Strategies and Explanations for Midterm Practice Exam
Strategies and Explanations for Midterm Practice Exam Chem 1220(Chemistry, Dr. Clark, General Chemistry)
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Chem 1220(Chemistry, Dr. Clark, General Chemistry)
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This 7 page Study Guide was uploaded by Phillip Fishbein on Friday April 1, 2016. The Study Guide belongs to Chem 1220(Chemistry, Dr. Clark, General Chemistry) at 1 MDSS-SGSLM-Langley AFB Advanced Education in General Dentistry 12 Months taught by Dr. Clark in Winter 2016. Since its upload, it has received 16 views. For similar materials see General Chemistry in Chemistry at 1 MDSS-SGSLM-Langley AFB Advanced Education in General Dentistry 12 Months.
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Date Created: 04/01/16
1. Acid-Base Equilibrium and Common Ion Effect Step 1: Find the value of the Equilibrium constant from given information (pK ia this question) Step 2: Find the expression for the equilibrium constant. Step 3: Set up BCA table with relevant information + - HA H A B 0.100 M 0.080 M (from HCl) 0 C -x +x +x A 0.100 – x 0.080 + x x Step 4: Set value of constant equal to new expression using terms from row A of table 2. Acid-Base Neutralization: Determine pH + - Step 1: Determine Reaction (H + OH H O) 2 Step 2: Determine moles of acid and base in problem + - Step 3: Find which is in excess, H or OH Step 4: Use remaining moles and total volume in the problem to get concentration of remaining reactant Step 5: Find pH or pOH, whichever is most applicable (pH for excess H , pOH for - excess OH ) Step 6 (optional): If you found pOH in Step 5, find pH using pOH and pK (wK isw14 unless a change in K ws mentioned in the problem) 3. Acid-Base Neutralization and Buffers: Determining pH Step 1: Determine first reaction (weak acid + strong base or strong acid + weak base) Step 2: Use BCA table to find excess, acid or base Step 3: If strong acid/base is in excess, use steps 4-6 from problem 2 to find pH. Otherwise, use step 4 from problem 2 for excess weak acid/base and its conjugate and move to step 4. Step 4: Use concentrations of weak acid/base and its conjugate and K oraK in b Henderson-Hasselbalch equation 4. Acid-Base Neutralization and Buffers: Determining Acidity of Final Solution Use methodology from question 3 to find pH relative to 7. Usually only need steps 1 and 2. 5. Titrations: Quantity Relation - Step 1: Determine relationship between number HA molecules and A molecules shown in figure Step 2: If more A than HA, pH > pK . If lass A than HA, pH < pK . Else, pHa= pK a *Not enough information given to determine if the equivalence point is reached. Knocks off 2 of the 5 possible answers. Also, solutions with diagrams like this will usually be buffers, as it shows there is a significant amount of conjugate base in solution. 6. Titrations: pH at Equivalence If the titration is between a strong acid and strong base, the pH at equivalence is 7. If it’s between a strong acid and weak base, the pH at equivalence is less than 7. If it’s between a weak acid and strong base, the pH at equivalence is greater than 7. 7. Titrations and Acid-Base Equilibria: pH of Two Titrations Since the two titrations involve a weak acid and strong base, the pH will be determined using Henderson-Hasselbalch’s equation until it reaches equivalence. At equivalence, K wibl be used for the pOH- which can be used to find pH. After equivalence, the newly added OH will determine the pOH, and thus, the pH. Since all the substances are equal molar and the same amount of acid is used in each titration, Henderson-Hasselbalch’s equation says the pH cannot be the same unless the pK isathe same, which is never true for two different acids. Through this logic, the K bon’t be the same, either. Therefore, the only time that pH is the same is past the equivalence point. Since everything is equal molar, it is easy to find the answer that is past the equivalence point. *This isn’t a general problem, so there’s no step-by-step solution that will work here that can be used for other problems. 8. K spand Molar Solubility: Find K sp from Molar Solubility Step 1: Find the expression for K forsphe substance in the problem Step 2: Using BCA table, fill in K spexpression in terms of x. Step 3: Replace x with molar solubility of the substance and you’ll get K spat that temperature. 