chem final

  1. CHAPTER 19
    Acid-base buffer
    • Solution that lessens the impact on pH of the addition of acid or base. Add a small amount of H3O+ or OH- to an UNBUFFERED solution, and the pH changes by several units.
    • Same addition of strong acid or base to a BUFFERED solution causes only a minor change in pH.
  2. What are the components of a buffer?
    • A conjugate acid-base pair.
    • Weak acid and conjugate base or weak base and conjugate acid
  3. What is the common ion effect?
    • This is how buffers work.
    • When you dissolve acetic acid in water, the acid dissociates slightly:
    • CH3COOH(aq) + H2O(l) <--> H3O1(aq) + CH3COO2(aq)
    • Adding the soluble salt sodium acetate solution instead of water, CH3COO-, will shift the equilibrium position to the left, thus, [H3O+] decreases, which lowers the extent of acid dissociation.
    • The same thing happens when we add acetic acid to a sodium acetate solution.
  4. Why is the acetate ion called the common ion?
    Because it is "common" to both the acetic acid and sodium acetate solutions
  5. When does the common ion effect occur?
    The common ion effect occurs when a given ion is added to an equilibrium mixture that already contains that ion, and the position of the equilibrium shifts AWAY from forming it.
  6. Why does a buffer work?
    because large amounts of the acidic (HA) and basic (A-) components consume small amounts of added OH- or H3O+, respectively
  7. If the ratio [HA]/[A-] goes up (more acid and less conjugate base), then [H3O+]goes up.
    If the ratio [HA]/[A-] goes down (less acid and more conjugate base), then [H3O+]goes down.
  8. Henderson Hasselbach equation
    • pH = pKa + log([base]/[acid])
    • pKa = -log(Ka)
    • Ka = antilog(-pKa)
    • Ka = [H3O+][A-]/[HA]
    • Ka = Kw/Kb
    • Kb = 10-pKb
  9. Buffer Capacity
    • 1 of the 2 key aspects of a buffer
    • measure of the "Strength" of the buffer, its ability to maintain the pH following addition of strong acid or base.
    • depends ultimately on component concentrations, both absolute and relative concentrations
    • absolute concentration: the more concentrated the buffer components, the greater the capacity.
    • relative concentration: the closer the component concentrations are to each other, the greater the capacity
  10. Buffer Range
    • pH range over which the buffer is effective and is also related to the relative buffer-component concentrations.
    • the further the concentration ratio is from 1, the less effective the buffer (the lower the buffer capacity)
    • buffers have a usable range within +/-1 pH unit of the pKa of the acid component
  11. Preparing a buffer
    • 1. Choose the conjugate acid/base pair
    • 2. Calculate the ratio of buffer component concentrations
    • 3. Determine the buffer concentration
    • 4. Mix the solution and correct the pH
  12. Acid-base titration curves
    plot of pH vs. volume of titrant added.
  13. acid-base indicator
    weak organic acid whose color differs from the color of its conjugate base
  14. Equivalence point
    • number of moles of OH- added equals number of moles of H3O+ originally present.
    • solution consists of anion(-) of strong acid and cation(+) of strong base
  15. End point
    occurs when indicator, which is added before the titration, changes color
  16. Ion-Product expression (Qsp) and Solubility-Product constant (Ksp)
    • Equilibrium exists between solid solute and aqueous ions
    • Ksp depends ONLY on temperature
    • Ksp value indicates how far the dissolution proceeds at equilibrum (saturation)
    • Qsp = [Mn+]p[Xz-]q = Ksp
  17. Special case of metal sulfides (S2-)
    • sulfide ion is so basic that it reacts completely with water to form the hydrogen sulfide ion (HS-) and (OH-)
    • (make sure u do the extra equations when ur solving Ksp for a compound with sulfide!!!)
