CHEMISTRY 59-320CHEMISTRY 59-320ANALYTICAL CHEMISTRYANALYTICAL CHEMISTRY

Fall - 2010Fall - 2010

Lecture 13

8-4 Systematic treatment of equilibrium

• Step 1: Write the pertinent reactions.• Step 2: Write the charge balance equation.• Step 3: Write mass balance equations.• Step 4: Write equilibrium constant

expression for each chemical reaction.• Step 5: Count the equations and

unknowns. The numbers should be the same.

• Step 6: Solve those equations.

Example 1

Example 2

8-5 Applying the systematic treatment of equilibrium

Solubility of calcium sulfate

Step 1

Step 2

Step 3

Step 4

Step 6

Step 5: counting

Solubility of magnesium hydroxide

Step 6

Step 4

Step 5

Example 3

9.1 Strong Acids and Bases • For a strong acid, the following reaction goes to completion.

• Calculating the pH of 0.10 M HBr, using activity coefficient

• Calculating the pH of 0.10 M HBr

• The above calculations illustrate the impact of activity coefficient.• Calculating the pH of 0.10 M KOH (a strong base):

Raman spectrum

• Strong acids in dilute solution are essentially completely dissociated.

• As concentration increases, the degree of dissociation decreases.

• The figure shows a Raman spectrum of solutions of nitric acid of increasing concentration. The spectrum measures scattering of light whose energy corresponds to vibrational energies of molecules.

• The sharp signal at 1049 cm−1 in the spectrum of 5.1 M NaNO3 is characteristic of free NO3

-.

The graph shows the fraction of dissociation deduced from spectroscopic measurements. It is instructive to realize that, in 20 M HNO3, there are fewer H2O molecules than there are molecules of HNO3 and dissociation decreases.

The dilemma with very dilute solutions

• What is the pH of 1.0 × 10−8 M KOH? • Using only Kw; you will find that the

pH=6.0 – wrong answer!!

• It is because one did not consider OH- from H2O.

• Taking into account the charge balance equation:

pH = 7.2

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 H2OIf 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].

• 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.

Notice that large changes in concentrationLead to rather small change in pH!

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-/H2PO4

2- 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?• The concentrations of HA

and A− are not what we mixed.

Neglecting OH-