ch4 – types of chemical reaction and solutions ch4.1 – 4.3 water, electrolytes, and...

Ch4.1 – Types of Chemical Reactions and Solutions

This is the process of hydration.

Dissociation equation: NaCl(s) Na+(aq) + Cl–(ag)

H2O

Solubility varies greatly. For ions - depends on who has the greater attraction.For molecules – depends on polarity

Solvent – the substance doing the dissolving (water)Solute – the substance that gets dissolved (salt)

ElectrolytesStrong electrolytes

- conduct current very efficiently- soluble salts, acids, bases

ElectrolytesStrong electrolytes

- conduct current very efficiently- soluble salts, acids, bases

Arrhenius Theory of Acids- produces H+ ions (protons) in water

HCl H2O H+(aq) + Cl–(aq)

- strong acid – vitrually every H ionizes

H2SO4 H2O H+(aq) + HSO4

–(aq)

Weak electrolytes – only a small degree of ionization Weak Acid Weak Base

HC2H3O2 H2O H+(aq) + C2H3O2

–(aq) NH3 H2O NH4

+(aq) + OH–

(aq)

Only about 1 in 100 H’s ionize. The rest stay in their molecules.Nonelectrolytes – dissolve in water, but don’t produce ions.

- like sugar (nonpolar substances)

Molarity

Units: Molar or M

Ex1) Calc the molarity of a solution prepared by dissolving 11.5g NaOH in enough water to make 1.50L soln.

soln of liters

solute of molesM

Concentration of Ions

Ex2) Give the concentration of ions in a 0.50M Co(NO3)2 soln.

Ex3) Calculate the number of moles of Cl– ions in

1.75L of 1.0x10-3M ZnCl2 soln.

Ch4 HW#1 p180+ 11(d-g),13,15(a,b),17(a,b)

Ch4 HW#1 p180+ 11(d-g),13,15(a,b),17(a,b)11. Show how each of the following strong electrolytes “ breaks up” into its

component ions dissolving in water.

d. (NH4)2SO4

e. HI f. FeSO4

g. KMnO4

h. HClO4

13. Calcium chloride is a strong electrolyte and is used to “salt” streets in the winter to melt ice and snow. Write a reaction to show how this substance breaks apart when it dissolves in water.



d. (NH4)2SO4 H2O 2NH4+ + SO4

-2

e. HI H2O H+ + I-

f. FeSO4 H2O Fe+ + SO4

-

g. KMnO4 H2O K+ + MnO4

-

h. HClO4 H2O H+ + ClO4

-




d. (NH4)2SO4 H2O 2NH4+ + SO4

-2

e. HI H2O H+ + I-

f. FeSO4 H2O Fe+ + SO4

-

g. KMnO4 H2O K+ + MnO4

-

h. HClO4 H2O H+ + ClO4

-


CaCl2 H2O Ca2+ + 2Cl-

15.Calculate the molarity of each of these solutions. a. A 5.623g sample of NaHCO3 is dissolved in enough water

to make 250 ml of solution.

b. A 184.6mg sample of K2Cr2O7 is dissolved in enough water to make 500 ml of solution.

15.Calculate the molarity of each of these solutions. a. A 5.623g sample of NaHCO3 is dissolved in enough water

to make 250 ml of solution.

b. A 184.6mg sample of K2Cr2O7 is dissolved in enough water to make 500 ml of solution.

M268.0NaHCO 84.0g soln 0.250L

NaHCO 1mol NaHCO 5.623g

3

33

17. Calculate the concentration of all ions present in each of the following solutions of strong electrolyte

a. 0.15M CaCl2

b. 0.26M Al(NO3)3

17. Calculate the concentration of all ions present in each of the following solutions of strong electrolyte

a. 0.15M CaCl2

0.15M Ca+ ions

0.30M Cl- ions

b. 0.26M Al(NO3)3

Ch4.2 More Concentration CalculationsEx1) Blood serum is 0.14M NaCl. What volume of blood contains

1.0mg NaCl?

Ex2) How would I prepare 500 mls of a 0.100M ______ solution,given solid solute?

Dilutions

Ex3) How would I prepare 500 mls of a 0.100M H2SO4 solution,

given concentrated stock solution of 18M?

Ch4 HW#2 p181 21,23(a,b),25

Ch4 HW#2 p181 21,23(a,b),2521. What volume of a 0.100 M solution of NaHCO3 contains

0.350 g of NaHCO3?

Ch4 HW#2 p181 21,23(a,b),2521. What volume of a 0.100 M solution of NaHCO3 contains

0.350 g of NaHCO3?

NaHCO 0.100mol NaHCO 84.0g

soln 1L NaHCO 1mol NaHCO 0.350g

33

33

23. Describe how you would prepare 2.00 L of each of the following solutions.a. 0.250M NaOH from solid NaOHb. 0.250M NaOH from 1.00M NaOH stock solution


soln 0.20 NaOH 1mol soln 1L

NaOH 40.0g NaOH 0.250molsoln 2Lg


b. M1.V1 = M2

.V2

NaOH 1mol soln 1L

NaOH 40.0g NaOH 0.250molsoln 2L

2

2

V

Vsoln 1L

1.00mol

1

soln 2L

soln 1L

0.250mol

25. A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 200.0 mL of stock solution.

A 10.0 mL sample of this stock solution is added to 50.0 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final

solution.

soln M391.0 SO)(NH 138.1g soln 0.200L

SO)(NH 1mol SO)(NH 10.8g

424

424424

25. A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 200.0 mL of stock solution.

A 10.0 mL sample of this stock solution is added to 50.0 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final

solution.

soln M391.0 SO)(NH 138.1g soln 0.200L

SO)(NH 1mol SO)(NH 10.8g

424

424424

2

2

M 1

50ml

1

soln 10mL

soln 1L

0.391molM

Ch4.3 Types of Solution Reactions

1. Precipitation Reactions2. Acid-base Reactions3. Oxidation-Reduction Reactions (Redox)

1. Precipitation Reactions solns mixed and an insoluble substance forms, and separates,

called a precipitate.

