Chem 113 Exam 3 – Flashcards

yields a WEAK conjugate base ionization equilibrium lies far to the RIGHT. HCl HBr HI HClO₄ HNO₃ H₂SO₄

large K_a value.

smaller pK_a value

will have H⁺ bound to F, N, or O atoms that can be donated

LiOH NaOH KOH RbOH CsOH Ca(OH)₂ Sr(OH)₂

Ba(OH)₂

larger K_b

smaller pK_b

will have N or O atoms that have lone pair electrons that can attract H⁺

if [H+] > [OH-] = solution is acidic if [H+] < [OH-] = solution is basic if [H+] = [OH-] = solution is neutral

K_w=[OH^-][H⁺]

K_w=1.0x10^-14

pH=-log[H⁺] or -log[H₃O⁺]

pH decreases as [H⁺] increases 

pH<7 = acidic solution

pH>7 = basic solution

pH=7 = neutral solution 

pH=-log[H⁺]=4.5

log[H⁺]=-4.5

10^log[H+]=10^-4.5

[H⁺]=3.16x10^-5 M

the solution is acidic 

[H₃O⁺]_{from HA}

             ______________ x 100 %

[HA]_initial

K_a x K_b = K_w

pKa + pKb = pKw

• cations may act as acids in water except:

Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺, Ca²⁺, Sr²⁺, Ba⁺²

• these are pH neutral
• these are the conjugate acids of strong bases and are such super weak acids that they will not act as an acid in water 
• (strong bases without OH^-)

• anions may act as base in water except:

Cl^-, Br^-, I^-, NO₃^-, HSO₄^-, ClO₄^-, BrO₄^-, IO₄^-

• these are pH neutral
• these are the conjugate bases of strong acids and are such super weak bases that they will not act as a base in water
• (strong acids without H⁺)

• contains the group H-O-X
• for a given series the acid strength increases with an increase in the number of oxygen atoms attached to the central atom
• the greater the ability of X to draw electrons toward itself, the greater the acidity of the molecule

the suppression of the ionization of the ionization of a weak electrolyte caused by the addition of an ion that is also a product of the ionization equilibrium of the weak electrolyte

• shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction.

the amount of acid or base that a buffer can neutralize before its pH changes is appreciable

• when the ratio n_base/n_acid is close to 1, the buffer has its maximum buffer capacity 

the equilibrium constant expression for a salt dissolving in water

CaF_2(s) - Ca²⁺_(aq) +2F^-_(aq)

Ksp= [Ca²⁺][F^-]²

the larger the Ksp, the more solid that will dissolve

• Q > K_sp ; precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy K_sp. 
• Q < K_sp ; no precipitation occurs

thermodynamic property related to the degree of disorder in a system

tends to increase if:

• liquids are formed from solids 
• gases are formed from either solids or liquids
• the number of molecules of gas increases as a result of a chemical reaction
• the temperature of a substance increases 
• the more complex the molecule

the difference in entropy between two states

• nature tends toward disorder. ie: a driving force for a spontaneous process is an increase in the entropy of the universe

to determine the sign of ΔS°, look at the coefficients of the gases

a gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system

therefore;

S_solid < S_liquid << S_gas

greater volume, the greater the entropy 

in any spontaneous process there is always an increase in the entropy of the universe

ΔS_univ>0

where ΔS_univ = ΔS_sys + ΔS_surr

• the sign of ΔS_surr depends on the direction of the heat flow
• the magnitude of ΔS_surr depends on the temperature
• ΔS_surr =q_rev/T = ΔH/T

q_rev: heat gained in a reversible process (joules)

T: temperature (Kelvin)

• since the change to the system is the reverse of the change to the surroundings 

• ΔG = ΔH - TΔS (from the standpoint of the system)
• a process (at constant T, P) is spontaneous in the direction in which free energy decreases 
• a "-" ΔG means "+" ΔS_univ
• test for spontaneity
  • ΔG < 0 (negative) = spontaneous
  • ΔG >0 (positive) = nonspontaneous
  • ΔG = 0 = equilibrium
• chemical equilibrium occurs at the lowest value of energy available to the reaction system
• ΔG° = RTln(K) = ΔH° - TΔS°

ΔG = ΔG° + RT ln(Q)

•  
  • R: gas law constant; 8.3145 J/k*mol
  • T: temperature (kelvin)
  • Q: reaction quotient (in partial pressures)
  • ΔG°: the free energy change at the standard state

• maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy 
  • W_max = ΔG
• first law: you can't win, you can only break even.
• second law: you can't break even. 

• write separate reduction, oxidation reactions.
• for each half reaction:
  • balance elements (except H,O).
  • balance O using H₂O.
  • balance H using H⁺.
  • balance charge using electrons.
• If necessary, multiply by integer to equalize electron count.
• Add half reactions.
• check that elements and charges are balanced.

• balance as in acid
• add OH^- that equals H⁺ions (both sides!)
• form water by combining H⁺, OH^-
• check elements and charges for balance

a device in which chemical energy is changed into electrical energy.

this is done with a oxidation-reduction (redox) reaction.

when a half reaction is reversed, the sign of E° is reversed.

when a half-reaction is multiplied by an integer,E° remains the same. 

a galvanic cell runs spontaneously in the direction that gives a positive value for E°_cell

• anode components are listed to the LEFT
• cathode components are listed to the RIGHT
• anode and cathode are separated by double vertical lines.
• a phase difference is indicated by a single vertical line.

directly related to the free energy difference between the reactants and the products in the cell

ΔG°= -nFE°

n=numbers of moles of electrons

F=Faraday = 96,485 coulombs per mole of electrons 

used to calculate the potential of a cell in which some or all of the components are not in their standard states.

E = E° - (RT/nF)*(ln(Q))

at 25° C, the nernst equation becomes.....

E= E° - (0.0591/n)*(log(Q))