

+ H₂O + 3 e

-+ H₂O + 2 H+ + 2 e

-2{ }

3{ }

Balance the following reactions occurring in aqueous solution.

_{Cr(OH)3 + OCl}- _ CrO

42- + Cl- + H2O

_{Cr(OH)3  CrO4}2-

_OCl- _ Cl

Practice

 Balance the following reactions occurring in aqueous solution.



_{Cu + NO3}- _ Cu2+ + NO (g)



_{Pb + PbO2 + SO4}2-  PbSO

4



_Mn2+ + NaBiO

-Now for a tough one

Fe(CN)₆4- + MnO

-Basic Solution



Do everything you would with acid, but add one

more step:



Add enough OH- to both sides to neutralize the

H+



H+ and OH- add together to give H₂O.



Try these:



Ag + Zn2+  Ag

2O + Zn



Cr₂O₇2- + C

Answers:

Ag + Zn2+ _{ Ag}

2O + Zn …?  Ag  Ag₂O

Zn2+ _{ Zn} 2 Ag + Zn2+ _{+ H}

2O  Ag2O + Zn + 2H+

Cr₂O₇2- _{+ 3C}

2H6O + 8OH- + H2O  2Cr3+ + 3C2H4O … Not finished!!

+ H₂O

2 + 2H+ + 2e

-Red-ox Titrations



Same as any other titration.



The permanganate ion is often used

because it is its own indicator. MnO₄- is

purple, Mn2+ is colorless. When reaction

solution remains clear, MnO₄- is gone.



Chromate ion is also useful, but its color

Example



The iron content of iron ore can be determined

by titration with standard KMnO₄ solution. The

iron ore is dissolved in excess HCl, and the

iron reduced to Fe+2 ions. This solution is then

titrated with KMnO₄ solution, producing Fe+3

and Mn+2 ions in acidic solution. If it requires

41.95 mL of 0.205 M KMnO₄ to titrate a

Other random information…

Solut

e

A solute is the dissolved substance in a

A solute is the dissolved substance in a

solution.

solution.

A solvent is the dissolving medium in a

A solvent is the dissolving medium in a

solution.

solution.

Solvent

Solvent

Salt in salt water Sugar in soda drinks

Carbon dioxide in soda drinks

“Like Dissolves Like”

FatsFats BenzeneBenzene

SteroidsSteroids HexaneHexane

WaxesWaxes TolueneToluene

Inorganic Salts

Inorganic Salts WaterWater

SugarsSugars Small alcoholsSmall alcohols

Acetic acidAcetic acid

Polar and ionic solutes dissolve best in polar solvents

Solubility Trends



The solubility of MOST solids increases with temperature.



The rate at which solids dissolve

increases with increasing surface area of the solid.



The solubility of gases decreases with increases in temperature.

Therefore…

Solids tend to dissolve best when:

o

Heated

o

Stirred

o

Ground into small particles

Gases tend to dissolve best when:

o

The solution is cold

Solubility Chart

Saturation of Solutions

 A solution that contains the maximum amount of solute that may be dissolved under existing conditions is saturated.

 A solution that contains less solute than a

saturated solution under existing conditions is

unsaturated.

 A solution that contains more dissolved solute than a saturated solution under the same

An electrolyte is:

An electrolyte is:

A substance whose aqueous solution

conducts an electric current.

A nonelectrolyte is:

A nonelectrolyte is:

A substance whose aqueous solution

does not conduct an electric current.

The ammeter measures the flow of electrons (current) through the circuit.

If the ammeter measures a current, and the bulb glows, then the solution conducts.

If the ammeter fails to measure a current, and the bulb does not glow, the solution is non-conducting.

1.

Pure water

2.

Tap water

3.

Sugar solution

4.

Sodium chloride solution

5.

Hydrochloric acid solution

6.

Lactic acid solution

7.

Ethyl alcohol solution

8.

Pure sodium chloride

1.

Pure water

2.

Tap water

3.

Sugar solution

4.

Sodium chloride solution

5.

Hydrochloric acid solution

6.

Lactic acid solution

7.

Ethyl alcohol solution

8.

Pure sodium chloride

ELECTROLYTES: NONELECTROLYTES: Tap water (weak)

NaCl solution HCl solution

Lactate solution (weak)

Pure water

Sugar solution Ethanol solution Pure NaCl

The reason for this is the polar nature of the water molecule…

Positive ions associate with the negative end of the water dipole (oxygen).

Negative ions associate with the positive end of the water dipole (hydrogen).

Covalent acids form ions in solution, with the help of the water molecules.

For instance, hydrogen chloride molecules, which are polar, give up their hydrogens to

water, forming chloride ions (Cl-) and

hydronium ions (H₃O+).

Other examples of strong acids include:



Sulfuric acid, H

₂

SO

₄



Nitric acid, HNO

₃



Hydriodic acid, HI



Perchloric acid, HClO

₄

Many of these weaker acids

are “organic” acids that contain a “carboxyl” group.

The carboxyl group does not easily give up its hydrogen.

Other organic acids and their sources include:

o

Citric acid – citrus fruit

o

Malic acid – apples

o

Butyric acid – rancid butter

o

Amino acids – protein

o

Nucleic acids – DNA and RNA

o

Ascorbic acid – Vitamin C

This is an enormous group of compounds; these are only a few examples.