AgNO3(aq) and NaCl(ag)

1. Precipitation Reactions solns mixed and an insoluble substance forms, and separates,

called a precipitate.

K2CrO4(aq) + Ba(NO3)2(aq)

Simple solubility rules for salts in water:

1. Most nitrates NO3-1 are soluble.

2. Most salts of alkali metals (Li+, Na+, K+, Cs+, Rb+ )are soluble. Same for ammonium, NH4

+ .

3. Most chloride, bromide, and iodide salts are soluble,

but not when with Ag+, Pb2+, Hg22+.

4. Most sulfates are soluble,

except BaSO4, PbSO4, Hg2SO4, CaSO4.

5. Most hydroxides are only slightly soluble,

except NaOH and KOH very soluble.

Ba(OH)2, Ca(OH)2, Sr(OH)2 only slightly soluble.

6. Most sulfides (S2–), carbonates (CO32–),

phosphates (PO43–), and chromtes (CrO4

2–)

are only slightly soluble.

Ex1) Use rules to predict what will happen if following solutions are mixed:

a) KNO3(aq) and BaCl2(aq) b) Na2SO4(aq) and Pb(NO3)2(aq) c) KOH(aq) and Fe(NO3)3(aq)

3 types of equations are used to describe reactions in solution:1. molecular eqn – great for doing stoichiometry.2. complete ionic eqn – all ions listed,

great for seeing strong electrolytes.3. net ionic eqn – only those soln components

that undergo change. Spectators not included.

Ex2) Aqueous potassium hydroxide is mixed with aqueous iron(III) nitrate to form a ppt of iron(III) hydroxide and aqueous potassium nitrate.Write all 3 eqns.

Ch4 HW#3 p181+ 29,31,33,35(a,b)

Ch4 HW#3 p181+ 29,31,33,35(a,b)29. When the following solutions are mixed together,

what precipitation (if any) will form?

a. BaCl2(aq) + Na2SO4(aq)

b. Pb(NO3)2(aq) + KCl(aq)

c. AgNO3(aq) + Na3PO4(aq)

d. NaOH(aq) + Fe(NO3)3(aq)

31. For the reactions in Exercise 29, write the balanced molecular equation, complete ionic equation, and net ionic equation. If no precipitate forms, write “No reaction.”



a. BaCl2(aq) + Na2SO4(aq) BaSO4(s)

b. Pb(NO3)2(aq) + KCl(aq) PbCl2(c)

c. AgNO3(aq) + Na3PO4(aq) Ag3PO4(s)

d. NaOH(aq) + Fe(NO3)3(aq) Fe(OH)3(s)




NIE: a. BaCl2(aq) + Na2SO4(aq) BaSO4(s)

NIE: b. Pb(NO3)2(aq) + KCl(aq) PbCl2(c)

NIE: c. AgNO3(aq) + Na3PO4(aq) Ag3PO4(s)

NIE: d. NaOH(aq) + Fe(NO3)3(aq) Fe(OH)3(s)


BME: a. BaCl2(aq) + Na2SO4(aq) 2Cl-(aq)+2Na+

(aq)+ BaSO4(s)

b. Pb(NO3)2(aq) + 2KCl(aq) 2NO3-(aq)+2K+

(aq)+ PbCl2(c)

c. 3AgNO3(aq)+Na3PO4(aq) 3NO3-(aq)+3Na+

(aq)+Ag3PO4(s)

d. 3NaOH(aq)+Fe(NO3)3(aq) 3NO3-(aq)+3Na+

(aq)+ Fe(OH)3(s)

33. Write net ionic equations for each of the following.

a. AgNO3(aq) + KI(aq)

b. CuSO4(aq) + Na2S(aq)

c. CoCl2(aq) + NaOH(aq)

d. NiCI2(aq) + KNO3(aq)

33. Write net ionic equations for each of the following.

a. AgNO3(aq) + KI(aq)

Ag+(aq) + I-

(aq) AgI(s)

b. CuSO4(aq) + Na2S(aq)

Cu2+(aq) + S2-

(aq) CuS(s) (only slightly)

c. CoCl2(aq) + NaOH(aq)

Co2+

(aq) + 2OH-(aq) Co(OH)2(aq) (only slightly)

d. NiCI2(aq) + KNO3(aq) No ppt

35. Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed.

a. Ammonium sulfate and barium nitrate

(NH4)2(SO4)(aq) + Ba(NO3)2(aq)

b. Lead(II) nitrate and sodium chloride

Pb(NO3)2(aq) + NaCl(aq)

35. Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed.

a. Ammonium sulfate and barium nitrate

SO42-

(aq) + Ba2+(aq) BaSO4(s)

b. Lead(II) nitrate and sodium chloride

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

Ch4.4 Mass of Precipitate

Ex1) Calculate the mass of solid NaCl that must be added to 1.50L of a 0.100M AgNO3 solution to precipitate all the Ag+ ions.

