Keys to the Study of Chemistry

Keys to the Study of Chemistry
AP Chemistry 1

SI System of Measurement
Prefixes—fill out chart with partner Name Giga hecto (BASE) micro Symbol M da xxxx m n 10^ 103 100 10-1 10-2 Mega kilo deka deci centi milli nano G k h d c  109 106 102 101 10-3 10-6 10-9 2

Base Units Measurement Units Symbol for Unit time seconds m mass mole
Kelvin s length meter kilogram kg amount mol K temperature 3

Volume Equivalents to Memorize
Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4

What is the difference between mass and weight?
Mass depends only on the amount of matter in an object Weight depends on the force of gravity 5

Density Back to the Future

Temperature Conversions
Celsius   Kelvin O C = K - 273 K = O C + 273 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7

Temperature Conversions
Celsius   Fahrenheit oF + 40 oC + 40 9 oF = 5 oC T (in oF) = 9/5 T (in oC) + 32 T (in oC) = [ T (in oF) - 32 ] 5/9 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8

Factor Label Method Multiply by a ratio = 1 to convert units
See note packet for sample problems 9

Calculator 2nd EE Mode Sci / Flo 10

Measurement Practice 3.0 ml 5.75 ml 0.35 ml 14
14

Uncertainty in Measurement
Rule: Measurements have a quantity (number) and a unit (label) The last number in all measurements is estimated Can only have ONE estimated value 15

Practice Measure the length of a 3x5 card with each ruler 16

Accuracy, Precision, and Bias
closeness to actual value Precision closeness to each other Bias a non-random, directional error 19

Significant Figures Significant = Meaningful
Definition: Used because there is uncertainty in all measurements 20

Exact Numbers Not Measured Have infinite sig figs
Ex: amount, on the Periodic Table, NA , conversion factors 21

Dot Right; If Not, Left If there is a DOT start left, move right
do not count leading zeros X 103 X 10-9 22

Dot Right; If Not, Left If there is NO DOT start right, move left
do not count leading zeros X 103 ,000 23

How many sig figs in the following?
X 102 X 102 1 1 3 3 1 1 2 2 24

Addition / Subtraction with Sig Figs
? ? 2.2 g cm 1.50 g cm g 6.98 g ? ? ? ? ? cm ? ? 7.21 g 7.0 g 25

Make sure the units are the same
Rule: Make sure the units are the same Line up the numbers by the decimal points Do the math Determine which measurement is the least accurate (has the least number of digits after the decimal point)—Round to its furthest place 26

Sample Problems (5.4 X 108 m) (1.29 X 10-3 m) = 6.966 X 105 m2
2 sf 3 sf 7.0 X 105 m2 27

(4.90 X 103 kg) ÷ (130.0 L) = 3.7692 X 101 3.77 X 101 kg/L 3 sf 4 sf
28

Multiplication/Division with SigFigs
Answer will have the same # of sig figs as the number with the least number of sig figs Example: 430 L = ? gal Answer has ___ sf 29

Mixed Operations Round after each distinct rule
1) Average (10.2 g g g) 3 = g = 30.6 =10.2 g (tenths) 30

Percent Error (2.9 g/cm3 – 2.7 g/cm3) 2.7 g/cm3 X 100 0.2 g/cm3 X 100
7 % 31

Matter Physical Properties Physical Changes Chemical Properties
Chemical Changes Green + yellow food coloring Burn Mg Silo gas Tear paper Rip Mg KI + NaCrO4 32

Energy The ability to do “work” Potential Energy Kinetic Energy
Due to the POSITION of the object Kinetic Energy Due to the MOTION of the object Tendency to favor situations of lower energy 33

+ + Potential Energy + + + + + + 34

Chapter 2: The Components of Matter
AP Chemistry 35 35

START MEMORIZING NAMES, SYMBOLS, CHARGES OF ELEMENTS AND POLYATOMIC IONS

Definitions for Components of Matter
Element simplest type of substance with unique physical & chemical properties consists of only one type of atom cannot be broken down into any simpler substances Molecule - consists of two or more atoms chemically bound together and behaves as an independent unit. Figure 2.1 37

Definitions for Components of Matter
Compound - a substance composed of two or more elements which are chemically combined. Figure 2.1 Mixture - a group of two or more elements and/or compounds that are physically intermingled.

Video of reaction

Mixtures and Compounds
Figure 2.21 S Fe Physically mixed therefore can be separated by physical means. Allowed to react chemically therefore cannot be separated by physical means.

Heterogeneous mixtures :
Homogeneous mixtures : Solutions : has one or more visible boundaries between the components. has no visible boundaries because the components are mixed as individual atoms, ions, and molecules. A homogeneous mixture is also called a solution. Solutions in water are called aqueous solutions. solutions can exist in all three physical states. 42

Basic Separation Techniques
Tools of the Laboratory Basic Separation Techniques Filtration : Separates components of a mixture based upon differences in particle size. Normally separating a precipitate from a solution, or particles from an air stream. Crystallization : Separation is based upon differences in solubility of components in a mixture. Distillation : separation is based upon differences in volatility. Extraction : Separation is based upon differences in solubility in different solvents (major material). Chromatography : Separation is based upon differences in solubility in a solvent versus a stationary phase. Tools Tools Tools 43

Modern Periodic Table Label the following on your periodic table:
Label the following on your periodic table: metals/non-metals/metalloids families periods alkali metals, alkaline earth metals, halogens, noble gases, transition metals Charge of ions in each group Diatomic Molecules 45 45

How is the modern Periodic Table arranged?
Group / Family Figure 2.11 How is the modern Periodic Table arranged? 46

Dmitri Mendeleev (Russia)
First draft of Periodic Table –1869 47 47

Mendeleev’s Periodic Law
If arranged according to their atomic mass*, the elements exhibit a distinct periodicity (repeatable variation) in their chemical properties” * Note today we use atomic number to order elements instead of atomic mass 48 48

Metals, Metalloids, and Nonmetals
Figure 2.12 49

Metalloids Semimetals 51 51
51 51

Diatomic Molecules There are 7 atoms that exist in pairs start here!
52 52

Bonds Electrostatic attraction between atoms
Protons are attracted to their own and electrons AND other atoms’ electrons + e -

“Ionic Bonds” Note: No bond is 100% ionic
Na Cl Electrons are TRANSFERRED from a metal to a non-metal Na Cl +1 -1 electrostatic attraction Note: No bond is 100% ionic 54 54

Cation: positive ion, gets smaller
Na +1 Na Anion: negative ion, gets bigger Cl -1 Cl Conductivity Demo 55 55

Figure 2.13 Video of reaction 57

Coulomb’s Law directly
Energy of attraction/repulsion is __________related to the charge of the particles and ___________related to their distance inversely Energy  distance Charge 1 X Charge 2 58 58

HAVE YOU STARTED MEMORIZING NAMES, SYMBOLS, CHARGES OF ELEMENTS AND POLYATOMIC IONS???

Figure 2.15 60

CovalentBonds Electrons are SHARED between 2 or more non-metals
Can be shared equally or unequally 61 61

HAVE YOU STARTED MEMORIZING NAMES, SYMBOLS, CHARGES OF ELEMENTS AND POLYATOMIC IONS???

Covalent Bonds 63 63

HAVE YOU MEMORIZED NAMES, SYMBOLS, CHARGES OF ELEMENTS AND POLYATOMIC IONS????

Elements That Are Polyatomic
Figure 2.18 A Polyatomic Ion 65

Chemical Formulas Empirical Formula – Molecular Formula –
Shows the relative number of atoms of each element in the compound. The simplest formula Derived from masses of the elements. Molecular Formula – Shows the actual number of atoms of each element in the compound. Structural Formula – Shows the actual number of atoms, and the bonds between them Shows the arrangement of atoms in the molecule. 66

Chemical Formulas—Example: Glucose
Empirical Formula - Molecular Formula - Structural Formula -

YOU BETTER HAVE THE NAMES, SYMBOLS, CHARGES OF ELEMENTS AND POLYATOMIC IONS MEMORIZED BY NOW!

