CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 2010

1. CHEMISTRY 59-320ANALYTICAL CHEMISTRYFall - 2010 Lecture 13

2. 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.

3. Example 1 Example 2

4. 8-5 Applying the systematic treatment of equilibrium Solubility of calcium sulfate Step 1 Step 2 Step 3 Step 4 Step 5: counting Step 6

5. Solubility of magnesium hydroxide Step 4 Step 5 Step 6

6. Example 3

7. For a strong acid, the following reaction goes to completion. 9.1 Strong Acids and Bases • Calculating the pH of 0.10 M HBr • Calculating the pH of 0.10 M HBr, using activity coefficient • The above calculations illustrate the impact of activity coefficient. • Calculating the pH of 0.10 M KOH (a strong base):

8. 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-. Raman spectrum 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.

9. 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: The dilemma with very dilute solutions pH = 7.2

10. 9.2 Weak acids and bases

11. 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.

12. 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.

13. 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

14. 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. The fraction of dissociation, α, is defined as the fraction of the acid in the form A−

15. 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−]. 9-4 Weak base equilibria

16. 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. 9-5 Buffers

17. 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. Henderson-Hasselbalch Equation Notice that large changes in concentration Lead to rather small change in pH!

18. 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 properties

19. 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.

20. Tris

21. 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-