Ex2) What mass of PbSO4 precipitates when 2.00L of 0.025M aqueous Na2SO4 and 1.25L of 0.0500M aqueous Pb(NO3)2 are mixed?

Ch4 HW#4 p182 39,41,43

Ch4 HW#4 p182 39,41,4339. What mass of NaCl is required to precipitate all the silver ions from 50.0 mL of a 0.0500 M solution of AgNO3?

NaCl + AgNO3 ?g 0.0500M

50.0ml

Ch4 HW#4 p182 39,41,4339. What mass of NaCl is required to precipitate all the silver ions from 50.0 mL of a 0.0500 M solution of AgNO3?

NaCl + AgNO3 ?g 0.0500M

50.0ml

Cl- + Ag+ AgCl(s)

NaCl 1mol Ag 1mol soln 1L

NaCl 58.5g NaCl 1mol Ag 0.050molsoln 0.050L

41. What mass of solid aluminum hydroxide is produced when 50.0 mL of 0.200 M Al(NO3)3 is added to 200.0 mL of 0.100M KOH?

Al(NO3)3 + 3KOH 0.200M 0.100M 50.0ml 200.0ml

41. What mass of solid aluminum hydroxide is produced when 50.0 mL of 0.200 M Al(NO3)3 is added to 200.0 mL of 0.100M KOH?

Al(NO3)3 + 3KOH K+(aq) + NO3

-(aq) + Al(OH)3(s)

0.200M 0.100M ?g 50.0ml 200.0ml

KOH is the limiting reactant and ____ g of aluminum hydroxide is produced

A1 mol 1 soln 1L

Al(OH) mol 1 A1 0.200molsoln 0.050L3

33

OH mol 3 soln 1L

Al(OH) mol 1 OH 0.100molsoln 0.200L 3

43. A 100.0-mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 mL of 0.200 M aqueous magnesium nitrate.a. Write a balanced chemical equation for any reaction that occurs.b. What precipitate forms?c. What mass of precipitate is produced?d. Calculate the concentration of each ion remaining in solution

after precipitation is complete.

43. A 100.0mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 mL of 0.200 M aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution


2KOH + Mg(NO3)2 2K+ + 2NO3- + Mg2+ + 2OH-

2KOH + Mg(NO3)2 Mg(OH)2(s) 0.200M 0.200M100ml 100ml

LR

Mg(OH) 1mol Mg 1mol soln 1L

Mg(OH) 58.3 Mg(OH) 1mol Mg 0.200molsoln 0.100L

22

222

43. A 100.0mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 mL of 0.200 M aqueous magnesium nitrate. d. Calculate the concentration of each ion remaining in solution


2KOH + Mg(NO3)2 2K+ + 2NO3- + Mg(OH)2(s)

start:end:

0.02mol K+

0.02mol OH–

0.02mol Mg+

0.04mol NO3

–

0.1 L 0.1 L

0.02mol Mg+ 0.04mol

NO3–

0.2 L

0.02mol K+ 0.02mol

OH–

Ch4.4B – Ion ConcentrationEx1) A 100.0mL aliquot of 0.100 M aqueous barium chloride is mixed with 100.0 mL of 0.300 M aqueous sodium sulfate. Barium sulfate precipitates out. a. Write a balanced chemical equation for any reaction that occurs. b. What mass of precipitate is produced? c. Calculate the concentration of each ion remaining in solution


c. Calculate the concentration of each ion remaining in solution after precipitation is complete.

BaCl2 + Na3SO4 3Na+ + 2Cl- + BaSO4(s)

start:end:

___ mol Ba2+

___ mol Cl–

___ mol Na+

___ mol SO4

–2

0.1 L 0.1 L0.2 L

___ mol Ba2+

___ mol Cl–

___ mol Na+

___ mol SO4

–2

c. Calculate the concentration of each ion remaining in solution after precipitation is complete.

BaCl2 + Na3SO4 3Na+ + 2Cl- + BaSO4(s) + SO4–2

start:end:

Ch4 HW#5 p182 44

0.01mol Ba2+

0.02mol Cl–

0.09mol Na+

0.03mol SO4

–2

0.1 L 0.1 L0.2 L

0.0 mol Ba2+

0.02mol Cl–

0.09mol Na+

0.02mol SO4

–2

Ch4 HW#5 p182 44 44. How many grams of silver chloride can be prepared by the reaction of 100.0mL of 0.20M silver nitrate with 100.0mL of 0.15M calcium chloride? Calculate the concentration of each ion remaining in solution after precipitation is complete.

AgNO3 + CaCl2 Ca+2 + NO3-1 + AgCl(s)

0.20M 0.15M ?g 0.1000L 0.1000L Don’t forget to balance!


