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. Kc = 2.0 x 10-10

Drill: 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. Kc = 2.0 x 10-12

Exp # [A] [B] [C] Rate 1) 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 Exp # [A] [B] [C] Rate 1) 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

([B3]/[B2])b = rate3/rate2 Exp # [A] [B] [C] Rate 1) 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

([C2]/[C1])c = rate2/rate1 Exp # [A] [B] [C] Rate 1) 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] k = Rate [A]2[B]3[C]

1 P2 + 2 SO2 + 1 O2  2 PO2 + 2 SO 2 PO2  1 P2O4 1 P2O4 + 2 SO  2 PO + 2 SO2 4 PO + 3 SO2  2 P2O5 + 3 S 3 P2O5 + 6 S  1 P6S6O15

1 P2 + 2 SO2 + 1 O2  2 PO2 + 2 SO 2 PO2  1 P2O4 1 P2O4 + 2 SO  2 PO + 2 SO2 4 PO + 3 SO2  2 P2O5 + 3 S 3 P2O5 + 6 S  1 P6S6O15

1 P2 + 2 SO2 + 1 O2  2 PO2 + 2 SO 2 PO2  1 P2O4 2 P2O4 + 4 SO  4 PO + 4 SO2 4 PO + 3 SO2  2 P2O5 + 3 S 3 P2O5 + 6 S  1 P6S6O15

1 P2 + 2 SO2 + 1 O2  2 PO2 + 2 SO 4 PO2  2 P2O4 2 P2O4 + 4 SO  4 PO + 4 SO2 4 PO + 3 SO2  2 P2O5 + 3 S 3 P2O5 + 6 S  1 P6S6O15

2 P2 + 4 SO2 + 2 O2  4 PO2 + 4 SO 4 PO2  2 P2O4 2 P2O4 + 4 SO  4 PO + 4 SO2 4 PO + 3 SO2  2 P2O5 + 3 S 3 P2O5 + 6 S  1 P6S6O15

2 P2 + 4 SO2 + 2 O2  4 PO2 + 4 SO 4 PO2  2 P2O4 2 P2O4 + 4 SO  4 PO + 4 SO2 4 PO + 3 SO2  2 P2O5 + 3 S 3 P2O5 + 6 S  1 P6S6O15

2 P2 + 4 SO2 + 2 O2  4 PO2 + 4 SO 4 PO2  2 P2O4 2 P2O4 + 4 SO  4 PO + 4 SO2 4 PO + 3 SO2  2 P2O5 + 3 S 2 P2 + 3 SO2 + 2 O2  2 P2O5 + 3 S

2 P2 + 3 SO2 + 2 O2  2 P2O5 + 3 S Rate = k[P2 ]2[SO2]3[O2]2

__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/Base H3N: + BF3 --> H3N-BF3 Base Acid Neutral

Common Names H+ Hydrogen ion H3O+ Hydronium ion H- Hydride ion OH- Hydroxide ion NH3 Ammonia NH4+ Ammonium ion

Amphiprotism Can act like an acid or a base Can donate or accept protons

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

Strong Acids or Bases Strong acids or bases ionize 100 % in solution Weak acids or bases ionize <100 % in solution

Drill: Name each of the following: KOH HBr Al(OH)3 H2CO3 HClO4 NH3

Review Drill & Check HW

CHM II HW Review PP-21 Complete the attached assignment & turn it in tomorrow.

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

Strong Acid/Base Ionizes 100 % (1 M) HA H+ + A- 1 M – all 1 1

Binary Acids Acids containing only 2 elements HCl Hydrochloric acid H2S Hydrosulfuric acid

Ternary Acids H2SO4 Sulfuric acid HNO3 Nitric acid Acids containing 3 elements H2SO4 Sulfuric acid HNO3 Nitric acid

Monoprotic Acids Acids containing only one ionizable hydrogen HBr Hydrobromic acid HC2H3O2 Acetic acid

Diprotic Acids Acids containing 2 ionizable hydrogens H2SO4 Sulfuric acid H2CO3 Carbonic acid

Triprotic Acids Acids containing 3 ionizable hydrogens H3PO4 Phosphoric acid H3AsO4 Arsenic acid

Polyprotic Acids H4SiO4 Silicic acid H2CO2 Carbonous acid Acids containing more than one ionizable hydrogens H4SiO4 Silicic acid H2CO2 Carbonous acid

Monohydroxic Base A base containing only one ionizable hydroxide NaOH Sodium hydroxide LiOH Lithium hydroxide

Neutralization Rxn HA(aq) + MOH(aq)  MA(aq) + H2O(l) 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)

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 Calculate the pH of each of the following: 1) [H+] = 0.040 M 2) [HCl] = 0.0025 M 3) [HBr] = 0.080 M

Calculate the pOH of each of the following: 1) [OH-] = 0.030 M 2) [KOH] = 0.0025 M 3) [NaOH] = 4.0 x 10-7 M

A solution with known concentration Standard Solution A solution with known concentration

Drill: Identify: acid, base, CA, & CB HCO3- + H2O H2CO3 + OH-

Review Drill & Check HW

CHM II HW Review PP-21 Complete the attached assignment & turn it in tomorrow.

Titration A method of determining the concentration of one solution by reacting it with a standard solution

Titration Formula for monoprotic solutions MAVA = MBVB

Equivalence Point The point where the concentrations of the two solutions in the titration are equal

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

Acid/Base Equivalence Point The point where the H+ concentration is equal to the OH- concentration

Titration Fact 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

Titration Formula for monoprotic solutions MAVA = MBVB

Dilution Formula M1V1 = M2V2

Calculate the mL of 16.0 M HNO3 it takes to make 4.0 L of 0.100 M HNO3

Calculate the mL of 12.5 M HCl required to make 2.5 L of 0.200 M HCl Make Calculations Calculate the mL of 12.5 M HCl required to make 2.5 L of 0.200 M HCl

Moles of solute per liter of solution (M) 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 for Acid/Base 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 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 Calculate the molarity 25.0 mL H3PO4 that neutralizes 50.00 mL 0.200 M Ca(OH)2 to its equivalence point

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

3.2 g HI is dissolved in a 1250 mL aqueous solution. Calculate its pH.

Calculate the volume of 0. 10 M H3PO4 that neutralizes 50. 00 mL 0 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 Drill: Calculate the molarity of 25.00 mL of H3PO4 that was titrated to its equivalence point with 75.00 mL of 0.125 M Ba(OH)2.