Chem. 31 – 4/28 Lecture

1. Chem. 31 – 4/28 Lecture

2. Announcements I • IC Lab Report – now due 5/5 • Returning AA and Soda Ash Lab Reports, test nearly graded • Homework Set Additional Problem 3.1 due Wed. and Quiz on Wednesday • AA Lab Comments: • Many students over diluted unknowns • Best region is in upper 2/3rds part of calibration plot (e.g. ~4 to 9 ppm for typical Ca Standards)

3. Announcements II • Today’s Lecture • Chromatography • Resolution and improving separations • Acid – Base Chemistry

4. ChromatographyOptimization – Resolution Equation How to improve resolution Increase N (increase column length, use more efficient column) Increase a (use more selective column or mobile phase) Increase k values (increase retention) Which way works best? Increase in k requires no new column (try first) but it will require more time and will not work if kB is large to begin with Increase N requires a new column Increasing a is the best but often requires a new column. What if resolution is very good (e.g. = 5)? Can decrease k to have faster chromatogram not in most recent version of text B for 2nd component

5. ChromatographyGraphical Representation Smaller H (narrower peaks) Initial Separation Increased alpha (more retention of 2nd peak) Larger k or longer column – Dt increases more than width

6. ChromatographySome Questions A GC is operated close to the maximum column temperature and for a desired analyte, k = 20. Is this good? Two columns are tried for a GC separation of compounds X and Y. Both give initial resolution values of 1.2. Column A has a kB value of 0.8 while column B has a kB value of 5.0 (B for 2nd eluting compound). Which column looks more promising?

7. Acid – Base Equilibria (Ch. 8) Weak Acid Problems: e.g. What is pH and the concentration of major species in a 2.0 x 10-4 M HCO2H (formic acid, Ka = 1.80 x 10-4) solution ? Can use either systematic method or ICE method. Systematic method will give correct answers, but full solution results in cubic equation ICE method works most of the time Use of systematic method with assumptions allows determining when ICE method can be used

8. Acid – Base Equilibria Weak Acid Problem – cont.: Systematic Approach (HCO2H = HA to make problem more general where HA = weak acid) Step 1 (Equations) HA ↔ H+ + A- H2O ↔ H+ + OH- Step 2: Charge Balance Equation: [H+] = [A-] + [OH-] 2 assumptions possible: ([A-] >> [OH-] or [A-] << [OH-]) Step 3: Mass Balance Equation: [HA]o = 2.0 x 10-4 M = [HA] + [A-] Step 4: Kw = [H+][OH-] and Ka = [A-][H+]/[HA] Step 5: 4 equations (1 ea. steps 2 + 3, 2 equa. step 4), unk.: [HA], [A-] [H+], [OH-]

9. Acid – Base Equilibria Weak Acid Problem – cont.: Assumption #1: [A-] >> [OH-] so [A-] = [H+] Discussion: this assumption means that we expect that there will be more H+ from formic acid than from water. This assumption makes sense when [HA]o is large and Ka is not that small (valid for [HA]o>10-6 M for formic acid) ICE approach (Gives same result as systematic method if assumption #1 is made) (Equations) HA ↔ H+ + A- Initital 2.0 x 10-4 0 0 Change - x +x +x Equil. 2.0 x 10-4 – x x x

10. Acid – Base Equilibria Weak Acid Problem – Using ICE Approach Ka = [H+][A-]/[HA] = x2/(2.0 x 10-4 – x) x = 1.2 x 10-4 M (using quadratic equation) Note: sometimes (but not in this case), a 2nd assumption can be made that x << 2.0 x 10-4 to avoid needing to use the quadratic equation [H+] = [A-] = 1.2 x 10-4 M; pH = 3.92 [HA] = 2.0 x 10-4 – 1.2 x 10-4 = 8 x 10-5 M Note: a = fraction of dissociation = [A-]/[HA]total a = 1.2 x 10-4 /2.0 x 10-4 = 0.60

11. Acid – Base Equilibria Weak Acid Problem – cont.: When is Assumption #1 valid (in general)? When both [HA]o and Ka are high or so long as [H+] > 10-6 M More precisely, when [HA]o > 10-6 M and Ka[HA]o > 10-12 See chart (shows region where error < 1%) Failure also can give [H+] < 1.0 x 10-7 M Assupmption #1 Works Fails

12. Acid – Base Equilibria Weak Base Problem: As with weak acid problem, ICE approach can generally be used (except when [OH-] from base is not much more than [OH-] from water) Note: when using ICE method, must have correct reaction Example: Determine pH of 0.010 M NH3 solution (Ka(NH4+) = 5.7 x 10-10, so Kb = Kw/Ka = 1.75 x 10-5) Reaction NH3 + H2O  NH4+ + OH- You can do the rest For bases, a = fraction of dissociation = [BH+]/[B]o

13. Acid – Base Equilibria Weak Acid/Weak Base Questions: A solution is prepared by dissolving 0.10 moles of NH4NO3 into water to make 1.00 L of solution. Show how to set up this problem for determining the pH using the ICE method. A student is solving a weak base problem for a weak base initially at 1.00 x 10-4 M using the ICE method and calculates that [OH-] = 2.4 x 10-8 M. Was the ICE method appropriate? The pH of an unknown weak acid prepared to a concentration 0.0100 M is measured and found to be 3.77. Calculate a and Ka.

14. Acid – Base Equilibria Buffer Solutions: A buffer solution is designed so that a small addition of acid or base will only slightly change the pH Most buffer solutions have a weak acid and its conjugate base both present Example: Determine pH of a mix of 0.010 M HCO2H and 0.025 M Na+HCO2- solution (ignoring activity) Go to board to show if ICE approach is needed