Chem 1332 Exam 3 Review

University of Houston Fundamentals of Chemistry

Get Started. It's Free
or sign up with your email address
Rocket clouds
Chem 1332 Exam 3 Review by Mind Map: Chem 1332 Exam 3 Review

1. Chapter 18: Acids and Bases

1.1. All pH calculations

1.1.1. 1. All correct starting concentration M = n/L M1V1 = M2V2

1.1.2. 2. Remove spectator ions Cations: Li+, Na+ K+, Rb+, Cs+, Ba2+, Sr2+ Anions: Cl-, Br-, I-, No3-, ClO4-

1.1.3. 3. Neutralization H+ + OH- --> H2O(l) SA + SB w.a. + OH- --> c.b. w.b. + H+ --> c.a. SCAN chart: Start Change (limiting reactant) After Neutralization

1.1.4. 4. Do the calculation a) H+ --> pH (SA) b) OH- --> pOH --> pH (SB) c) HA + H2O <=> H3O+ + A- Eqm calc. using Ka(HA) --> pH (WA) d) B + H2O <=> OH- + BH+ Eqm. calc. using Kb(B) --> pOH --> pH (WB) e) A- + H2O <=> OH- + HA Eqm. calc. w/ Kb(A-) = Kw/Ka(HA) --> pOH --> pH f) BH+ + H2O <=> H3O+ + B Eqm. calc w/ Ka(BH+) = Kw/Kb(B) --> pH

2. Chapter 19: Ionic Equilibria in Aqueous Solutions

2.1. Acid-Base Buffers

2.1.1. Buffer Solution: a conjugate acid-base pair within a 10:1 concentration ratio of each other (WA w/ conj. base or WB w/ conj. acid

2.1.2. Added H3O+ or OH- causes only a small change in the buffer-component concentration ratio & a small change in pH

2.1.3. Ideal Buffer: 1:1 concentration ratio

2.1.4. Common-ion effect: an ion is added to a solution that already contains it, and the equilibrium position shifts away from forming it (Le Chatelier's principle)

2.1.5. Ka = [H3O+] [A-] / [HA]

2.1.6. Henderson Hasselbalch Equation: solve directly for pH pH = pKa + log([A-]/[HA]) Ideal buffer: pH = pKa because [base] = [acid]

2.2. Titrations

2.2.1. Monitoring pH w/ Acid-Base Indicators An indicator changes color over a range of ~2 pH units

2.2.2. SA-SB Titration Curve pH = 7 at equivalence (end) point

2.2.3. WA-SB Titration Curve Buffer region Half-way equivalence point: at midpoint of the buffer region; half of the initial WA has reacted, so [HA] = [A-] and pH = pKa Equivalence point: pH > 7.00 The sol'n contains A-, which acts as a WB

2.2.4. WB-SA Titration Curve Halfway equivalence point: pOH = pKb Equivalence point: pH < 7.00

2.2.5. Polyprotic Acid Titration Curve 2 equiv. pts. & 2 buffer regions The pH at the midpoint of each buffer region is equal to the pKa of that acidic species

2.3. Solubility Equilibria

2.3.1. "Slightly soluble" ionic compounds

2.3.2. Qsp: ion-product expression For PbF2(s) <=> Pb(2+)(aq) + 2F-(aq): Qsp = Qc[PbF2(s)] = [Pb2+(aq)][F-(aq)]^2

2.3.3. Ksp: solubility-product constant Saturated sol'n: Qsp = Ksp (at equilibrium) Ksp depends on temperature only

2.3.4. Special case: sulfide ion, S(2-) reacts completely w/ water : S(2-)(aq) + H2O(l) --> HS-(aq) + OH-(aq)

2.3.5. Determining Solubility (S) from Ksp: Ksp = [C]^p * [A]^q Set up SCE chart: Use "S" instead of "x" Pay attention to stoich. ratios... # moles S = molar solubility Higher Ksp --> greater solubility

2.3.6. Adding a common ion decr. the solubility of a slightly soluble ionic compound

2.3.7. Effect of pH on Solubility In acidic sol'n, most ionic compounds are more soluble NO EFFECT FOR: Cl-, Br-, I-, ClO2-, NO3-

2.3.8. Prediction the formation of a precipitate If Qsp > Ksp, sol'n is saturated; no change If Qsp > Ksp, a precipitate will form until saturation If Qsp < Ksp, no precipitate will form (sol'n is unsaturated)

