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Strong Acid-Weak Base and Weak Acid - Strong Base. CH 3 COOH(aq) + OH - (aq) -> CH 3 COO - (aq) + H 2 O(l). Slow change in pH before equivalence point; solution is a buffer CH 3 COOH(aq)/CH 3 COO - (aq) At halfway point [HA] = [A - ] pH = pK a

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Strong Acid-Weak Base and Weak Acid - Strong Base

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CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)

Slow change in pH before equivalence point; solution is a buffer


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) + OH-

determines the pH at equivalence

Answer: 8.26


Polyprotic acid and bases

Titration of H2CO3 with a strong base


A compound whose color changes noticeably over a short range of pH.

pH = 7.0 8.5 9.4 9.8 12.0


[In- (aq)]


KIn =

[HIn(aq) ]

The indicator is a weak acid itself (HIn)

HIn(aq) + H2O(l)  In-(aq) + H3O+ (aq)

The acid form, HIn, has a different color from the base form In-

The 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


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
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


Ksp =

[Ag+ (aq)] [Cl- (aq)]

AgCl(s)  Ag+ (aq) + Cl- (aq)

Define an equilibrium constant for this process: solubility product, Ksp

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
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