General Chemistry 3 Advanced ReKaps Advanced CTE 1) Within a period acidity increases from left to right, so it is easier to break the H-F bond because the electrons were already spending a disproportionate amount of time with the F atom! On the other hand, looking at acidity down a group we find that HCl, HBr, and HI are strong acids but HF is not…polarity differences would favor HF being the strongest acid so another factor that didn’t matter when we compared atoms in the same period must have come into play. 2) The factor must be atomic radius, and as the atomic radius gets larger the bond strength between the halogen and the hydrogen weakens. Advanced CTE 1. pH of a 2.3 x 10-13 M H+ solution is actually 12.64. Via the shortcut it is 12.77. 2. If the pH is 7.18, then the approximate concentration is 8.2 x 10-8 (actual concentration is 6.7 x 10-8) 3. If the pH is 10.10, then the approximate concentration is 9.0 x 10-11 (actual concentration is 8.0 x 10-11) A1) B: H2S is a weak acid, thus immediately ruling out answer choices C and D. For answer A – via the rule that KA * KB (of conjugate base) = KW – if the ionization constant of a weak acid is near zero that implies that the KB of the conjugate base, which in this case is HS-, is near infinity. HS- is a weak base, thus it’s KB must also be less than one. A2) B: The addition of Mg3N2 to water resulted in the production of hydroxide ions, thus you can conclude that Mg3N2 is acting as a base. The Lewis definition of a base is a species that donates electrons, thus answer choice B fits best. A3) D: The Ksp of a solid only changes if the temperature is varied thus answer choice A can be eliminated quickly. From the dissociation reaction CaCO3 -> Ca2+ + CO32- it would appear that decreasing the concentration of H+ would have no effect on this equilibrium. However, the AAMC expects students to recognize that common weak acids and bases…and the carbonate ion CO32- is indeed a well-known weak base: CO32- + H2O -> HCO3- + OHIncreasing the pH will result in a rise of OH- ions, which via Le Chatelier’s principle will result in an increase of carbonate ions. An increase in carbonate ions will result in a Le Chateliere’s principle shift of the dissociation reaction of calcium carbonate. Specifically, it will shift the equilibrium to the left, i.e. the molar solubility of the solid will decrease. Advanced titration questions 1) Equivalents acid = (40.0 mL HCl)(10-1 mmol HCl/mL HCl)(1 meq HCl/1 mmol HCl) = 4 meq At equivalence point, must have 4 meq NaOH. Determine number of milligrams in 4 meq by using MW of NaOH:

(4 meq NaOH)(1 mmol NaOH/1 meq NaOH)(40.0 mg NaOH/mmol NaOH) = 160 mg NaOH 2) No math is necessary here. Acid and base strength is irrelevant to determining the number of equivalents required for neutralization. 4 meq of ANY acid will neutralize 4 meq of ANY base. Both titrations started with the same # of equivalents of acid, thus the answer to this question is the same as the first: 160 mg. 3) Solving for the pH of a strong acid solution or a strong base solution is fairly easy because in aqueous solution the acid and base react completely with water. Thus, in this case, all of the HCl will react with water to form H3O+ and Cl- ions. Therefore, the final concentration of H+ will be equal to the initial concentration of HCl. The initial concentration of HCl is 10-1 M, thus the final concentration of H+ will also be 10-1 M. The pH is therefore 1. 4) Ka = [H+][CH3COO-]/[CH3COOH – x] 10-5 = (x)(x)/[CH3COOH – x] 10-5 = (x)(x)/[10-1 – x] (10-1 – x ≈ 10^-1) 10-5 = (x)(x)/(10^-1) X^2 = 10^-6 x = 10^-3 = [H+] pH = 3 5) In all strong acid-strong base titrations the only product that produces H+ ions is H2O, which we know produces H+ at a concentration of 10-7 M, thus the pH will always be 7. 6) The acetate ion, seen at equilibrium, is a weak base that will react with water to produce some hydroxide ions, therefore the pH at the equivalence point will be above 7. 7) No Henderson-Hasselbach here, for this is not a buffer solution. The conjugate base of HCl is Cl-. Since Cl- is inert we do not have a buffer solution. At the half-equivalence point, the major player in determining the pH remains HCl. The initial concentration was 10-1 M, thus at the half-equivalence point the concentration will be half that number*: 10-1 M/2 = 5 * 10-2 M. The pH is therefore approximately 1.5. 8) pH = pKA + log (A- / HA) pH = 5 + log (A- / HA) At the half-equivalence point, half of the initial HA has reacted, thus [HA] = [A-], which means that the above simplifies to: pH = 5 + log (1) Since log (1) = 0, pH = 5. A4) Stop: higher, lower, same

