Review Guide, Sample Problems-- now with solutions

• Assume that 1.0 % of acetic acid vapor dimerizes at 2.5 atm pressures.
  2 CH₃CO₂H (g) ---> {CH₃CO₂H}₂ (g)
  Evaluate Keq
  • we start with P(CH₃CO₂H) = 2.5 atm
  • we lose 1% by reactions (2.475 atm remains)
  • this 0.025 atm reacts,but becomes (1/2)(0.025 atm) of the dimer
  • K = P(dimer) / P(acid)²
• K _p= (0.0125) / (2.475)²
  ------------------------------------
  A 0.050 M solution of a weak acid has a pH of 4.55. What is Ka for the acid?
  • [H3O+] = 10-4.55= 2.82 x 10^-5
  • [anion] = same
  • [acid] = 0.050 - 2.82 x 10^-5 which is nearly the same as 0.050
  • ka = (2.82 x 10^-5)2 / 0.050 = ...
• -------------------------------------
  The salt QF₃ --> Q³⁺ (aq) + 3F^- (ag) and has a solubility of 2.1 x 10^-4 moles per liter.
  Evaluate Ksp
• typo, may have read QMF₃ by mistake, sorry)
• [Q3+] =2.1 x 10^-4
• [F-] = 3 x 2.1 x 10-4 =6.3 x 10^-4
• Ksp = (2.1 x 10^-4) * (6.3 x 10^-4)³
  -------------------------------------
  Ksp for Ba(OH)₂ is 3 x 10^-3. If we mix equal volumes of 0.05M BaCl₂ and 0.020 M NaOH, will we get a precipitate to form?
  • remember, simply mixing solutions dilutes each
  • start with 0.025 M Ba²⁺ and 0.010 M OH-
  • Q = [Ba²⁺] [OH-]² = 0.025 * (0.010)² = 2.5 x 10^-6
• Q << Ksp so no precipitate
  -----------------------------------
  NH₄+ (or NH₄Cl ) acts like a weak acid
  Kb for ammonia is 1.75 x 10-5
  what is Ka for ammonium ion?
• Ka * Kb = Kw
• Ka = 1.0 x 10^-14 / 1.75 x10^-5 = 5.71 x 10^-10
  what is the pH of a 0.10 M solution of ammonium chloride
• NH₄+ + H₂O ---> H₃O+ + NH₃
• [H₃O+] = x = [NH₃]
• [NH₄+] = 0.10 -x
• x²/ (0.1-x) = 5.71 x 10^-10
• can ignore x in the 0.1-x term
• x = sqrt( 5.71 x 10^-11) = 7.56 x 10-6
• pH = -log (ans) = 5.12
  -----------------------------------
  What is the pH of a solution that is 0.10 M Acetic acid; Ka = 1.8 x 10^-5
• really the same as the NH4+ problem above, with a different Ka
  ------------------------------------
  Kc = 138 for the reaction Fe³⁺ + SCN- ----> Fe(SCN)²⁺
  If we start with 2.0 x 10-3M Fe³⁺ and 0.050 M SCN-, how much Fe(SCN)²⁺ forms?
• let x = [Fe(SCN)²⁺] at equilibrium (also change in the amount)
• [Fe³⁺ ] = 0.0020 -x
• [SCN-] = 0.0500 -x
• at equil K= 138 = x / {(0.0020-x)*(0.050-x)}
• this is a quadratic
• but x can't exceed 0.002 (all Fe3+ used) so [SCN-] can't be less than 0.048
• approximate 138 = x /(0.0020-x)*(0.050 ___)
• this is easy to solve for X
• ----------------------------------
  What is the pH of a solution that is 0.030M Acidic acid and 0.020 M potassium acetate?
  Ka for acetic acid is 1.8 x 10^-5.
• pka = - log(1.8 x 10^-5) = 4.75
• pH = pKa + log [(Ac)/(HAc)] = 4.75 + log( .003/0.002)
  ----------------------------------
  

Concepts--

• Why is sodium oxalate more soluble in 0.1 M HCl than it is in water?
• oxalate ion reacts with H+ forming the waek acid
• therefore, reduces oxalte ion concentration and more salt dissolves.
• Why is PbCl₂ less soluble in 0.1M HCl than it is in water?
• Cl- is a common ion
• Cl- is not accociated with a weak acid, so it ignores the H+

• Will adding a drop of HCl (1M) to 5 ml of 1 M solution of acetic acid
  • a. raise the pH significantly (The HCl alone now fixes the pH)
  • b. lower the pH significantly
  • c, have relatively little effect on the pH
• Will adding a drop of 1M Acetic acid to 5 ml of 1 M HCl
  • (same three choices)
  • (c.-- no change, the HCl fixes the H+
• If you add 1 drop (1/20 ml) of 1 M NaOH to 5.0 ml of water, what will be the pH?
  • [OH-] = 1.0 M x (0.05/5) where the ( ) is the dilution ratio
  • [OH-] = 0.010
  • pOH = 2
  • pH = 14 - pOH = 14.0-2.0 = 12.00
• If you added that NaOH to a solution that is 0.1M Acetic acid and 0.10 Sodium acetate,
  • why would the pH change be very small
  • this is a buffer mixture
  • the OH- is used up converting Acetic acid to acetate ion
  • the pH still stays close to pKa (see an ealier problem)
  • Adding sodium thiosulfate to water forms a solution that is several degrees cooler than the starting water.
    • Is sodium thiosulfate more soluble in hot water or in cold water?
    • the reaction (dissolving) is endothermic
    • Le Chatlier says equil constant is increased if temp rises
    • so solubility is greaster in hot water.
  • If you slowly add 0.1M NaOH to a solution of HCl and acetic acid could you...
    • produce a solution that is mainly NaCl and Acetic acid?
      • yes.. the NaOH removes H3O+ (HCl) but won't attack weak acid ntil the HCl is gone.
    • produce a solution that is mainly HCl and sodium acetate?
      • no (see above) H3O+ + acetate ---> acetic acid
    • a solution that has lost half of the HCl and half of the acetic acid?
      • no, see above
    • a solution where OH- and acetate ion both exist at modest concentrations?
      • sure, the OH- will eventually convert the acetic acid t acetate ion
      • just keep adding more OH- until it matches
    • a solution that is pH 7.00 ?
      • sure, we start acidic (pH 1.0) and gradually increase pH
      • at some point we will pass through pH 7.00
    • ---------
• The freezing point of a 1.0 M solution of glucose (a sugar) is -1.86 ^oC
  • The freezing point of a 1.0 M solution of a weak acid is 2.01 ^oC
  • why the difference?
    • since the acid partly reacts, the total concentration of ions > 1.0 M
    • here it acts like a (2.01/1.860 = 1.081molal
    • so we get 0.016 molal H3O+, 0.081 molal anion and we keep 0.919 M acid
    • Ka = 0.081 x 0.081 / 0.919
    • HA --> H+ + A-
  • how could this information be used to find Ka for the acid
• ----------------
• CaCO ₃ (solid limestone) can be heated in air to completely decompose to CaO + CO₂ (g)
  • If CaO is left exposed to normal air, it picks up CO₂ and becomes CaCO₃ again.
  • can you explain and use this information to tell something about the reaction
  • basically, the equiibrium shifts with temperature
  • at high temperature, we favor gas and CaO
  • at lower temepratures, even the 0.01% CO2 in air will react.---------------
  • so the reaction must be endothermic