Exp # [A] [B] [C] Rate 1) 27 C 0.10 0.10 0.10 4.0

Post on 10-Feb-2016

41 views 1 download

Tags:

description

1A + 2B  1C + 1D Calculate the equilibrium concentrations of each species when 150 mL 2.5 M A is mixed with 100.0 mL 2.5 M B. K c = 2.0 x 10 -10. 1A + 2B  1C + 1D Calculate the equilibrium concentrations of each species when a solution is made with 1.0 M A & 1.0 M B. K c = 2.0 x 10 -12. - PowerPoint PPT Presentation

transcript

1A + 2B 1C + 1DCalculate the equilibrium

concentrations of each species when 150 mL 2.5 M A is mixed with 100.0 mL 2.5 M B. Kc = 2.0 x 10-10

1A + 2B 1C + 1DCalculate the equilibrium

concentrations of each species when a solution is

made with 1.0 M A & 1.0 M B. Kc = 2.0 x 10-12

Exp # [A] [B] [C] Rate1) 27 C 0.10 0.10 0.10 4.0 2) 27 C 0.10 0.10 0.20 8.0 3) 27 C 0.10 0.20 0.20 64 4) 27 C 0.30 0.10 0.10 36 5)127 C 0.10 0.10 0.10 400.0

Exp # [A] [B] [C] Rate1) 27 C 0.10 0.10 0.10 4.0 2) 27 C 0.10 0.10 0.20 8.0 3) 27 C 0.10 0.20 0.20 64 4) 27 C 0.30 0.10 0.10 36 5)127 C 0.10 0.10 0.10 400.0

([A4]/[A1])a = rate4/rate1

3a = 9; thus, a = 2

Exp # [A] [B] [C] Rate1) 27 C 0.10 0.10 0.10 4.0 2) 27 C 0.10 0.10 0.20 8.0 3) 27 C 0.10 0.20 0.20 64 4) 27 C 0.30 0.10 0.10 36 5)127 C 0.10 0.10 0.10 400.0

([B3]/[B2])b = rate3/rate2

2b = 8; thus b = 3

Exp # [A] [B] [C] Rate1) 27 C 0.10 0.10 0.10 4.0 2) 27 C 0.10 0.10 0.20 8.0 3) 27 C 0.10 0.20 0.20 64 4) 27 C 0.30 0.10 0.10 36 5)127 C 0.10 0.10 0.10 400.0

([C2]/[C1])c = rate2/rate1

2c = 2; thus c = 1

Rate = k[A]2[B]3[C]

