Aqueous solution reactions1 Aqueous-solution Reactions Classify a reaction by Homogeneous chemical...

aqueous solution reactions 1

Aqueous-solution Reactions

Classify a reaction by

Homogeneous chemical reactions:

gas phasesolutions

aqueous-solution (common occurrence)non-aqueous-solution

Heterogeneous (more than one phase) chemical reactions:

gas / liquidgas / solidliquid / solid

Know the meaning of terms


Nature of Aqueous Solutions

Nature of compounds

molecular substances (polar, non-polar, H-bonding)non-electrolytes

ionic substances (acids, bases, salts)strong electrolytes (completely ionized in solution)week electrolytes (not completely ionized in solution)

Know your terms and species (in the solution)


Dissolving a Strong Electrolyte

See them in your imagination

http://www.chem.ufl.edu/~chm2040/cgi-bin/image_eval.cgi?Op=display&Chapter=4&File=solutions&Image=c4f10.gif


http://www.chem.ufl.edu/~chm2040/cgi-bin/image_eval.cgi?Op=display&Chapter=4&File=Coolstuff/ionic&Image=c4f8.gif


Strong ElectrolytesStrong acids: HNO3, H2SO4, HCl, HClO4

Strong bases: MOH (M = Na, K, Cs, Rb etc)

Salts: All salts dissolving in water are completely ionized. The ions may react with water (to be discussed in Chem 123)

Stoichiometry & concentration relationship

NaCl (s) Na+ (aq) + Cl– (aq)

Ca(OH)2 (s) Ca+(aq) + 2 OH– (aq)

AlCl3 (s) Al3+ (aq) + 3 Cl– (aq)

(NH4)2SO4 (s) 2 NH4 + (aq) + SO42– (aq)


Concentration of Ions

A bottle labeled as 0.100 M Al2(SO4)3.

[Al3+] = _____ M (mol / L)

[SO42–] = _____ M

Assume sea water is 0.438 M NaCl, 0.0512 M MgCl2, and 0.001 M CaCl2

[Na+] = _____ M

[Mg2+] = _____ M

[Ca2+] = _____ M

[Cl–] = _____ M Know how to calculate your quantities


Precipitations

When ions form a solid that is not very soluble, a solid is formed. Such a phenomenon is called precipitation.

The formation of a precipitation is also an equilibrium phenomenon (a subject to be covered in Chem123)

Ag+ (aq) + Cl– (aq) AgCl(s)or AgCl(s) Ag+ (aq) + Cl– (aq) Ksp = [Ag+][Cl–] is a constant

the solubility product


Precipitation Reactions

Ag+ (aq) + NO3– (aq) + Cs+ (aq) + I– (aq) AgI (s) + NO3

– (aq) + Cs+ (aq)

Ag+ (aq) + I– (aq) AgI (s) (net reaction)or

Ag+ + I– AgI (s)

Heterogeneous Reactions

Spectator ions or bystander ions

Soluble ions

Alkali metals, NH4+

nitrates, ClO4-,

acetate

Mostly soluble ions

Halides, sulfates

Mostly insoluble

Silver halidesMetal sulfides, hydroxidescarbonates, phosphates


Acid-base Reactions HCl (g) H+ (aq) + Cl– (aq)

NaOH (s) Na+ (aq) + OH– (aq)

neutralization reaction: H+ (aq) + OH– (aq) H2O (l)

Explain these reactions

Mg(OH)2 (s) + 2 H+ Mg2+ (aq) + 2 H2O (l)

CaCO3 (s) + 2 H+ Ca2+ (aq) + H2O (l) + CO2 (g)

Mg(OH)2 (s) + 2 HC2H3O2 Mg2+ (aq) + 2 H2O (l) + 2 C2H3O2 – (aq)

acetic acid


Oxidation-reduction reactions

Oxidation reaction must be accompanied by reductions – redox reactions

Increasing oxidation state is oxidation (loss e–, LEO)Decreasing oxidation state is reduction (gain e–, GER)

What elements are oxidized and reduced in each reaction? Work out the oxidation state changes for them as well!

2 KClO3 2 KCl + 3 O2

Fe2O3 + 3 CO 2 Fe + 3 CO2

MnO2 + 4 H+ + 2 Cl– Mn2+ + 2 H2O + Cl2


Review Oxidation StatesOxidation states is an assigned number. Formal charge concept may be used to assign oxidation states.

