Chapter 6 - Acid/Base Equilibria
- Acid/Base theories
- Arrhenius
- Bronsted - Lowry
- Lewis
- Solvern/System
- pH and pOH
- -log = p function
- based on activities of species (pH = -log aH+)
- For 'working' definition assume activities = concentrations
- -log[H+] = pH
- -log[OH-] = pOH
- -logKa = pKa
- neutral - pH = 7.00 ([H+] = 1.0 x 10-14)
- pH + pOH = 14.00
- pH < 7, acidic
- pH > 7, basic
- [H+] = 10^-pH
- If acidic pH greater than 6.5 OR basic pH less than 7.5 we must consider H+ or
OH- from water!!
- Calculation of pH
- Strong acid
- assume complete dissociation, [H+] = [strong acid]
- Strong base - calculate hydroxide ion concentration from formula of salt
- soluble hydroxide - 1 OH- for every hydroxide
- soluble oxide - 2 OH- for every O2-
- If temp different from 25C, Kw may change - change 'neutral' pH.
Problems 6-13
- Dissociation of weak acids and bases
- HA H+ + A-
- Ka =[H+][A-]/[HA]
- Ka - acid dissociation constant
- B + H2O HB+ + OH-
- Kb = [HB+][OH-] / [B]
- Kb - basic equilibrium constant (hydrolysis constant)
- NOTE - B can be 1 of 2 types of base
- 'Molecular' base - look up Kb in table
- Conjugate base of weak acid
- note OH- comes from reaction with water.
- Calculation of pH
- Acid
- Strong acid
- assume complete dissociation, [H+] = [strong acid]
- Weak acid
- Calculate CHA / Ka (CHA = original concentration of acid)
- If CHA / Ka 100, then [H+] = the square root of (Ka*CHA)
- If CHA / Ka < 100, then use the equation based on the
quadratic formula
- Base
- Strong base - calculate hydroxide ion concentration from formula of salt
- soluble hydroxide - 1 OH- for every hydroxide
- soluble oxide - 2 OH- for every O2-
- Weak base - Calculate CB / Kb (CB = original concentration of base)
- If CB / Kb 100, then [OH-] = SQRT(Kb*Cb)
- If CB / Kb < 100, then use formula based on the quadratic
equation
- Conjugate acids/bases
- Calculate "missing" K value from Kw = Ka * Kb
- Treat like any other weak a/b
Problems 14-32 even
- Common Ion Effect
- Common ion - common to the equilibrium system
- causes a shift away from the ionic side of the equilibrium
- For HA H+ + A-
- Addition of H+ or A- causes shift back to HA
- 'Suppresses ionization'
- Buffers
- Typically mixture of weak conjugate a/b pair
- Weak acid reacts with strong base
- Weak base reacts with strong acid
- prepared by:
- addition of weak acid and soluble salt of the weak acid
- HA + NaA (NaA -> Na+ + A-)
- partial neutralization of weak acid with strong base
- base is limiting reactant
- reverse is true for weak base / conj. acid systems
- Resistant to a change in pH
- Derive Henderson-Hasselbalch eqn
- pH dependent on ratio of base to acid
Problems 42-49
- Buffer capacity - amount of material to get ratio of base/acid outside limits
- 10 base/acid 0.1
- ideal buffer - b/a = 1
- Selection of buffer
- pH < 7, weak acid/conj. base
- pH > 7, weak base/conj. acid
- Choose one that pH pKa (or [H+] Ka)
Stepwise equilibria
- Essentially all first proton comes off before any second
- H2A H+ + HA- (K1)
- HA- H+ + A2- (K2)
- Solution of weak acid - treat like monoprotic acid
- Use K1 - describes equilibrium for proton donation
- Solution of 'stripped' anion (A2-) treat like conjugate base of monoprotic acid
- must have all donatable protons removed
- use 'last' K value
- equilibrium w/formation of stripped base
- Acid salt
- partially neutralized polyprotic acid (HA-)
- Amphiprotic species
- conjugate base of polyprotic acid
- weak acid w/protons left to donate
- Use K value for dissociation of proton (K2 for diprotic) and K value for
formation (K1 for diprotic)
- Same types of calculations true for triprotic acids
- Caution choosing K values for acid salt
- Buffers from polyprotic acids
- Choose Kx that has both buffering species in equation
- Treat like other buffer
Problems 37-41; 50-52
Questions? E-mail me.
Return to my home page.