Transcript K b

CHEMISTRY 59-320
ANALYTICAL CHEMISTRY
Fall - 2010
Lecture 14
9.2 Weak acids and bases
• The relationship between Ka and Kb for a
conjugate acid-base pair: Ka*Kb = Kw.
• Weak is conjugate to weak. The conjugate
base of a weak acid is a weak base and
vice verse.
9-3 Weak acid equilibria
• Why is the ortho isomer 30 times more acidic than the para isomer?
The product of the acid dissociation reaction forms
a strong, internal hydrogen bond, that drives the
reaction forward.
Calculate the pH of a weak acid
For any respectable weak acid, [H+] from HA will be much greater than [H+] from H2O
If dissociation of HA is much greater than H2O dissociation, [A-] >>[OH-]
Example: Calculation the pH of 0.0500 M o-hydroxybenzoic acid solution. The
Ka = 1.0 x 10-3.
Solution: plug into the above equation with F= 0.0500 and Ka = 1.0 x 10-3
The fraction of dissociation, α, is defined as the fraction
of the acid in the form A−
• Fraction of dissociation of
a weak electrolyte
increases as electrolyte is
diluted. The stronger acid
is more dissociated than
the weaker acid at all
concentrations.
• Weak electrolytes
(compounds that are only
partially dissociated)
dissociate more as they
are diluted.
9-4 Weak base equilibria
• Calling the formal concentration of base F (= [B] + [BH+]),
•
Notice that approximation in the last equation looks like a weak-acid
problem, except that now x = [OH−].
9-5 Buffers
• A buffered solution resists
changes in pH when acids or
bases are added or when
dilution occurs.
• The buffer is a mixture of an
acid and its conjugate base.
There must be comparable
amounts of the conjugate acid
and base (say, within a factor
of 10) to exert significant
buffering.
Henderson-Hasselbalch Equation
• Regardless of how
complex a solution
may be, whenever
pH = pKa, [A−] must
equal [HA].
Notice that large changes in concentration
Lead to rather small change in pH!
• This relation is true
because all equilibria
must be satisfied
simultaneously in any
solution at equilibrium.
If there are 10 different
acids and bases in the
solution, there is only
one pH, because there
can be only one
concentration of H+ in
a solution.
Buffer properties
• Buffer capacity, β, is a
measure of how well a solution
resists changes in pH when
strong acid or base is added.
• where Ca and Cb are the
number of moles of strong acid
and strong base per liter
needed to produce a unit
change in pH.
• (a) Cb versus pH for a solution
containing 0.100 F HA with
pKa = 5.00. (b) Buffer capacity
versus pH for the same system
reaches a maximum when pH
= pKa.
• The lower curve is the
derivative of the upper curve.
Buffer pH Depends on Ionic Strength and
Temperature
•
The correct Henderson-Hasselbalch equation, includes activity coefficients. Failure
to include activity coefficients is the principal reason why calculated pH values do
not agree with measured values.
•
The H2PO4-/H2PO42- buffer has a pKa of 7.20 at 0 ionic strength. At 0.1 M
ionic strength, the pH of a 1:1 mole mixture of this buffer is 6.86. Molecular
biology lab manuals list pKa for phosphoric acid as 6.86, which is
representative for ionic strengths employed in the lab.
As another example of ionic strength effects, when a 0.5 M stock solution of
phosphate buffer at pH 6.6 is diluted to 0.05 M, the pH rises to 6.9—a rather
significant effect.
Changing ionic strength changes pH.
Buffer pKa depends on temperature. Tris has an exceptionally large
dependence, −0.028 pKa units per degree, near room temperature. A
solution of tris with pH 8.07 at 25°C will have pH ≈ 8.7 at 4°C and pH ≈
7.7 at 37°C.
Changing temperature changes pH.
•
•
•
•
Tris
• A Dilute Buffer Prepared
from a Moderately Strong
Acid.
• What will be the pH if 0.010 0
mol of HA (with pKa = 2.00)
and 0.010 0 mol of A− are
dissolved in water to make
1.00 L of solution?
Neglecting OH-
• The concentrations of HA
and A− are not what we
mixed.
Chapter 10: Polyprotic acids and
bases
• Polyprotic systems
are acids or bases
that can donate or
accept more than one
proton.
• A molecule that can
both donate and accept
a proton is said to be
amphiprotic.
•
•
•
(a) How many grams of NaHCO3 (FM 84.007) must be added to 4.00 g of
K2CO3 (FM 138.206) to give a pH of 10.80 in 500 mL of water?
(b) What will be the pH if 100 mL of 0.100 M HCl are added to the solution in
part (a)?
(c) How many milliliters of 0.320 M HNO3 should be added to 4.00 g of K2CO3
to give a pH of 10.00 in 250 mL?
(a)
(b)
(c)
pH in polyprotic acid solutions
The Acidic Form, H2L+ - diprotic acid
Example: calculate the pH and composition of individual
solutions of 0.050 0 M H2L+, 0.050 0 M HL, and 0.050 0 ML−.
Leucine hydrochloride contains the protonated
species H2L+
HL is an even
weaker acid,
because
K2 = 1.80 × 10−10.
Monobasic species
• We can treat L− as a monobasic species,
with Kb = Kb1. Great approximation!
Amphiprotic character
Solving for H+