2AgNO3(aq) + CaCl2(aq) 2AgCl(s) + Ca+2(aq) + 2NO3

-(aq)

0.20M 0.15M ?g 0.1000L 0.1000L

AgCl 1mol AgNO 2mol soln 1L

AgCl 143.4 AgCl 2mol AgNO 0.2molsoln 0.100L

3

3

AgCl 1mol CaCl 1mol soln 1L

AgCl 143.4 AgCl 2mol CaCl 0.15molsoln 0.100L

2

2



-(aq)

___ mol Ag+

___ mol NO3

–

___ mol Ca+2

___ mol Cl–

0.1 L 0.1 L0.2 L

___ mol Ag+

___ mol NO3

–

___ mol Ca+2

___ mol Cl–

44. How many grams of silver chloride can be prepared by the reaction of 100.0ml of 0.20M silver nitrate with 100.0ml of 0.15M calcium chloride.How much leftover of each ion? Balanced Molecular Equation:


-(aq)

0.20M 0.15M ?g 0.1000L 0.1000L

ions: Ag+ NO3- Ca+2

Cl- start:end:

44. How many grams of silver chloride can be prepared by the reaction of 100.0ml of 0.20M silver nitrate with 100.0ml of 0.15M calcium chloride.How much leftover of each ion? Balanced Molecular Equation:


-(aq)

0.20M 0.15M ?g 0.1000L 0.1000L

ions: Ag+ NO3- Ca+2

Cl- start:end:

AgCl 1mol AgNO 2mol soln 1L

AgCl 143.4 AgCl 2mol AgNO 0.2molsoln 0.100L

3

3

AgCl 1mol CaCl 1mol soln 1L

AgCl 143.4 AgCl 2mol CaCl 0.15molsoln 0.100L

2

2

44. How many grams of silver chloride can be prepared by the reaction of 100.0ml of 0.20M silver nitrate with 100.0ml of 0.15M calcium chloride.How much leftover of each ion? Complete Ionic Equation: Look at the CIE, not the molecular equation:

2Ag+(aq)+ NO3

–(aq)+Ca2++2Cl–(aq)2AgCl(s)+Ca+2

(aq)+2NO3-(aq)

0.20M 0.20M 0.15M 0.30M 2.87g 0.1000L 0.1000L 0.1000L 0.1000L

ions: Ag+ NO3- Ca+2

Cl- start: 0.02mol 0.02mol 0.015mol 0.030molend:

44. How many grams of silver chloride can be prepared by the reaction of 100.0ml of 0.20M silver nitrate with 100.0ml of 0.15M calcium chloride.How much leftover of each ion? Complete Ionic Equation:

2Ag+(aq)+NO3

–(aq)+Ca2++2Cl–(aq)2AgCl(s)+Ca+2

(aq)+2NO3–

(aq)

0.20M 0.20M 0.15M 0.30M 2.87g 0.1000L 0.1000L 0.1000L 0.1000L

ions: Ag+ NO3- Ca+2

Cl- start: 0.02mol 0.02mol 0.015mol 0.030molend: 0mol 0.02mol 0.015mol ?mol

All the silver ions get used up. Ag+: 0MNo NO3

- are used, only spectator ions: NO3-: 0.02mol/.200L = 0.1M

No Ca2+ are used, only spectator ions: Ca2+:0.015mol/.20L=0.075MBut what about the chloride ions?

44. How many grams of silver chloride can be prepared by the reaction of 100.0ml of 0.20M silver nitrate with 100.0ml of 0.15M calcium chloride.How much leftover of each ion? Net Ionic Equation:

Ag+(aq) + Cl–(aq) AgCl(s)

ions: Ag+ NO3- Ca+2

Cl- start: 0.02mol 0.02mol 0.015mol 0.030molend: 0mol 0.02mol 0.015mol 0.010mol

All the silver ions get used up. Ag+: 0MNo NO3

- are used, only spectator ions: NO3-: 0.02mol/.200L = 0.1M

No Ca2+ are used, only spectator ions: Ca2+:0.015mol/.20L=0.075MBut what about the chloride ions?

They were consumed at a rate of one-to-one with Ag+.So if 0.02mol Ag+ is consumed, same with Cl-.That leaves 0.01 moles leftover.

Cl–: 0.010mol/.20L=

Ch4.5 – Acid-Base ReactionsBronsted-Lowry Theory of acids and bases:

Acid – proton donorBase – proton acceptor

HCl(aq) + NaOH(aq)

HC2H3O2(aq) + KOH(aq)

Ex1) What volume of a 0.100M HCl solution is needed to neutralize

25.0mL of 0.350M NaOH?

HCl(aq) + NaOH(aq)

Ex2) In a certain experiment, 28.0ml of 0.250M HNO3 and 53.0ml of 0.320M KOH are mixed.

Calculate the amount of water formed and the amount

of excess H+ or OH-.

Acid-Base Titrations - delivery of a measured volume of soln (usually from a buret)

of know concentration into a soln being analyzed (the analyte). The point where enough titrant has been added to react

exactly is called the equivalence point, often indicated by an indicator (substance that changes color.)This point is often called the endpoint.

Ex3) Benzoic acid, HC7H5O2, is a component of waste effluent released

in some industrial processes. If 10.59 ml of 0.1546M NaOH is required to neutralize it, what volume of benzoic acid was present?

(next slide)

Ex3) Benzoic acid, HC7H5O2, is a component of waste effluent released

in some industrial processes. If 10.59 ml of 0.1546M NaOH is required to neutralize it, what volume of benzoic acid was present?