9. pH and Solubility Either determine which anion is the conjugate of a strong acid or determine three anions that are weak bases. 10. Molar Comparison of Dissolution of Weak Acid/Base If a weak acid is dissolved in the solution, its concentration will be greater than its conjugate. If the salt being dissolved includes a weak base, the concentration of the base will be greater than its conjugate acid. 11. K spand Point of Precipitation Step 1: Find the expression of each K spinvolved in the problem. Step 2: Fill in concentration of known ions. Step 3: Fill in concentration of unknown with its molar multiple of x. Step 4: Solve for x in both expressions. Step 5: The equation with the smaller x will precipitate out first. X is the minimum amount of the added ion needed to cause the formation of a precipitate. 12. Relating Spontaneity, ΔG, ΔH, and ΔS Using ΔG = ΔH – TΔS, remember that endothermic reactions are nonspontaneous unless entropy is positive and the reaction occurs at a high temperature. Exothermic reactions are spontaneous at low temperatures where entropy is negative. Exothermic reactions with positive entropy are always spontaneous. 13. Heat, Entropy, Enthalpy, and Temperature Step 1: Convert all temperature values given to Kelvin for easier comparison Step 2: Determine sign of q in entropy formula (S = q/T) Step 3: Compare relative magnitude based on temperatures in Kelvin 14. Ranking Molar Entropy Step 1: Organize based on phase (solid, liquid, gas) with solid as the lowest and gas as the highest Step 2: In each phase with multiple substances, organize based on number of atoms in each particle, with less particles being lower than more particles. Step 3: Rank any remaining ties based on molar mass. Lower molar mass are less than higher molar mass. 15. Explaining Difference in Molar Entropy Use methodology from problem 14: more ways to vibrate relates to number of atoms in each particle, more ways to rotate relates to phases. 16. Entropy as an Extensive Property Entropy only relies on amount, not which microstate the system is currently in. Go with the largest number of whatever is talked about in the problem. 17. Relating Different Phase Changes to Entropy and Enthalpy Remember that deposition is when a substance changes from gas to solid directly. This decreases the overall internal potential energy of the substance, therefore it is exothermic. Remember that adding heat to a system/surroundings increases the entropy of that system/surroundings. Therefore, an exothermic system adds heat to the surroundings, thus increasing entropy. 18. Find Temperature of Boiling/Freezing Points with Enthalpy and Entropy Data Remember that when freezing, boiling, melting, condensing a substance, the process is actually in equilibrium. This means that ΔG = 0 for these processes. From the provided formulas, you known that ΔG = ΔH – TΔS. Use the enthalpy and entropy data to find ΔH and ΔS. Then solve for T. 19. Equilibrium: Equilibrium Constant and Temperature for Gases Δn Formula: K p K (cT) , Δn = change in moles of gas Determine if Δn is positive or negative based on chemical equation. If it is positive,pK increases with increased temperature. If it is negative, p decreases with increased temperature. 20. Solubility: Calculating pH of Semi-Soluble Strong Bases Use the methods from problem 8, except solving K spfor x using the given values for each substance Use x and molar constants to find molarity of OH in each solution. The solution with - the highest concentration of OH has the highest pH. 21. Spontaneity and ΔG, ΔH, and ΔS Reactions are nonspontaneous if ΔG is positive. That eliminates two of the answers. 0 Remember that ΔG is for reactions that occur at stp (standard temperature and pressure). Since the pressure of at least one of the gases is not 1 atm, the system 0 does not satisfy the requirements to use ΔG . Therefore, the answer is ΔG. 22. Finding Entropy with Temperature and Enthalpy Data: Boiling/Freezing Points Use the method from problem 18 solving for ΔS instead of T. You don’t have to find ΔH with enthalpy data. If the temperature is given to you in Celsius, make sure to convert it to Kelvin. 