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K >>1 Forward rxn dominates (rxn lies to the right). Mostly products at equilibrium, [products] >> [reactants]
Cl2(g) + 2 NO(g) 2 NOCl(g) K = 6.2 x 104
K <<1 Reverse rxn dominates (rxn lies to
the left). Mostly reactants at equilibrium,
[products] << [reactants]
COCl2(g) CO(g) + Cl2(g) K = 2.0 x 10-10
H2(g) + F2(g) 2 HF(g) K = 10
K 1 Forward and reverse rxn occur to roughly
the same extent, [products] [reactants]
2.0 moles of NH3 gas are introduced into a previously
evacuated 1.0 L container. At a certain temperature the
NH3 partially dissociates by the following equation.
At equilibrium 1.0 mol of NH3 remains. Calculate the
equilibrium constant for this reaction.
2 NH3(g) N2(g) + 3 H2(g)
Reaction Quotient (Q)
Q = K: The rxn is at equilibrium. No shift.
Q < K: The rxn shifts right to produce
products to increase Q.
Q > K: The rxn shifts left to produce
reactants to decrease Q.
If a change (stress) is imposed on a system
at equilibrium, the position of the equilibrium
will shift in a direction that tends to reduce
that change (stress).
Le Chatelier’s Principle
Le Chatelier’s Principle
1. SO2(g) is removed.
2. O2(g) is added.
3. SO3(g) is added.4. The volume of the reaction container is
halved.5. An inert gas like Ar is added.6. A catalyst is added.7. Temperature is increased.
2 SO2(g) + O2(g) 2 SO3(g) H = 198 kJ
Le Chatelier’s Principle
1. CO2(g) is added.
2. CaCO3(s) is added.
3. The volume is increased.
4. The temperature is decreased.
CaCO3(s) CaO(s) + CO2(g) H = 556 kJ
N2O4(g) 2 NO2(g)
1.0 mol of N2O4(g) is placed in a 10.0 L
vessel and then reacts to reach equilibrium.
Calculate the equilibrium concentrations of
N2O4 and NO2. K = 4.0 x 10-7
Comments on the Conjugates of Acids and Bases.
• The weaker the acid the stronger its conjugate base.
• The weaker the base the stronger its conjugate acid.
• The conjugate base of a weak acid is a WEAK base.
• The conjugate base of a strong acid is worthless.
• The conjugate acid of a weak base is a WEAK acid.
Acid Ka Realative acid
strength
Conjugate base
Kb Relative base
strength
HCl ~106
HF 7.2 x 10-4
HC2H3O2 1.8 x 10-5
HOCl 3.5 x 10-8
NH4+ 5.6 x 10-10
Acid Ka Realative acid
strength
Conjugate base
Kb Relative base
strength
HCl ~106 Cl- ~10-20
HF 7.2 x 10-4 F- 1.4 x 10-11
HC2H3O2 1.8 x 10-5 C2H3O2- 5.6 x 10-10
HOCl 3.5 x 10-8 OCl- 2.9 x 10-7
NH4+ 5.6 x 10-10 NH3 1.8 x 10-5
Stuff you should now know.
1. Ka value is directly related to acid strength.
2. Weak acids vs. strong acids (Ka’s and % dissociation.3. Conjugate acid-base pairs.
4. KaKb=Kw
5. Kb value is directly related to base strength.
6. How to write out Ka and Kb rxns and expressions.7. The weaker the acid the stronger the conjugate base
(and vice versa).8. Conjugate bases of strong acids have no basic
properties whatsoever! (Kb << Kw)
Acidic, Basic, or Neutral?
1. NaCN2. NH4NO3
3. KI4. LiC2H3O2
5. C6H5NH3Cl6. KF7. NaNO3
8. HClO4
9. Ca(OH)2
10. NH4CN11. NH4C2H3O12. CaO13. SO3
Acidic, Basic, or Neutral?
1. NaCN Na+ - worthless, CN- - weak base, basic2. NH4NO3 NO3
- - worthless, NH4+ - weak acid, acidic
3. KI K+ - worthless, I- - worthless, neutral4. LiC2H3O2 Li+ - worthless, C2H3O2
- - weak base, basic5. C6H5NH3Cl Cl- - worthless, C6H5NH3
+ - weak acid, acidic6. KF K+ - worthless, F- - weak base, basic7. NaNO3 Na+ - worthless, NO3
- - worthless, neutral8. HClO4 HClO4 – strong acid, acidic9. Ca(OH)2 Ca(OH)2 – strong base, basic10. NH4CN KaNH4 < KbCN- - basic11. NH4C2H3O KaNH4 = KbC2H3O- - neutral12. CaO metal oxide - basic13. SO3 nonmetal oxide - acidic
Buffers
Buffer – A solution where a weak acid and
its conjugate base are both present
in solution.
• Buffers resist changes in pH
Good Buffers
• Good buffers will have the following:– EQUAL concentrations of the weak acid and
its conjugate base.
– LARGE concentrations of the weak acid and its conjugate base.
– pKa = pH of desired pH.
Examples of Buffers
• HCN/CN-
• NH4+/NH3
• H2PO4-/HPO4
2- - intracellular fluid buffer
• H2CO3/HCO3- - blood buffer
Calculate the pH when 0.10 mol of HCl is
Added to a 1.00 L solution containing 1.00 M
HNO2 and 1.00 M NaNO2.
KaHNO2 = 4.0 x 10-4
Calculate the pH when 0.10 mol of NaOH
are added to a 1.0 L solution containing
1.00 M HNO2 and 1.00 M NaNO2.
KaHNO2 = 4.0 x 10-4
Calculate the pH of a solution formed by
Mixing 500.0 mL of 0.100 M NH3 and 500.0
mL of 0.0500 M HCl. KbNH3 = 1.8 x 10-5
You want to prepare a HOCl buffer of pH
8.00. You want to make a 500. mL solution
and use all of the 0.75 mol of HOCl you
have on hand. How many mol of KOCl must
you add? KaHOCl = 3.5 x 10-8
Calculate the pH of a solution formed by
mixing 500. mL of 1.50 M HCN with 250. mL
of 1.00 M NaOH. KaHCN = 6.2 x 10-10
Proposed Study Plan
• Thursday: HE III Material (finish Lon Capa)• Friday: HE I Material• Saturday: HE II Material• Sunday: HE III Material• Monday: HE III Material• Tuesday: He III Material• Wednesday: HE I, II Material• Thursday: HE I, II, III Material