H+ + C7H5O2- + Na+ + OH-

Ch4 HW#5 p182 45(a,b),47(a,b),49(a,b)

Ch4 HW#6 p182 45(a,b),47(a,b),49(a,b)45. Write the balanced molecular, complete ionic, and net ionic equations for each of the following acid-base reactions.

a. HClO4(aq) + Mg(OH)2(s) →

2H+(aq) +2ClO4

-(aq)+Mg+2

(aq) + 2(OH)-(aq)→

b. HCN(aq) + NaOH(aq) →

H+(aq) +CN-

(aq)+Na+(aq) + (OH)-

(aq)→

45. Write the balanced molecular, complete ionic, and net ionic equations for each of the following acid-base reactions.


2H+(aq) +2ClO4

-(aq)+Mg+2

(aq) + 2(OH)-(aq)→H(OH)(l) +Mg+2

(aq)+ ClO4-(aq)

H+(aq) + (OH)-

(aq)→H(OH)(l)


H+(aq) +CN-


(aq)→

45. Write the balanced molecular, complete ionic, and net ionic equations for each of the following acid-base reactions.


2H+(aq) +2ClO4

-(aq)+Mg+2

(aq) + 2(OH)-(aq)→H(OH)(l) +Mg+2

(aq)+ ClO4-(aq)

H+(aq) + (OH)-

(aq)→H(OH)(l)


H+(aq) +CN-


(aq)→H(OH)(l) +Na+(aq)+ CN-

(aq)

H+(aq) + (OH)-

(aq)→H(OH)(l)

47. Write the balanced molecular, complete ionic, and net ionic equations for the reactions that occur when the following are mixed.

a. potassium hydroxide (aqueous) and nitric acid

KOH(aq) + HNO3(aq)

b. barium hydroxide (aqueous) and hydrochloric acid

Ba(OH)2(aq) + HCl(aq)

47. Write the balanced molecular, complete ionic, and net ionic equations for the reactions that occur when the following are mixed.

a. potassium hydroxide (aqueous) and nitric acid

K+(aq)+OH-

(aq)+H+(aq)+NO3

-(aq)H+

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

(aq)+NO3-(aq)

H+(aq)+ OH-

(aq) H(OH)(l)

b. barium hydroxide (aqueous) and hydrochloric acid

Ba2+(aq)+2(OH)-

(aq)+2H+(aq)+2Cl-(aq) 2H+

(aq)+2(OH)-(aq)+ Ba2+

(aq)+ 2Cl-

H+(aq)+ OH-

(aq) H(OH)(l)

49. What volume of each of the following acids will react completely with 50.00 mL of 0.200 M NaOH?a. 0.100 M HCl

HCl(aq) + NaOH(aq)→H(OH)(l) +Na+(aq)+ Cl-(aq)

0.100M 0.200M ?L 0.05000L

b. 0.150 M HNO3

49. What volume of each of the following acids will react completely with 50.00 mL of 0.200 M NaOH?a. 0.100 M HCl

HCl(aq) + NaOH(aq)→H(OH)(l) +Na+(aq)+ Cl-(aq)

0.100M 0.200M ?L 0.05000L

b. 0.150 M HNO3

HNO3(aq) + NaOH(aq)→H(OH)(l) +Na+(aq)+ NO3

-(aq)

0.150M 0.200M ?L 0.05000L

HCl 0.100L HCl 0.100mol NaOH 1mol soln 1L

HCl 1L HCl 1mol NaOH 0.200molsoln 0.0500L

Lab 4.2 Pre-lab QuestionsThe following data was obtained from a regular chemistry lab group

conducting this lab:

Trial 1 NaOH Initial Volume

Final Volume

Volume Used

Molarity 0.100M 1mL 12mL 11mL

HCl Initial Volume

Final Volume

Volume Used

Molarity 0.0157M0.016M0.02M

2mL 9mL 7mL

3. Knowing my regular chem classes, this trial may have lacked accuracy, but it definitely lacked precision. Please list the obvious non-precise measurements.4. Look at your calculator answer. Round the answer to the correct number of significant digits. If you look at the first 3 digits on your calculator screen, how different do your 2 answers look?


Final Volume

Volume Used


HCl Initial Volume

Final Volume

Volume Used


2mL 9mL 7mL

5. If the actual molarity of HCl was 0.150M, please find the % error for the answer with the correct sig digs.6. Can you see how precise measuring can be as important as accurate measuring?


Final Volume

Volume Used


HCl Initial Volume

Final Volume

Volume Used


2mL 9mL 7mL

Ch4 HW#7 Mid Chapter Review p180+16c,18c,24b,54

Lab4.2 NeutralizationVolume of NaOH: ____ (measured)Molarity of NaOH: 0.200Volume of HCl: ____ (measured)Molarity of HCl: ?

HClMHClLNaOHmol

HClmol

NaOHL

NaOHmolNaOHL ____

____ 1

1

1

2.0 ___

Ch4 HW#7 Mid Chapter Review p180+ 16c,18c,24b,5416c. A 0.4508-g sample of iron is dissolved in a small amount of concentrated nitric acid forming Fe3+ ions in solution and is diluted to a total volume of 500.0ml. Calculate the molarity of Fe3+.

16c. A 0.4508-g sample of iron is dissolved in a small amount of concentrated nitric acid forming Fe3+ ions in solution and is diluted to a total volume of 500.0ml. Calculate the molarity of Fe3+.

