Strong Acid-Weak Base and
Weak Acid - Strong Base
CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)
Slow change in pH before
equivalence point; solution
is a buffer
CH3COOH(aq)/CH3COO-(aq)
At halfway point
[HA] = [A-]
pH = pKa
At equivalence, pH
determined by CH3COO-(aq)
Changes in pH during a titration of a weak acid/base with a
strong base/acid:
Halfway to the stoichiometric point, the pH = pKa of the acid
The pH is greater than 7 at the equivalence point of the
titration of a weak acid and strong base
The pH is less that 7 at the equivalence point of the titration
of a weak base and strong acid
Beyond the equivalence point, the excess strong acid or base
will determine the pH of the solution
Titration of 100.0 mL of 0.1000 M CH3COOH(aq) with 0.1000 M
NaOH
Before addition of NaOH: pH determined by CH3COOH(aq)
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)
Answer: pH = 2.88
Before the equivalence point: determine pH for a buffer
Addition of 30.00 mL of NaOH(aq)
The OH-(aq) reacts with the CH3COOH(aq). Determine
concentration of CH3COOH(aq) and CH3COO- (aq) in solution
after addition of the base. Answer: pH = 4.38
At half equivalence: [CH3COOH(aq)] = [CH3COO-(aq)]
pH = pKa
At equivalence: enough OH-(aq) added to react with all
CH3COOH(aq).
For this problem, equivalence is reached when 100.0mL of
OH- is added; i.e. 0.01000 moles of OH-(aq) added
Solution contains 0.01000 moles CH3COO-(aq) in 200.0 mL
solution; [CH3COO-(aq)] = 0.05000 M
pH determined by
CH3COO-(aq) + H2O(l)  CH3COOH(aq) + OH- (aq)
pH = 8.72
(note greater than 7.0)
Beyond equivalence: pH determined by excess OH-(aq)
Estimate the pH at the equivalence point of the titration of
25.00 mL of 0.100 M HCOOH(aq) with 0.150 M NaOH(aq)
(Ka(HCOOH) = 1.8 x 10-4)
At the equivalence point, enough NaOH(aq) has been added
to react with all the HCOOH(aq) forming HCOO-(aq)
The reaction:
HCOO-(aq) + H2O(l)  HCOOH(aq) + OHdetermines the pH at equivalence
Answer: 8.26
Polyprotic acid and bases
Titration of H2CO3 with a strong base
Indicators
A compound whose color changes noticeably over a short
range of pH.
pH = 7.0
8.5
9.4
9.8
12.0
The indicator is a weak acid itself (HIn)
HIn(aq) + H2O(l)  In- (aq) + H3O+ (aq)
[In- (aq)] [H3O+(aq)]
KIn =
[HIn (aq) ]
The acid form, HIn, has a different color from the base form
InThe end point of a titration is the point at which the
concentrations of the acid and base forms of the indicator are
equal; [HIn(aq)] = [In-(aq)]
Color change occurs when pH = pKIn
In choosing an indicator:
pKIn ≈ pH(equivalence point) ± 1
Applications
Atmospheric CO2(g)
CO2(g) + H2O(l)  H2CO3(aq)
H2CO3(aq) + H2O(l)  HCO3-(aq) + H3O+(aq)
Acid rain: When pH < 5.5
due to pollutants in the air like SO2, SO3, NO2, which
dissolve in water to form strong acids
Many lakes have water too acidic to sustain life, forests
have also been damaged.
In the ground, acidic rain water can be neutralized by ions
in the soil
The ocean is buffered to a pH of 8.4 by buffering that
depends on the presence of hydrogen carbonates and
silicates
Physiological Buffers:
Body fluids such as blood function over a very narrow pH
range, maintained by buffers
Blood contains two buffering systems:
1) Phosphate buffer (H2PO4-/HPO42-)
H2PO4-(aq) + H2O(l)  H3O+(aq) + HPO42-(aq)
Ka2 = 6.2 x 10-8, pKa2 = 7.21
Average pH of blood is 7.40 indicates that
[HPO42-(aq)] / [H2PO4-(aq) ] = 1.55
2) HCO3-/H2CO3 buffer
CO2(g) + H2O(l)  H2CO3(aq)
H2CO3(aq) + H2O(l)  HCO3-(aq) + H3O+(aq)
pKa1 = 6.36
[HCO3-(aq)] / [H2CO3(aq) ] = 11.0
Build up of H2CO3 would destroy this balance.