Rate[A]2[B]3[C]k =

__N2 +__CO2 +__O2 __NO2 +__CO

__NO2 __N2O4

__N2O4 + __CO __NO +__CO2

__ NO + __ CO2 __ N2O5 + __ C

__N2O5 + __C __C2N2O5

__N2 +__CO2 +__O2 __NO2 +__CO

__NO2 __N2O4

__N2O4 + __CO __NO +__CO2

4 NO + 3 CO2 2 N2O5 + 3 C

__N2O5 + __C __C2N2O5

__N2 +__CO2 +__O2 __NO2 +__CO

__NO2 __N2O4

2 N2O4 + 4 CO 4 NO +4 CO2

4 NO + 3 CO2 2 N2O5 + 3 C

__N2O5 + __C __C2N2O5

__N2 +__CO2 +__O2 __NO2 +__CO

__NO2 __N2O4

2 N2O4 + 4 CO 4 NO +4 CO2

4 NO + 3 CO2 2 N2O5 + 3 C

__N2O5 + __C __C2N2O5

__N2 +__CO2 +__O2 __NO2 +__CO

4 NO2 2 N2O4

2 N2O4 + 4 CO 4 NO +4 CO2

4 NO + 3 CO2 2 N2O5 + 3 C

__N2O5 + __C __C2N2O5

__N2 +__CO2 +__O2 __NO2 +__CO

4 NO2 2 N2O4

2 N2O4 + 4 CO 4 NO +4 CO2

4 NO + 3 CO2 2 N2O5 + 3 C

__N2O5 + __C __C2N2O5

2 N2 + 4 CO2 + 2 O2 4 NO2 + 4 CO

4 NO2 2 N2O4

2 N2O4 + 4 CO 4 NO +4 CO2

4 NO + 3 CO2 2 N2O5 + 3 C

__N2O5 + __C __C2N2O5

2 N2 + 4 CO2 + 2 O2 4 NO2 + 4 CO

4 NO2 2 N2O4

2 N2O4 + 4 CO 4 NO +4 CO2

4 NO + 3 CO2 2 N2O5 + 3 C

2 N2 + 3 CO2 + 2 O2 2 N2O5 + 3 C __N2O5 + __C __C2N2O5

2 N2 + 4 CO2 + 2 O2 4 NO2 + 4 CO

4 NO2 2 N2O4

2 N2O4 + 4 CO 4 NO +4 CO2

4 NO + 3 CO2 2 N2O5 + 3 C

2 N2 + 3 CO2 + 2 O2 2 N2O5 + 3 C 2 N2O5 + 4 C 2 C2N2O5

2 N2 + 4 CO2 + 2 O2 4 NO2 + 4 CO

4 NO2 2 N2O4

2 N2O4 + 4 CO 4 NO +4 CO2

4 NO + 3 CO2 2 N2O5 + 3 C

2 N2 + 3 CO2 + 2 O2 2 N2O5 + 3 C 2 N2O5 + 1 4 C 2 C2N2O5

Acid/Base

Properties of Acids·Sour taste, Change color of dyes, Conduct electricity in solution, React with many metals, React with bases to form salts

Properties of Bases·Bitter taste, Feel slippery, Change color of dyes, Conduct electricity in solution, React with acids to form salts

Arrhenius·Acids: release H+ or H3O+ in solution

·Bases: release OH- in solution

Arrhenius·Acid: HA --> H+ + A- ·HCl --> H+ + Cl-

·Base: MOH --> M+ + OH-

·NaOH -->Na+ + OH-

Bronsted-Lowry·Acid: Proton donor ·Base: Proton Acceptor

Bronsted-Lowry· HA + H2O --> H3O+ + A-

· HI + H2O --> H3O+ + I- · Acid Base CA CB· NH3 + H2O --> NH4

+ + OH-

· Base Acid CA CB

Lewis Acid/Base·Acid: Electron Acceptor ·Base: Electron Donor

Lewis Acid/BaseH3N: + BF3 --> H3N-BF3

Base Acid Neutral

Drill:•List 3 properties

each of both acids & bases

Common Names· H+ Hydrogen ion· H3O+ Hydronium ion· H- Hydride ion· OH- Hydroxide ion· NH3 Ammonia

· NH4+ Ammonium ion

Amphoterism·Can act like an acid or a base

·Can donate or accept protons

Define acids & bases by each of

the three methods

Naming Acids· All acids are H-anion· If the anion is:· -ides hydro___ic acids· -ates ___ic acids· -ites ___ous acids

Naming Bases·Almost all bases are metal hydroxides

·Name by normal method·Ammonia (NH3) as well as many amines are bases

Drill: Name each of the following:

NaOH HIBa(OH)2 H2SO4

HMnO4 H3PO3

Strong Acids or Bases·Strong acids or bases ionize 100 % in solution

·Weak acids or bases ionize <100 % in solution

Strong Acids· HClO4 Perchloric acid

· H2SO4 Sulfuric acid

· HNO3 Nitric acid· HCl Hydrochloric acid· HBr Hydrobromic acid· HI Hydroiodic acid

