Acid-Base Models Conjugate Acids and
Acidity and Ka Bases
pH, pOH, and pKa Structure and
Dominant Sources Oxoacids
Polyprotic Acids Lewis Acids
% dissociation Acid-Base Solution
Bases and Kb Al Gore Rhythms
Definition Example Definition Example
of Acid of Acid of Base of Base
Svante Donates Donates
Arrhenius H+ OH–
Brøsted- Donates Accepts NH3 or
Lowry H+ H+ H2O
G.N. Accepts Donates
•3 : NH3
Lewis electron electron
Greek: kaustikos kaiein “to burn”
Acids & bases corrode nearly everything
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
HA “acid” and A– “conjugate base”
H2O “base” and H3O+ “conjugate acid”
H3O+ “hydronium ion” no free protons!
Ka = [H+] [A–] / [HA] “acid dissociation contant”
Acidity and Ka
Strong acid: Ka means [HA]eq = 0
Weak acid: Ka 0 or [HA]eq [HA]0
A–(aq) + H3O+(aq) HA(aq) + H2O(l)
K=Ka–1 or weak acid is associated with a
strong conjugate base and vice versa.
Acids HSO4– HC2H3O2 HCN
Ka 1.210–2 1.810–5 6.210–10
pH, pOH, and pKa
pX – log10(X)
Reduces Kw = 1.01210–14 to pKw = 13.995
to a manageable entity for comparison
Actually pKw = 14 is usually memorized.
2 H2O(l) H3O+(aq) + OH–(aq) K = Kw
pH+pOH=pKw because Kw = [H+][OH–]
Acid (pH<7), Neutral (pH=7), Base (pH>7)
E.g., pH(0.1 M acetic acid) = 2.87
Strong acids Weak Acids
[H+]eq = [HA]0 Ka = [H+][A–] / [HA]
pH = p[HA]0 Ka = x2 / ( [HA]0 – x)
Strong bases Ka x2 / [HA]0
[OH–]eq = [BOH]0 x ( Ka [HA]0 )½
pOH = p[BOH]0 pH = – log10(x)
pH = 14 – pOH Weak Bases
Don’t be fooled by x ( Kb [MOH]0 )½
[HA]0 < 10–7 pH = 14 - px
If more than one H+ source (multiple acids
or a polyprotic acid), pKa will usually differ by
a few units, and the largest will
determine [H+]eq. E.g., 0.01M H3PO4
pKa: 2.12, 7.21, and 12.32, respectively
Use Ka for 1st H only: pH = 2.24 (quadratic formula)
Determines H3PO4, H2PO4–, and H+
Use other two pKa to find HPO42– and PO43–
Weak electrolyte present in molecular
form mostly with few ions & will have a
small % dissociation unless diluted.
100% [ionic form] / [initial amount]
E.g., 0.1 M acetic acid = [initial amount]
[acetate ion] (0.10 Ka)½ = 0.0013 M
% dissoc. 100% (Ka/0.1)½ = 1.3%
0.01 M gives 4.2% shows more ionization
upon dilution (but pushes the “5% approx.”)
Bases and Kb
Strong base: [OH–]eq = [MOH]0
where M=metal and pH = 14 - pOH
Weak bases often amines, :NR3
where R = H and/or organic groups
Amine steals H+ from H2O to leave OH–
Kb = [NR3H+] [OH–] / [:NR3]
For NH3 (ammonia, not an amine), Kb = 1.810–5
Same pH rules & approximations apply.
Conjugate Acids and Bases
“Salt of a weak (acid/base) is itself a
Example of a weak base, NH3
NH4+ + OH– NH3 + H2O K = Kb–1
H2O H+ + OH– K = Kw (add eqns)
NH4+ NH3 + H+ Kconjugate = Kw / Kb
Kacid(NH4+) = 10–14/1.810–5 = 5.510–10
Weaker parent yields stronger conjugate!
Conjugate pH Calculations
Salts of strong parents are neutral;
those of very weak bases, very acidic:
Fe(H2O)63+ Fe(H2O)5OH2+ + H+
pKa1 = 2.2 or Ka1 = 6.310–3 (pKb = 11.8)
pH of 3 M Fe(NO3)3 ? (HNO3 strong; ignore!)
Ka1 = [FeOH2+] [H+] / [Fe3+] = x2 / (3-x)
x [3Ka1]½ = 0.14 (5% OK) pH = 0.86
pKa2 = 3.5, pKa3 = 6, pKa4 = 10
In that Fe complex, pKa1 and pKa2 are
very close (2.2 and 3.5), so the second
conjugate may be important:
Fe(H2O)5OH2+ Fe(H2O)4(OH)2+ + H+
This will certainly be the case at lower
formal iron concentrations, F, since %
dissociation is greater there. Try F=0.01
Even at F=0.01, we can safely ignore
pKa3, pKa4, and pKw, since they are
dominated by pKa1 and pKa2.