In the body
H2CO3(aq)  H2O(l) + CO2 (g)
CO2 is exhaled from the lungs to prevent buildup of H2CO3
Solubility Equilibria
Many ionic solids dissociate into their ions in water:
NaCl(aq) -> Na+(aq) + Cl- (aq).
Compounds such as NaCl exist completely as Na+ (aq) and
Cl- (aq) in aqueous solutions unless the amount of NaCl
exceeds the solubility of NaCl in water
Other compounds such as CaCO3 dissolve to a very small
extent in water - sparingly soluble
The Earth’s crust consists largely of sparingly soluble salts;
e.g. gypsum (CaSO4.2H2O), calcite (CaCO3), dolomite
(xCaCO3.yMgCO3), oxides and sulfides of metals such as Fe,
aluminosilicates (XAlSi3O8 or XAlSi2O8, X = Na+, K+, Ca2+)
Hard-water contains high levels of Ca2+ and Mg2+
Ca2+ forms soap scum with detergents.
Add soluble Na2CO3 (washing soda) to precipitate CaCO3
which washes off.
Chemical weathering includes the dissolving of sparingly
soluble salts.
CaCO3(s) + CO2(g) + H2O(l)  Ca2+(aq) + 2 HCO3-(aq)
Highly soluble compounds: several grams of the compound
dissolves per 100 g of water
At 298 K: 36 g of NaCl, 122 g of AgNO3
Sparingly soluble compounds: less than one gram dissolves
per 100 g of water
At 298 K: 2.4 x 10-4 g of AgCl; 9.3 x 10-4 g of CaCO3,
4.4 x 10-14 g of PbS
AgCl(s)  Ag+ (aq) + Cl- (aq)
Define an equilibrium constant for this process: solubility
product, Ksp
Ksp =
[Ag+ (aq)] [Cl- (aq)]
The solubility product, Ksp, is the equilibrium constant for the
equilibrium between an undissolved salt and its ions in a
saturated solution.
The molar solubility of Ag2Cr2O4 is 6.5 x 10-5 mol/L at 25o C.
Determine the value of Ksp.
Ag2Cr2O4(s)  2 Ag+ (aq) + Cr2O42- (aq)
Ksp = [Ag+ (aq) ]2 [Cr2O42- (aq)]
[Ag+ (aq) ] = 2 x 6.5 x 10-5 mol/L
[Cr2O42- (aq) ] = 6.5 x 10-5 mol/L
Ksp = 1.1 x 10-12
Determine the solubility of BaSO4(s) in pure water at 298 K
in moles/liter and grams/liter. Ksp (BaSO4) = 1.1 x 10-10
BaSO4(s)  Ba2+ (aq) + SO42- (aq)
If x is the solubility in moles/liter
[Ba2+(aq)] [SO42-(aq)] = x2 = 1.1 x 10-10
[Ba2+(aq)] = [SO42-(aq)] = 1.0 x 10-5 M
Solubility of BaSO4 is 1.0 x 10-5 M or 2.3 x 10-3 g/L
Precipitation from Solution
If equal volumes of aqueous solutions of 0.2 M Pb(NO3)2 and
KI are mixed will PbI2(s) precipitate out? Ksp of PbI2 is 1.4 x
10-8
Use the reaction quotient, Q, to predict whether precipitation
will occur
Pb(NO3)2 (aq) + KI(aq) -> PbI2 (s) + KNO3 (aq)
Net ionic equation: Pb2+ (aq) + 2I- (aq) -> PbI2 (s)
The reverse of this reaction defines Ksp
PbI2 (s)  Pb2+ (aq) + 2I- (aq)
Ksp = [Pb2+ (aq)] [I- (aq)]2
If Q > Ksp precipitation; if Q < Ksp no precipitation
Equal volumes of Pb(NO3)2 and KI are mixed
On mixing, volume of mixed solution is twice initial volume
[Pb2+ (aq)] = 0.2M / 2 = 0.1 M
[I- (aq)] = 0.1 M
Q = [Pb2+(aq)] [I- (aq)]2 = (0.1)(0.1)2 = 0.001 M
Q > Ksp; PbI2(s) precipitates