Acids and Bases

A 1 M solution of which of the following compounds would have the lowest pOH? Question 8 Answer Choices A. KNH2 B. HF C. NH4I D. NaBr

A. A 1 M solution of KNH2 would have the lowest pOH of the four salts listed. As pH = 14 - pOH, very basic solutions will have low values of pOH (high pH), so this question is asking for the most basic compound. HF is a weak acid, and will yield a solution of relatively low pH and high pOH. NH4I is an acidic salt, and gives a solution of relatively high pOH. NaBr is a neutral salt, and therefore a 1 M solution will have both pH and pOH near 7. The anion of KNH2, NH2-, is a strongly basic anion, which will cause KNH2 solutions to be highly basic with low values of pOH.

Addition of sodium acetate to a solution of acetic acid will cause the pH to: Question 15 Answer Choices A. increase due to the common ion effect. B. remain constant because sodium acetate is a buffer. C. remain constant because sodium acetate is neither acidic nor basic. D. decrease due to the common ion effect.

A. Addition of sodium acetate to a solution of acetic acid will cause the pH to increase due to the common ion effect. Sodium acetate is a basic compound, because acetate is the conjugate base of acetic acid, a weak acid ("the conjugate base of a weak acid acts as a base in water"). The addition of a base to any solution, whether it is buffered or not, will increase the pH. The answer is "increase due to the common ion effect."

Because it dissolves carbon dioxide from the atmosphere to make mild carbonic acid, natural rain has a pH of around 5.5. However, due to pollutants such as sulfates from coal-fired power plants and nitrates from car exhausts, the pH of rain can drop to as low as 2. This decrease in pH represents a change in H+ concentration by approximately what factor? Question 1 Answer Choices A. 3000 B. 300 C. 3.5 D. 35

A. Because it dissolves carbon dioxide from the atmosphere to make mild carbonic acid, natural rain has a pH of around 5.5. However, due to pollutants such as sulfates from coal-fired power plants and nitrates from car exhausts, the pH of rain can drop to as low as 2. This decrease in pH represents a change in H+ concentration by approximately a factor of 3000. Each change by 1 in the pH value corresponds to a change by a factor of 10 in [H+] since pH is a logarithmic scale. Since 5.5 - 2 = 3.5, the factor change in [H+] must be greater than 103 = 1000. More precisely, the change in [H+] is 103.5 = 103 × 100.5 ≈ 1000 × 3 = 3000.

If a solution with a pH of 11 is diluted with pure water until the volume is increased by a factor of 10, the pH will then equal: Question 9 Answer Choices A. 10 B. 9 C. 12 D. 1

A. If a solution with a pH of 11 is diluted with pure water until the volume is increased by a factor of 10, the pH will then equal 10. Each pH unit corresponds to a 10-fold change in the concentration of H+ (or OH-). Therefore, diluting the basic solution by a factor of ten would result in a solution one pH unit closer to neutrality.

Red wines are generally acidic, with proper pH values near 3.6. At this pH, which of the following organic acids commonly found in wine have the highest proportion of molecules in the -1 charge-state? (note: pKa1 denotes the pKa of the most acidic proton on the compound; the pKa values of all subsequent protons on the compounds of interest are all near pH = 6) Question 13 Answer Choices A. Maleic acid (pKa1 = 1.90) B. Citric acid (pKa1 = 3.09) C. Succinic acid (pKa1 = 4.05) D. Tartaric acid (pKa1 = 2.95)

A. Red wines are generally acidic, with proper pH values near 3.6. At this pH, maleic acid will have the highest proportion of molecules in the -1 charge-state. As this is a question asking for extremes, it is safe to eliminate tartaric and citric acids for having the non-extreme pKa values. The Henderson-Hasselbalch equation states that pH = pKa + log ([A-]/[HA]), where A- is the molecule in the -1 charge state, and HA is the protonated, neutral molecule. Rearranging this equation and substituting the given pH yields: log ([A-]/[HA]) = 3.6 - pKa. The acid that gives the largest positive number for the right side of this equation is the one with the smallest pKa, maleic acid.

Which of the following statements is most accurate regarding the titration shown below? A. A weak base is being titrated by a strong acid, and this base has a pKb of 4.2. B. A weak acid is being titrated by a strong base, and this acid has a pKa of 4.2. C. A weak base is being titrated by a strong acid, and this base has a pKa of 4.2. D. A weak acid is being titrated by a strong base, and this acid has a pKb of 4.2.