Work out the oxidation states of all elements in these species:

NH3 N2H4 NH2OH N2 N2O NO NO2– NO2 NO3

– Cl– Cl2 ClO– ClO2

– ClO2 ClO2

– ClO3–

ClO4–

CO H2C2O4 C2O42–

C2H6 CH4 CO2 CO32–

PH3 P4 H3PO4 PO43–

H2S HS– S2– S6 SO2 SO3– SO3 S2O3

2– SO42–

H2O2

MnO2 KMnO4 MnO4– K2CrO4 CrO4

– K2Cr2O7 Cr2O7–

Review stoichiometry


Half ReactionsThese reactions explained during the lecture:

Zn = Zn2+ + 2 e–

Cu2+ + 2 e– = Cunet (electron transfer) Zn (s) + Cu2+ (aq) = Zn2+ (aq) + Cu (s)

For these half reactions,

Zn = Zn2+ + 2 e–

2 H+ + 2 e– = H2 get and explain the net reaction yourself

Explain how reactions proceed


Balance Half Equations

1. Identify the key element that undergoes an oxidation state change.

2. Balance the number of atoms of the key element on both sides.

3. Add the appropriate number of electrons to compensate for the change of oxidation state.

4. Add H+ (in acid medium), or OH- (in basic medium), to balance the charge on both sides of the half-reactions; and H2O, if necessary, to balance the equations.

Balance

MnO4 – + ____ + _____ Mn2+ + __ H2O

Page 9 of handout


Balance these half equations

Zn (s) Zn2+(aq)

Cu2+ (aq) Cu (s)

H2(g) H+ (aq)

I –(aq) I2

Fe2+ (aq) Fe3+ (aq)

SO32– (aq) SO4

2– (aq)

2 S2O32–(aq) S4O6

2–(aq) + 2e–1

Cr2O72–(aq) Cr3+ (aq)

I will illustrate the balance of these half reaction equations and those equations in other slides during the lecture. If you are not at my lecture, you should practice their balance to acquire your skills.

The textbook gives a slight different method to balance redox equations and please find out the difference. Both ways wok. You may use either method.

Work on these from time to time


Balance Redox Reaction equations

Add two half reaction equations so that you can cancel all electrons to obtain a balanced redox reaction equation

H2O2 + I – I2 + H2O

MnO4 – + H2O2 Mn2+ + O2

MnO4 – + SO3

2– Mn2+ + SO42–

MnO4 – + Fe2+ Mn2+ + Fe3+

Cr2O7 2– + UO2+ Cr3+ + UO2

2+

Work on these from time to time to refresh your skills.


Disproportionation ReactionsBalance disproportionation reaction (the same substance is both oxidized and reduced)

H2O2 2 H2O + O2

S2O32– SO4

2– + S (s)

S2O32– SO2 + S (s)

Work on these from time to time


Analyze and Learn the SkillsAnalyze this example and learn the skills to help you overcome any difficulty.

Task: Balance the equation: S2O32– SO4

2– + S (s)

Identify the element oxidized and reduced: oxidation states of S in S2O32– SO4

2– and S are +4, +6, and 0 respectively. S is both oxidized and reduced.

The oxidation half reaction: S2O32– SO4

2– + 8 e– (2 S from +2 +6, (2×4=8))The reduction half reaction: S2O3

2– + 4 e – (2S from +2 0, (2×2=4)) S

Balance the charge with H+ S2O32–SO4

2– + 8 e– + 10 H+ (both sides have 2-) S2O3

2– + 4 e – + 6 H+ S (both sides have 0)

Add water to balance: S2O32– + 5 H2O SO4

2– + 8 e– + 10 H+ S2O3

2– + 4 e – + 6 H+ S + 3 H2O

Make # of e the same: S2O32– + 5 H2O SO4

2– + 8 e– + 10 H+ 2 S2O3

2– + 8 e – + 12 H+ S + 6 H2O

The balanced equation: 3 S2O32– + 2 H+ SO4

2– + 4 S + 10 H+

Make sure you fill in the details too much to be included here!!!


Balance Redox Reaction in Basic Solutions

Redox reactions may have different products depends on the acidity (pH) of the solutions.

In basic solutions, there are more OH– ions than H+ ions. Thus, it is sensible to have OH– appearing in the equations than to have H+ ions.

Balance these reactions in a basic solution:

MnO4 – + CN – MnO2 + OCN –

Any one of several ways to assign oxidation states for CN works.

