Schedule1 Schedule2 Schedule3 Schedule4
Schedule to Exam 3
| Date | Day | Topics |
|---|---|---|
| 3/9 | Mon | Talked a bit about Exam 2 - 29 questions. Then I covered a bit about acid strengths in certain series. Trends: the more electronegative the atom is that the acidic hydrogen is attached to, the weaker the bond, and therefore the stronger the acid. Also, for oxy-acids in a series, the more oxygens on the central atom, the strong the acid. I then started on Chapter 8 - Type 2 problems. I showed the basic set up on the ICE table, then I switched over to the web site and showed the help sheet for Type 2 - a weak acid and its salt. I finished by showing a fraction of species diagram and pointed out the RATIOS of the conjugates throughout the diagram. Only in the CROSSOVER region (pKa±1) are the ratios such that both components must be considered reasonable. |
| 3/11 | Wed | Chapter 8. Defined acid/base buffers. Buffers resist pH change - they do not stop it though. Must have both an acid and base present in order to have a buffer. Only way to do this is to use conjugate pairs like HA and A-, or B and BH+. The "sweet spot" pH for any buffer is the pKa for the acid. At that spot the ratio of HA/A- (or B/BH+) is 1/1. To shift the pH away from that spot (say to the nearest integer value) you change the ratio of the conjugates to a maximum of 10/1 or 1/10. A "TRUE" buffer protects both against both acid attacks and base attacks. These "attacks" are the same as "stresses" when using LeChatlier's lingo. The crossover region on a fraction of species diagram is the buffer zone. |
| 3/13 | Fri | READ Chapter 8. Read more. Celebrate by NOT having class today. |
| 3/16 | Mon | Spring Break - no class |
| 3/18 | Wed | Spring Break - no class |
| 3/20 | Fri | Spring Break - no class |
| 3/23 | Mon | Rehashed much of what I did on Wednesday, 3/18. Showed the importance of getting that conjugate pair ratio. Also calculated the pH of many points during the titration of HCl with NaOH. I then showed a plot of pH vs mL of NaOH added. The curve is an "s" shaped curve with the equivalence point in the center of the vertical rise part. I also showed how just 1 drop of the titrant (OH-) will send the pH from about 4 to 10 when you are right at the equivalence point. |
| 3/25 | Wed | Looked closer at titrations, and in particular, acid/base titrations. Pointed out the 4 different parts of a titration (pH) curve and how to calculate the pH at those points or regions. The start point and the equivalence points are both single species (Type 1) calculations. The start point (0% titrated) is all HA, and the equivalence point (100% titrated) is all A- --- calculate just like we did in chapter 7 for type 1 problems. Between those two points (10-90% titrated) is the buffer region of the curve and you have a definite conjugate pair ratio to use to calculate pH. This is a type 2 problem is just like solving for any buffer system. The last zone is the overshoot area past the equivalence point (>100% titrated) and it is easily calculated by just knowing the excess hydroxide in the solution. The excess hydroxide SETS the pOH and therefore pH. |
| 3/26 | Thu | H09 due by noon |
| 3/27 | Fri | Chapter 8 |
| 3/30 | Mon | UT Class DROP DEADLINE! Pointed out equivalent points on a fraction of species diagram and a pH titration curve. This is especially interesting and helpful for polyprotic acids. A titration curve for a triprotic acid like phosphoric acid will have 3 equivalence points. I then introduced solubility equilibria and Ksp. You MUST have a saturated solution in order to have equilibrium between the solid state and the dissolved state. Showed how an ICE table looks for solubility equilibria. You get distinct answers for various ratioed salts (like 1:1 salts, 1:2 salts, 1:3, 2:3, etc...). The "x" in these problems is the amount that dissolves and is the molar solubility. You MUST be able to calculate the molar solubility for any Ksp value and vice versa. |
| 3/31 | Tue | H10 due by noon |
| 4/1 | Wed | Went over fractional precipitation. Showed you how to predict the order of precipitation. Then showed a precipitation diagram where each precipitate comes out at different titrant concentration onsets. Those onsets are calculated via the concentration given and the Ksp's of each ion in solution. My example was with iodide, bromide, and chloride ions. All of those ions will precipitate with silver ion. The more divergent the Ksp's the better the separation will be and the closer the Ksp's the worse the separation. Remember that Ksp's alone do not determine when a precipitation occurs - different salt ratios will have much different Ksp's even though their solubilities might be identical. |
| 4/3 | Fri | Told you about review sheet for exam 3. Discussed the process of dissolving precipitates. Le Chatlier's principle helps here... remove a product ion and the solid will dissolve some more to replace it. Keep up the removal and you can completely dissolve an otherwise insoluble solid. Acid will help dissolve any salts with a relatively strong conjugate base as the anion - see book for listing. Complex ions: metal cation + ligand will give a complex ion. Many complex ions have strong color and therefore are great for qualitative analysis. Discussed the different in formation constants (Kf's) and dissociation constansts (Kd's). We will tend to use Kd's because they are more like Ksp's. |
| 4/5 | Sun | H11 due by 3pm | 4/6 | Mon | talk about exam - introduced a little electrochemistry. Electric potential is measured in volts (voltage). |
| 4/7 | Tue | EXAM 3, 7-9 PM |