Fe(s) + HNO3(aq) Fe3+ (aq) + NO3-(aq) + H2(g)

Fe 55.8gsoln 0.500L

Fe 1mol Fe 0.4508g

18c.Calculate the concentration of all ions present in 5.00g of NH4Cl in 500.0mL of solution.

NH4Cl H2O NH4+ + Cl-

18c.Calculate the concentration of all ions present in 5.00g of NH4Cl in 500.0mL of solution.

NH4Cl H2O NH4+ + Cl-

ClNH 53.5gsoln 0.500L

ClNH 1mol ClNH 5.00g

4

44

24b. How would you prepare 1.00 L of a 0.50M HCl soln from “concentrated” (12M) reagent?

24b. How would you prepare 1.00 L of a 0.50M HCl soln from “concentrated” (12M) reagent?

2

2

V

Vsoln 1L

12.00mol

1

soln 1L

soln 1L

0.500mol

54. What volume of 0.0200 M calcium hydroxide is required to neutralize 35.00 mL of 0.0500 M nitric acid?

2HNO3(aq) + Ca(OH)2(aq)→ 2H(OH)(l) +Na2+(aq)+ NO3

-(aq)

0.050M 0.200M 0.03500L ?L

54. What volume of 0.0200 M calcium hydroxide is required to neutralize 35.00 mL of 0.0500 M nitric acid?

2HNO3(aq) + Ca(OH)2(aq)→ 2H(OH)(l) +Na2+(aq)+ NO3

-(aq)

0.050M 0.200M 0.03500L ?L

Ca(OH) 0.200mol HNO 2mol soln 1L

Ca(OH) 1L Ca(OH) 1mol HNO 0.050molsoln 0.03500L

23

223

Ch4.6 Oxidation-Reduction Reactions (Redox)Redox reactions involve the transfer of electrons.

Ex1) Na(s) + Cl2(g)

In redox reactions, one element is oxidized (loses electrons),one element is reduced (gains electrons).

OIL RIG

(Oxidation Is Losing, Reducing Is Gaining)

(reduced in charge)

Oxidation numbers – in covalent bonds, where electrons are shared,they are rarely shared equally. So we assign a charge to each.

Here’s the rules:

Oxidation numbers – in covalent bonds, where electrons are shared,they are rarely shared equally. So we assign a charge to each.

Here’s the rules:

1. Free elements are 0. Exs: Na(s), O2(g) 2. Monatomic ions take ox #’s

based on their column: Exs: Na: +1, Cl: -13. Fluorine is always -1.

4. Oxygen is usually -2. Exs: CO, CO2, SO2, SO3

Exceptions: the O22- group is -1, like in H2O2

when bonded to fluorine it +2, OF2.

5. With nonmetals, H is +1 Exs: HCl, NH3, H2O, CH4

6. Sum in a compound adds up to 0, oradds up to the sum of the whole polyatomic ion charge.

Exs: NH4+

Ex2) Assign oxidation states to all the atoms:

a) CO2 b) SF6

c) NO3- d) Fe2O3

Ex3) In the following reaction, identify which element is oxidizedand which is reduced:

2Al(s) + 3I2(s) 2AlI3(s) 0 0 +3 -1

Al: oxidized

I2: reduced

I2 is referred to as the oxidizing agent (takes electrons).

Al is the reducing agent (furnishes electrons).

Ex4) In the following reaction, identify which element is oxidizedand which is reduced:

2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)

Ch4 HW#8 p182+ 57,59,61(a,b)

Ch4 HW#8 p182+ 57,59,61(a,b) (Go over fast!)57. Assign oxidation states for all atoms in each of the following compounds.

You have one of these on ur test!

a. KMnO4 b. NiO2

c. K4Fe(CN)6 (Fe only) d. (NH4)2HPO4

e. P4O6 f. Fe3O4

g. XeOF4 h. SF4

i. CO j. Na2C2O4

57. Assign oxidation states for all atoms in each of the following compounds.

a. KMnO4 b. NiO2

(+1)___(-8) ___(-4)


(+4)___(-6) (+2)(+1)___(-8)

e. P4O6 f. Fe3O4

___(-12) ___(-8)

g. XeOF4 h. SF4

___(-2)(-4) ___(-4)

i. CO j. Na2C2O4

___(-2) (+2)___(-8)

57. Assign oxidation states for all atoms in each of the following compounds.

a. KMnO4 b. NiO2

(+1)+7(-8) +4(-4)


(+4)+2(-6) (+2)(+1)+5(-8)

e. P4O6 f. Fe3O4

g. XeOF4 h. SF4

i. CO j. Na2C2O4

59. Assign the oxidation for chlorine in each of the following anion: You have one of these on ur test!

OCl-

ClO2-

CLO3-

CLO4-

59. Assign the oxidation for chlorine in each of the following anion:

OCl- [(-2)___](-1)

ClO2-

[___ (-4)](-1)

ClO3-

[___(-6)](-1)

ClO4-

[___(-8)](-1)

59. Assign the oxidation for chlorine in each of the following anion:

OCl- [(-2)+1](-1)

ClO2-

[+3 (-4)](-1)

ClO3-

[+5(-6)](-1)

ClO4-

[+7(-8)](-1)

61(a,b). Specify which of the following are oxidation-reduction reactions, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced.

a. CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

b. Zn(s) + 2HCI(aq) ZnCI2(aq) + H2(g)


a. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) [+4(-4) ] (0) [+4(-4)] [+2(-2)]

-1each -2each +1eachOIL: H lost 2e- so is oxidized, making it the reducing agent.RIG: O gained 2e- so is reduced, making it the ox agent.

b. Zn(s) + 2HCI(aq) ZnCI2(aq) + H2(g) (0) [+1(-1)] [+2(-1each)] (0)


a. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) [+4(-4) ] (0) [+4(-4)] [+2(-2)]

-1each -2each +1eachOIL: H lost 2e- so is oxidized, making it the reducing agent.RIG: O gained 2e- so is reduced, making it the ox agent.

b. Zn(s) + 2HCI(aq) ZnCI2(aq) + H2(g) (0) [+1(-1)] [+2(-1each)] (0)

OIL: Zn lost 2e- so is oxidized, making it the reducing agent.RIG: H gained 1e- each so is reduced, making it the ox agent.

