Aqueous Equilibria
Chapter 16
pH
 The pH of aqueous solutions not only
blood plasma, but seawater, detergents,
sap and reaction mixtures is controlled, by
the transfer of protons between ions and
water molecules.
Ions as acids and bases
 Any cation that is the conjugate acid of a weak
base functions as an acid and lowers the pH of
the solution. Ammonium ion the conjugate acid
of weak base ammonia, is an acid.
 NH4 +(aq) +H2O(l) H3O+(aq) +NH3(aq)
 Small highly charged metal cations that can act
as Lewis acids in water, such as Al3+ and Ti3+
also produce acidic solutions, even though the
cations don’t have any protons to donate. The
protons come from the water molecules that
hydrate the ions in solution.
 Salts that contain the conjugate acids of
weak bases produce acidic aqueous
solutions; so do salts that contain small,
highly charged metal cations. Salts that
contain the conjugate bases of weak acids
produce basic aqueous solutions.
pH of a salt solution
 Aquoeus solutions of salts with acidic
cations have a pH lower than 7; salts with
basic anions produce a pH higher than 7
in aqueous solutions.
Class Practice
 Estimate the pH of 0.15M
Ca(CH3COO)2(aq).
pH of mixed solutions
 The pH of a solution of a weak acid
increases when a salt containing its
conjugate base is added. The pH of a
solution of a weak base decreases when a
salt containing its conjugate acid is added.
Class Practice
 Calculate the pH of a solution that is 0.500
M HNO2(aq) and 0.100M KNO2(aq)
 Ka=4.3x10-4 for HNO2
Titrations
 Strong acid strong base titrations
 Strong base and strong acid
 In the titration of a strong acid with a
strong base (or a strong base with a
strong acid), the pH increases (or
decreases) slowly initially, increases (or
decreases) rapidly through pH=7 at the
stoichiometric point, and then increases
(or decreases) slowly again.
Class Practice
 Suppose we carry out a titration in which
0.34M HCl (aq) is the titrant and the
analyte initially consists of 25 ml of 0.25M
NaOH (aq). What is the pH of the analyte
solution after the addition of 5 ml of titrant?
Calculating the pH at the stochiometric point
of the titration of weak acid with strong base
 Calculate the pH of the solution resulting
when 5 ml of 0.150 M NaOH (aq) is added
to 25 ml of 0.100M HCOOH (aq) . Use
Ka=1.8x10-4 for HCOOH (aq).
Buffer solutions
 Solutions that resist the change in pH
when small amounts of strong acids or
base are added are called as buffers.
Acid Buffer
 An acid buffer solution consists of a weak
acid and its conjugate base; it has pH<7.
 CH₃COOH (aq) + H₂O(l)↔H₃O+(aq) + CH₃CO₂(aq)
 Acid Buffer action: The weak acid transfers
protons to the OH- ions from added strong
base.
 The conjugate base of the weak acid accepts
protons from the H₃O+ ions supplied by a
strong acid.
Base Buffer
 A base buffer solution consists of a weak
base and its conjugate acid ; it has pH >7.
 NH₃(aq) + H₂O(l)↔NH₄+ (aq) + OH- (aq)
 Base Buffer action: The weak base accepts
protons from the H₃O+ ions supplied by a
strong acid.
 The conjugate acid of the weak base transfers
protons to the OH- ions from added strong
base.
Calculate the pH of a buffer
solution
 For a solution containing a weak acid HA




and a salt that provides the conjugate
base anion, A- , the proton transfer
equilibrium is;
HA(aq) +H2O(l)↔H3O+ (aq) +A- (aq)
Ka=[H3O+ ][A-]/[HA]
Rearranging the equation
1/[H3O+ ] =1/Ka x[A-]/[HA]
 Taking logarithm of both sides:
 log1/[H₃O+ ] = log1/ Ka + log[A-]/[HA]
 -log[H₃O+ ] = -log[Ka] + log[A-]/[HA]
 That is pH = pka + log[A-]/[HA]
 pH =pKa + log ([base]initial/[acid]initial)
 This relation is called the Henderson Hassel
balch equation.
