Lectures 3,4
Water and pH
Water is essential for life; and it has unusual physical properties related to the weak
forces that connect atoms--these affect thermodynamic properties of solutes
•
•
•
•
•
covalent: strong (142 –O:O- to 946 N:::N, -C:C- 343 kJ/mol; 0.15 nm)
Vanderwaal’s: weak (0.4-4.0 kJ/mol ; 0.1 nm)
hydrogen: moderate (12-30 kJ/mol ; 0.3 nm; combine two electronegative
atoms, O,N, e.g. O...H-O: water, note 10-ps “flickering clusters”)
ionic: moderate (20 kJ/mol ; 0.25 nm in crystal; but moderated by H2O: F =
Q1Q2/εr2 and ε [dielectric constant] = 1 in vacuum, 80 in H2O)
hydrophobic: moderate (<40 kJ/mol; depend on hydrocarbons disrupting the
H-bonds “flickering clusters” of H2O)
Forming covalent, Vanderwaal's, hydrogen, ionic bonds reduces ΔE; forming
hydrophobic increases ΔS
Dissociation of H-O bonds in water
H2O + H2O ↔ H3O+ + OHH
H
O
H
H
O
O
H
+
H
H
H
O
O
H
H
H
O
H
[H3O+] = [OH-] = 10-7 in pure H2O
as with hydrogen bonds, note “flickering” as H is traded around H2O molecules
Note the convention: H2O ↔ H+ + OH- and
[H3O+] = [H+]
Keq = [H+][OH-]/[H2O]
Because [H2O] is approx. constant, define Kw = Keq *[H2O]
Kw = [H+][OH-] = (10-7)( 10-7) = 10-14 M2
Define the p function: p[X] = - log10[X]
pKw = pH + pOH = 7 + 7 = 14 (because –log(10-14) = 14)
Strong acids and bases
pH of 0.1 M HCl = 1, etc.
pOH of 0.1 M NaOH = 1; pH of 0.1 M NaOH = 13; etc
([H+] = 10-3M)
([H+] = 10-7+ 10-9M)
(pH 0: - log10(X) = 0, X = exp10 (-0) = 1 M)
Note that a base can be either an OH- donor or an H+ acceptor.
What is the pH of 1 mM HCl?
What is the pH of 1 nM HCl?
Can you made a solution of pH 0?
Weak acids and bases
Weak acids
HA↔ H+ + AHA is “conjugate acid”; A- is conjugate base
Ka = [H+][ A-]/[HA] = [H+]*[ A-]/[HA]
pKa = pH – log ([ A-]/[HA])
Example: CH3COOH ↔ H+ + CH3COO-
Ka = 1.74 x 10-5; pKa = 4.76
What is the pH of a 0.1 M CH3COOH solution? What is [H+]?
CH3COOH ↔
0.1 – X
H+ + CH3COOX
X
Ka = [H+]*[CH3COO-]/[CH3COOH] = 1.74 x 10-5
X*X = (0.1-X)* 1.74 x 10-5 Assume (0.1-X) = 0.1 if error < 5%
X = sqrt (1.74 x 10-6) = 1.32 x 10-3 = [H+]
pH = - log10[H+] = 2.88
Weak bases
A- + H2O ↔ HA + OHKeq = [HA][OH-]/[ A-] [H2O]
HA is “conjugate acid”; A- is conjugate base
Kb = Keq * [H2O] = [HA][OH-]/[ A-]
pKb = pOH – log ([HA]/[ A-])
Show that Kw = Ka * Kb
pKw = pKa + pKb = 14 Useful!
Example: imidazole C3N2H4 + H2O ↔ C3N2H5+ + OHKb = 0.98 x 10-7; pKb = 7.01
What is the pH of a 0.1 M imidazole solution in water? What is [H+]?
H
C
C3N2H4 + H2O ↔ C3N2H5+ + OH0.1 – X
X
X
HN
Kb = [C3N2H5+][OH-]/[C3N2H4] = 0.98 x 10-7
X*X = (0.1-X)* 0.98 x 10-7 Assume (0.1-X) = 0.1
X = sqrt (0.98 x 10-8) = 1.0 x 10-4 = [OH-]
pOH = 4; pH = 14 – 4 = 10; [H+] = 10-10 M
C === C
H
H
__________________________________________________________
Buffers
Henderson-Hasselbach equation
General solution when you know the concentrations of conjugate acid and base
Recall:
Ka = [H+][ A-]/[HA] = [H+]*[ A-]/[HA]
pKa = pH – log ([ A-]/[HA])
Rearrange:
pH = pKa + log ([ A-]/[HA]) = pKa + log ([conjugate base]/[conjugate acid])
Example: What pH do you get when to 0.1 M HA, you add 0.02 M NaOH?
HA
0.1-0.02
↔
H+ + A0.02
HA
0.1-0.02
NaOH
0.02
pH = pKa + log [(0.02)/(0.1-0.02)]
Apply to acetic acid, pKa = 4.76:
A0.02
N
H+
0.02
OH0.02
pH = 4.76 + log [1/4]
= 4.76 + (- 0.60) = 4.16
H2O
Na+
0.02
Example: How much HCl must be added to 0.1 M Tris to give pH 7.8? pKa (Tris) = 8.1
(Tris is tris-hydroxy-amino-methane and generally is the base form, e.g. A-.)
A- + H+ ↔ HA
0.1-X
X
pH = pKa + log ([ A-]/[HA])
7.8 = 8.1 + log ([ A-]/[HA])
log ([ A-]/[HA]) = -0.3
[ A-]/[HA] = 10-0.3 = 0.5
[ A-] = 0.5[HA]
0.1-X = 0.5X
X = [HA] = 0.1/1.5 = 0.067 M
A0.1−X
HCl
X
H+
X
ClX
HA
X
(HCl sufficient to make this concentration)
Titrations
Example: Add NaOH to 0.1 M CH3COOH
pH = pKa + log([ A-]/[HA])
when [ A-] ~ [HA], then [ A-]/[HA] ~ 1, and log([ A-]/[HA]) ~ 0 and pH ~ pKa
Also, when log([ A-]/[HA]) ~ 0, it is most resistant to changes in HA
i.e., when pH ~ pKa: buffers!
So expect most "resistance," lowest d(pH)/d(NaOH) at 0.05 M
[CH3COO-] / [CH3COOH] = 0.05 / (0.1 – 0.05) = 1
pH
pKa
4.76
2.88
0
NaOH →
0.1 M
pKa for ammonium = 9.25, imidazole = 6.99, acetate = 4.76 (note the shapes
are all the same)
Phosphate dissociation and disproportionation:
H3PO4 ↔ H2PO4- ↔ HPO4-2 ↔ PO4-3
pK1 = 2.15 pK2 = 7.2 pK3 = 12.4
Put H2PO4- into water:
H2PO4- ↔ Η+ + HPO4-2
H2PO4- + Η+ ↔ H3PO4
so [HPO4-2] = [H3PO4]
pH = pK1 + log[H2PO4-1]/[H3PO4] = pK1 + log[H2PO4-] - log[H3PO4]
pH = pK2 + log[HPO4-2]/[H2PO4-1] = pK2 + log[HPO4-2] - log[H2PO4-]
Add equations
2pH = pK1 + pK2
pH = ½( pK1 + pK2)
Carbonated beverage
CO2 + H2O ↔ Η+ + ΗCO3pK1 = 3.77
The taste of CO2 (for instance, in Coke) is the taste of H+ (Frommer, Science 327:275
15 Jan 2010)