Determine the empirical and molecular formulas for each of the following:
M: C6H12O6 formaldehyde -D-galactose E: CH2O M: CH2O E: CH2O -D-glucose M: C2H4O2 -D-fructose acetic acid M: C6H12O6 E: CH2O E: CH2O M: C6H12O6 E: CH2O 69 69

Naming Compounds Pre-Test
Name: P4O7 Fe2O3 H2SO4 NO3 HBr K2ClO4 Write the Formula for: hydrofluoric acid ammonium carbonate copper (II) nitride calcium hydroxide phosphoric acid dinitrogen tetroxide tetraphosphorus heptoxide HF iron (III) oxide (NH4)2CO3 sulfuric acid Cu3N2 nitrogen trioxide Ca(OH)2 hydrobromic acid H3PO4 potassium perchlorate N2O4 70 70

Oxoanions ClO4- ClO3- ClO2- ClO- “hyper” perchlorate chlorate chlorite
hypochlorite 72 72

Naming oxoanions “hyper” Prefixes Root Suffixes Examples per root ate
Figure 2.20 “hyper” Prefixes Root Suffixes Examples per root ate ClO4- perchlorate No. of O atoms root ate ClO3- chlorate root ite ClO2- chlorite hypo root ite ClO- hypochlorite

Sample Problem 2.8 Determining Names and Formulas of Ionic Compounds Containing Polyatomic Ions PROBLEM: Give the systematic names or the formula or the formulas for the names of the following compounds: (a) Fe(ClO4)2 (b) sodium sulfite (c) Ba(OH)2 8H2O SOLUTION: (a) ClO4- is perchlorate; iron must have a 2+ charge. This is iron(II) perchlorate. (b) The anion sulfite is SO32- therefore you need 2 sodiums per sulfite. The formula is Na2SO3. (c) Hydroxide is OH- and barium is a 2+ ion. When water is included in the formula, we use the term “hydrate” and a prefix which indicates the number of waters. So it is barium hydroxide octahydrate.

Sample Problem 2.9 Recognizing Incorrect Names and Fromulas of Ionic Compounds PROBLEM: Something is wrong with the second part of each statement. Provide the correct name or formula. (a) Ba(C2H3O2)2 is called barium diacetate. (b) Sodium sulfide has the formula (Na)2SO3. (c) Iron(II) sulfate has the formula Fe2(SO4)3. (d) Cesium carbonate has the formula Cs2(CO3). SOLUTION: (a) Barium is always a +2 ion and acetate is -1. The “di-” is unnecessary. (b) An ion of a single element does not need parentheses. Sulfide is S2-, not SO32-. The correct formula is Na2S. (c) Since sulfate has a 2- charge, only 1 Fe2+ is needed. The formula should be FeSO4. (d) The parentheses are unnecessary. The correct formula is Cs2CO3.

Sample Problem 2.11 Determining Names and Formulas of Binary Covalent Compounds SOLUTION: PROBLEM: (a) What is the formula of carbon disulfide? (b) What is the name of PCl5? (c) Give the name and formula of the compound whose molecules each consist of two N atoms and four O atoms. (a) Carbon is C, sulfide is sulfur S and di-means 2 - CS2. (b) P is phosphorous, Cl is chloride, the prefix for 5 is penta-. Phosphorous pentachloride. (c) N is nitrogen and is in a lower group number than O (oxygen). Therefore the formula is N2O4 - dinitrogen tetraoxide.

Sample Problem 2.12 Recognizing Incorrect Names and Formulas of Binary Covalent Compounds PROBLEM: Explain what is wrong with the name of formula in the second part of each statement and correct it: (a) SF4 is monosulfur pentafluoride. (b) Dichlorine heptaoxide is Cl2O6. (c) N2O3 is dinitrotrioxide. SOLUTION: (a) The prefix mono- is not needed for one atom; the prefix for four is tetra-. So the name is sulfur tetrafluoride. (b) Hepta- means 7; the formula should be Cl2O7. (c) The first element is given its elemental name so this is dinitrogen trioxide.

Anion “-ate” suffix becomes an “-ic” suffix in the acid. Anion “-ite”
Naming Acids 1) Binary acids solutions form when certain gaseous compounds dissolve in water. For example, when gaseous hydrogen chloride(HCl) dissolves in water, it forms a solution called hydrochloric acid. Prefix hydro- + anion nonmetal root + suffix -ic + the word acid - hydrochloric acid 2) Oxoacid names are similar to those of the oxoanions, except for two suffix changes: Anion “-ate” suffix becomes an “-ic” suffix in the acid. Anion “-ite” suffix becomes an “-ous” suffix in the acid. The oxoanion prefixes “hypo-” and “per-” are retained. Thus, BrO4- is perbromate, and HBrO4 is perbromic acid; IO2- is iodite, and HIO2 is iodous acid. 80

Law of Conservation of Mass:
The total mass of substances does not change during a chemical reaction. total mass total mass = reactant reactant 2 product calcium oxide carbon dioxide calcium carbonate CaO CO2 CaCO3 56.08g g 100.08g 81

Law of Definite (or Constant) Composition:
No matter what its source, a particular chemical compound is composed of the same elements in the same parts (fractions) by mass. CaCO3 1 atom of Ca 40.08 amu 1 atom of C 12.00 amu 3 atoms of O 3 x amu amu 40.08 amu amu = parts Ca 12.00 amu amu = parts C 48.00 amu amu = parts O

Sample Problem 2.2 Calculating the Mass of an Element in a Compound PROBLEM: Pitchblende is the most commercially important compound of uranium Analysis shows that 84.2g of pitchblende contains 71.4g of uranium, with oxygen as the only other element. How many grams of uranium can be obtained from 102kg of pitchblende?

Law of Multiple Proportions
If elements A and B react to form two compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers: Carbon Oxide I Carbon Oxide II g oxygen / 100 g compound 57.1 72.7 g carbon / 100 g compound 42.9 27.3 g oxygen/g carbon 57.1 42.9 = 1.33 72.7 27.3 = 2.66 How many more oxygen is in compound II than in compound I ? = Formulas??? 84 84

The Atomic Basis of the Law of Multiple Proportions
Figure 2.4 85

John Dalton’s Atomic Theory— (England, 1766-1844)
1. All matter consists of atoms. 2. Atoms of one element cannot be converted into atoms of another element. 3. Atoms of an element are identical in mass and other properties and are different from atoms of any other element. 4. Compounds result from the chemical combination of a specific ratio of atoms of different elements. 86

J.J. Thompson (England) 1856-1940
Cathode Ray Tube Determined the charge/mass ratio of an electron. Suggested plum pudding model for atom 87 87

Experiments to Determine the Properties of Cathode Rays
Figure 2.5 88

November 1899 Scientific American
“At the recent meeting of the British Association for the Advancement of Science Prof. J.J.Thomson gave an interesting account of recent researches on the existence of masses smaller than atoms…It would appear that electrification seems to consist in the removal from an atom of a small corpuscle,…consisting of a very small portion of the mass with a negative charge, while the remainder of the atom possesses a positive charge.” 89 89

Millikan’s Oil-Drop Experiment for Measuring an Electron’s Charge
Figure 2.7 Mass of electron = 9.1 X kg 90

Ernest Rutherford (New Zealand) 1871-1937
1911- Ernst Rutherford- Gold Foil Experiment (Nobel Prize 1923) Rutherford & Geiger Suggested presence of a positively charged nucleus Video 91 91

Rutherford’s -Scattering Experiment
and Discovery of the Atomic Nucleus Figure 2.8 92

“As if you fired a 15 inch artillery shell at a piece of tissue paper and it came back and hit you”
Fun Fact: If you took all of the atoms on earth and smushed them, they would all fit into one sugar cube 93 93

General Features of the Atom
Figure 2.9 95

Properties of the Three Key Subatomic Particles
Charge Mass Location in the Atom Nucleus Outside Name(Symbol) Electron (e-) Neutron (n0) Proton (p+) Relative 1+ 1- Absolute(C)* x10-19 x10-19 Relative(amu)† Absolute(g) x10- 24 x10-24 x10-28 Table 2.2 96

Atomic Symbols, Isotopes, Numbers
J The Symbol of the Atom or Isotope Z J = Atomic symbol of the element A = mass number; A = Z + N Z = atomic number (the number of protons in the nucleus) N = number of neutrons in the nucleus Isotope = atoms of an element with the same number of protons, but a different number of neutrons Figure 2.10 Tools 97

Isotopes Atoms with the same number of protons, but different numbers of neutrons Symbols: C C A, Mass Number #p + #n Z, Atomic Number #p 98 98

Atomic Masses C-12 is assigned a mass of exactly 12 amu and other atoms’ masses are given relative to this standard A is used to compare the masses of atoms mass spectrometer 99 99

The Modern Reassessment of the Atomic Theory
1. All matter is composed of atoms. The atom is the smallest body that retains the unique identity of the element. 2. Atoms of one element cannot be converted into atoms of another element in a chemical reaction. Elements can only be converted into other elements in nuclear reactions. 3. All atoms of an element have the same number of protons and electrons, which determines the chemical behavior of the element. Isotopes of an element differ in the number of neutrons, and thus in mass number. A sample of the element is treated as though its atoms have an average mass. 4. Compounds are formed by the chemical combination of two or more elements in specific ratios. 100

Formation of a Positively Charged Neon Particle
Tools of the Laboratory Formation of a Positively Charged Neon Particle in a Mass Spectrometer Figure B2.1 Tools 103

