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77 Cards in this Set
- Front
- Back
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equilibrium
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[reactants] and [products] doesn't change; rate of forward reaction is equal to the rate of the reverse reaction
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Keq
given aA +bB -><- cC + dD |
Keq = [C]^c [D]^d/[A]^a [B}^b
brackets = molar concentraion at equilibrium solids and liquids are not included (because concentraion doesnt change) If gaseous, then can use partial pressure mass action ration |
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K eq is constant
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at given temp regardless of concentrations of product or reactant. Reaction will go in direction to equilibrium
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If Keq < 1
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reaction favors the reactant
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If Keq = 1
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reaction ba;ances the reacant and product
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Keq > 1
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products favored
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Equilirium reached if
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ration of products and reactants remains constant & soulution is saturated with solute (system no longer changes with time)
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Q = reaction quotient
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aA =bB -><- cC + dD
where Q = [C]^c [D]^d/[A]^a [B}^b |
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Q = Keq
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no change
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Q < Keq
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Reaction proceed forward to produce more product
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Q > Keq
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Reaction proceeds in the reverse direction to attain more reactant
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Le Chatliers
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a system at equilibrium will try to neutralize any imposed change or stress (e.g., adding reactant) to reestablish equilibrium
Eg., add more reactant, system will react by favoring forward reaction to consume reactant |
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Changing volume of reaction container which is at equilibrium
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If gas, and decrease volume (increase pressure), then reaction favors the soulte with least number of moles and vice cersa
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+ change H
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endothermic
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-chnage H
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exothermic
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changing heat in a system at equilibrium
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only effects endo thermic/ exothermic reaction
If add heat, and the process is exothermic, reaction proceeds left If cool it down, reaction proceeds to right |
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Keq and changes in temp
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changing temp will effect Keq; increases with endothermic, decreases with exothermic
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addition of inert or unreactive gas to system in equilibrium
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has no effect
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adding a catalyst
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increases reaction rate but does not affect the quilibrium; adding catalyst to a system that is already in quilibrium will have n oeffect
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change in H is negative
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exothermic and heat is on the product side
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Change in H is positive
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endothermic
heat on reactant side |
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solubility product constant
Ksp |
the extent to which a salt will disolve in water can be determined from its solunility product constant Ksp
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molar solubility
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number of moles of a salt that will saturate a liter of water (asked to find out just how much of a solid can disosle in H20
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given Ksp = 2.7 x 10 ^-9
for Li3PO4, how many moles of this salt would be required to form a saturated 1 liter aq solution |
equil = li3PO4 -><- 3Li + PO4
let x = [PO4] and [Li] = 3x Ksp = (3x)^2x = 27x^4= 2.7 x 10^-9 -> x = 3.2^-3 ol needed |
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Qsp < Ksp
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more salt can be dissolved
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Qsp = Ksp
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solution is saturated
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Qsp> Ksp
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excess salt will precipitate
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Common ion effect
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reduce the solubility of ion is solution already has common ion; adding more ions that are in product will drive reaction to the left
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Arrenhius acid
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acids ionize in water to produce hydrogen H+ ions
HCL -> H+ + Cl- |
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Arrenhius base
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ionize in H20 to produce OH- ions
NaOH -> Na+ +OH- |
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Bronset Lowry Acid
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acids are proton donor
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Bronsted Lowry Base
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Bases are proton acceptors
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Lewis Acid
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electron pair acceptor
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Lewis Base
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electron pair donor
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strong acid
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on e that dissociates completly ( or nearly so ) in H20
E.g., HCl (aq) + H2) (l) -> H30 (aq) + Cl- (aq) (HF is a example of weak acid) |
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Ka (equilirbium expression for acid dissocation)
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"acid ionization or dissocation constant"
strength of acid is directly related to how much the products are favored over the reactants Ka = [H30] [A-]/[HA] |
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Ka > 1