  18. Determining Ksp from solubility:
    • 1. convert solubility to molar solubility (ex. 3.2x10^-4 g of Ag3PO4 is soluble in 50 mL solution)
    • 2. convert the g to mL by multiplying by 1000 mL, then multiple 1 mol/molar mass of the compound
    • 3. determine the correctly balanced equation, then determine Ksp
    • 4. Fill in molarity values and solve for Ksp. (ex. Ag3PO4 Ksp is [Ag+1]3[PO4-3] so you do 3 times the molarity for Ag, because there's 3 moles of Ag ions in 1 mol of Ag3PO4)
  19. Determining solubility from Ksp:
    • 1. set up ICE table, where "S" is the unknown molar solubilty of the product ions. DO NOT put anything under the reactant for any I, C, or E.
    • 2. make sure it's a balanced equation and determine the Ksp equation and make sure it's the same in the table. (ex. 3Ca+2 would be "3S") and I, as in initial on the table, is always 0.
    • 3. solve for Ksp equation by inputting the E, as in equilibrium, constants for it. (ex. 3Ca+2 and 2PO4-3 from Ca3(PO4)2 would be Ksp = [Ca+2]3[PO4-3]2
    • 4. filling in the E values in the Ksp equation would then be Ksp = (3s)3(2s)2
    • 5. Then solve the equation so 108s5 = 1.2x10-29 (which is the given Ksp value)
  20. Using Ksp values to compare solubilites
    • as long as we compare compounds with the same total number of ions in their formulas, Ksp values indicate relative solubility.
    • the higher the Ksp, the greater the solubility.
  21. Effect of common ion on solubility
    adding a common ion decreases the solubility of a slightly soluble ionic compound
  22. Effect of pH on solubility
    if a slightly soluble ionic compound contains the anion of a weak acid, addition of H3O+ (from a strong acid) increases its solubility
  23. Predicting formation of a precipitate: Qsp vs Ksp
    • If Qsp = Ksp, the solution is saturated and no change will occur.
    • If Qsp > Ksp, a precipitate will form until the remaining solution is saturated.
    • If Qsp < Ksp, no precipitate will form because the solution is unsaturated
  24. selective precipitation
    a solution of one precipitating ion is added to a solution of two ionic compounds until the Qsp of the MORE soluble compound is almost equal to its Ksp.
  25. complex ion
    • consists of a central metal ion covalently bonded to 2 or more anions or molecules, called ligands.
    • hydroxide, chloride, and cyanide ions are examples of ionic ligands.
    • water, carbon monoxide, and ammonia are examples of molecular ligands.
    • EX) Cr(NH3)6+3 is a complex ion because the central Cr+3 is surrounded by six NH3 ligands.
    • WHEN A SALT DISSOLVES IN WATER, A COMPLEX ION FORMS with water as ligands around the metal ion.
  26. simple ion
    like Na+ or CH3COO-, consists of one or a few bonded atoms, with an excess or deficit of electrons
  27. Formation constant (new equilibrium constant): Kf
    • The sum of the equations gives the overall equation, so the product of the individual formation constants gives the overall formation constant
    • basically products/reactants in the same way the Ksp is formed. don't include water in the Kf equation.
  28. to find initial concentrations:
    Mf = MiVi/Vf
  29. Calculating concentration of a complex ion
  30. Ligand
    increases the solubility of a slightly soluble ionic compound if it forms a complex ion with the cation
  31. Calculating the effect of complex-ion formation on solubility
  32. Difference between Qsp and Ksp?
    • Qsp = ion product is a constant for an unsaturated or supersaturated solution NOT AT EQUILIBRIUM.
    • Ksp = solubility product constant for a saturated solution AT EQUILIBRIUM.
  33. Koverall
    • Koverall = KspKf
  34. CHAPTER 20
    Spontaneous change
    • a spontaneous change of system is one that occurs under specified conditions WITHOUT A CONTINUOUS input of energy from outside the system.
    • does NOT mean it is instantaneous. Just means that given enough time, the process will happen by itself.
    • ex) ripening, rusting, aging.
  35. Non-spontaneous change
    • only occurs if the surroundings CONTINUOUSLY SUPPLY the system with an input of energy
    • if a change is spontaneous in one direction, IT IS NON-SPONTANEOUS IN THE OTHER DIRECTION.