Strong Bases· All column I hydroxides· Ca(OH)2 Calcium hydroxide

· Sr(OH)2 Strontium hydroxide

· Ba(OH)2 Barium hydroxide

Binary Acids·Acids containing only 2 elements

·HCl Hydrochloric acid·H2S Hydrosulfuric acid

Ternary Acids· Acids containing 3 elements

·H2SO4 Sulfuric acid

·H2SO3 Sulfurous acid

·HNO3 Nitric acid

Drill: Name & give the formula for at

least 4 each of strong acids & strong bases

Strong Acid/BaseIonizes 100 % (1 M) HA H+ + A-

1 M – all 1 1

Monoprotic Acids·Acids containing only one ionizable hydrogen

·HBr Hydrobromic acid·HCN Hydrocyanic acid·HC2H3O2 Acetic acid

Diprotic Acids·Acids containing 2 ionizable hydrogens

·H2SO4 Sulfuric acid

·H2SO3 Sulfurous acid

·H2CO3 Carbonic acid

Triprotic Acids·Acids containing 3 ionizable hydrogens

·H3PO4 Phosphoric acid

·H3PO3 Phosphorus acid

·H3AsO4 Arsenic acid

Polyprotic Acids· Acids containing more than one ionizable hydrogens

·H2SO4 Sulfuric acid

·H4SiO4 Silicic acid

·H2CO2 Carbonous acid

Monohydroxic Base·A base containing only one ionizable hydroxide

·NaOH Sodium hydroxide·KOH Potassium hydro.·LiOHLithium hydroxide

AP CHM HW•Read: Chapter 13•Problems: 1 - 3•Page: 395

CHM II HW•Read: Chapter 18•Problems: 3 & 5•Page: 787

Neutralization Rxn· A reaction between an acid

& a base making salt & H2O

·HA(aq) + MOH(aq) MA(aq) + H2O(l)

Neutralization Rxn

HCl(aq) + NaOH(aq)

NaCl(aq) + H2O(l)

Drill: Identify: acid, base, CA, & CB HCO3

- + H2O

H2CO3 + OH-

pH·The negative log of the hydrogen or hydronium ion concentration

·pH = -log[H+]·pOH = -log[OH-]

Calculate the pH of each of the following:1) [H+] = 0.040 M2) [HCl] = 0.0025 M 3) [HBr] = 0.080 M

Calculate the pOH of each of the following:

1) [OH-] = 0.030 M2) [KOH] = 0.0025 M3) [NaOH] = 4.0 x 10-12 M

AP CHM HW•Read Chapter 13•Work problems 17 & 19 on page 395

Standard Solution

·A solution with known

concentration

Titration· A method of determining the

concentration of one solution by reacting it with a standard solution

· MAVA = MBVB for monoprotics

Titration·When titrating acids against bases, the end point of the titration is at the equivalence point

Equivalence Point·The point where the H+

concentration is equal to the OH-

concentration

Titration No changes will be observed when titrating acids against bases; thus, one must use an indicator to see changes

Indicator·An organic dye that changes color when the pH changes

Drill:·Calculate the molarity of 25.0 mL HCl when it’s titrated to its equivalence point with 50.0 mL 0.200 M NaOH

Dilution Formula

M1V1 = M2V2

·Calculate the mL of 16.0 M HNO3 it takes

to make 4.0 L of 0.100 M HNO3

Make Calculations·Calculate the mL of 12.5 M HCl required

to make 2.5 L of 0.200 M HCl

Molarity·Moles of solute per

liter of solution (M)

Normality·Number of moles of hydrogen or hydroxide ions per liter of solution (N)

Titration Formula·NAVA = NBVB ·Elliott’s Rule:·#HMAVA = #OHMBVB

Make Calculations·Calculate the molarity of 30.0 mL H2CO3 when it’s titrated to its equivalence point with 75.0 mL 0.200 M NaOH

Make Calculations·Calculate the molarity of 40.0 mL H3PO4 when it’s titrated to its equivalence point with 30.0 mL 0.20 M Ba(OH)2

Calculate the volume of 0.250 M HCl

needed to titrate 50.00 mL 0.200 M NaOH to its equivalence point

Calculate the molarity 25.0 mL H3PO4 that

neutralizes 50.00 mL 0.200 M Ca(OH)2 to its equivalence point

Drill: Calculate the volume of 0.10 M

H3PO4 that neutralizes 50.00 mL 0.200 M

Ca(OH)2 to its equivalence point

AP CHM HW•Read: Chapter 13•Problems: 7 & 9•Page: 395

CHM II HW•Read: Chapter 18•Problems: 27•Page: 787

Drill: Calculate the molarity of 25.00 mL of H3PO4 that was titrated to its equivalence point

with 75.00 mL of0.125 M Ba(OH)2.

Drill: 3.2 g HI is dissolved in a

125 mL aqueous solution.

Calculate its pH.

Titration Curve:Strong acid vs

strong base

Titration Curve:Strong acid vs

strong base; then weak acid vs strong base

Titration Curve:Strong base vs

strong acid; then weak base vs strong acid