Transcript The Equilibrium Constant - RIT - People

The Equilibrium Constant
We have the general reaction
aA  bB  cC  dD
Equilibrium Constant
• From this we can write an expression for
our equilibrium
C  D 
a
b
 A  B 
c
K 
d
Where we have the products over the
reactants and each compound is represented
by its concentration
A Word on Units
• What are the units on K?
• From the formula we might guess that they
would depend on the specific equilibrium
involved.
• This is not the case however.
• Let us visit the concept of standard state.
Standard States
• The terms and unit that we will use are
dictated by convention
• Solutions
M
• Gas
bar
• The values used in our equilibrium
expression will be based on some reference
value
Standard States
• For solutions the standard state is 1.00M
• Gas 1.00 bar
• Pure solids or solvents the standard state is
the substance.
• So if a solution is 2.31 x 10-3 M then the
value we use in the equilibrium expression
is....
Standard State
• 2.31 x 10-3 M / 1.00 M = 2.31 x 10-3 with
the units factoring out.
• Since gas values are usually expressed as
pressure instead of seeing [X], you will see
a gas expressed as PX
• Different standard states could be picked
but convention is well established on these
terms in Analytical Chemistry.
Equilibrium Constants
• The equilibrium expression will depend on
how the expression is written.
• As a dissociation
HA = H+ + A
K1
[H ][A

[HA]

]
Equilibrium
• But written as an association
H+ + A- = HA
K2
[HA]



[H ][A ]
• Which we can see that K1 = 1/ K2
• So if we write an equation in reverse we
must take the reciprocal of the constant
Equilibrium
What about added equilibria.
+
H2S = H + HS
K a1
[ H  ][ HS  ]

[H 2S ]
then HS- = H+ + S2K a2
[ H  ][ S 2  ]

[ HS  ]
Equilibrium
• If we add the two reactions we get.
+
2H 2S = 2 H + A
Giving us the product of the equilibria.
K aT
[ H  ]2 [ S 2  ]