Think: titration used to determine unknown; measured out grams of KHP and dissolved in H2O Predict: equivalents (mol as measured in g) remain unchanged so calculated concentration irrelevant matter. You would use the formula (mass/molec weight)(eq/mol) = VbNb since the source of the acid was solid, not aqueous. Match: C A5) Stop: mg sodium carbonate Think: 2 HCl neutralize 1 Na2CO3 Predict: mL HCL x mmol HCl/mL HCl x mmol Na2CO3/mmol HCl x mg Na2CO3/mmol Na2CO3 = 25.1 x 0.07520/1 x 1/2 x 106/1 = 25 x 3/40 x 1/2 x 100 = 75/80 x 100 = 100 Match: B ReDox CTEs 1) standard EMF of the cell: 1.82 – (-2.71) = 4.53 V 2) standard EMF for this electrolytic cell: -2.71 – 1.82 = -4.53 V 3) Solution: (6*10-3 mol e- )*(1 mol Zn/2 mol e-)*(65.39g Zn/1 mol Zn) = 1.96*10-1 g of Zn A6) B: First, determine the anode and cathode half-reactions. For a spontaneous cell, the reduction half-reaction that with the most positive reduction potential will occur at the cathode; therefore the relevant half-reactions are: Cathode: Ag+ + e- --> Ag (s) + 0.799 V Anode: Cu (s) --> Cu2+ + 2e- 0.337 V To balance the reaction, the cathode half-reaction must be multiplied by two: 2Ag+ + 2e- --> 2Ag (s) + 0.799 V Cu (s) --> Cu2+ + 2e- 0.337 V 2Ag+ + Cu (s) --> 2Ag (s) + Cu2+ + 0.462 V The reaction quotient is the concentration of products over reactants. A7) C: The strongest oxidant (i.e. oxidizing agent) is the species that is most likely to get reduced, thus the strongest oxidant will have the most positive reduction potential. According to Table 1, that corresponds to answer choice C. A8) B: Alkali Metal or Alkaline Earth metal (s) + water (l) metal-hydroxide (aq) + H2 (g) MINI MCAT A1) A: Choice A: You should know from your outside knowledge that conductivity is a property of metals. So if the precipitate were a metal, you would expect it to be an excellent conductor. Choice B: : For hydrogen gas to form, the following half-reaction would have to take place: 2H+ (aq) + 2e--> H2 (g). Since the following reaction is a redox reaction, you might be tempted to select choice B as the correct answer. However the reaction as written (i.e., proceeding to the right) is a reduction –

hydrogen is reduced. Hence, this reaction cannot complete an oxidation-reduction pair with the reduction of Ag+(s). Nothing gets oxidized. Choice B is incorrect. Choice C: this statement supports Student 2’s hypothesis. HPO4- having a mole fraction of 1 during the spill means the fully protonated phosphoric acid does not exist in solution – it has become deprotonated, as per Student 2’s hypothesis. Choice D: even adding 10 mL of a strong base such as NaOH, let alone a salt like AgNO3, would not significantly raise the pH of a strong acid such as sulfuric acid, H2SO4. This statement does nothing to support or detract from either hypothesis. A2) C: Excess AgNO3 – and consequently excess Ag+(aq) – would drive the deprotonation of H3PO4 to the right endpoint, leaving only H+ and NO3- ions, changing the original solution to a strong solution of nitric acid with a silver phosphate precipitate. A3) A> From your outside knowledge, you should recognize that this is the point where pH = pKa1. From the Henderson-Hasselbalch equation, pH = pKs + log( [A-]/[HA] ), you should know that at this point, [A-] = [HA]. So when the solution pH = pKa1, the acid is the fully protonated phosphoric acid, and the base is phosphoric acid’s conjugate base, H2PO4-. A4) B. At the first equivalence point, enough base has been added to neutralize the protons released from the complete reaction H3PO4(aq) -> H2PO4-(aq) + H+(aq). In 50 mL of 0.1 M H3PO4(aq), the number of protons released is 0.050 L -> ( 0.1 mol / L ) = 5.0 -> 10-3 mol. Now you can figure out the concentration of the NaOH solution, since you know there must be 5.0 -> 10-3 mol of OH- released to neutralize the acid. If the first equivalence point was reached with the addition of 10 mL of solution, then the concentration is ( 5.0 -> 10-3 mol OH- ) / 0.01 L = 0.5 M. A5) B: pH = 6.1 + log (0.14 M/1.4 M) pH = 6.1 + log (0.1) = 6.1 – 1.0 = 5.1 A6) A: In order to answer this question correctly you need to know that the quantity of charge--Q --is equal to the current in Amps--i--times the time--t--in seconds. This will give an answer in coulombs. You now have to convert from coulombs into moles