Download Acid K a

CO(g) + H2O(g)
CO2(g) + H2(g)
CO(g) + H2O(g)
CO2(g) + H2(g)
CO(g) + H2O( g)
CO2(g) + H2(g)
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
K  1 Forward and reverse rxn occur to roughly
the same extent, [products]  [reactants]
H2(g) + F2(g)
2 HF(g)
K = 10
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.
2 NH3(g)
N2(g) + 3 H2(g)
At equilibrium 1.0 mol of NH3 remains. Calculate the
equilibrium constant for this reaction.
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.
Le Chatelier’s Principle
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
2 SO2(g) + O2(g)
1.
2.
3.
4.
2 SO3(g) H = 198 kJ
SO2(g) is removed.
O2(g) is added.
SO3(g) is added.
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.
Le Chatelier’s Principle
CaCO3(s)
1.
2.
3.
4.
CaO(s) + CO2(g) H = 556 kJ
CO2(g) is added.
CaCO3(s) is added.
The volume is increased.
The temperature is decreased.
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
Brønsted-Lowry Model
Acids – are proton donors
Bases – are proton acceptors
HC2H3O2 is a stronger acid then HCN which
Has the stronger conjugate base?
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
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
Realative
acid
strength
Conjugate
base
Kb
Relative
base
strength
Acid
Ka
Realative
acid
strength
Conjugate
base
Kb
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
Relative
base
strength
Stuff you should now know.
1.
2.
3.
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7.
8.
Ka value is directly related to acid strength.
Weak acids vs. strong acids (Ka’s and % dissociation.
Conjugate acid-base pairs.
KaKb=Kw
Kb value is directly related to base strength.
How to write out Ka and Kb rxns and expressions.
The weaker the acid the stronger the conjugate base
(and vice versa).
Conjugate bases of strong acids have no basic
properties whatsoever! (Kb << Kw)
Calculate the pH of a 0.10 M HBr solution.
Calculate the pH of a 0.10 M HOCl solution.
KaHOCl = 3.5 x 10-8
Calculate the pH of a 0.10 M NaF.
KaHF = 7.2 x 10-4
Calculate the pH of a 0.10 M Ca(OH)2
solution.
Calculate the pH of a solution containing
0.10 M HOCl and 0.02 M NaOCl.
KaHOCl = 3.5 x 10-8
Calculate the pOH of 0.05 M Ba(OH)2.
Calculate the pOH of 0.50 M KOCl.
KaHOCl = 3.5 x 10-8
Calculate the pOH of 1.00 M HI.
Calculate the pOH of 0.25 M NH4Cl.
Ka NH4+ = 5.6 x 10-10
Calculate the pOH of a solution containing
0.25 M NH4Cl and 0.10 M NH3.
Ka NH4+ = 5.6 x 10-10
Calculate the pH of 1.6 x 10-13 M HNO3.
A solution of 8.00 M HCOOH is 0.47%
Ionized. What is the Ka for the acid? pH?
Acid
HF
C6H5NH3+
HC2H3O2
HCN
NH4+
Ka
7.2 x 10-4
2.6 x 10-5
1.8 x 10-5
6.2 x 10-10
Acidic, Basic, or Neutral?
1.
2.
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NaCN
NH4NO3
KI
LiC2H3O2
C6H5NH3Cl
KF
NaNO3
HClO4
Ca(OH)2
NH4CN
NH4C2H3O
CaO
SO3
Acidic, Basic, or Neutral?
1.
2.
3.
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6.
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10.
11.
12.
13.
NaCN
NH4NO3
KI
LiC2H3O2
C6H5NH3Cl
KF
NaNO3
HClO4
Ca(OH)2
NH4CN
NH4C2H3O
CaO
SO3
Na+ - worthless, CN- - weak base, basic
NO3- - worthless, NH4+ - weak acid, acidic
K+ - worthless, I- - worthless, neutral
Li+ - worthless, C2H3O2- - weak base, basic
Cl- - worthless, C6H5NH3+ - weak acid, acidic
K+ - worthless, F- - weak base, basic
Na+ - worthless, NO3- - worthless, neutral
HClO4 – strong acid, acidic
Ca(OH)2 – strong base, basic
KaNH < KbCN - basic
KaNH = KbC H O - neutral
metal oxide - basic
nonmetal oxide - acidic
-
4
4
2
3
-
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-/HPO42- - intracellular fluid buffer
• H2CO3/HCO3- - blood buffer
Calculate the pH of a solution that is 1.00 M
HNO2 and 1.00 M NaNO2.
KaHNO = 4.0 x 10-4
2
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.
KaHNO = 4.0 x 10-4
2
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.
KaHNO = 4.0 x 10-4
2
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. KbNH = 1.8 x 10-5
3
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
• Total Points in course: 800
• Points to be decided next week: ~415
Proposed Study Plan
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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
A 100. mL solution of 0.10 M HF is titrated
by 0.10 M NaOH. Calculate the pH when
0.0, 25.0, 50.0, 100.0, and 125.0 mL of
NaOH have been added. KaHF = 7.4 x 10-4