Mass Spectrometer 104 104

Mass Spectrogram 105 105

We use the mass spectrogram to calculate average atomic mass
90.92% Ne-20 0.26 % Ne-21 8.82% Ne-22 ( x 20) + ( x 21) + ( x 22) = amu = amu Back 106 106

Chapter 17: Equilibrium AP Chemistry 107

Label the Forward Rate and Reverse Rate on the graph below
108

Reaching equilibrium on the macroscopic and molecular levels
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 17.1 N2O4(g) NO2(g) 109

N2O NO2 K E K E K E Label the region in which the study of kinetics is concerned Label the region that indicates the region has reached equilibrium 110

Equilibrium The state in which there are no observable changes as time goes by Rate of fwd rxn = Rate of rev rxn [R] and [P] are constant is dynamic on the molecular level; no further net change is observed because changes in one direction are balanced by changes in the other. Examples: acetone, NaCl(s), NO2 tubes 111

Which of these are at equilibrium?
A lit Bunsen burner An Alka Seltzer tablet in water Vapor pressure in a sealed flask A saturated solution of PbCl2 Student population of OHS over a year Salt dissolving in water 114

Law of Mass Action aA + bB cC + dD No units!! Initial Conc’ns
Equilibrium Conc’ns No units!! 116

N2O4 2NO2 Table 17.1 Initial and Equilibrium Concentration Ratios for
the N2O4-NO2 System at 1000C Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. N2O NO2 Ratio(Q) Ratio(K) Initial Equilibrium [NO2]2 [N2O4] [NO2]eq2 [N2O4]eq Experiment [N2O4] [NO2] [N2O4]eq [NO2]eq 1 0.1000 0.0000 0.0000 0.0491 0.1018 0.211 2 0.0000 0.1000 ∞ 0.0185 0.0627 0.212 3 0.0500 0.0500 0.0500 0.0332 0.0837 0.231 4 0.0750 0.0250 0.0411 0.0930 0.210 117

At equilibrium, Q = K The change in Q during the N2O4- NO2 reaction.
Figure 17.3 The change in Q during the N2O4- NO2 reaction. At equilibrium, Q = K 119

aA + bB cC + dD If K = 1 If K >> 1 If K << 1
Concentration of products = concentration of Reactants Concentration of products > concentration of Reactants “Equilibrium lies far to the right” Concentration of products < concentration of Reactants “Equilibrium lies far to the left” 121

The range of equilibrium constants
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 17.2 small K large K intermediate K 122

Write the reaction quotient, Qc, for each of the following reactions:
Sample Problem 17.1 Writing the Reaction Quotient from the Balanced Equation PROBLEM: Write the reaction quotient, Qc, for each of the following reactions: (a) The decomposition of dinitrogen pentoxide, N2O5(g) NO2(g) + O2(g) (b) The combustion of propane gas, C3H8(g) + O2(g) CO2(g) + H2O(g) (a) N2O5(g) NO2(g) + O2(g) 2 4 Qc = [NO2]4[O2] [N2O5]2 Qc = [CO2]3[H2O]4 [C3H8][O2]5 (b) C3H8(g) O2(g) CO2(g) H2O(g) 5 3 4

Form of Q or K for an Overall Rxn

Sample Problem 17.2 Writing the Reaction Quotient for an Overall Reaction PROBLEM: At very high temperatures reached during the explosive combustion of gasoline within the cylinders of an auto engine, some of the N2 and O2 present form nitric oxide which combines with more O2 to form nitrogen dioxide, a toxic pollutant that contributes to photochemical smog. (1) N2(g) + O2(g) NO(g) Kc1 = 4.3 x 10-25 (2) 2NO(g) + O2(g) NO2(g) Kc2 = 6.4 x 109 (a) Show that the Qc for the overall reaction sequence is the same as the product of the Qcs of the individual reactions. (b) Calculate the Kc for the overall reaction.

Sample Problem 17.2 Writing the Reaction Quotient for an Overall Reaction SOLUTION: Qc1 = [NO]2 [N2][O2] (1) N2(g) + O2(g) NO(g) (a) (2) 2NO(g) + O2(g) NO2(g) Qc2 = [NO2]2 [NO]2[O2] Qc = [NO2]2 [N2][O2]2 N2(g) + 2O2(g) NO2(g) [NO2]2 [NO]2[O2] = [NO2]2 [N2] [O2] 2 [NO]2 [N2][O2] Qc1x Qc2 = (b) Kc = Kc1 x Kc2 = (4.3 x 10-25) x (6.4 x 109) = 2.8 x 10-15

Modifying the Balanced Eqn
Original rxn Reverse rxn N2O4 (g) NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4] K = [N2O4] [NO2]2 ‘ = 1 K = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 127

Modifying the Rxn con’t
Original rxn New rxn N2O4 (g) NO2 (g) ½N2O4 (g) NO2 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4] = K1/2 K = [NO2] [N2O4]1/2 = (4.63 X 10-3)1/2 New rxn 2N2O4 (g) NO2 (g) = K2 K = [NO2]4 [N2O4]2 = (4.63 X 10-3)2 128

Sample Problem 17.3 Determining the Equilibrium Constant for an Equation Multiplied by a Common Factor PROBLEM: For the ammonia formation reaction N2(g) + 3H2(g) NH3(g) the equilibrium constant, Kc, is 2.4x10-3 at 1000K. If we change the coefficients of the equation, which we’ll call the reference (ref) equation, what are the values of Kc for the following balanced equations? (a) 1/3N2(g) + H2(g) /3NH3(g) (b) NH3(g) /2N2(g) + 3/2H2(g) SOLUTION: (a) The reference equation is multiplied by 1/3, so Kc(ref) will be to the 1/3 power. Kc = (2.4x10-3)1/3 = 0.13 (b) The reference equation is reversed and halved, so Kc(ref) is to the -1/2 power. Kc = (2.4x10-3)-1/2 = 20.

Heterogeneous Equilibria
Reactants and products are in different phases CaCO3 (s) CaO (s) CO2 (g) How are the 2 pictures different from each other? How are they the same? PCO2 130

Kc = [CO2] Kp = PCO2 CaCO3 (s) CaO (s) + CO2 (g)
Solids and liquids do not affect K; they have a constant concentration (i.e. density) 131

Water Softeners Problem: “Hard Water” contains lots of Ca2+
Solution: “Soften” water by replacing Ca2+ with Na+ 132

Homogeneous Equilibrium
All species are in the same phase (i.e. all gases) N2O4 (g) NO2(g) But…with gases Kc ≠ Kp 133

Sample Problem 17.4 Converting Between Kc and Kp PROBLEM: Calculate Kc for the following, if CO2 pressure is in atmospheres. CaCO3(s) CaO(s) + CO2(g) Kp = 2.1x10-4 at 1000K PLAN: We know Kp and can calculate Kc after finding ngas. R = L*atm/mol*K. SOLUTION: ngas = since there is only a gaseous product and no gaseous reactants. Kp = Kc(RT)n Kc = Kp/(RT)n = (2.1x10-4)( x 1000)-1 = 2.6x10-6

system proceeds from right to left to reach equilibrium
Reactants Products IF Qc > Kc system proceeds from right to left to reach equilibrium (reverse reaction is favored) Qc = Kc system is at equilibrium Qc < Kc system proceeds from left to right to reach equilibrium (forward reaction is favored) 136

Reaction direction and the relative sizes of Q and K
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 17.5 reactants products reactants products Equilibrium: no net change Reaction Progress Reaction Progress

Comparing Q and K to Determine Reaction Direction
Sample Problem 17.6 Comparing Q and K to Determine Reaction Direction PROBLEM: For the reaction N2O4(g) NO2(g), Kc = 0.21 at 1000C. At a point during the reaction, [N2O4] = 0.12M and [NO2] = 0.55M. Is the reaction at equilibrium. If not, in which direction is it progressing? PLAN: Write an expression for Qc, substitute with the values given, and compare the Qc with the given Kc. SOLUTION: Qc = [NO2]2 [N2O4] = (0.55)2 (0.12) = 2.5 Qc is > Kc, therefore the reaction is not at equilibrium and will proceed from right to left, from products to reactants, until Qc = Kc. 139

Let x be the change in concentration of Br2
At a given temp, (Kc) = 1.10 x for the reaction Br2 (g) 2 Br (g) If initially [Br2] i = M and [Br] i = M, calculate the concentrations of these species at equilibrium. Are we at equilibrium????? Br2 (g) Br (g) ICE Let x be the change in concentration of Br2 Initial (M) 0.163 0.0120 Change (M) -x +2x Equilibrium (M) x x [Br]2 [Br2] Kc = Kc = ( x)2 x = 1.1 x 10-2 Solve for x 140