Abbreviate those equilibria as
A3+ B2+ + H+ K1 = B H / A
B2+ C+ + H+ K2 = C H / A
But those are 2 eqns in 4 unknowns!
2 Additional Conditions!
Conservation of iron requires that
Conservation of charge must include a
spectator anion for A3+ such as 3 Cl–
3 A3+ + 2 B2+ + C+ + H+ = 3 F (conc. Cl–)
Notice that higher charges get scaled by
their charge for proper balance.
Solve 4 Nonlinear Equations
3F = 3A + 2B + C + H
B H = K1 A
C H = K2 B
Remove A by substituting from 1st eqn
Solve 3 Nonlinear Equations
Remaining after substitution
3F = 3F – 3B – 3C + 2B + C + H
B H = K1 ( F – B – C )
C H = K2 B
H = B + 2C
B H = K1 F – K1 B – K1 C
C H = K2 B substitute C = K2 B / H
Solve 2 Nonlinear Equations
H = B + 2 K2 B / H
B H = K1 F – K1 B – K1 K2 B / H
H2 = B H + 2 K2 B B = H2 / (H + 2K2)
B H2 = K1 F H – K1 B H – K1 K2 B
Substitute for B
H3 + K1 H2 – K1 (F – K2)H – 2 K1 K2 F = 0
Only H remains as an unknown!
Solve 1 Cubic Equation
Use the dominant solution as first guess
in an iteration to the non-dominant one.
A3+ B2+ + H+ K1 = B H / A
K1 = x2 / (F – x)
x2 + K1 x – F K1 = 0
x = ( – K1 + [K12 + 4F K1]½) / 2
x = 5.310–3 H
F = 0.01, K1 = 610–3, and K2 = 310– 4
H3 + 610–3 H2 – 5.810–3 H – 3.610–8 = 0
Hout = (3.6 10–8 + 5.810–3 Hin – 610–3 Hin2)1/3
Hin (the guess) Hout (the refinement)
[H+] = 5.5810– 3 M and by B = H2 / (H + 2K2),
[B2+] = 5.0410– 3 M and by C = K2 B / H,
[C+] = 2.7 10– 3 M which makes [A3+] = 4.710– 3 M
More Conjugate Calculations
HCO3– H+ + CO32– pKa2 = 10.25
pH of 0.10 M Na2CO3 ? (NaOH strong; ignore)
CO32– + H2O HCO3– + OH– K = Kw/Ka2
K = [HCO3–] [OH–] / [CO32–] = x2 / (0.10-x)
x [0.10 1.810–4 ]½ = 4.210–3 (5% OK)
pOH = 2.37 pH = 14.00 – 2.37 = 11.63
Sodium carbonate is in laundry soap.
Acid Structure and Strength
Higher polarity liberates protons.
While all hydrogen halides are strong acids
(save HF), acidity lessens with halide size.
Even non-adjacent electronegativity
drains e– density, liberating protons.
HClO is weak but HClO4 is strong.
(the H is attached to only 1 of the oxygens)
Cl3CCO2H is much stronger than CH3CO2H
German: O2 = sauerstoff or the stuff
that makes sour oxoacids, –X–O–H
True when X is a fairly electronegative
non-metal (halogen, N, P, S) or a metal in a
high oxidation state (e.g., H2CrO4)
True as nonadjacent electronegativity
aids polarity (e.g., CH3CO2H)
Metals in lower oxidation states tend to
give up the OH– and not the H+
Sr(OH)2 Sr2+ + 2 OH–
Al(OH)3 is basic; Al3+ is acidic.
Water is amphoteric too, acting as a
base by becoming H3O+, for example.
– Brønsted definition: water accepts the proton.
Lewis Acids and Bases
Water is both acid & base by Lewis too.
Base via H2O: + H+ H3O+
donates lone pair e– to proton
Acid via H2O + S2– OH– + HS–
accepts the one of sulfide’s electrons
Lewis’ is the most general definition
E.g., Cu2+ + 6 :NH3 Cu(NH3)62+
Cu’s the acid and :NH3’s the base.
Know the significant Solve controlling K
species for the Write K expression.
conditions. Set initial conditions
Ignore the rest. for smallest change.
Define all species in
terms of x (change).
reactions, keep only
Find x solving K exp.
Check any approx.
Find acids & bases. Find pH & [species]