MnO4 – + SO3

2– MnO2 + SO42–

Practice balance these from time to time to polish you skills!


Oxidizing and Reducing Agents

Oxidizing agent or oxidant such as O2 or F2 is a substance that is able to oxidize other substances. It is reduced in the process (gains electron or decreases oxidation state).

Describe reducing agent or reductant

NH3 N2H4 NH2OH N2 N2O NO NO2 – NO2 NO3

–

Species Cannot be reduced further

Species Cannot be oxidized further

O x I d a n t

R e d u c t a n t


Titrations

Titration is a method used in volumetric analysis. The addition of a solution is carefully controlled so that stoichiometric amounts can be read from a burette.

Titration can be carried out for instantaneous or rapid reactions such as neutralization and oxidation reactions.

Explain these terms: neutralization,

titration,quivalence point,

half equivalence point, indicators


Conductance in Titration

conductance

V of NaOH added

Cl– Na+

H+ OH–

Conductance measurement of a HCl solution titrated by NaOH is shown:measured (total) conductancearea due to the ions labeled

Do all ions have the same conductance? Why or why not?

Why does the total conductance vary?

Is conductance of an ion depends on the concentration?

Explain physical properties of chemicals


Stoichiometry in Solution Chemistry

For titration calculations, the amount of reactant m is evaluated from the concentration C and volume V by

m = C * V

For example, when m1 amount of acid is neutralized by m2 amount of base, m1 = m1. For redox reactions, similar relationship can also be used, but the stoichiometric relationship should be kept in mind.

Amount in mmol = C in M * V in mL


Volumetric AnalysisA 5.00-mL sample of vinegar requires 38.08 mL of 0.1000 M NaOH solution to reach the equivalence point. What is the concentration of acetic acid in the vinegar by weight percent if its density is 1.01 g / mL?

Solution:

Net reaction: OH– + HC2H3O2 H2O + C2H3O2 – (aq)

38.08 mL OH–

5 mL vinegar0.1000 mol1000 mL

1 mol HC2H3O2 1 mol OH–

60.05 g HC2H3O2 1 mol HC2H3O2

1 mL vinegar1.01 g vinegar

= 0.0453 HC2H3O2 in vinegar

= 4.53 % HC2H3O2 in vinegarThe vinegar has 4.53 % of acetic acid by mass.

What’s the concentration of a 7% vinegar?

If you find this solution difficult to understand, use your own method to solve it


38.08 mL OH–

5 mL vinegar

0.1000 mol1000 mL

1 mol HC2H3O2 1 mol OH–

60.05 g HC2H3O2 1 mol HC2H3O2

1.01 g vinegar1 mL vinegar

= 0.0453 HC2H3O2 in vinegar

= 4.53 % HC2H3O2 in vinegar

What the concentration of a 7% vinegar?

A 5.00-mL sample of vinegar requires 38.08 mL of 0.1000 M NaOH solution to reach the equivalence point. What is the concentration of acetic acid in the vinegar by weight percent if its density is 1.01 g / mL?Solution: a slight variation and hope you find it easier to followNet reaction: OH– + HC2H3O2 H2O + C2H3O2

– (aq)

Mass of acetic acid

Mass of sample4.53 % ofHC2H3O2 in vinegar =

Another way of thinking


Chemical Analysis Application

How much 0.1000 M KMnO4 solution is required to reach the equivalence point in a titration of 1.00 g oxidize oxalic acid (oa = H2C2O4.2H2O)?

Solution: Redox reaction:2 MnO4

– + 5 H2C2O4 + 6 H+ = 2 Mn2+ + 8 H2O + 10 CO2 (balanced? Chk pls)

1.00 g oa1 mol oa

126 g oa

2 mol MnO4

5 mol oa

1000 mL

0.1000 mol= 31.75 mL MnO4

–

Do problem 5.97

Molar mass of H2C2O4.2H2O = 126 g mol-1

Note mole ratio in the balanced equation.


Back TitrationA 1.00-g sample containing MnO2 dissolved in solution is treated with 2.00 g of oxalic acid (oa = H2C2O4.2H2O). Then 50.00 mL 0.1000 M KMnO4 is required for the titration of the excess oa. What is the % MnO2 by mass?