 Kc = (0.0120 + 2x)2 0.163 - x = 1.10 x 10-2 ax2 + bx + c =0
-b ± b2 – 4ac  2a x = Recall the quadratic equation gives 2 solutions! We must determine which is correct to find the equilibrium concentrations x = x = “MATH” “0: Solver” Up Arrow to edit the equation Enter the equation “Enter” Type “0” after the “X=“ Push “SOLVE” (ALPHA Enter) to get the value of “X” “QUIT” “X,T,Theta,n” Br2 (g) Br (g) Initial (M) Change (M) Equilibrium (M) 0.163 0.0120 -x +2x x x [Br2] eq = M [Br] eq= M 141

Are we at equilibrium????? 2CO2 (g) 2 CO (g) + O2 (g) 0.40 Initial (M)
At a particular temperature, K = 2.0 X 10-6 for the reaction: 2CO2 (g) CO(g) + O2 (g) If 2.0 mol of CO2 is initially placed into a 5.0 L vessel, calculate the equilibrium concentrations of all species Are we at equilibrium????? 2CO2 (g) CO (g) + O2 (g) 0.40 Initial (M) +2x +x -2x Change (M) x Equilibrium (M) x 2x [CO]2[O2] [CO2]2 Kc = Kc = (2x)2 (x) ( x)2 = 2.0 x 10-6 Solve for x 142

≈0.4 Kc = (2x)2 (x) (0.4 - 2x)2 = 2.0 x 10-6 Kc = (4x3) (0.4 - 2x)2
LOOK! K IS REALLY SMALL YUCK!!! Kc = (4x3) ( x)2 = 2.0 x 10-6 2CO2 (g) CO (g) + O2 (g) 0.40 Initial (M) +2x +x -2x Change (M) x Equilibrium (M) x 2x ≈0.4 Kc = 4x3 ( x)2 = 2.0 x 10-6 Kc = 4x3 (0.4)2 = 2.0 x 10-6 x = 4.3 x 10-3 M = [O2]eq [CO] eq= 8.6 X 10-3 M [CO2]eq= M 143

Calculator MATH “0: Solver” 0= AX2 + BX + C ENTER After X = , type 0
“QUIT” “X,t,theta,n” button 144

Check Assumption ≈0.4 2CO2 (g) 2 CO (g) + O2 (g) 0.40 Initial (M) +2x
0.40 Initial (M) +2x +x -2x Change (M) x Equilibrium (M) x 2x ≈0.4 0.40 – 2x = 0.40 – 2(4.3 X 10-3) = 0.3914 145

When can we use the assumption?
K is very small (K < 10-3) K is very big (K > 103) 146

Sample Problem 17.6 Calculating Kc from Concentration Data PROBLEM: In a study of hydrogen halide decomposition, a researcher fills an evacuated 2.00-L flask with 0.200mol of HI gas and allows the reaction to proceed at 4530C. 2HI(g) H2(g) + I2(g) At equilibrium, [HI] = 0.078M. Calculate Kc. PLAN: First find the molar concentration of the starting material and then find the amount of each component, reactants and products, at equilibrium. SOLUTION: [HI] = 0.200 mol 2.00 L = 0.100M Let x be the amount of [H2] at equilibrium. Then x will also be the concentration of [I2] and the amount of [HI] is 2x or 0.078M.

Sample Problem 17.6 Calculating Kc from Concentration Data continued concentration (M) 2HI(g) H2(g) I2(g) initial 0.100 change -2x + x + x equilibrium x x x [HI] = = x ; x = 0.011M Qc = [H2] [I2] [HI]2 = [0.011][0.011] (0.078)2 = = Kc

Henri Louis Le Chatlier (1850-1936)
Inventor of acetylene torch Professor of Industrial Chemistry and Metallurgy Instrumental in the development of cement and Plaster of Paris 157

LeChatlier’s Principle
When a stress is applied to a system at equilibrium, the system will respond to partially undo the stress Add Reactant, Add Product, Remove Reactant, Remove Product, Add Heat, Increase Pressure,… 158

2 H+ + 2 CrO42- Cr2O72- + H2O produced used produced used [Cr2O72-]
Add HCl System wants? Shift? Color? [H+] Add NaOH Add a catalyst Use H+ ORANGE (Add H+) [H+] [CrO42-] [Cr2O72-] [H2O] (Use H+) Produce H+ YELLOW Reaches equilibrium faster---no change! H+ H+ H+ Na+ 161

2 NO2 N2O4 + energy produced used produced used [NO2] [N2O4] Add Heat
System wants? Shift? Color? Remove Heat Increase Pressure Decrease Pressure Shift DARKER Use Heat Produce Heat LIGHTER Darker; then lighter Decrease Pr. Lighter; then Darker Increase Pr. 163

Predicting the Effect of a Change in Temperature on
Sample Problem 17.13 Predicting the Effect of a Change in Temperature on the Equilibrium Position PROBLEM: How does an increase in temperature affect the concentration of the underlined substance and Kc for the following reactions? (a) CaO(s) + H2O(l) Ca(OH)2(aq) H0 = -82kJ (b) CaCO3(s) CaO(s) + CO2(g) H0 = 178kJ (c) SO2(g) S(s) + O2(g) H0 = 297kJ PLAN: Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on Kc. SOLUTION: (a) CaO(s) + H2O(l) Ca(OH)2(aq) heat An increase in temperature will shift the reaction to the left, decrease [Ca(OH)2], and decrease Kc. (b) CaCO3(s) + heat CaO(s) + CO2(g) The reaction will shift right resulting in an increase in [CO2] and increase in Kc. (c) SO2(g) + heat S(s) + O2(g) The reaction will shift right resulting in an decrease in [SO2] and increase in Kc. 164

Table B17.1 Effect of Temperature on Kc for Ammonia Synthesis
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. T (K) Kc 200. 7.17 x 1015 300. 2.69 x 108 400. 3.94 x 104 500. 1.72 x 102 600. 4.53 x 100 700. 2.96 x 10-1 800. 3.96 x 10-2 165

Le Chatelier Explains Tooth Decay!
Ca5 (PO4)3OH (s)  5 Ca+2 (aq) + 3 PO4–3 (aq) + OH- (aq) 166

Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Changes in Concentration N2 (g) + 3H2 (g) NH3 (g) Add NH3 Equilibrium shifts left to offset stress 14.5 168

Changes in Concentration continued aA + bB cC + dD Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left 14.5 169

Changes in Volume and Pressure A (g) + B (g) C (g) Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas 14.5 170

The effect of added Cl2 on the
Figure 17.7 The effect of added Cl2 on the PCl3-Cl2-PCl5 System 181

Predicting the Effect of a Change in Concentration on
Sample Problem 17.11 Predicting the Effect of a Change in Concentration on the Equilibrium Position PROBLEM: To improve air quality and obtain a useful product, sulfur is often removed from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O2; 2H2S(g) + O2(g) S(s) + 2H2O(g) What happens to (b) [H2S] if O2 is added? (a) [H2O] if O2 is added? (d) [H2S] if sulfur is added? (c) [O2] if H2S is removed? PLAN: Write an expression for Q and compare it to K when the system is disturbed to see in which direction the reaction will progress. Q = [H2O]2 [H2S]2[O2] SOLUTION: (a) When O2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H2O] increases. 182

Predicting the Effect of a Change in Concentration on
Sample Problem 17.11 Predicting the Effect of a Change in Concentration on the Equilibrium Position continued Q = [H2O]2 [H2S]2[O2] (b) When O2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H2S] decreases. (c) When H2S is removed, Q increases and the reaction progresses to the left to come back to K. So [O2] decreases. (d) Sulfur is not part of the Q (K) expression because it is a solid. Therefore, as long as some sulfur is present the reaction is unaffected. [H2S] is unchanged. 183

The effect of pressure (volume) on an equilibrium system.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 17.8 + lower P (higher V) more moles of gas higher P (lower V) fewer moles of gas 184

(a) CaCO3(s) CaO(s) + CO2(g)
Sample Problem 17.12 Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position PROBLEM: How would you change the volume of each of the following reactions to increase the yield of products. (a) CaCO3(s) CaO(s) + CO2(g) (b) S(s) + 3F2(g) SF6(g) (c) Cl2(g) + I2(g) ICl(g) PLAN: When gases are present a change in volume will affect the concentration of the gas. If the volume decreases (pressure increases), the reaction will shift to fewer moles of gas and vice versa. SOLUTION: (a) CO2 is the only gas present. To increase its yield, we should increase the volume (decrease the pressure). (b) There are more moles of gaseous reactants than products, so we should decrease the volume (increase the pressure) to shift the reaction to the right. (c) There are an equal number of moles of gases on both sides of the reaction, therefore a change in volume will have no effect. 185

Chapter 19 Ionic Equilibria in Aqueous Systems

Ionic Equilibria in Aqueous Systems
19.1 Equilibria of Acid-Base Buffer Systems 19.2 Acid-Base Titration Curves 19.3 Equilibria of Slightly Soluble Ionic Compounds 19.4 Equilibria Involving Complex Ions 19.5 Application of Ionic Equilibria to Chemical Analysis