Class Practice
 Calculate the pH of a buffer solution that is
0.40 M NaCH₃CO₂ (aq) and 0.80M
CH₃COOH(aq) at 25⁰C.
 Calculate the ratio of the molarities of CO₃2and HCO₃- ions required to achieve buffering
at pH =9.50. The pka of H₂CO₃ is 10.25.
Solubility product
 The solubility product is the equilibrium
constant for the equilibrium between an un
dissolved salt and its ions in a saturated
solution.
Class Practice
 The molar solubility of silver chromate
Ag2CrO4 is 6.5x10-5 mol/L. determine the
value of Ksp.
 The Ksp for lead(II)iodide in water is 1.4x10-8 .
Estimate its molar solubility.
The common ion effect
 The common-ion effect is a term used to
describe the effect on a solution of two
dissolved solutes that contain the same
ion.The presence of a common ion
suppresses the ionization of a weak acid
or a weak base.
 If both sodium acetate and acetic acid are
dissolved in the same solution they both
dissociate and ionize to produce acetate
ions. Sodium acetate is a strong
electrolyte so it dissociates completely in
solution. Acetic acid is a weak acid so it
only ionizes slightly.
 NaC2H3O2(s) → Na+(aq) + C2H3O2-(aq)
HC2H3O2(l) ↔ H+(aq) + C2H3O2-(aq)
 According to Le Chatelier's principle, the
addition of acetate ions from sodium
acetate will suppress the ionization of
acetic acid and shift its equilibrium to the
left. This will decrease the hydrogen ion
concentration and thus the common-ion
solution will be less acidic than a solution
containing only acetic acid.
Class Practice
 What is the approximate molar solubility of
silver chloride in 0.10M NaCl (aq)?
Predicting precipitation
 In the Ksp expression the right hand side of
the expression is known as the "ion
product". At saturation when the ions in
solution are in equilibrium with the solid
slightly soluble salt, the ion product is
equal to a fixed value called the solubility
product constant:
 Ksp = ion product
 Two further cases exist:
 ion product < Ksp
 ion product > Ksp
 If the ion product < Ksp then no
precipitation will occur even though the
salt may be insoluble according to the
solubility rules. If, on the other hand, the
ion product > Ksp the ion concentration will
be large enough for precipitation to occur
 The concentration of Calcium ion in blood
plasma is 0.0025 M. If the concentration of
Oxalate ion is 1 X 10-8 M. Will Calcium
Oxalate, CaC2O4 precipitate? Ksp = 2.3 X
10-9.
 Write the slightly soluble salt equilibria.





CaC2O4 = Ca+2 + C2O4-2
Write the ion product ion product = [Ca+2]
[C2O4-2]
Identify the given ion concentrations
[Ca+2] = .0025 = 2.5 X 10-3 M
[C2O4-2] = 1 X 10-8 M
Plug in the given ion concentrations into
the ion product equation
 ion product = [Ca+2] [C2O4-2] = (2.5 X 10-3)
( 1 X 10-8)
 ion product = 2.5 X 10-11
 Compare the ion product value with Ksp
value and make a conclusion. Since the
ion product value (2.5 X 10-11) is less than
the Ksp (2.3 X 10-9) we have to conclude
that no precipitation will take place.
Class Practice
 Lead II Chromate, PbCrO4, is used in
yellow paint pigment("chrome yellow").
When the solution is 5.0 X 10-4 M in Pb+2
ion and 5.0 X 10-5 M in Chromate ion.
Would you expect some of the Lead
Chromate to precipitate. Ksp PbCrO4 = 1.8
X 10-14.
Homework
 Page 744
 16.22 all
 16.30 all
 16.55 all
 16.85