The equilibrium constant, K:
aA + bB
cC + dD
At equilibrium, the concentrations of reactants and
products are constant. An important relationship in
thermodynamics is called the equilibrium constant,
K:
[
C ] [D ]
=
[A]a [B ]b
c
K
d
where products are in the
numerator and reactants are in
the denominator
Let’s look at:
H2O
+
H3O+
H2O
+
OH-
[
H O ][OH ]
K=
+
−
3
[H 2O ]2
Since [H3O+] and [OH-] are 10-7 M in a neutral
solution, and [H2O] is roughly 55.6 M, the
concentration of water is essentially constant. For
this reason, it has become the convention not to
include this value in the equilibrium constants.
Hence,
[
][
]
K w = H 3O + OH − = 10-14 M2 at 25 ºC
The same convention is used for any expression with
pure liquid, a pure solid or a solvent in great excess.
The condition of equilibrium: Where forward and reverse
reaction rates are equal. In this case, no net changes in
concentrations are observed, despite the fact that both reactions
continue to proceed. There is a dynamic equilibrium.
Le Châtelier’s Principle: The position of an equilibrium
will shift in a direction to relieve an applied stress.
aA + bB
cC + dD
The rate for the forward reaction is given by:
Ratef = kf[A]w[B]x
where kf is a constant, [] = concentration
The rate for the reverse reaction is given by:
Rater = kr[C]y[D]z
where kr is a constant
At equilibrium, Ratef = Rater. The addition or removal of A,B,C,
or D will initially change either Ratef or Rater but eventually these
two rates will become equal again, but with different values.
[
][
] [ ][
]
K w = H 3O + OH − = H + OH − therefore,
[ ]
[
− log K w = − log H + + − log OH −
]
or, 14 = pH + pOH
Ka and Kb represent dissociation constants for weak
acids and bases, respectively:
e.g., in H2O:
Ka
[A ][H O ]
=
Kb
[
HA][OH − ]
=
−
A- + H3O+
HA + H2O
A- + H2O
Ka × Kb
HA + OH-
+
3
[HA]
[A ]
−
[A ][H O ]× [HA][OH ] = [OH ][H O ] = K
=
[A ]
[HA]
−
+
−
3
−
−
+
3
K a Kb = K w
Therefore, if you know the Ka of the acid, the Kb of the
conjugate base can be readily determined. (note: Ka and
Kb are in units of M)
w
Ex. What is the Kb of the acetate anion, CH3COO-?
We recognize that the acetate anion is the conjugate
base of acetic acid CH3COOH.
CH3COOH + H2O
[A ][H O ] = 1.75 x10
=
−
Ka
CH3COO- + H3O+
+
3
[HA]
−5
see appendix 3
We are interested in the reaction:
CH3COO- + H2O
Kb
CH3COOH + OH-
[
HA][OH − ]
=
[A ]
−
We know that:
K a Kb = K w
Therefore,
K w 1.01x10 −14
=
= 5.77 x10 −10
Kb =
−5
K a 1.75 x10
Be aware that the use of equilibrium expressions
based on concentrations alone in ionic equilibria can
lead to error at high electrolyte concentrations:
Electrolytes can lead to shielding of ions in solution
resulting in lower “effective” concentrations of the
ions.
Under conditions of high electrolyte concentrations,
chemists refer to the “activities” of ions.
Activity, a, is essentially the “effective”
concentration of the ion and is given by
aX = [X]γX where
γX is called the “activity coefficient” (it is always ≤ 1)
The activity coefficient is dependent upon the total
“ionic strength”, μ, of the solution.
(
)
1
μ = [A]Z A2 + [B ]Z B2 + [C ]Z C2 + ...
2
where [A] is the concentration of ion A and ZA is its
charge
The “thermodynamic” equilibrium constant is based
on activities:
aA + bB
cC + dD
K'
c
d
(
aC ) (a D )
=
(a A )a (aB )b
Whereas the “formal” equilibrium constant is based
upon molar concentrations:
K
c
d
[
C ] [D ]
=
[A]a [B ]b
They approach one another at low concentration.
We leave this subject here…and simply point out that
all of our discussions about equilibria of ionic
systems apply to solutions of low enough ionic
strength that we can assume all ions have activity
coefficients of 1.
Acid/base calculations for common cases:
Strong acids/bases in water
Weak acids/bases in water
Buffers
Salts
Concentration symbols:
Equilibrium concentration
[BH], [H3O+], [OH-], [B-]…
[Z] (real)
Analytical concentration
CBH, CH+, COH-, CB-…
CZ (conceptual)
If we dissolve 5 moles of HCl in a liter of water, the
analytical concentration, CHCl, is 5 M. The real
concentration, [HCl], is essentially zero because all
of the HCl dissociates into solvated H+ and Cl-.
Therefore,
HB + H2O
H3O+ + B[H+] ≈ [Cl-] ≈ CHCl
H2O + H2O
H3O+ + OHKa < 10-3 M very weak acid
10-3 < Ka < 103 intermediate
Ka > 103 M very strong acid
i)
ii)
how to handle autoprotolysis (CHBKa or CBKb)
how to handle [acid or base] (CHB/Ka or CB/Kb)
Strong acid in water:
Add 1x10-3 mol of HCl to water and dilute to 1 L at
25 ºC. What is the pH?
HCl + H2O → H3O+ + ClH2O + H2O
H3O+ + OH[H3O+] = [Cl-] + [OH-]
[Cl-] = CHCl
[OH-] = Kw/[H3O+]
charge balance
So, [H3O+] = CHCl + Kw/[H3O+]
If CHCl ≥ 5x10-7 M, Kw/[H3O+] < 0.05 CHCl (see below)
(Kw/[H3O+] < ((10-14/5x10-7)) = 2 x 10-8
(2x10-8/5x10-7)=0.04)
Therefore, [H3O+] ≈ CHCl
CHCl
[HCl]
[H3O+]
[Cl-]
[OH-]
=
=
=
=
=
10-3 mol/L = 10-3 M
0M
10-3 M (10-3 >> 10-7 M) pH = 3
10-3 M
Kw/[H3O+] ≈ 10-14/10-3 ≈ 10-11 M, pOH=11
If CHCl < 5x10-7 M,
[H3O+] = CHCl + Kw/[H3O+] (= [Cl-] + [OH-])
or
0 = [H3O+]2 - CHCl[H3O+] - Kw
ax2 + bx + c = 0 a quadratic equation
solutions to a quadratic equation:
− b ± b 2 − 4ac
x=
2a
[ H 3O + ] =
2
C HCl + C HCl
+ 4K w
+
2
When we solve for [H3O ] or [OH-], only positive solutions have any
physical significance.
E.g., if CHCl = 1.3x10-7 M,
+
[ H 3O ] =
1.3x10− 7 M +
-log 1.8x10-7 = 6.74
(1.3x10 )
−7 2
(
+ 4 1x10−14
2
-log 1.3x10-7 = 6.89
) = 1.8x10
−7
M
Summary of strong acid case: Ka > 103 M
[HB] = 0 ; [B-] = CHB
Can autoprotolysis be ignored?
Yes:
CHB > 5 x 10-7 M
[H3O]+ = CHB
No:
CHB < 5 x 10-7 M
[ H 3O + ] =
2
C HB + C HB
+ 4K w
2