The effect of addition of acid or base to …
Figure 19.1 The effect of addition of acid or base to … an unbuffered solution acid added base added or a buffered solution acid added base added Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

The Effect of Added Acetate Ion on the Dissociation of Acetic Acid
Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH3COOH]initial [CH3COO-]added % Dissociation* pH 0.10 0.00 0.050 0.15 1.3 2.89 0.036 4.44 0.018 4.74 0.012 4.92 * % Dissociation = [CH3COOH]dissoc [CH3COOH]initial x 100 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

How a buffer works Figure 19.2 Buffer after addition of H3O+
H2O + CH3COOH H3O+ + CH3COO- Buffer with equal concentrations of conjugate base and acid Buffer after addition of OH- CH3COOH + OH H2O + CH3COO- H3O+ OH- Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Terms: Common Ion Effect: HC2H3O2 H+ + C2H3O2- Add NaC2H3O2?
A response to the addition of an ion already involved in the equilibrium 191

A sol’n that RESISTS a change in the pH Made up of:
Buffered Solution: A sol’n that RESISTS a change in the pH Made up of: A WEAK ACID and its SALT OR A WEAK BASE and its SALT i.e. HA and NaA i.e. B and BHCl NH3 and NH4Cl 192

Why WEAK acids and bases?
do not 100% dissociate HA + H2O H3O A- add H+? Shift makes more acid (HA) add OH-? Shift makes more base (A-) Acid Base C.Acid C. Base 193

Why WEAK acids and bases?
do not 100% dissociate B + H2O BH OH- add H+? Shift makes more acid (BH+) add OH-? Shift makes more base (B) Acid Base C.Acid C. Base 194

Sample Problem 19.1 Calculating the Effect of Added H3O+ and OH- on Buffer pH PROBLEM: Calculate the pH: (a) of a buffer solution consisting of 0.50M CH3COOH and 0.50M CH3COONa (b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a) (c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a) Ka of CH3COOH = 1.8x10-5. (Assume the additions cause negligible volume changes. PLAN: We know Ka and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed step-wise through changes in the system. SOLUTION: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (M) (a) Initial Change Equilibrium 0.50 - 0.50 - x - + x + x 0.50-x - 0.50 +x x

Sample Problem 19.1 Calculating the Effect of Added H3O+ and OH- on Buffer pH continued (2 of 4) [H3O+] = x [CH3COOH]equil ≈ 0.50M [CH3COO-]initial ≈ 0.50M [H3O+][CH3COO-] [CH3COOH] [H3O+] = x = Ka [CH3COO-] [CH3COOH] Ka = = 1.8x10-5M Check the assumption: 1.8x10-5/0.50 X 100 = 3.6x10-3 % (b) [OH-]added = 0.020 mol 1.0L soln = 0.020M NaOH CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O (l) Concentration (M) Before addition 0.50 - 0.50 - Addition - 0.020 - - After addition 0.48 0.52 -

Sample Problem 19.1 Calculating the Effect of Added H3O+ and OH- on Buffer pH continued (3 of 4) Set up a reaction table with the new values. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (M) Initial 0.48 - 0.52 Change - x - + x + x Equilibrium 0.48 -x - 0.52 +x x [H3O+] = 1.8x10-5 0.48 0.52 = 1.7x10-5 pH = 4.77 0.020 mol 1.0L soln (c) [H3O+]added = = 0.020M H3O+ CH3COO-(aq) + H3O+(aq) CH3COOH(aq) + H2O (l) Concentration (M) Before addition 0.50 - 0.50 - Addition - 0.020 - - After addition 0.48 0.52 -

Sample Problem 19.1 Calculating the Effect of Added H3O+ and OH- on Buffer pH continued (4 of 4) Set up a reaction table with the new values. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (M) Initial 0.52 - 0.48 Change - x - + x + x Equilibrium 0.52 -x - 0.48 +x x [H3O+] = 1.8x10-5 0.48 0.52 = 2.0x10-5 pH = 4.70

Henderson-Hasselbalch Equation
199

Buffer capacity The amount of H+ / OH- a buffer can absorb without a significant change in pH Depends on [HA] and [A-]

The relation between buffer capacity and pH change
Figure 19.3 The relation between buffer capacity and pH change

Buffer Range The pH range over which the buffer acts effectively
Within [HA] / [A-] = 10 to = 0.1

Preparing a Buffer Phosphate buffers are important in regulating the pH of intracellular fluids at pH values generally between 7.1 and 7.2. What is the concentration ratio of H2PO4- to HPO42- in intracellular fluid at pH = 7.15? H2PO4- (aq)  H+ (aq) + PHPO42- (aq) Ka = 6.2 X 10-8 zO.87 it]

Colors and approximate pH range of some common acid-base indicators
Figure 19.4

STRONG ACID—STRONG BASE
Titrations and pH Curves How does the pH of 50.0 ml M HCl change as M NaOH is added? STRONG ACID—STRONG BASE pH before any NaOH is added? major species? HCl  H Cl- [0.100 M] [0.100 M] [0.100 M] H+ Cl- H2O pH = -log (0.100) = 1.000 207

2.pH after some NaOH is added
Add 10.0 ml M NaOH major species? H+ Cl- Na+ OH- H2O 50.0 ml 0.100 M HCl 10.0 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(60) MT = M For OH- (0.2)(10) = MT(60) MT = M 60.0 ml total 209

Step 2: Neutralization Banana Chart: Before, Neutralization, After
H+ + OH  H2O B N A ≈ major species? H+ H2O pH = -log (0.0500) = 1.300 210

pH after some NaOH is added
ii. Add 20.0 ml M NaOH major species? H+ Cl- Na+ OH- H2O 50.0 ml 0.100 M HCl 20.0 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(70) MT = M For OH- (0.2)(20) = MT(70) MT = M 70.0 ml total 212

H+ + OH- H2O B N A ≈ major species? H+ H2O pH = -log ( ) = 1.845 213

iii. Add 24.0 ml M NaOH major species? H+ Cl- Na+ OH- H2O 50.0 ml 0.100 M HCl 24.0 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(74) MT = M For OH- (0.2)(25) = MT(74) MT = M 74.0 ml total 215

H+ + OH- H2O B N A ≈ major species? H+ H2O pH = -log (0.0027) = 2.569 216

iv. Add 24.9 ml M NaOH major species? H+ Cl- Na+ OH- H2O 50.0 ml 0.100 M HCl 24.9 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(74.9) MT = M For OH- (0.2)(25) = MT(74.9) MT = M 74.9 ml total 218

H+ + OH- H2O B N A ≈ major species? H+ H2O pH = -log (0.0003) = 3.523 219

vi. Add 25.0 ml M NaOH major species? H+ Cl- Na+ OH- H2O 50.0 ml 0.100 M HCl 25.0 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(75) MT = M For OH- (0.2)(25) = MT(75) MT = M 75.0 ml total 221

H+ + OH- H2O B N A ≈ ≈ major species? H2O pH = -log(1 X 10-7) = 7 222

equivalence point (mol H+ = mol OH-) 223

Step 1: DILUTION 3. pH after 30 ml NaOH added major species?
H+ Cl- Na+ OH- H2O 50.0 ml 0.100 M HCl 30.0 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(80) MT = M For OH- (0.2)(25) = MT(80) MT = M 80.0 ml total 224

H+ + OH- H2O B N A ≈ major species? OH- H2O pOH = -log (0.0125) = 1.903 pH = 14 – pOH = 14 – = 225

Figure 15. 1 The pH Curve for the Titration of 50. 0 mL of 0
Figure 15.1 The pH Curve for the Titration of 50.0 mL of M HNO3 with M NaOH

Curve for a strong acid-strong base titration
Figure 19.6 Curve for a strong acid-strong base titration Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

WEAK ACID—STRONG BASE How does the pH of 50.0 ml M acetic acid sol’n change as M NaOH is added? pH before any NaOH is added? major species? HC2H3O2 H2O 229

ICE, ICE, BABY Eqn? HC2H3O2 H+ + C2H3O2- I C 0.100 ≈0 ≈0 E -x +x +x
≈ ≈0 -x +x x 0.100-x x x 230

x = 0.00134 pH = -log (0.00134) = 2.872 HC2H3O2 H+ + C2H3O2- I C
≈ ≈0 ≈ HC2H3O2 H+ + C2H3O2- I C E pH = -log ( ) = 2.872 231

Add 5.0 ml M NaOH major species? HC2H3O2 Na+ OH- H2O 50.0 ml 0.100 M HAc 5.0 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(55) MT = M For OH- (0.2)(5) = MT(55) MT = M 55.0 ml total 233

Step 2: Neutralization major species: HC2H3O2 OH- H2O Eqn?
HC2H3O2 + OH- H2O + C2H3O2- B N A ≈0 ≈ major species? HC2H3O2 C2H3O2- H2O Eqn? HC2H3O2 H+ + C2H3O2- 234

