Equilibrium
Equilibrium is a macroscopic event,
regardless of the direction of approach
the forward rate and the reverse rate are
the same and changes in concentration
cannot be observed.
Products favored 103 > K > 10-3 Reactants favored
[C]c[D]d
Kc = [A]a[B]b
Mathematical Relationships
Reverse the reaction equation K’ = 1/K
Scale the reaction equation K’’ = Kn
Add equations Krxn = K1∙ K2 ∙ K3 ...
Relationship to Kp = K(RT)Dn
To solve for concentrations using K you
will need a balanced reaction equation and
an ICE table.
Heterogeneous Solutions
For mixtures the pure solids, pure liquids, and
components that represent large amounts of
solvents can be removed from the
equilibrium expression and rolled into the
equilibrium constant.
Predicting reaction direction from nonequilibrium conditions – Q
The equilibrium expression for Q is the same as
for K. Compare the values of Q and K (the
ratios of products to reactants)
Q=K the system is at equilibrium
Q>K favors reactants
Q<K favors products
Factors that affect Equilibrium – [ ], P&V, T
For [ ] check Q or use the teeter totter,
For gases – P&V, [ ] is to 1/V which is to P, ↑P, ↓V, or ↑[ ] will push to fewer moles of gas.
For T check DH is q product or reactant
Acid
HCl
Base
NaOH
Arrhenius
Produces H+
Produces OH-
BrönstedLowery
Donates H+
Accepts H+
Lewis
Accepts e- pair Donates e- pair
pH = -log[H+] pOH = -log[OH-] pKa = -logKa etc.
Kw = 1E-14 = [H+][OH-] = 1E-7∙1E-7
pKw = pH + POH = 14 = 7 + 7
Henderson-Hasselbach equation
pH = pKa + log([salt]) – log([acid])
pOH = pKb + log([salt]) – log([base])
Determination of strength –
or who wins the tug of war for the proton
Strong acids Ka > 1
Weak acids Ka < 1
Strong bases Kb > 1
Weak bases Kb < 1
Buffer
Common Ion Effect
Titrant
Equivalence Point
Primary Standard
End Point
Secondary Standard
Titration
Solubility
Molar solubility
Ksp - Solubility Product Constant
Ka – Acid dissociation constant
Kb – Base dissociation constant
Kw – water dissociation constant
Le Chatelier’s Principle
Lewis Acid
Lewis Base
Bronsted-Lowry Acid
Bronsted-Lowry Base
Arrenhius Acid
Arrhenius Base
Henderson-Hasselbach equation
Conjugate acid
Conjugate Base
Titration
1. Add solution from the
buret.
2. Reagent (base) reacts with
compound (acid) in solution
in the flask.
3. Indicator shows when exact
stoichiometric reaction has
occurred.
4. Net ionic equation
H+ + OH- --> H2O
5. At equivalence point
moles H+ = moles OH-
http://www.chem1.com/acad/webtext/abcon/abcon-2.html
In General How to Solve Acid/Base Equilibria and Buffer Problems
1. Are the compounds involved strong or weak electrolytes
a. If strong then there is a 100% dissociation and no way back use a single
headed arrow and when adding the simultaneous equations together only and
add the products.
i.e. NaOH → Na+ + OHb. if a weak electrolyte use a double headed arrow and when adding
simultaneous equations together add both reactants and products
i.e. CH3COOH ↔ CH3COO- + H+
2. Determine if reactions are neutralization, dissociation, or equilibrium reactions
I. Neutralization
a. Class I – strong electrolyte/strong electrolyte: produces salt and
water. Use ICE table with units of amount (mmol or moles), cancel
terms, determine amounts of products. Stop.
b. Class II – weak electrolyte/weak electrolyte: Compare Kas or Kbs to
determine which predominates products or reactants. Use the equlibrium
equation and ICE table with units of concentration (mmol/mL or mol/L) to
determine equilibrium concentrations. Stop.
c. Class III – strong electrolyte/weak electrolyte: produces salt of the
conjugate and water and further actions, Dissociation or Equilibrium
reactions must be determined continue to 2 II
II. Dissociate any products and begin again at repeat step 1 then continue to 2 III.
Products of the dissociation though by name appear to be the same compounds
they are not identical in that they have a different source and are treated separately.
III. Equilibrium – write new stoichiometrically balanced chemical equilibrium
equations.
3. Write the equilibrium equation and solve for x and determine
[H+] or [OH-]
4. Find pH or pOH
5. Determine the pOH from pH or pH from pOH
6. Determine the [H+] or [OH-] not found in step 3
Ksp
Write a stoichiometrically balanced equation for the limited dissociation of the solid
If provided with the Ksp and looking for molar solubility of products
Write the stoichiometric ratio, i.e. 1:2
Multiply the ratio through by x, i.e. x:2x
Write the equilibrium equation, i.e. Ksp = [M2+][Y-]2
Substitute in the stoichiometrically determined x values for the [ ]s, Ksp = (x)(2x)2
Solve for x
If provided a mass determined to be in solution and looking for Ksp
convert mass to moles
convert moles to molarity
Write the equilibrium equation, i.e. Ksp = [M2+][Y-]2
Solve for Ksp