A. Regarding the titration shown below, it is most accurate that a weak base is being titrated by a strong acid, and this base has a pKb of 4.2.This is a two-by-two question. Since pOH starts low and increases (i.e., pH decreases) as titrant is added, the unknown is a base and the titrant is an acid (choices stating the titrant is a base can be eliminated). Since it is pOH that is measured, the pKb can be determined at the half-equivalence point of the graph. This is the buffer region occurring at a pOH of approximately 4, making "a weak base is being titrated by a strong acid, and this base has a pKb of 4.2" the correct answer.

What is the Brønsted-Lowry base in the following reaction? HCl(g) + H2O(l) H3O+(aq) + Cl-(aq) Question 15 Answer Choices A. H2O B. Cl- C. HCl D. H3O+

A. The Brønsted-Lowry base in the following reaction is H2O. HCl(g) + H2O(l) H3O+(aq) + Cl-(aq) A Brønsted-Lowry base is a molecule or ion that is a H+ acceptor. Based on the reaction given in the question, H2O accepts H+. (Note that if there were reversible arrows—which there are not—the answer choice Cl- could have also been correct.)

The resulting solution made from the combination of 50 mL of 1.0 M LiOH with 50 mL of 1.0 M HBr will be identical in all respects to 100 mL of: Question 13 Answer Choices A. 0.5 M LiBr. B. 1 M LiBr. C. a saturated solution of LiBr(s). D. 2 M LiBr.

A. The resulting solution made from the combination of 50 mL of 1.0 M LiOH with 50 mL of 1.0 M HBr will be identical in all respects to 100 mL of 0.5 M LiBr. The best way to approach this question is to ignore OH- and H+, because they will neutralize each other when mixed. Thus, in this question we are mixing 50 mL of 1.0 M Li+ and 50 mL of 1.0 M Br-. Since the solution volume is doubling, the concentration of each ion is expected to decrease by a factor of 2. Therefore, the correct choice is "0.5 M LiBr".

Which of the following best accounts for the difference in acid strength between H2S and H2Se? Question 6 Answer Choices A. Increased stability is observed in HSe- compared to HS-. B. The H—S bond is weaker than the H—Se bond. C. Reactivity of HSe- is increased relative to HS-. D. The bond length observed in H—Se is shorter than that of H—S.

A. At equilibrium, the side of the reaction with the greatest stability predominates. In this instance, if HSe- is more stable than HS-, the dissociation of H2Se will occur to a greater extent and helps explain why H2Se is a stronger acid (choice A is correct, choice C is not correct). A weaker bond between hydrogen and sulfur would not help explain how H2Se is a stronger acid (eliminate choice B), and bond length would have limited impact on acid strength (eliminate choice D).

Thallium-201 decays by electron capture. What are the atomic and mass numbers of the daughter nucleus, respectively? Question 14 Answer Choices A. 80, 201 Correct Answer (Blank) B. 82, 201 C. 81, 200 D. 81, 202

A. Electron capture involves the addition of an electron to the nucleus with the conversion of a proton to a neutron. This results in mercury-201 as the product, making choice A correct.?

In the span of two hundred minutes, 30 grams of copper is plated on a sample of iron with a constant current of 3.8 A. How many moles of electrons are transferred per mole of plated copper? Note: F = 96,500 C/mol e- Question 31 Answer Choices A. 1 mole B. 2 moles C. 3 moles D. 4 moles

A. Given I = C/t, we can rearrange to get C = I/t where I is current (in amps), C is a measure of charge (in coulombs), and t is time (in seconds). (Choice A is correct)

Acetylsalicylic acid, also known as aspirin, possesses a pKa of 3.5. What is the pH of a 1 M solution of acetylsalicylic acid? Question 21 Answer Choices A. 1.8 Correct Answer (Blank) B. 3.5 C. 5.3 D. 7

A. Here we are asked to find the pH of a solution containing a weak acid (HA). To do this, we can use an ICE table to find the solution. Initially, there is only 1 M HA present and some unknown quantity 'x' will dissociate and form equal quantities 'x' of H+ and A-. This results in equilibrium concentrations of 1 - x, x, and x for HA, H+, and A-, respectively. We will make the standard assumption that 1 is substantially larger than x and find that Ka = 10-3.5 = x2/1. Thus we find that x = (10-3.5)0.5 = 10-1.75 = [H+] and the pH = 1.75, making choice A correct.