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products favored
acid is strong |
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Ka < 1
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reactants favored and acis is weak
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Comparing the Ka strengths
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The larger Ka, the stronger the acid
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Common strong acids
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HI, HBr, HCl, HClO3, HClO4, H2SO4, HNO3
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Weak acid
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any acid that is not a strong acid from the list
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Kb
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Base ionization constant
Kb = [HB][OH]/[B] Larger Kb, stronger base |
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Strong base
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Group 1 hydroxides (LiOH, NaOH, KOH, RbOH, CsOH)
CaOH2 CH3NH2 Sr(OH)2 Ba(OH)2 Metal amids (NaNH2 |
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Weak base
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NH3 (ammonia) and amines
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Conjugate base of a strong acid
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has no basic properties (Cl- will not tend to accept proton, and doesnt act like a base
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The conjugate base of a weak acis
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weak base (F- has some tendency to accept proton)
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polyprotic
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has more than 1 proton to donatel; generally, everytime a polyprotic acid donates a proton, the resulting species will be a weaker acid than its predecessor
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amphoteric
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can act like a base or acid
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autoionization (self-ionization)
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of water, reacts with self in a Bronster Lowry acid-base reaction, 1 molecules acting as the acid and the other as athe base
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Kw
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=1.0*10^-14 = [H3O][OH]
varies with temp (like other equillibrium constants), this is at 25 degrees C Kw is constatnt at a given temp regardless of the H3O concentration |
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If the introduction of an acid increases the concentration of H3O ions, then equilibrium is disturbed
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the reverse reaction is favored, decreasing the concentration of OH- ions
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If introduce base to increase its concentration, equilibium is disturbed
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reverse reaction is favored, decreasing the concentration of H3O ions, however the produce to [H3O] and [OH ]will remain equal to Kw
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Kw = [OH][H3O]
Increase one concentration |
decreases the other
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pH = -log[H+]
means that [H+] = |
10^-pH
E.g., pH = 7 , [H] = 10^-7 |
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trend for acidity
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down and to the left
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pOH = -log[OH]
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[OH] = 10^-pOH
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pH + pOH
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=14
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[H] = 6.2x10^-5
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pH is between 4 and 5
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lowest concentration of H3O
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has highes pH
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solution has pH of 3.5, what is [H]
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[H] = 10^-3.5 = 10^.5 x 10^-4 = sqrt(10) x 10^-4 = 3.2 *10^-4
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KaxKb=
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Kw
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pH calculation for strong acid
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pH comes directly from molarity of solution; so .01 M solution of HCl will have [H} = .01 = pH 2
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weak acid
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concentration of a H ions will be much less than the undissociated ions
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when K < 10 ^-4
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x added to or subtracted from a number is neglible
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with weak acids , use ICE
HCN -><- H+ + CN- Initial: .2 0 0 at equil: (.2-x)M x x |
know Ka = 4.9 x 10^-10
x^2/(.2-x) equivl to: x^2/.2 = 4.9 x 10^-10 |
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"p"
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take negative log
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Ka1 < Ka2
So pKa1 > pKa2 |
Ka2 is more acidic
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pKa + pKb
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14
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weak acid
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may have a weak or strong conj base
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10^x = 3.16
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x = log(3.16)
x between 0 and 1 (10^0 or 10^1) so x equal .5 |
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log (AB) = log (A) + log (B)
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log (10^2) = 2
log (10^3) = 3 log (10^2 x 10^3) = 5 |
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in a series of oxyacids
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more acids means stronger acid
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rate constant (kinetics) K and equilibrium (thermodynamics) are not related formally
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are not related formally
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If H2) is the solvent
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do NOT include it in the equillibrium expression
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condensation
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is exothermic
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there's no net consumption or production of H+ in an acetal reaction
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H is a catalysit??
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If the reacion is does not involve gases
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ignore the effect of pressure (if it does include gases, do include pressure such that reaction will favor products or reactants that have least gas moles
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