Ch4.7 - Balancing Redox EquationsHalf reactions: break up a redox reaction into 2 parts,

one involving oxidation, one involving reduction. Slightly different methods if the solution is acidic or basic,

so determine that first. (We will only do acidic.)Ex1) Aluminum metal is placed in a 2.0M soln of sodium hydroxide.

Hydrogen gas is liberated.

Al(s) + NaOH Al(OH)4–

(aq) + H2(g) + Na+(aq)

a. Which substance is oxidized?b. Which substance is reduced?c. Which substance is the reducing agent?d. Which substance is the oxidizing agent?e. Balance the 2 half reactions.f. Balance the complete reaction.

(next slide) Ch4 HW#9 p183 63(a,b), 65(a) (modified)

The Procedure: (These are the official steps, copy these online.)Step1: Write ½ rxns. (Most FRQ’s ask intermediate questions 1st.)Step2: Balance ½ rxns. a. Balance all elements except H and O. b. Balance O using H2O. c. Balance H using H+. d. Balance charge using electrons.Step3: Equalize electrons in both ½ rxns.Step4: Add the ½ rxns, cancel identical items.Step5: Check that elements and charges balance.

Ex1) Aluminum metal is placed in a 2.0M soln of sodium hydroxide.Hydrogen gas is liberated.


(aq) + H2(g) + Na+(aq)

a. oxidized: (OIL)

b. reduced: (RIG)

c. reducing agent:

d. oxidizing agent:

e. Balance the 2 half reactions. f. Balance the complete reaction.

Ex1) Aluminum metal is placed in a 2.0M soln of sodium hydroxide.Hydrogen gas is liberated. 0 +1 -2 -1 +3 -2 -1 0 +1


(aq) + H2(g) + Na+(aq)

a. oxidized: (OIL) Al(s) b. reduced: (RIG) H c. reducing agent: Al(s) d. oxidizing agent: OH-

e. ox ½ rxn:

red ½ rxn:

f. complete rxn:

Ex2) Potassium dichromate, K2Cr2O7, reacts with ethyl alcohol, C2H5OH, in the following reaction:

H+(aq) + Cr2O7

2-(aq) + C2H5OH Cr+3

(aq) + CO2(aq) + H2O(l) (yellow) (violet)

Ex2) Potassium dichromate, K2Cr2O7, reacts with ethyl alcohol, C2H5OH, in the following reaction: (Acidic)

H+(aq) + Cr2O7

2-(aq) + C2H5OH(l) Cr+3

(aq) + CO2(aq) + H2O(l) (yellow) (violet)OIL:RIG:

Step1: Step2:

a.b.c.d.

Step3:Step4:Step5:Reduction ½ rxn:

Oxidation ½ rxn:

H+(aq) + Cr2O7

2-(aq) + C2H5OH(l) Cr+3

(aq) + CO2(aq) + H2O(l) Step1: Write ½ rxns.Step2: Balance ½ rxns. (Start with 1st.)

a. Balance all except H n O.b. c. d.

Step3:Step4:Step5:

Reduction ½ rxn: Cr2O72-

(aq) Cr+3(aq)

Oxidation ½ rxn: C2H5OH(l) CO2(aq)

(+6) (+3) (–2) (+4)

H+(aq) + Cr2O7

2-(aq) + C2H5OH(l) Cr+3


a. Balance all except H n O.b. Balance O using H2O.c. d.

Step3:Step4:Step5:


(aq) 2Cr+3(aq)


(+6) (+3) (–2) (+4)

H+(aq) + Cr2O7

2-(aq) + C2H5OH(l) Cr+3


a. Balance all except H n O.b. Balance O using H2O.c. Balance H using H+.d.

Step3:Step4:Step5:


(aq) 2Cr+3(aq) + 7H2O(l)


(+6) (+3) (–2) (+4)

H+(aq) + Cr2O7

2-(aq) + C2H5OH(l) Cr+3


a. Balance all except H n O.b. Balance O using H2O.c. Balance H using H+.d. Balance charge using electrons.

Step3:Step4:Step5:

Reduction ½ rxn: 14H+(aq) + Cr2O7

2-(aq) 2Cr+3

(aq) + 7H2O(l)


(+6) (+3) (–2) (+4)

H+(aq) + Cr2O7

2-(aq) + C2H5OH(l) Cr+3

(aq) + CO2(aq) + H2O(l) Step1: Write ½ rxns.Step2: Balance ½ rxns. (Start with 1st.) (Repeat w 2nd.)