What percentage of HF (pKa = 3.17) dissociates in one liter of solution containing two moles of the acid? Question 8 Answer Choices A. 1.8% B. 32% C. 72% D. 98%

A. Here we are asked to find the percentage of a weak acid (HF) which has dissociated. To do this, we can use an ICE table. Thus choice B is correct.

If the equation below represents an overall reaction in an electrochemical cell, which of the following materials likely constitutes the anode? 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) Question 18 Answer Choices A. Cu(s) B. CuNO3(s) C. NO3-(aq) D. H+(aq)

A. The anode is the site of oxidation (an ox) and in the above reaction, copper is oxidized (loses an electron) from Cu(s) to Cu2+(aq) making Cu(s) the anode (choice A is correct). While nitrate is the species being reduced to nitrogen monoxide, this would be taking place at the cathode (eliminate choice C). Hydrogen is neither oxidized nor reduced in the reaction (eliminate choice D) and CuNO3 is the mono-oxidized product of the Cu anode (eliminate choice B).

Which of the following pairs will generate an electrolytic cell with the least stable products? Al3+ + 3 e− → Al(s) E° = -1.68 V Cd2+ + 2 e− → Cd(s) E° = -0.40 V Ag+ + e− → Ag(s) E° = 0.80 V Br2(l) + 2 e− → 2Br- E° = 1.08 V Question 5 Answer Choices A. Al3+ and Ag(s) B. Br2(l) and Al(s) C. Cd2+ and Br- D. Ag+ and Cd(s)

A. The electrolytic cell with the most negative electrical potential will have the least stable products (choice A is correct). To determine the overall potential of any given cell, the reduction and oxidation half-reactions are added. To find the oxidation half-reactions, the reduction half-reaction in question can be reversed and its potential reversed in sign. In this instance to find the oxidation half-reaction, the reduction of silver above can be reversed to Ag(s) → Ag+ + e− E° = -0.80 V which can then be add to the reduction half-reaction Al3+ + 3 e− → Al(s) E° = -1.68 V. Note that while the oxidation half-reaction is multiplied 3 to appropriately balance the reaction, this does not impact its electrical potential. The overall electrical potential can be found by adding the two half-reaction potentials: -1.68 V + (-0.80 V) = -2.49 V.

Equal moles of HCl and NaOH are mixed together in an aqueous solution. Which of the following is a true statement if solid NaCl is added to the subsequent solution? Question 6 Answer Choices A. The pH of the solution remains at 7. B. The pH of the solution decreases below 7. C. A buffer solution will be formed. D. The pH of the solution increases above 7.

A. The pH of the solution remains at 7.

Compared to a 50 mL solution of a 1 M CH3CO2H, the complete neutralization of 50 mL of a 1 M HCl solution will require: Question 3 Answer Choices A. the same amount of base. B. less base. C. an indeterminable amount of base given the information provided. D. more base.

A. the same amount of base.

4-Nitrophenol (pKa = 7.08) is an indicator that is yellow when deprotonated and colorless when protonated in solution. If a solution has a pOH of 10, its color will be: Question 3 Answer Choices A. indeterminable from the information given. B. colorless. C. light yellow because both acid and conjugate base will be present in equal amounts. D. yellow.

B. 4-Nitrophenol (pKa = 7.08) is an indicator that is yellow when deprotonated and colorless when protonated in solution. If a solution has a pOH of 10, its color will be colorless. Since pH + pOH = 14, we know that the pH of the solution in question is 4. Since this pH is less than the pKa of the indicator, the 4-nitrophenol will be protonated. Therefore, the solution will be colorless.