  36. ΔEsys = -ΔEsurr
    • (q+w)sys = -(q+w)surr
    • The change in energy, and therefore heat and/or work absorbed by the system is released by the surroundings.
  37. ΔEsys + ΔEsurr = -ΔEsurr + ΔEsurr = 0 = ΔEuniv
    • The total energy of the universe is CONSTANT. Change in energy of the universe is 0.
    • The first law by itself does not predict the DIRECTION of a spontaneous change.
  38. ΔH < 0
    Spontaneous and exothermic
  39. ΔH > 0
    • Non-spontaneous and endothermic
    • (even some endothermic reactions ARE SPONTANEOUS)
    • The sign of ΔH by itself does not predict the DIRECTION of a spontaneous change.
  40. So what predicts the direction of a spontaneous change then?
    A change in the freedom of motion of particles in a system, as in the dispersal of their energy of motion, is a key factor for predicting the direction of a spontaneous change.
  41. Why do freedom of motion and dispersal of energy relate to spontaneous change?
    • Quantization of energy: the complete quantum state of the molecule at any instant consists of its electronic states and these transitional, rotational, and vibrational states.
    • Number of microstates: each quantized state of the system is called a microstate, and at any instant, the total energy of the system is dispersed throughout one microstate.
    • Dispersal of energy: at a given set of conditions, each microstate has the same total energy as any other.
  42. S = klnW
    • W: number of microstates
    • S: entropy
    • k: Boltzmann constant; universal gas constant (R) divided by avogadro's number = 1.38x10-23 J/K
  43. A system with lower microstates (smaller W) has lower entropy (lower S)
    A system with higher microstates (larger W) has higher entropy (higher S)
  44. ΔSsys = Sfinal - Sinital
    ΔSsys depends only on the difference between its final and initial values
  45. Second Law of Thermodynamics
    • When we consider BOTH the system and its surroundings, we find that all real processes occur spontaneously in the direction that increases the entropy of the universe.
    • ΔSuniv = ΔSsys + ΔSsurr > 0
  46. Enthalpy
    There is NO ZERO POINT, so we can measure only changes.
  47. Entropy
    • There IS A ZERO POINT, and we can determine absolute values by applying the Third Law of Thermodynamics
    • increases when a solid or liquid solute dissolves in a solvent
  48. Third Law of Thermodynamics
    • A perfect crystal has zero entropy at absolute zero.
    • ΔSsys = 0 at 0o K
  49. ΔSorxn
    • Standard entropy of reaction
    • Entropy change that occurs when all reactants and products are in their standard states.
    • #moles of gas increases = ΔSorxn is POSITIVE.
    • #moles of gas decreases = ΔSorxn is NEGATIVE.
    • ΔSorxn = ΣmSoproducts - ΣnSoreactants
  50. Exothermic change
  51. Endothermic change
  52. ΔSsurr
    • directly related to an opposite change in the heat of the system (qsys) and inversely related to the temperature at which heat is transferred
    • ΔSsurr = -qsys/T
    • For a process at constant pressure: 
    • ΔSsurr = -ΔHsys/T
  53. ΔSuniv = ΔSorxn + ΔSsurr
    ΔSsys = -ΔSsurr
    • When ΔSuniv is positive, reaction is SPONTANEOUS
    • When ΔSuniv is negative, reaction is NON-SPONTANEOUS
    • ΔSuniv >0 for a spontaneous process
    • ΔSuniv < 0 for a nonspontaneous process
    • ΔSuniv = 0 for a process at equilibrium
  54. Gibbs free energy
    • Free energy (G) combines the system's enthalpy and entropy:
    • G = H - TS
    • Measure of the spontaneity of a process and of the useful energy available from it.