ICE, ICE, BABY x = 7.19 X 10-5 HC2H3O2 H+ + C2H3O2- I 0.0727 ≈0 0.0182
≈ -x +x x x x x x = 7.19 X 10-5 235

pH = -log (7.19 X 10-5) = 4.143 HC2H3O2 H+ + C2H3O2- I C
E ≈ 7.19 X X X 10-5 ≈ X ≈0.0182 pH = -log (7.19 X 10-5) = 4.143 236

Terms: Common Ion Effect: HC2H3O2 H+ + C2H3O2- Add NaC2H3O2?
A response to the addition of an ion already involved in the equilibrium 238

A sol’n that RESISTS a change in the pH Made up of:
Buffered Solution: A sol’n that RESISTS a change in the pH Made up of: A WEAK ACID and its SALT OR A WEAK BASE and its SALT i.e. HA and NaA i.e. B and BHCl NH3 and NH4Cl 239

ii. Add 10.0 ml M NaOH major species? HC2H3O2 Na+ OH- H2O 50.0 ml 0.100 M HAc 10.0 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(60) MT = M For OH- (0.2)(10) = MT(60) MT = M 60.0 ml total 240

Henderson-Hasselbalch Eqn
ICE, ICE, BABY Or……….. Henderson-Hasselbalch Eqn 242

ICE, ICE, BABY x = 2.70 X 10-5 HC2H3O2 H+ + C2H3O2- I 0.0500 ≈0 0.0333
≈ -x +x x 0.05-x x x x = 2.70 X 10-5 243

pH = -log (2.70 X 10-5) = 4.568 HC2H3O2 H+ + C2H3O2- I
E ≈ 2.70 X X X 10-5 ≈ X ≈0.0333 pH = -log (2.70 X 10-5) = 4.568 244

Or…H-H Eqn 4.74 245

iii. Add 12.5 ml M NaOH major species? HC2H3O2 Na+ OH- H2O 50.0 ml 0.100 M HAc 12.5 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(62.5) MT = M For OH- (0.2)(12.5) = MT(62.5) MT = M 62.5 ml total 247

ICE, ICE, BABY x =1.80 X 10-5 HC2H3O2 H+ + C2H3O2- I 0.0400 ≈0 0.0400
≈ -x +x x 0.04-x x x x =1.80 X 10-5 251

E ≈ 1.80 X X X 10-5 ≈ X ≈0.0400 pH = -log (1.80 X 10-5) = 4.745 252

Or…H-H Eqn 4.74 =1!!! 253

iv. Add 20 ml M NaOH major species? HC2H3O2 Na+ OH- H2O 50.0 ml 0.100 M HAc 20 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(70) MT = M For OH- (0.2)(20) = MT(70) MT = M 70 ml total 255

ICE, ICE, BABY x =4.51 X 10-5 HC2H3O2 H+ + C2H3O2- I 0.0143 ≈0 0.0571
≈ -x +x x x x x x =4.51 X 10-5 258

E ≈ 4.51 X X X 10-5 ≈ X ≈ pH = -log (4.51 X 10-5) = 5.346 259

Or…H-H Eqn 4.74 260

Add 25 ml M NaOH major species? HC2H3O2 Na+ OH- H2O 50.0 ml 0.100 M HAc 25 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(75) MT = M For OH- (0.2)(20) = MT(75) MT = M 75 ml total 262

HC2H3O2 + OH- H2O + C2H3O2- B N A ≈0 ≈ ≈ major species? C2H3O2- H2O Eqn? C2H3O2- + H2O HC2H3O2 + OH- 263

ICE, ICE, BABY Kb!!! x =6.08 X 10-6 C2H3O2- + H2O HC2H3O2 + OH- I
-x x x Kb!!! x x x x =6.08 X 10-6 264

pOH = -log (6.08 X 10-6) = 5.215 pH = 14 – pOH = 14 – 5.215 = 8.784
C2H3O H2O HC2H3O2 + OH- I C E -6.08 X X X 10-6 X X 10-6 pOH = -log (6.08 X 10-6) = 5.215 pH = 14 – pOH = 14 – = 8.784 265

4. Add 50 ml M NaOH major species? HC2H3O2 Na+ OH- H2O 50.0 ml 0.100 M HAc 50 ml 0.200 M NaOH MoVo = MTVT Step 1: DILUTION For H+ (0.1)(50) = MT(100) MT = M For OH- (0.2)(20) = MT(100) MT = M 100 ml total 267

HC2H3O2 + OH- H2O + C2H3O2- B N A ≈0 ≈ major species? C2H3O2- OH- H2O pH depends on? 268

pOH = -log (0.0500) = 1.301 pH = 14 – pOH = 14 –1.301 = 269

Buffering region Eq pt > 7; pH depends on A-
pH = pKa =half-way to eq pt Buffering region 270

Figure The pH Curve for the Titration of 50.0 mL of M HC2H3O2 with M NaOH

Curve for a weak acid- strong base titration
Titration of 40.00mL of M HPr with M NaOH Figure 19.7 Curve for a weak acid- strong base titration pH = 8.80 at equivalence point pKa of HPr = 4.89 methyl red [HPr] = [Pr-] Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

WEAK BASE—STRONG ACID How does the pH of ml 0.05 M ammonium hydroxide change as M HCl is added? pH before any HCl is added? major species? NH4OH H2O Comes from dissolving NH3 in water 275

ICE, ICE, BABY Eqn? x = 9.5 X 10-4 NH3 + H2O NH4+ + OH- I C
≈ ≈0 -x x x 0.05-x x x x = 9.5 X 10-4 276

pH = -log (9.5X10-4) = 10.98 NH3 + H2O NH4+ + OH- I 0.05 --- ≈0 ≈0 C E
≈ ≈0 -9.5 X X X 10-4 ≈ X X 10-4 pH = -log (9.5X10-4) = 10.98 277

2.pH after some HCl is added
Add 5.0 ml M HCl major species? NH3 H+ Cl- H2O 5.0 ml 0.200 M HCl 100 ml 0.05 M NH3 MoVo = MTVT Step 1: DILUTION For H+ (0.2)(5) = MT(105) MT = M For OH- (0.05)(100) = MT(105) MT = M 105.0 ml total 279

Step 2: Neutralization major species: NH3 H+(H3O+) H2O Eqn?
H3O NH3 H2O NH4+ B N A ≈0 ≈ major species? NH3 NH4+ H2O Eqn? NH3 + H2O NH OH- 280

ICE, ICE, BABY x = 7.22X 10-5 NH3 + H2O NH4+ + OH- I
≈0 -x x x x x x x = 7.22X 10-5 282

pOH = -log (7.22X10-5) = 4.14 pH = 14 - pOH = 14 - 4.14 = 9.86
NH H2O NH OH- I C E ≈0 -7.22 X X X 10-5 ≈ X 10-5 pOH = -log (7.22X10-5) = 4.14 pH = 14 - pOH = = 9.86 283

Or…H-H Eqn 4.74 284

ii. Add 10.0 ml M HCl major species? NH3 H+ Cl- H2O 10.0 ml 0.200 M HCl 100 ml 0.05 M NH3 MoVo = MTVT Step 1: DILUTION For H+ (0.2)(10) = MT(110) MT = M For OH- (0.05)(100) = MT(110) MT = M 110.0 ml total 286

ICE, ICE, BABY x = 2.70 X 10-5 NH3 + H2O NH4+ + OH- I
≈0 -x x x x x x x = 2.70 X 10-5 289

pOH = -log (1.8 X10-5) = 4.74 pH = 14 - pOH = 14 - 4.74 = 9.26
NH H2O NH OH- I C E ≈0 -1.8 X X X 10-5 ≈ X 10-5 pOH = -log (1.8 X10-5) = 4.74 pH = 14 - pOH = = 9.26 290

Or…H-H Eqn 4.74 291

iii. Add 12.5 ml M HCl major species? NH3 H+ Cl- H2O 12.5 ml 0.200 M HCl 100 ml 0.05 M NH3 MoVo = MTVT Step 1: DILUTION For H+ (0.2)(12.5) = MT(112.5) MT = M For OH- (0.05)(100) = MT(112.5) MT = M 112.5 ml total 293

ICE, ICE, BABY x = 1.8 X 10-5 NH3 + H2O NH4+ + OH- I C
≈0 -x x x x x x x = 1.8 X 10-5 296

pOH = -log (2.70 X10-5) = 4.57 pH = 14 - pOH = 14 - 4.57 = 9.43
NH H2O NH OH- I C E ≈ ≈0 -2.7 X X X 10-5 ≈ X X 10-5 pOH = -log (2.70 X10-5) = 4.57 pH = 14 - pOH = = 9.43 297