A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. Which of the following best approximates the pH of the resultant solution? (pKb of SCH3- = 3.6) Question 18 Answer Choices A. -log (1 × 10-3.6) B. -log (1 × 10-5.2) C. -log (2 × 10-5.2) D. -log (1 × 10-1.8)

B. A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. -log (1 × 10-5.2) best approximates the pH of the resultant solution. (pKb of SCH3- = 3.6) The result of the titration will be a 1 M solution of methylthiol, HSCH3 since the initial 2 M solution has been doubled in volume during the titration. Since we know the pKb of its conjugate base SCH3- is 3.6, then we know the pKa of HSCH3 is 10.4. Therefore, at equilibrium: 10-10.4 = (x)(x)/(1-x) Where x = [H+] = [SCH3-], at equilibrium. As the resultant value of x is negligibly small compared to 1, we ignore x in the denominator and say that x = [H+] = [SCH3-] = (10-10.4)(0.5) giving 1 × 10-5.2. The answer -log (2 × 10-5.2) would be correct if the resultant solution was 2 M in concentration. The answer -log (1 × 10-1.8) is the result of using the pKb of the methylthiolate anion in the calculation rather than the pKa of HSCH3, and -log (1 × 10-3.6) would be the result of a similar a calculation using this pKb value, while not taking the square root.

A salt solution contains ammonium cation (Ka = 5.7 × 10-10) and nitrite anion (Kb = 1.4 × 10-11) in equal proportions. The pH of the solution will be: Question 6 Answer Choices A. acidic because nitrite has a higher Kb value than the Ka of ammonium. B. acidic because ammonium has a higher Ka value than the Kb of nitrite. C. basic because ammonium has a higher Ka value than the Kb of nitrite. D. basic because nitrite has a higher Kb value than the Ka of ammonium.

B. A salt solution contains ammonium cation (Ka = 5.7 × 10-10) and nitrite anion (Kb = 1.4 × 10-11) in equal proportions. The pH of the solution will be acidic because ammonium has a higher Ka value than the Kb of nitrite. The Ka of ammonium is greater than the Kb of nitrite. Eliminate choices stating "nitrite has a higher Kb value than the Ka of ammonium." The fact that ammonium has a higher Ka means that it, in general, is in a more dissociated state than the nitrite is in a protonated state. Therefore, the solution will be acidic.

Ethylenediaminetetraacetic acid (EDTA) is a commonly used scavenger of heavy metals. The molecule has 4 carboxylic acid moieties, with successive pKa values of 1.99, 2.67, 6.16, and 10.26. At neutral pH, what is the predominant charge state of EDTA? Question 17 Answer Choices A. -1 B. -3 C. -2 D. -4

B. At neutral pH, the predominant charge state of EDTA is -3. The Henderson-Hasselbalch equation (pH = pKa + log ([A-]/[HA])) shows that the deprotonated form of a compound is favored when pH > pKa. At neutral pH this is true for three of the four acidic protons on EDTA, giving a charge state of -3.

Enough HF (Ka = 7.4 × 10-4) is added to water to create a pH = 2.1 solution. The addition of which of the following would have the least impact on the pH of this solution? Question 2 Answer Choices A. NH4F B. PbF2 C. Na2CO3 D. NaF

B. Enough HF (Ka = 7.4 × 10-4) is added to water to create a pH = 2.1 solution. The addition of PbF2 would have the least impact on the pH of this solution. Both NaF and NH4F are soluble salts, and will decrease the solubility (acidity) of HF due to the common ion effect, so both of these answer choices can be eliminated. Na2CO3 is a soluble salt and CO32- is the conjugate base of a weak acid, so it will act to increase the pH. PbF2 is essentially insoluble in water, and therefore will have the least effect on the pH, making it the best answer.

Which of the following acids (in conjunction with an equal amount of its conjugate base) would be most appropriate to use in order to maintain a buffer solution of pH 7.4 at 25°C? Question 1 Answer Choices A. H2CO3 (pKa = 6.35) B. H2PO4- (pKa = 7.21) C. HNO2 (pKa = 3.35) D. HCO3- (pKa = 10.33)

B. H2PO4- (pKa = 7.21) would be the most appropriate acid (in conjunction with an equal amount of its conjugate base) to use in order to maintain a buffer solution of pH 7.4 at 25°C. In general, it is best to choose a buffering system for which the pKa of the acid is as close as possible to the desired pH. Among the given choices, the best acid to use is the dihydrogen phosphate ion, H2PO4-, since its pKa is closest to 7.4.