Step3:Step4:Step5:

Reduction ½ rxn: 6e- + 14H+(aq) + Cr2O7

2-(aq) 2Cr+3

(aq) + 7H2O(l)


(+6) (+3) (–2) (+4)

H+(aq) + Cr2O7

2-(aq) + C2H5OH(l) Cr+3



Step3: Equalize electronsStep4:Step5:

Reduction ½ rxn: 6e- + 14H+(aq) + Cr2O7

2-(aq) 2Cr+3

(aq) + 7H2O(l)

Oxidation ½ rxn: C2H5OH(l) + 3H2O(l) 2CO2(aq) + 12H+(aq)+12e-

(+6) (+3) (–2) (+4)

H+(aq) + Cr2O7

2-(aq) + C2H5OH(l) Cr+3



Step3: Equalize electronsStep4: Add the ½ reactions togetherStep5:

Reduction ½ rxn:2(6e- +14H+(aq) +Cr2O7

2-(aq) 2Cr+3

(aq) + 7H2O(l)) 12 28 2 4 14


(+6) (+3) (–2) (+4)

H+(aq) + Cr2O7

2-(aq) + C2H5OH(l) Cr+3



Step3: Equalize electrons.Step4: Add the ½ rxns and cancel identical items.Step5: Check that elements and charges balance.

Reduction ½ rxn:12e- +28H+(aq)+2Cr2O7

2-(aq) 4Cr+3

(aq) +14H2O(l)


C2H5OH(l)+16H+(aq)+2Cr2O7

2-(aq) 2CO2(aq)+4Cr+3

(aq)+11H2O(l)

(+6) (+3) (–2) (+4)

Ex3) An aqueous soln of cyanide ion is often used to extract silver from ore. It occurs in a basic soln in the following reaction:

Ag(s) + CN-(aq) + O2(g) Ag(CN)2

-(aq)

Balance using ½ rxn method.

HW#63b) Balance using ½ rxn method:

I-(aq) + ClO-

(aq) I3-(aq) + Cl-(aq)

Ch4 HW#9 p183 63(a,b), 65(a) (modified)

Ch4 HW#9 p183 63(a,b), 65(a) (modified)63. Balance the following oxidation-reduction reactions that occur in acidic solutions.

a. Zn(s) + HCl(aq) Zn2+(aq) + H2(g)

63. Balance the following oxidation-reduction reactions that occur in acidic solutions.

b. I-(aq) + ClO-(aq) I3-(aq) + Cl-(aq)

65. Balance the following oxidation-reductions reactions that occur in acidic solutions.

a. Al(s) + MnO4-(aq) MnO2(s) + Al(OH)4

-(aq)

65. Balance the following oxidation-reductions reactions that occur in acidic solutions.

b. Cl2(g) Cl-(aq) + OCl-(aq)

Ch4 HW#10 p183 64(a,b),66(a) (modified)64. Balance the following oxidation-reduction reactions that occur in acid solution using the half-reactions method.

a. Cu(s) + NO3-(aq) Cu2+(aq) + NO(aq)

64. Balance the following oxidation-reduction reactions that occur in acid solution using the half-reactions method.

b. Cr2O72-(aq) + Cl-(aq) Cr3+(aq) + Cl2(g)

66. Balance the following oxidation-reduction reactions the occur in acidic solution.

a. Cr(s) + CrO42(aq) Cr(OH)3(s)

66. Balance the following oxidation-reduction reactions the occur in basic solution.

b. MnO4-(aq) + S2-(aq) MnS(s) + S(s)

Lab 4.3 Redox

Ch4 Rev#1 p180+ 12c,16b,18d, 22,24e, 36(a,b)

12c. Show how Ca(OH)2 “breaks up” into its component ions

dissolving in water.

16b. An 853.5-mg sample of KIO3 is dissolving in enough water

to make 250.0 mL of solution. Calculate the molarity.

18d. Calculate the concentration of all ions present of 1.00 g K3PO4

in 250.0 mL of solution.

22. How many grams of NaOH are contained in 250.0 mL of a 0.400 M sodium hydroxide solution.

24e. How would you prepare 1.00 L of a 0.50 solution of sodium carbonate from the pure solid.

36. Write net ionic equations for the reaction, if any, when aqueous solutions of the following are mixed.

a) Cobalt(III) chloride and sodium hydroxide.

b)Silver nitrate and ammonium carbonate.

Ch4 Rev#2 p180+ 42, 53, 62a, 67

Ch4 Rev#2 p180+ 42, 53, 62a, 6742. What mass of barium sulfate is produced when 100.0 mL

of a 0.100M solution of barium chloride is mixed with 100.0 mL of a 0.100M solution of iron(III) sulfate?

Before After

FeBa

ClSO4

Cl

SO4 SO4

53. A 25.00 mL sample of hydrochloric acid solution requires 24.16 mL of 0.106 M sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric acid solution?

62a. Specify if Cu(s) + 2Ag+(aq) 2Ag(s) +Cu2+(aq) is a oxidation-reduction reaction, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced.

67. Chlorine gas was first prepared in 1774 by C.W. Scheele by oxidizing sodium chlorine with manganese(IV) oxide. The reaction is:

NaCl(aq) + H2SO4(aq) + MnO2(s) Na2SO4(aq) + MnCl2(aq) + H2O(l) + Cl2(g)

ch4 – types of chemical reaction and solutions ch4.1 – 4.3 water, electrolytes, and...

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