[H 2S ]
Thermodynamics
• How does this all relate to “thermo”
• Two contributions
– Enthalpy DH, a measure of the heat of a
reaction. + endothermic, - exothermic
– Entropy DS, a measure of disorder in a system.
+ for more disorder, - for less disorder
Thermodynamics
When HCl (which is a gas) is bubbled into
water the following reaction occurs
H2O + HCl(g) = H+(aq) + Cl-(aq)
o
Which has a DH of -75.15 kJ/mole @ 25 C
Heat can be viewed as ending up in bonds.
What bonds are in this reaction?
The naught (o) means that the product and
reactants are in the standard state.
Thermodynamics
• When we dissolve KCl, a salt similar to
NaCl then we get an entropy effect. Since
the ions in solution are more disordered
than the ions in the solid.
• H2O + KCl(s) = K+ (aq) + Cl-(aq)
o
DS is + 76 J/(K.mole) @ 25C and entropy
favors this reaction
Thermodynamics
• But for the reaction of HCl with water the
DS value is negative. Why would that be
the case?
• What happens if the terms work against
each other.
Thermodynamics
• This is where Gibbs free energy comes in.
DG = DH - TDS
• For our reaction with HCl in water what do we
get?
DG = -75.15 kJ/mole - (298K)(-131.5J/K.mole)
and calculating we get DG = -35.94 kJ/mole
• If DG is negative then the reaction is favored.
Exergonic
• If DG is positive then the reaction is disfavored
Endergonic
Thermodynamics
• If DG is zero then the reaction is at
equilibrium.
• Remember that the values we have been
looking at are in their standard state and
these values will vary with concentration.
• What is the relationship between DGo and K
K e
 DG o
RT
La Chatelier’s Principle
• If a change disturbs a equilibrium system
that system will proceed back to
equilibrium to partially offset the change.
• Use Q, same form as the equilibrium
expression. If Q is less than K then the
system must proceed to the right.
• If Q is more than K then the system must go
to the left.
La Chatelier’s Principle
• Acetic acid in water
HAc = H+ + AcKa 
[ H  ][ Ac  ]
[ HAc ]
 1.75 x10  5
• If you double the hydrogen ion
concentration by adding a strong acid you
will cause the amount of HAc to increase
Effect of heat on K
• The equilibrium constant of an endothermic
reaction increases if temperature is raised.
• The equilibrium constant of an exothermic
reaction decreases if temperature is raised.
• Look at it this way
reactants = products + heat
Solubility
• A very important parameter in many areas
of chemistry.
– Likes dissolve likes
– Polar compounds will dissolve in polar solvents
• sugar in water
• alcohols in water
– Non polar compounds will dissolve in non
polar solvents.
• oils in carbon tetrachloride
• flavor in fats
Salts in water
• Some salts are ‘soluble’ in water and some
are not.
– Soluble
• NaCl
KCl
Na3PO4
LaF3
Cu(OH)2
– Not Soluble
• AgCl
• Yet all lead to varying amounts of ions in
solution that are very polar.
Solubility Products
• Let us look at an example
AgCl(s) = Ag+ + Cl-
• Ksp = [Ag+][Cl-]
• If we look up the value for this Ksp in
appendix F we find that the value listed is
1.8 x 10-10, what is [Ag+] for a saturated
solution?
• The Ksp for NaCl would be 49. What does
that mean?
Ksp Problem
• What will be the concentration of La3+ if
1.00 grams of the solid LaF3 is added to 10
mL of water and allowed to come to
equilibrium?
• Ksp = 2 x 10-19
• Best way to solve. Set up a table!
Ksp solutions
LaF3 = La3+
Initial Conc. solid
0
Final Conc. solid
x
3F0
3x
Where x is the concentration of La3+ at
equilibrium
Ksp = [La3+][F-]3
Ksp Solutions
Ksp = [La3+][F-]3
From last slide [La3+] = x and [F-] = 3x
so 2 x 10-19 = x (3x)3 = 27x4
so x = 9 x 10-6 M La3+ and
what value for F- ? ____________________
Common Ion Effect
• We can figure out how much AgCl would
dissolve in pure water. How much would
dissolve if the solution already had some of
either ion already in it?
• Let’s look at the more complex example
however.
Common Ion Solutions
Let the initial Fluoride concentration be 0.001 M
Initial Conc
Final Conc
LaF3 =
solid
solid
La3+
3F0
0.001
x
3x + 0.001
Where x is the concentration of La3+ at
equilibrium
Common Ion
Ksp = [La3+][F-]3
From last slide[La3+] = x and [F-] = 3x + 0.001
so 2 x 10-19 = x (3x + 0.001)3 = quadric equation
to solve. Fun!!!!!
What do we do??????
Common Ion
We know that the amount of F- we are going
to get will be small. We will make an
approximation. We will assume that x is
small in the F- term. Giving
2 x 10-19 = x (0.001)3
Which gives us x = 2 x 10-10 M for La3+
Now what is F-? Does this check out?
Approximations
• Knowing when to make approximations will be
very helpful to us.
• When approximations are made then we must
check our answer to make sure that this
approximation make some sense.
• Is the final F- concentration close to the 0.001 M
as we assumed? If not then we will need to solve
without making the approximation which can be
difficult.
Separation by Pcpt
• In theory this will work great.
• Co-precipitation will lead to significant
error.
Complex Formation
• Many different equilibria can be going on at
the same time. The second type is the
formation of complexes. The example in
the book is the formation of iodides of
lead(II).
Complexes
• With lead and iodide we can have solid PbI2
formed. Ksp = [Pb2+][I-]2
• With extra added iodide we have complex
formed.
• As iodide is added we form the mono, di, tri
and tetra iodide species. Each step has and
equilibrium constant. These are formation
constants.
Complexes
• Pb2+ + I- = PbI-
K1
• Pb2+ +2I- = PbI2
b2
• Pb2+ +3I- = PbI32-
b3
• Pb2+ + 4I- = PbI42-
b4
• Each step also has a stepwise function also.
• PbI3 + I- = PbI42K4
Complexes
• All equilibria must be satisfied at the same
time. In this case here we can control the
solubility of the PbI2 by added iodide ion in
increasing amounts giving a mixture of the
species that were mentioned.
Protic Acid and Bases
• Bronsted and Lowry
– Acids are proton donors
– Bases are proton acceptors
• We increase or decrease the amount of
hydronium ion (H3O+) in solution with these
acids and bases.
• We also have the Lewis acid base definition.
Deals with electron pairs.
Salts
• When acids and bases are brought together
they will neutralize each other. This
produces a salt. Most salts added to water
might not dissolve completely but will
dissociate completely.
• NaCl
Na2SO4 Na3PO4 NH4Cl
Conjugates
• Products of acid/base reactions are also
classified as acid/base.
• Acetic acid + Ammonia = Acetate Ion + Ammonium ion
Acetic acid
acid
Acetate ion
conjugate base
Ammonia
base
Ammonium ion
conjugate acid
Water
• Water is our most common solvent. Its
behavior and properties are important to
world as we know it.
• Water can either gain a proton or lose a
proton.
• We can use the hydronium ion H3O+ in our
reactions or we often get lazy and just use
H+
Water
• Water can react with itself. One water could
donate a proton and another water could lose that
proton. This is called autoprotolysis.
• H2O + H2O = H+3O + OH• Other protic solvents can do this ie. Acetic acid
• 2CH3COOH = CH3COOH2+ + CH3COO• There are solvents that are called aprotic. These
solvents do not give up or accept the proton
• Diethyether, hexane, acetonitrile, toluene.
Water
• Autoprotolysis Constant
• H2O = H+ + OH- has a special
equilibrium constant. Kw
• Kw = [H+][OH-] Kw = 1.01X 10-14 @25C
• Kw varies with temp. We will often see this
expressed on the log scale.
• pKw = -log Kw
Water
• What is pH?
• pH = -log[H+]
• What is the pH of pure water.
For each H2O that dissociates we get one H+ and one OH-,
so [H+] = [OH-]
Kw = [H+]2 = 1 x 10-14
H+ = 1.00 x 10-7
pH = 7.0
Water will usually have traces of CO2 and other ions
that change the pH
Acid and Base Strength
• Some acids and base are considered strong! This
means that they fully dissociate in water.
– Acids
– Bases
HCl, H2SO4, HNO3, HClO4
NaOH, KOH
• Many other acid are weak. This only means that
they do not fully dissociate in water. Do not infer
that they are less hazardous.
– Acids
– Bases
Acetic, H3PO4, Benzoic etc.
Ammonia, other amines.
Weak Acids
• The dissociation of weak acid have and
equilibrium constant called Ka. If more that
one proton can dissociate, i.e. H3PO4 each
added loss will have its own equilibrium
constant.
Acid/Conjugate Relationship
• Acetic acid HA
HA = H+ + AKa = [H+][A-]/[HA]
• The conjugate base. NaA Sodium Acetate
NaA + H2O = HA + OH- + Na+
Kb = [HA][OH-]/[A-]
Acid/Conjugate Relationship
• What is Ka*Kb
Ka*Kb = [H+][A-]/[HA]*[HA][OH-]/[A-] = [H+][OH-] = Kw
So once you know the Ka for the acid form you
know the Kb for the conjugate base. It must be the
same conjugate pair.
Goodbye bases, You will note that there are no Kb
values listed in the Appendix in the text. This the
the way that biochemists look at the world.