Solution: Redox reaction:2 MnO4

– + 5 H2C2O4 + 6 H+ = 2 Mn2+ + 8 H2O + 10 CO2

H2C2O4 + MnO2 + 2 H+ = Mn2+ + 2 H2O + 2 CO2 (balanced? Chk pls)

50.00 mL MnO4

0.1000 mol MnO4

1000 mL

5 mol oa

2 mol MnO4

86.9 g MnO2

126 g oa

100 %

1.0 g Sample

126 g oa

1 mol oa= 1.575 g oa

(2.000 - 1.575) g oa = 29.0 % MnO2 by mass

Cool head 4 complicated problem, eh!


Review 1Sufficient amount of AgNO3 is placed in a 10.00 mL of tab water, and the AgCl solid is filtered and dried. The solid weighs 0.123 g. What is the concentration of chloride ion?

Solution: access and limiting reagent, and ppt

1 mol AgCl143.4 g AgCl

1 mol Cl-1 mol AgCl

0.123 g AgCl 10.00 mL

= 0.0858 mol / L of Cl–

1000 mL1 L

Express [Cl-] in mol and mass %.


Review 2A 0.1234-g sample (S) NaCl and sugar mixture contains 40% NaCl. When dissolved in water, the solution is treated with AgNO3. How much dry AgCl is the theoretical yield?

Solution: percentage analysis

40 g NaCl100 g S

1 mol AgCl58.5 g NaCl0.1234 g S

= 0.1323 g AgCl theoretical yield

143.4 g AgCl1 mol AgCl

Use the factors to help you think!


Review 3A 0.2345-g sample (S) NaCl and CaCl2 mixture contains 40% NaCl. When dissolved in water, the solution is treated with AgNO3. How much dry AgCl is the theoretical yield?

Solution:

0.2345*0.40 g NaCl = 0.2299 g AgCl 143.4 g AgCl58.5 g NaCl

Solve the same for a mixture of NaCl and KCl.

Find the percentage of NaCl if 0.5000 g AgCl is collected.

0.2345*0.60 g CaCl22*143.4 g AgCl

110 g CaCl2= 0.3668 g AgCl

(0.2299 + 0.3668) g AgCl = 0.5967 g AgCl


Review 4When 0.2345-g NaCl and CaCl2 mixture is dissolved in water, the solution is treated with AgNO3. The mass of dry AgCl collected is 0.5967 g. What is the percentage of NaCl in the mixture?

Solution: Assume the sample contains x fraction of NaCl0.2345 (x) g NaCl 143.4 g AgCl

58.5 g NaCl

Solve the same problem for 0.6000 instead of 0.5967 g AgCl.

In this problem, what are the min. and max. masses of AgCl? (0.5748 – 0.6114 g)

+ 0.2345 (1 – x) g CaCl22*143.4 g AgCl

110 g CaCl2= 0.5967 g AgCl

Solve for x in the equation: x = 0.40 = 40%


Review 5 - skills

obtain empirical formula from composition and combustion experiment

find molecular formula from empirical formula and molar mass

evaluate theoretical yield in a reaction, identify the limiting reagent

Identify limiting reagent and evaluation percent yield in a reaction

predict amounts of products or reagents required in reactions

find percent yield of a reaction but pay attention to limiting reagent

give concentration of common ions in a solution of several salts

calculate masses of reagents or products (AgCl) involving common ions

find percentage of a compound in a mixture by titration or precipitation


Review 6 – skill 2

Identify oxidation states and recognize reagent oxidized

balance redox reaction equations

evaluate quantities of reagents or products involving redox reactions

stoichiometry calculation in titration experiments paying attention to molar relationship such as H3PO4, Ca(OH)2, as well as redox reactions

evaluate quantities involving isotope composition of elements

calculate one of density, mass and volume in solid, liquid, and gas

name inorganic compounds

write proper net ionic equation of reaction (know your spectator ions)


TA Hours & Rooms – fall 03The Tutors handle the tutorial periods for CHEM 120/121 but they can also provide assistance to individuals or small groups on a drop-in basis according to the following schedule. (i.e. One of the Tutors will be in each room for the specified period. You can drop in at any time during that period to get some help.)

Day Time Room*Monday 12:30-1:20 PHY 313Monday 1:30-2:20 MC 4062Monday 2:30-3:20 PHY 150Wednesday 12:30-3:20 PHY 313Wednesday 1:30-2:20 MC 4060

* These rooms are for one-on-one tutorial, not tutors office.

Aqueous solution reactions1 Aqueous-solution Reactions Classify a reaction by Homogeneous chemical...

Documents

Transcript of Aqueous solution reactions1 Aqueous-solution Reactions Classify a reaction by Homogeneous chemical...