Or…H-H Eqn 4.74 298

iv. Add 24.5 ml M HCl major species? NH3 H+ Cl- H2O 12.5 ml 0.200 M HCl 100 ml 0.05 M NH3 MoVo = MTVT Step 1: DILUTION For H+ (0.2)(12.5) = MT(112.5) MT = M For OH- (0.05)(100) = MT(112.5) MT = M 112.5 ml total 300

Figure 15.5 The pH Curve for the Titration of mL of M NH3 with 0.10 M HCI

Curve for a weak base- strong acid titration
Titration of 40.00mL of M NH3 with M HCl Figure 19.8 pKa of NH4+ = 9.25 Curve for a weak base- strong acid titration pH = 5.27 at equivalence point Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Curve for the titration of a weak polyprotic acid.
Figure 19.9 Curve for the titration of a weak polyprotic acid. pKa = 7.19 pKa = 1.85 Titration of 40.00mL of M H2SO3 with M NaOH Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Review Quiz 1) [H3O+] in a soln of NaOH having a pH of 10.3?
a) 5X10-11M b) 5.0X10-11 M c) 5.01X10-11 M d) other 2) pH of neut body a) b) 7.00 c) 6.81 d) other 3) pH (at 25oC) of a solution of 1.0 X M HCl? a) b) c) 7.00 d) other 4) endpoint of titration b/w acid & a) <7.00 b) >7.00 c) 7.00 d) can’t say 5) equiv pt of titration b/w an acid & base at 25oC? 6) What is the conjugate base of NH3? a) NH b) NH2- c) OH- d) H3O+ 304

Solubility Equilibrium
A ___________ sol’n is at equilibrium Ex: PbCl2 Eqn? PbCl2 Pb Cl- Equilibrium Expression? K = [Pb2+] [Cl-]2 saturated Salts dissociate 100%; but the AMOUNT that dissolves varies sp Solubility Product 305

Ksp of Silver Halides AgBr AgCl
Silver Bromide grains on X-ray film emulsion - Kodak 306

Differences between solubility and solubility product
# moles solute dissolved / L solution or… # grams solute dissolved / 100 ml water Solubility Product Ksp—an equilibrium constant; only changes with temperature 307

Calculate the Ksp for Li2CO3 if its solubility is 7.4 X 10-2 mol/L
1st!!! Write the equation Li2CO3 2 Li CO32- 2nd!!! Write the equilibrium expression Ksp = [Li+]2[CO32-] Ksp = (1.48X10-1)2(7.4X10-2) Ksp = 1.6 X 10-3 1.48X10-1 7.4X10-2 7.4X10-2 308

Calculate the solubility of PbI2 (Ksp = 1.4 X 10-8)
1st!!! Write the equation PbI2 Pb I- 2nd!!! Write the equilibrium expression Ksp = [Pb2+][I-]2 Ksp = (x)(2x)2 1.4X10-8 = 4x3 x = 1.5 X 10-3 mol / L 2x x x 309

Common Ion Effect 2x x 3x 1st!!! Write the equation
Calc the solubility of Ca3(PO4)2 (Ksp = 1.3X10-32) 1st!!! Write the equation Ca3(PO4)2 3Ca PO43- 2nd!!! Write the equilibrium expression Ksp = [Ca2+]3[PO43-]2 Ksp = (3x)3(2x)2 1.3X10-32 = (27x3)(4x2) = 108x5 x = 1.6 X 10-7 mol / L 2x x 3x 310

Calculate the solubility of Ca3(PO4)2 in 0.20 M Na3PO4
1st!!! Write the equation Ca3(PO4)2 3Ca PO43- Major species? Prediction? or Na+ PO H2O Ca3(PO4)2 3Ca PO43- I C E --- ≈ x x x x 311

x = 2.3 X 10-11 mol / L (vs. 1.6 X 10-7 mol / L)
Ca3(PO4)2 3Ca PO43- I C E --- ≈ x x x x Ksp= (Ca2+)3(PO4)2 = (3x)3(0.2+2x)2 = (3x)3(0.2)2 1.3X10-32 = 1.08x3 x = 2.3 X mol / L (vs. 1.6 X 10-7 mol / L) 312

pH and solubility A + B C + D K = 2 C + D  E + F K = 3 A + B  E + F
Reaction Coupling (one reaction drags a slower one along) A + B C D K = 2 C D  E F K = 3 A + B  E F K = 2 X 3 = 6 Note: K is higher than each rxn individually 313

Limestone with acid rain
Dissociation of Calcium Carbonate in acid: CaCO Ca2+ + CO32- Ksp = 8.7X10-9 H+ + CO HCO3- K = 1/Ka2 = 1/5.6X10-11 = 1.8X1010 CaCO3 + H Ca2+ + HCO3- K = 155 314

Milk of Magnesia+Stomach Acid
Mg(OH) Mg2+ +2OH- Ksp=8.9X10-12 2H++ 2OH H2O 2(1/Kw) = 2X1014 Mg(OH)2 +2H H2O + Mg2+ K=1.8X103 315

Which salt’s solubility will depend on pH?
AgF AgCl AgBr AgF Ag F- H++ F HF AgF + H+ Ag+ + HF Why won’t this work with a strong acid? 316

Precipitation Ion product (Q) (Use initial concentrations!)
PbCl2 Pb Cl- Q = [Pb2+]i [Cl-]2i If Q > K precipitate forms If Q < K no precipitate forms 317

Step 1: DILUTION Pb2+ NO3- K+ Cl- H2O MoVo = MTVT
A sol’n is prepared by mixing 50.0 ml of 0.10 M Pb(NO3)2 with 50.0 ml of 1.0 M KCl. Calculate the concentrations of Pb2+ and Cl- at equilibrium. Ksp for PbCl2 = 1.6X10-5. Major species Pb2+ NO3- K+ Cl- H2O 50.0 ml 0.10 M Pb(NO3)2 1.0 M KCl MoVo = MTVT Step 1: DILUTION For Pb2+ (0.1)(50) = MT(100) MT = 0.05 M For Cl- (1.0)(50) = MT(100) MT = 0.50 M 100.0 ml total 318

Calculate Q Write equation and equilibrium expression
PbCl2 Pb Cl Ksp = [Pb2+][Cl-]2 Q =(0.05)(0.5)2 = Q>K 319

Precipitation major species? Cl- PbCl2 H2O Eqn? PbCl2 Pb2+ + 2 Cl-
≈ major species? Cl- PbCl2 H2O Eqn? PbCl2 Pb Cl- 320

ICE, ICE, BABY x = 1.0 X 10-4 PbCl2 Pb2+ + 2Cl- I C ----- ≈0 0.40 E
≈ x x x x x = 1.0 X 10-4 321

[Pb2+]eq = 1 X 10-4 M [Cl-]eq = 0.40 M
Calculate the concentrations of Pb2+ and Cl- at equilibrium. PbCl2 Pb Cl- I C E ≈ X X10-4 X X 10-4 [Pb2+]eq = 1 X 10-4 M [Cl-]eq = 0.40 M 322

Sample Problem 19.4 Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds PROBLEM: Write the ion-product expression for each of the following: (a) Magnesium carbonate (b) Iron (II) hydroxide (c) Calcium phosphate (d) Silver sulfide PLAN: Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d). SOLUTION: (a) MgCO3(s) Mg2+(aq) + CO32-(aq) Ksp = [Mg2+][CO32-] (b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq) Ksp = [Fe2+][OH-] 2 (c) Ca3(PO4)2(s) Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO43-]2 (d) Ag2S(s) Ag+(aq) + S2-(aq) S2-(aq) + H2O(l) HS-(aq) + OH-(aq) Ksp = [Ag+]2[HS-][OH-] Ag2S(s) + H2O(l) Ag+(aq) + HS-(aq) + OH-(aq)

Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic
Compounds at 250C Name, Formula Ksp Aluminum hydroxide, Al(OH)3 3 x 10-34 Cobalt (II) carbonate, CoCO3 1.0 x 10-10 Iron (II) hydroxide, Fe(OH)2 4.1 x 10-15 Lead (II) fluoride, PbF2 3.6 x 10-8 Lead (II) sulfate, PbSO4 1.6 x 10-8 Mercury (I) iodide, Hg2I2 4.7 x 10-29 Silver sulfide, Ag2S 8 x 10-48 Zinc iodate, Zn(IO3)2 3.9 x 10-6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Sample Problem 19.5 Determining Ksp from Solubility PROBLEM: (a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4? (b) When lead (II) fluoride (PbF2) is shaken with pure water at 250C, the solubility is found to be 0.64g/L. Calculate the Ksp of PbF2. PLAN: Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. SOLUTION: PbSO4(s) Pb2+(aq) + SO42-(aq) (a) Ksp = [Pb2+][SO42-] 4.25x10-3g 100mL soln 1000mL L 303.3g PbSO4 mol PbSO4 = 1.40x10-4M PbSO4 Ksp = [Pb2+][SO42-] = (1.40x10-4)2 = 1.96x10-8

