Students will be able too...
Outcome Topic | Directly Relates To... | |
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1. | Describe the relationship between free energy and equilibrium | Equilibrium: ∆G = 0, system minimizes the overall free energy |
2. | Convert ΔG to Q, as well as ΔG° to K and vice versa. | \(\Delta G = \Delta G^\circ + RT \ln Q\) \(\Delta G^\circ = -RT \ln K\) |
3. | Know the importance of the activity of a species and how it relates to concentration, pressure, and equilibrium. | \( a_{\rm gas} = P_{\rm gas}/1\;{\rm atm}\) \( a_{\rm X(aq)} = [{\rm X}]/1\;{\rm M}\) activities are used in TRUE equilibrium constants. |
4. | Write the mass action expression for homogeneous and heterogeneous equilibria. | \( {{{a_{\rm C}^c \; a_{\rm D}^d}\over a_{\rm A}^a \; a_{\rm B}^b} }\) |
5. | Determine new values for K when combining multiple reactions. | when you flip a reaction: \( K' = K^{-1}\) when you double a reaction: \(K'' = K^2\) when you ½ a reaction: \(K''' = \sqrt{K}\) |
6. | Determine if a system is at equilibrium and if not which direction the reaction will shift to achieve equilibrium. | Compare Q to K and decide |
7. | Know the difference between Kp and Kc and be able to convert between the two. | \(K_{\rm p} = K_{\rm c}(RT)^{\Delta n}\) |
8. | Set up and solve a RICE table. | This is a procedural thing... know how to create and use the table |
9. | Calculate the concentration/pressure of all species at equilibrium. | Solve for the x in the RICE table, then substitute x back in for all the equilibrium concentrations or pressures (the "E" line). |
10. | Show a complete understanding of Le Chatelier's principle. | Stresses are: concentration or pressure changes and/or temperature changes. System shifts (forward or reverse) so as to relieve the stress. |
11. | Predict the response of a reaction to an applied stress (concentration, pressure, volume, temperature) both qualitatively and quantitatively. | Same as above but you ADD an "S" line to your RICE table and get a new "I" line. Remember that the "I" line is meant to NOT be at equilibrium (Q≠K). filler filler filler filler filler filler |
12. | Calculate the new value of K when the temperature changes to a new value. | the van't Hoff equation: \( \ln\left({K_2\over K_1}\right) = {\Delta H\over R}\left({1\over T_1} - {1\over T_2}\right) \) |
Students will be able too...
Outcome Topic | Directly Relates To... | |
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1. | Understand the strength of an acid (or base) as determined by the percent of ionization in solution. | Percent ionization (%ion) leads to a different [H+] than the stated concentration except for the special "strong" cases where the ionizations are 100%. EVERYTHING else is less than 100% and are typically down below the 1% level. Every ionization below 100% is called "weak". |
2. | Identify strong and weak acids and bases. | Memorize the 7 Strong Acids and 8 Strong Bases |
3. | Identify acid/base conjugate pairs and their relative strengths. | Conjugate Pairs differ by one and only one proton. The relative strengths directly relate to the magnitudes of Ka for the acid strength and Kb for the base strength. And for ALL conjugate pairs, Ka × Kb = Kw. |
4. | Understand the process of auto-ionization of water and what is meant by acidic, basic, and neutral. | Water will split into protons and hydroxide ions in a very very very very small amount. [H+][OH-] = Kw = 10-14. Acidic is ANY solution where [H+]>[OH-], basic is ANY solution where [OH-]>[H+], and neutral is the special case of [H+]=[OH-]. |
5. | Know the value of Kw at 25°C, and the relationship between Ka and Kb for a conjugate pairs. | Ka × Kb = Kw |
6. | Convert between hydronium ion concentration, hydroxide ion concentration, pH and pOH for a given solution. | [H+][OH-] = Kw = 10-14 pH + pOH = pKw = 14 (at 25°C) |
7. | Determine the pH of a strong acid or base solution. | [strong acid] = [H+], [strong base] = [OH-] if only 1 OH- in formula |
8. | Determine the pH of a weak acid or weak base solution. | Set up RICE table and solve for H+ or OH-, then get pH. |
9. | Determine the pH of the solution made from the salt of a weak acid or the salt of a weak base. | Same as above, remember that salts of weak acids are bases and behave that way while salts of weak bases are acids and behave that way. Now Set up RICE table and solve for H+ or OH-, then get pH. |
10. | Recognize and predict the components of a buffer solution. | Always a specific ratio of the two conjugates of a pair. |
11. | Calculate the pH of a buffer solution, and a buffer solution after the addition of strong acid or strong base. filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler filler | Remember that the ratio of the two conjugates SETS the pH of the buffer - the Henderson-Hasselbalch equation comes in handy here. The addition of more acid or base is a "STRESS"... system will respond and shift left or right. The result is a NEW ratio of conjugates and hence, a new pH which is slightly shifted from the original pH. |
12. | Determine the majority species for acid/base solutions as well as the pH following neutralization. | You can think of the multitude of conjugate pair ratios and how they are governed by the given pH. The conjugates are FORCED into their new ratios based on the forced pH brought on by adding strong acid or base (a titration). |
13. | Interpret a titration curve plot including calculating the concentration and Ka or Kb for the analyte. | Know the inflection points (there are two of them) one is the equivalence point while the other is the 1/2 titration point. |
14. | Determine the protonation state (or overall charge) for a polyprotic species at a particular pH. | Understand how setting pH SETS the ratio of the conjugates... knowing that ratio also allows you to know the relative "protonation" state of the species. |
15. | Apply concepts from equilibria to acid/base problems | Apply Le Chatlier's Principle in the arena of acid/base reactions. |
Know (memorize) the value for Kw at 25°C which is 1.0 × 10-14.
Know ALL necessary formulas for answering all the questions that go with the outcomes listed above.