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Chemistry Review 3

The last of the chemistry reviews...

    Equilibria, continued:

  • Le Chatelier's Principle: if a system in equilibrium is disturbed, it will attempt to re-establish equilibrium by opposing the disturbance. Some examples: raising temperature shifts equilbrium in the endothermic direction; adding reactant drives the reaction in the forward direction, while adding product reverses the reaction; increased pressure in a gaseous equilibrium favors the side of the reaction with the fewest number of moles of gas, while decreased pressure shifts the equilibrium to increase gas moles; pressure has virtually no effect when only solids and liquids are involved, so give it up.
  • Catalysts don't change Kc or Kp... they get a reaction to equilibrium more quickly.
  • ...
    Acid-Base Equilibria
  • The ions of strong acids and bases dissociate completely in water, but "weak" acids and bases only dissociate partially in water. All this equilibrium stuff we've discussed previously can be applied to weak acids and bases by using equilibrium concepts to describe weak acid/base dissociation and strength.
  • H2O equilbrium: Kc = ([H3O+][OH-]) / [H2O]^2 = 3.2 x 10^-18
  • Well, that's so low that the concentration of H2O doesn't change much at all, so we consider it to be constant and cancel out the H2O concentration:
  • Kc * [H2O]^2 = [H+][OH-] = Kw
  • Kw is the ion product constant of water! And it varies with temperature!
  • Kw = [H3O+][OH-] = [H+][OH-] = 1.00 x 10^-14 @ 298K (25C)
  • [H+] > [OH-]: ACIDIC!
  • [H+] = [OH-]: NEUTRAL!
  • [H+] < [OH-]: BASIC!
  • yes, it's this easy, people - I can't make this stuff up!
  • The relative concentrations of H+ and OH- are usually very small in the case of weak acids and bases, and with stronger acids/bases one becomes very small and the other comparatively large, so we compare the concentrations using logarithms:
  • pH = -log [H+] . . . logarithm of H3O+ molarity, fool!
  • pOH = -log [OH-] . . . logarithm of OH- molarity!
  • [H+] = 10^(-pH) mol/L
  • [OH-] = 10^(-pOH) mol/L
  • -log (Kw) = 14
  • pH + pOH = 14 @ 298K
  • Weak Acid: HA and its conjugate base A-
    HA + H2O <--> H+ + A-
  • Weak Acid Kc = [H3O][A-] / [HA][H2O]
  • Well, like before, [H2O] remains nearly constant, so we cancel it out:
  • Kc * [H2O] = Ka
  • Ka: acid dissociation constant, which tells you how readily the acid dissociates from its conjugate base in water, and how strong the acid is! Lower the Ka, weaker the acid!
  • Ka = [H+][A-] / [HA]
  • Since Ka concentrations vary exponentially (just like H+ and OH-), we use a logarithmic relationship to compare dissociation constants:
  • pKa = -log (Ka)
  • Ka = 10^(-pKa)
  • Larger pKa values indicate weaker acids (inverse relationship, just like pH)
  • If Ka and [HA] are given and you're calculating [H+]:
    Ka < 1x10^(-4) . . . ignore x in denominator ([HA] - x)
    Ka > 1x10^(-4) . . . use quadratic formula or successive estimations of x
  • Repeat for bases:
  • Weak Base: B and its conjugate acid BH+
    B + H2O <--> BH+ + OH-
  • Kb = ([BH+][OH-]) / [B]
  • pKb = -log (Kb)
  • Kb = 10^(-pKb)
  • Large Kb values indicate strong bases
  • Large pKb values indicate weak bases
  • ...
    Buffers
  • Buffers resist change in pH ([H+]) when strong acid/base is added
  • Buffers come in two flavors:
  • Weak acid / conjugate base: HC2H3O2 & H2H3O2-
  • Weak base / conjugate acid: NH3 & NH4+
  • Henderson-Hasselbalch: pH = pKa + log ([A-]/[HA])
  • When preparing buffers, try to make [A-] = [HA]. Since log 1 = 0, weak acids are most effective where pH = pKa of the acid!
  • Weaker acids have stronger conjugate bases
  • Weaker bases have stronger conjugate acids
    Therefore:
  • Ka * Kb = Kw
  • pKa + pKb = pKw = 14.00
  • ...
    Titration
  • Strong monoprotic acid / strong base titrations equivalence point = pH 7.00
  • Titration calculation steps: calculate initial pH of solution; add titrant; calculate moles of titrant; subtract from initial moles of solute; calculate new solute M using remaining moles and new solution volume; calculate new pH
  • ...
    Polyprotic Acids
  • H2CO3 can donate more than one proton; each species has it's own Ka value:
    H2CO3 --> H+ + HCO3- . . . Ka1 = 4.5 x 10^-7
    HCO3- --> H+ + CO3(2-) . . . Ka2 = 4.7 x 10^-11
  • Each successive Ka is 10^4 - 10^5 times smaller than the one before it
  • When calculating [H+] of Ka1, ignore Ka2's [H+] contribution - it's way small

    :: Bryan Travis :: 11/03/2003 @ 20:49 :: [link] ::
    ...