23. Entropy: Atom Theory No help here. Just remember that Boltzmann was the main guy for the theory end of entropy and that he mainly advocated for atom theory. If you just remember the atom theory part, it should help lead you to the answer. 24. Coordinate Chemistry: Finding the Formula for the Coordinate Complex Experimentally Since one mole of AgCl precipitated for one mole of RhCl * 4NH , that means one 3 3 mole of Cl is one the outside of the coordinate complex per mole of RhCl * 4NH .3 3 Since only one answer has one Cl outside the coordinate complex formula, that answer is the correct answer. 25. Charge of a Complex You just have to know which ligands are charged and how charged they are to find the charge of a complex. Since platinum has a charge of +2, and Br has a charge of - -1, ammonia is neutral, and there are two Br ions per complex, the charge of the complex comes out to be 0. 26. Explanations of Monodentate vs Bidentate Ligands While water has two lone pairs available for coordination complexes, the two pairs are on the same point on the molecule. Therefore, it can only “bite” in one place with those two pairs. (en), however, also has two lone pairs, but the lone pairs are located at different ends of the molecule. Therefore, it can “bite” in two different places with one pair each. 27. Coordinate Complexes and Magnetism Diamagnetic substances have the no unpaired electrons in their orbitals. Remember that low spin metals fill the lower three of the five orbitals completely before filling the other two, whereas high spin metals fill the orbitals normally. That means that it must either have no electrons, ten electrons, or 6 electrons with low spin in the d- orbital to be diamagnetic. 28. Crystal-Field Theory: D-orbital Energy Explanation Since ligands introduce electrons at the d-orbitals, the potential energy of the whole substance increases in those orbitals. However, in an octahedral orientation, the 2 2 ligands only truly line up to the orbitals along the x, y, and z axes, or the d x -yand dzorbitals. 29. Chelating Agents: Comparison of Ligands to Find Relationship between Formation Constant and Enthalpy or Entropy This is a difficult one. Dr. Clark’s notes in the key explain it better than I could; however, I’ll try. As Dr. Clark points out in the key, the chelating effect is due to the increase in entropy when a bidentate or polydentate ligand coordinates to a metal. Therefore, in order to determine what affects it, you should compare a monodentate ligand to a bidentate ligand. However, to best determine if the entropy is affecting the constant or the enthalpy, you should pick two that have either similar entropy changes or enthalpy changes. Since the chelate effect is an increase in entropy with a bidentate ligand, look for the enthalpy to be similar. In order for the enthalpy to be similar, the two ligands should have similar atoms that act as the Lewis base for the molecule. Since carbonate and water are the only ones that share a Lewis base, they are the best choice for the experiment. 30. Coordinate Chemistry: Determining Facts about a Complex Step 1: Determine the charge of the complex by looking at the ions bonded to the whole complex Step 2: Determine the coordinate number of the complex based on the number of ligands coordinated to the metal multiplied by the number places it can take up (bidentate and polydentate) Step 3: Determine the charge of ligand(s) Step 4: Use the overall charge of the complex and the charge of the ligands to find the charge of the metal cation in the complex Step 5: Use the charge to determine the number of d-electrons in the complex Step 6: Look at your options and see if they match any of the details determined above 31. Coordinate Chemistry: Strong/Weak Field Ligands and the Splitting of the d-orbitals Remember that stronger ligands increase the difference in the energies of the d- orbitals. Since the gap in the energies of the orbitals are larger, it takes a photon with higher energy to jump to the orbitals of higher energy. If the energy gap gets large enough, it is more favorable to fill the lower energy orbitals completely before putting an electron in the higher energy orbitals. This is the definition of a low spin complex.
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