2.4. Complex Ion Equilibria

2.4.1. Complex ion: central metal ion covalently bonded to two or more anions/molecules (ligands) Hydrated metal ions: salt dissolves in water --> H2O ligands around metal ion Cr(NH3)6 (3+)

2.4.2. Formation constant, Kf = Kc / [H2O]^n (The concentration of water is constant in aqueous reactions)

2.4.3. A ligand incr. the solubility of a slightly soluble ionic compound if it forms a complex ion w/ the cation

3. Chapter 20: Thermodynamics

3.1. Spontaneity

3.1.1. Spontaneous change: occurs under specified conditions w/out a continuous input of energy from the surroundings

3.1.2. "Spontaneous" does NOT mean "instantaneous"

3.1.3. A chemical rxn proceeding toward equilibrium is spontaneous

3.1.4. First Law of Thermodynamics does NOT predict spontaneous change Law of conservation of energy ΔEsys + ΔE surr = 0 = ΔEuniv Change in internal energy: ΔE= q + w

3.1.5. The sign of ΔH does NOT predict spontaneous change Exothermic (ΔH < 0) and endothermic (ΔH > 0) processes can be spontaneous

3.2. Entropy

3.2.1. Definition Freedom of particle motion & dispersal of particle energy Entropy Incr. as: - solid --> liq. --> gas - A salt dissolves A change in the freedom of motion of particles in a system affects the direction of a spontaneous process Incr. # of microstates --> incr. entropy Entropy (S) is a state function ΔS(sys) = S(final) - S(initial) ΔS(sys) = q(rev) / T q = heat absorbed Extensive property: depends on amount Standard molar entropy (S°) (units of J/mol*K)

3.2.2. Second Law of Thermodynamics All processes occur spontaneously in the direction that incr. the entropy of the universe ΔS(univ) = ΔS(sys) + ΔS(surr) > 0 for any real spontaneous process

3.2.3. Third Law of Thermodynamics A perfect crystal has zero entropy at absolute zero (0 K) At equilibrium, ΔS(univ) = 0 (no net change)

3.2.4. Predicting Relative S° S° incr. as T incr. S° incr. as state changes from solid to liquid to gas ΔS°vap >> ΔS°fus S(sol'n) > S(solute) + S(solvent) Dissolving a gas in a liquid: entropy decr. Dissolve gas in another gas: entropy incr. Entropy incr. w/ size (down a group on the periodic table) & chemical complexity (for compounds)

3.2.5. Calculating Entropy Change of a Reaction Standard entropy of reaction, ΔS°rxn: entropy change that occurs when all reactants & products are in their standard states Predicting the sign of ΔS°rxn # mol gas incr. --> ΔS°rxn is positive # mol gas decr. --> ΔS°rxn is negative Hess's Law: ΔS°rxn = ΣmS°(products) - ΣnS°(reactants) The only way for an endothermic reaction (ΔHsys > 0) to occur spontaneously is if ΔSsys is positive & large enough to outweigh the negative ΔSsurr Solid ionic compound dissolved: ΔSsys >> 0 --> ΔSuniv > 0

3.3. Free Energy (G)

3.3.1. G = H - TS

3.3.2. Free energy change (ΔG) Measure of the spontaneity of a process & the useful energy available from it The sign of ΔG ΔSuniv > 0: spontaneous: ΔG < 0 ΔSuniv = 0: equilibrium: ΔG = 0 ΔSuniv < 0: nonspontaneous: ΔG > 0

3.3.3. Gibbs Equation ΔGsys = ΔHsys - TΔSsys Standard free energy change ΔG°sys = ΔH°sys - TΔS°sys

3.3.4. Standard free energy of formation, ΔG°f When 1 mol of a compound is made from its elements w/ all components in standard state ΔG°rxn = ΣmΔG°f(prod.) - ΣnΔG°f(reacts) ΔG°f of an element in its standard state is zero An eq. coefficient (m,n) multiplies ΔG°f by that number Reversing a reactin changes the sign of ΔG°f

3.3.5. Work ΔG = W(max) ΔG is the max. useful work done by a system during a spontaneous process Spont. rxn (ΔGsys is negative) does work on surroundings (-W) ΔG is the minimum work done to a system to make a nonspont. process occur Nonspont. rxns occur only if the surroundings do work on the system (+W)