  55. Gibbs Equation
    • ΔGsys = ΔHsys - TΔSsys
    • ΔG < 0 for a spontaneous process
    • ΔG > 0 for a nonspontaneous process
    • ΔG = 0 for a process at equilibrium
  56. ΔGorxn units are k/J
    ΔSorxn units are J/K
  57. Standard free energy of formation
    • ΔGof
    • ΔGorxn = ΔHorxn - TΔSorxn
    • ΔGorxn = ΣmΔGofproducts - ΣnΔGofreactants
  58. ΔG = ΔH - TΔS
    • finding the temperature at which a reaction becomes spontaneous
    • ΔGorxn = 0 = ΔHorxn - TΔSorxn
    • ΔHorxn = TΔSorxn
    • T = ΔHo/ΔSo
  59. CHAPTER 21
    Half Reaction Method
    • Divides the overall redox reaction into oxidation and reduction half-reactions.
    • Step 1: Divide the skeleton reaction into 2 half reactions. Each half reaction contains the oxidized and reduced forms of one of the species. Ex) if the oxidized form of a species is on the left side, then the REDUCED form MUST be on the RIGHT.
    • Step 2: balance the atoms and charges in each half-reaction. First you do atoms OTHER THAN O and H. THEN you do O and H LAST. Charge is balanced by adding electrons to the left side in the REDUCTION HALF REACTION because the reactant gains them and to the RIGHT SIDE IN THE OXIDATION HALF REACTION because the reactant loses them.
    • Step 3: if needed, multiply one or both half reactions by an integer so that #e- gained in reduction = #e- lost in oxidation
    • Step 4: Add the balanced half reactions, and include states of matter.
    • Step 5: Check atoms and charges are balanced.
  60. Half Reaction Method
    • 1. Divide into half-reactions. Group the reactants and products with similar atoms.
    • 2. For each half-reaction balance
    •     a) Atoms other than O and H
    •     b) O atoms with H2O
    •     c) H atoms with H+
    •     d) Charge with e-, get total charges on reactant and product side to be the same.
    • 3. Multiply each half-reaction by an integer that will make the number of electrons lost equal to the number of electrons gained on the reactant side and product side of the 2 reactions. 
    • 4. Add half-reactions and cancel substances appearing as both reactants and products from the reactant and product side of the 2 reactions. Get overall reaction.
    • 5. Add hydroxide ions to neutralize H+. The neutralization reaction produces water on the product side.
    • 6. Balance the reaction. Final reaction add in the spectator ion from the question.
  61. Voltaic Cell
    uses a spontaneous reaction (ΔG < 0) to generate electrical energy
  62. Electrolytic Cell
    uses electrical energy to drive a non-spontaneous reaction (ΔG > 0)
  63. Similarities between voltaic and electrolytic cells
    • Two electrodes which conduct the electricity between cell and surroundings
    • They are dipped into an electrolyte, a mixture of ions (usually aqueous solution) that are involved in the reaction or that carry the charge.
    • An electrode is identified either as anode or cathode depending on the half-reaction that takes place there.
    • Oxidation half reaction occurs at the ANODE. Electrons lost by the substance leave the oxidation half cell at the anode.
    • Reduction half reaction occurs at the CATHODE. Electrons gained by the substance being reduced enter the reduction half cell at the cathode.
  64. Differences between voltaic and electrolytic cells
    • Voltaic: energy is released from a spontaneous redox reaction.
    • Electrolytic: energy is absorbed to drive a non-spontaneous redox reaction.
    • Voltaic: System does work on surroundings (load)
    • Electrolytic: surroundings (power supply) does work on system (cell)
  65. Voltaic cell
    • Wire and salt bridge complete the circuit.
    • Electrons move left to right through the wire
    • Anions move right to left through the salt bridge
    • Cations move left to right through the salt bridge.
  66. Cell potential (Ecell)
    • difference in electrical potential between the two electrodes.
    • Also called VOLTAGE and EMF (electromotive force)
  67. Electrons flow spontaneously from negative to positive electrode.
    • Ecell > 0 for spontaneous rxn. More positive Ecell is, the more work it can do.
    • Ecell < 0 for non-spontaneous
    • Ecell = 0 for equilibrium
  68. Standard Electrode Potential (Eohalf-cell)
Card Set
chem final
final ch 19,20,21