Sample Problem 19.5 Determining Ksp from Solubility continued (b) PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]2 0.64g L soln 245.2g PbF2 mol PbF2 = 2.6x10-3 M Ksp = (2.6x10-3)(5.2x10-3)2 = 7.0x10-8

Sample Problem 19.6 Determining Solubility from Ksp PROBLEM: Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6. PLAN: Write out a dissociation equation and Ksp expression; Find the molar solubility (S) using a table. SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2 Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Concentration (M) Initial - Change - +S + 2S Equilibrium - S 2S Ksp = (S)(2S)2 S = = 1.2x10x-2M

Table 19.3 Relationship Between Ksp and Solubility at 250C
No. of Ions Formula Cation:Anion Ksp Solubility (M) 2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4 2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4 2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH)2 1:2 5.5 x 10-6 1.2 x 10-2 3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

The effect of a common ion on solubility
Figure 19.11 The effect of a common ion on solubility PbCrO4(s) Pb2+(aq) + CrO42-(aq) CrO42- added PbCrO4(s) Pb2+(aq) + CrO42-(aq) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Sample Problem 19.7 Calculating the Effect of a Common Ion on Solubility PROBLEM: In Sample Problem 19.6, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6. PLAN: Set up a reaction equation and table for the dissolution of Ca(OH)2. The Ca(NO3)2 will supply extra [Ca2+] and will relate to the molar solubility of the ions involved. SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Concentration(M) Initial - 0.10 Change - +S +2S Equilibrium - S 2S Ksp = 6.5x10-6 = ( S)(2S)2 = (0.10)(2S)2 S << 0.10 Check the assumption: S = = 4.0x10-3M 4.0x10-3M x 100 = 0.10M 4.0%

Test for the presence of a carbonate
Figure 19.12 Test for the presence of a carbonate

Sample Problem 19.8 Predicting the Effect on Solubility of Adding Strong Acid PROBLEM: Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds: (a) Lead (II) bromide (b) Copper (II) hydroxide (c) Iron (II) sulfide PLAN: Write dissolution equations and consider how strong acid would affect the anion component. SOLUTION: (a) PbBr2(s) Pb2+(aq) + 2Br-(aq) Br- is the anion of a strong acid. No effect. (b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) OH- is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility. (c) FeS(s) Fe2+(aq) + S2-(aq) Sr2- is the anion of a weak acid and will react with water to produce OH-. FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq) Both weak acids serve to increase the solubility of FeS.

Sample Problem 19.9 Predicting Whether a Precipitate Will Form PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF? PLAN: Write out a reaction equation to see which salt would be formed. Look up the Ksp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION: CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11 mol Ca2+ = L(0.30mol/L) = 0.030mol [Ca2+] = 0.030mol/0.300L = 0.10M mol F- = L(0.060mol/L) = 0.012mol [F-] = 0.012mol/0.300L = 0.040M Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6x10-4 Q is >> Ksp and the CaF2 WILL precipitate.

Cr(NH3)63+, a typical complex ion
Figure 19.13 Cr(NH3)63+, a typical complex ion

Sample Problem 19.10 Calculating the Concentration of a Complex Ion PROBLEM: An industrial chemist converts Zn(H2O)42+ to the more stable Zn(NH3)42+ by mixing 50.0L of M Zn (H2O)42+ and 25.0L of 0.15M NH3. What is the final [Zn (H2O)42+]? Kf of Zn(NH3)42+ is 7.8x108. PLAN: Write the reaction equation and Kf expression. Use a reaction table to list various concentrations. Remember that components will be diluted when mixed as you calculate final concentrations. It is obvious that there is a huge excess of NH3 and therefore it will drive the reaction to completion. SOLUTION: Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)42+(aq) + 4H2O(l) Kf = [Zn(NH3)42+] [Zn(H2O)42+][NH3]4 [Zn(H2O)42+]initial = (50.0L)(0.0020M) 75.0L = 1.3x10-3M [NH3]initial = (25.0L)(0.15M) 75.0L = 5.0x10-2M

Sample Problem 19.10 Calculating the Concentration of a Complex Ion continued Since we assume that all of the Zn(H2O)42+ has reacted, it would use 4 times its amount in NH3. [NH3]used = 4(1.3x10-3M) = 5.2x10-3M [Zn(H2O)42+]remaining = x(a very small amount) Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)42+(aq) + 4H2O(l) Concentration(M) Initial 1.3x10-3 5.0x10-2 - Change ~(-1.3x10-3) ~(-5.2x10-3) ~(+1.3x10-3) - Equilibrium x 4.5x10-2 1.3x10-3 - Kf = [Zn(NH3)42+] [Zn(H2O)42+][NH3]4 = 7.8x108 = (1.3x10-3) x(4.5x10-2) x = 4.1x10-7M

The stepwise exchange of NH3 for H2O in M(H2O)42+
Figure 19.14 M(H2O)42+ The stepwise exchange of NH3 for H2O in M(H2O)42+ NH3 M(H2O)3(NH3)2+ 3NH3 M(NH3)42+ Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Sample Problem 19.11 Calculating the Effect of Complex-Ion Formation on Solubility PROBLEM: In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na2S2O3), through formation of the complex ion Ag(S2O3)23-. Calculate the solubility of AgBr in (a) H2O; (b) 1.0M hypo. Kf of Ag(S2O3)23- is 4.7x1013 and Ksp AgBr is 5.0x10-13. PLAN: Write equations for the reactions involved. Use Ksp to find S, the molar solubility. Consider the shifts in equilibria upon the addition of the complexing agent. SOLUTION: AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.0x10-13 (a) S = [AgBr]dissolved = [Ag+] = [Br-] Ksp = S2 = 5.0x10-13 ; S = 7.1x10-7M (b) AgBr(s) Ag+(aq) + Br-(aq) Ag+(aq) + 2S2O32-(aq) Ag(S2O3)23-(aq) AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)23-(aq)

Sample Problem 19.11 Calculating the Effect of Complex-Ion Formation on Solubility continued [Br-][Ag(S2O3]23- [AgBr][S2O32-]2 Koverall = Ksp x Kf = = (5.0x10-13)(4.7x1013) = 24 AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)23-(aq) Concentration(M) Initial - 1.0 Change - -2S +S +S Equilibrium - 1.0-2S S S Koverall = S2 (1.0-2S)2 = 24 S 1.0-2S = (24)1/2 S = [Ag(S2O3)23-] = 0.45M

The amphoteric behavior of aluminum hydroxide
Figure 19.15 The amphoteric behavior of aluminum hydroxide

Sample Problem 19.12 Separating Ions by Selective Precipitation PROBLEM: A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20. PLAN: Both precipitates are of the same ion ratio, 1:2, so we can compare their Ksp values to determine which has the greater solubility. It is obvious that Cu(OH)2 will precipitate first so we calculate the [OH-] needed for a saturated solution of Mg(OH)2. This should ensure that we do not precipitate Mg(OH)2. Then we can check how much Cu2+ remains in solution. SOLUTION: Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10 Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = 2.2x10-20 [OH-] needed for a saturated Mg(OH)2 solution = = 5.6x10-5M

Sample Problem 19.12 Separating Ions by Selective Precipitation continued Use the Ksp for Cu(OH)2 to find the amount of Cu remaining. [Cu2+] = Ksp/[OH-]2 = 2.2x10-20/(5.6x10-5)2 = 7.0x10-12M Since the solution was 0.10M CuCl2, virtually none of the Cu2+ remains in solution.

The general procedure for separating ions in qualitative analysis
Figure 19.16 The general procedure for separating ions in qualitative analysis Add precipitating ion Add precipitating ion Centrifuge Centrifuge Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

A qualitative analysis scheme for separating cations into five ion groups
Figure 19.17 Acidify to pH 0.5; add H2S Centrifuge Add NH3/NH4+ buffer(pH 8) Centrifuge Add (NH4)2HPO4 Centrifuge Add 6M HCl Centrifuge Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Step 5 Dissolve in HCl and add KSCN
Figure 19.18 A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+ Step 1 Add NH3(aq) Centrifuge Centrifuge Step 2 Add HCl Step 3 Add NaOH Centrifuge Step 4 Add HCl, Na2HPO4 Step 5 Dissolve in HCl and add KSCN Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

A view inside Carlsbad Caverns, New Mexico
Figure B19.1 A view inside Carlsbad Caverns, New Mexico

Formation of acidic precipitation
Figure B19.2 Formation of acidic precipitation

Keys to the Study of Chemistry

Similar presentations

Presentation on theme: "Keys to the Study of Chemistry"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Keys to the Study of Chemistry

Similar presentations

Presentation on theme: "Keys to the Study of Chemistry"— Presentation transcript:

Similar presentations

About project

Feedback