3.3.6. Reaction Spontaneity Most exothermic rxns are spont. because the large negative ΔH makes ΔG negative ΔH ΔS -TΔS ΔG Rxn spontaneity - + - - Spont. at all T + - + + Nonspont. at all T + + - - Spont. at high T or + Nonspont. at low T - - + - Spont. at low T or + Nonsp. at high T The temperature at which a rxn becomes spont. Set ΔG = 0 and solve for T ΔH = TΔS T = ΔH/ΔS

3.3.7. Equilibrium & Rxn Direction ΔG = RT ln(Q/K) = RT lnQ - RT lnK ΔG° = -RT lnK ΔG = ΔG° + RT lnQ

4. Chapter 21: Electrochemistry

4.1. Redox Reactions

4.1.1. Oxidation: loss of e- Reducing agent is oxidized (gives e-) Oxidized substance ends up w/ a higher O.N.

4.1.2. Reduction: gain of e- Oxidizing agent is reduced (takes e-) Reduced substance ends up w/ a lower O.N.

4.1.3. Half-reaction method 1) Divide the skeleton ionic rxn into two half-rxns, each containing the oxidized & reduced forms of one of the species 2) Find the oxidation # of each atom 3) Balance the atoms other than O & H 4) Account for the change in O.N. by adding electrons on one side of each equation 5) Find the LCM of the electrons' coefficients, multiply each equation to match that # of e- 6) Sum the equations and cancel out the electrons 7) Balance the charges by adding some H+ ions to one side 8) Balance the # of O atoms by adding some H2O to one side 9) Check the # of H atoms on each side 10) In basic sol'n, add an OH- ion on both sides for each H+ ion present. The OH- on the same side as the H+ combine to form H2O. Excess H2O is canceled.

4.2. Electrochemical Cells

4.2.1. Two electrodes are dipped into an electrolyte Anode: Oxidation Cathode: Reduction

4.2.2. Voltaic (galvanic cell) Uses a spontaneous rxn (ΔG < 0) to generate electrical energy. The system does work. Electrons flow through the wire from anode to cathode Anode: negative charge Oxidation: X --> X+ + e- Cathode: positive charge Reduction: e- + Y+ --> Y Salt bridge: joins the half-cells; "liquid wire" Output of a Voltaic Cell Cell potential, Ecell: the voltage of the cell (electromotive force, emf); difference in electric potential between the two electrodes Ecell > for a spontaneous process If Ecell = 0, the rxn has reached equilibrium & the cell will do no more work Standard cell potential (E°cell)

4.2.3. Electrolytic cell Uses electrical energy to drive a nonspontaneous rxn (ΔG > 0). The surroundings do work on the system. Electrons flow from Cathode to Anode ΔG is (+) and Ecell is (-) Anode: positive charge Oxidation: A- --> A + e- Cathode: negative charge Reduction: e- + B+ --> B Electrolysis: splitting a substance w/ the input of electrical energy; decompose a compound into its elements Recall diatomic elements: H2, N2, O2, F2, Cl2, I2, Br2 Electrolysis of Pure Molten Salts Electrolysis of Mixed Molten Salts Electrolysis of Water Electrolysis of Aqueous Salt Sol'ns Stoichiometry of Electrolysis

4.3. Free Energy & Electrical Work

4.3.1. The signs of ΔG and Ecell are opposite for a spont. rxn.

4.3.2. Electrical work: Wmax = -Ecell x charge (in Joules)

4.3.3. The charge of 1 mol of electrons (Faraday constant): F = 96500 J/V*mol e-

4.3.4. ΔG = -nFEcell (n moles e-) ΔG° = -nFE°cell when all components are in standard state E°cell = (RT/nF) lnK (R = 8.314 J/mol K) Positive E°cell: rxn proceeds spont. to the right at standard state Reaction Parameters at Standard State ΔG° K E°cell Rxn at standard state < 0 >1 > 0 Spontaneous 0 1 0 At equilibrium >0 <1 < 0 Nonspontaneous Effect of Concentration on Cell Potential (under nonstandard conditions): ΔG = ΔG° RT lnQ Nernst Equation: Ecell = E°cell - (RT/nF) lnQ Q contains only species w/ concentrations/pressures that can vary. Solids do not appear Converting pressure to molarity: n/V = P/RT, R = 0.0821 atm L/mol K