of electrons using the faraday constant: 96,500 coulombs per mole of electrons, which is given to you in the question stem. Having moles of electrons, you need to convert to the number of moles of copper; this is done by looking at the second halfreaction in Table 1 and seeing that two moles of electrons give one mole of copper. You now need to divide your answer by two; this gives an answer in moles of copper. This number must now be divided by the molecular weight of copper, which is 63.5 grams and is given in the provided periodic table. Let’s stop here for a minute and look at the answer choices, hoping that we can simplify things. Choice A is the only answer choice less than one, all the others are quite a bit larger than one. Let’s take a look at our calculation: 2 amps times 10 minutes times sixty seconds divided by 96,500 coulombs divided by 2 moles of electrons times 63.5 grams. You should be able to quickly calculate that the numerator is about 72,000: 60--from the 60 seconds per minute term--times 60--the approximate atomic weight of

copper--gives 3,600; 3,600 times 20--from 2 amps times 10 minutes--gives 72,000. You can already see that the denominator is bigger than this--96,500 coulombs is down there. So, since the answer is less than one, choice A is the correct answer. A7) C: For this question, it’s now important that you know what is going on. Two important points: the applied voltage potential is 0.5 volts and the voltages of cells connected in series--which these are--are additive. So, the combined voltage of these two cells must be greater than –0.5 volts. (You should know that the potential is negative because this is an electrolytic cell, not a galvanic one.) Starting from the left, electrode number one has electrons coming into it, so a reduction is taking place. Copper is being reduced at this electrode. At electrode number 2, oxidation must, therefore, be taking place; the oxidation that is occurring is the reverse of the fourth reaction down in Table 1--oxygen gas is produced at this electrode. Because electrons are flowing into electrode number three, a reduction is taking place--the reduction of H+ to form hydrogen gas--reaction one in Table 1. Finally, at electrode number four, oxygen gas is formed by the same reaction that occurs at electrode 2. So, using the familiar Ecell = Ecathode – Eanode, the cell potential of the right-hand cell is 0.34 V--from the reduction of copper-minus 0.40--from the oxidation of hydroxide ion, giving a total of minus 0.06 V. For the left-hand cell, it’s 0 volts--from the reduction of H+--minus 0.40 V--again, from the oxidation of hydroxide ion, giving a total of minus 0.40 V. Adding the two cell voltages together, we get a total of minus 0.46 V, so the applied voltage of 0.5 V is enough to drive the cell forward. So, after all that, oxygen is formed at electrodes 2 and 4, making answer choice C the correct answer. A8) B: As was stated in the explanation to question A7, the reduction of H+ to form hydrogen gas is occurring at electrode number 3. Choice A is wrong because the applied voltage, 0.5 V, is not enough to drive the reduction of sodium ion. Choice C is wrong because this is not even a half-reaction. Choice D is wrong because this reaction is an oxidation. Remember: if electrons are coming into an electrode a reduction is occurring, not an oxidation. A9) B: In the previous two questions we determined that reduction of H+ to form hydrogen gas is occurring at electrode 3 and oxidation of hydroxide ion is occurring at electrode 4. What you need to do now is add the two half-reactions, remembering to balance the electrons of course. The reduction of H+ is the top reaction in Table 1, and hydroxide appears in the fourth reaction down, this reaction must be reversed before you add it to the hydrogen reaction. So, reversing the hydroxide equation, you can see that, in order to balance the electrons, the hydrogen equation must be multiplied by 2. Doing this and adding the two half-reactions gives 4H+ + 4OH- 2H2(g) + O2(g) + 2H2O. The ratio of hydrogen to oxygen is 2:1. A10) D: If electrons are going into an electrode, reduction is occurring; the electrode at which reduction is occurring is the cathode. From the arrows in Figure 1, electrons are entering electrodes 1 and 3; these electrodes are the cathodes A11) D: Pure water is not an electrolyte and will not conduct electricity, so an additional substance must be added to serve as an electrolyte.