The indicator bromothymol blue has a pKa near the pH of neutral water. Its basic form is a deep blue whereas its acidic form is a pale yellow. Near pH = 7 the indicator is green. If a solution of CsOH was slowly added to a solution of bromothymol blue in distilled water, what color change would be observed? Question 9 Answer Choices A. Yellow to blue B. Green to blue Correct Answer (Blank) C. Green to yellow D. Blue to green

B. If a solution of CsOH was slowly added to a solution of bromothymol blue in distilled water, the solution would change from green to blue. When bromothymol blue is initially added to distilled water (pH = 7), the solution is green. As the base CsOH is added to solution, the pH of the solution increases and the basic form of the indicator dominates, giving a blue color. A change to yellow, or a change from blue to green, would indicate a decrease in pH.

Which of the following statements is true about a solution of acetic acid in equilibrium? CH3CH2COOH (aq) H+ (aq) + CH3CH2COO- (aq) Question 7 Answer Choices A. When HI is added, CH3CH2COOH acts as a Brønsted-Lowry acid. B. When HCl is added, CH3CH2COO- acts as a Brønsted-Lowry base. C. When NaOH is added, CH3CH2COOH acts as a Lewis acid. D. When HF is added, CH3CH2COOH acts as an Arrhenius base.

B. It is true about a solution of acetic acid in equilibrium that when HCl is added, CH3CH2COO- acts as a Brønsted-Lowry base. CH3CH2COOH (aq) <--> H+ (aq) + CH3CH2COO- (aq) In the presence of a strong acid, CH3CH2COO- will behave as a Brønsted-Lowry base by accepting a proton and the equilibrium will shift to the left. In the presence of a strong base, H+ and not CH3CH2COOH will behave as a Lewis acid and accept electrons from OH-. CH3CH2COOH does not contain the hydroxide ion, even though the formula for a carboxylic acid (COOH) might suggest that. Therefore, it cannot behave as an Arrhenius base.

Which of the following best characterizes the ionization constant of a strong acid? Question 14 Answer Choices A. 0 < Ka < 0.1 B. Ka > 1 C. Ka < 0 D. 0.1 < Ka < 1

B. Ka > 1 best characterizes the ionization constant of a strong acid. A strong acid dissociates virtually completely in aqueous solution; therefore, the concentrations of A- and H3O+ should both be much greater than the concentration of undissociated HA, which implies a large value of Ka, certainly greater than 1.)

The acidic strength of HF is greatly enhanced when it is used in conjunction with SbF5. The resultant system, known as "magic acid", is one of the strongest acids known, and follows the dissociation equation below. HF + SbF5 → H+ + SbF6- Which of the following best describes the role of SbF5 in magic acid? Question 12 Answer Choices A. Lewis base B. Lewis acid C. Brønsted base D. Brønsted acid

B. SbF5 acts as a Lewis acid in the dissociation of magic acid. SbF5 accepts (not donates, eliminating Lewis base) an electron pair on F- from dissociating HF, thus driving this dissociation reaction toward nearly full dissociation. SbF5 neither accepts nor donates H+, thus it acts as neither a Brønsted acid nor base.

The Ka of formic acid (HCOOH) is 1.8 × 10-4. What is the pKb of the formate ion? Question 16 Answer Choices A. -14 - log (1.8 × 10-4) B. 14 + log (1.8 × 10-4) C. -14 + log (1.8 × 10-4) D. 14 - log (1.8 × 10-4)

B. The Ka of formic acid (HCOOH) is 1.8 × 10-4. The pKb of the formate ion is 14 + log (1.8 × 10-4). For a conjugate pair, such as formic acid and the formate ion, the product of the Ka of the acid and the Kb of the base is equal to 10-14. The equation KaKb = 10-14 is equivalent to the equation pKa + pKb = 14. Therefore, pKb = 14 - pKa = 14 - [-log (1.8 × 10-4)]=14 + log (1.8 × 10-4)

The conjugate bases of HSO4-, CH3OH, and H3O+ are, respectively: Question 18 Answer Choices A. SO4-, CH2OH-, H2O B. SO42-, CH3O-, H2O C. SO4-, CH3O-, OH- D. SO42-, CH2OH-, OH-

B. The conjugate bases of HSO4-, CH3OH, and H3O+ are, respectively SO42-, CH3O-, H2O. The conjugate base of a chemical species is simply that species after it has lost an H+. Therefore, the conjugate base of HSO4- is SO42- (eliminating "SO4-, CH2OH-, H2O" and "SO4-, CH3O-, OH-"), and the conjugate base of H3O+ is just H2O (eliminating "SO42-, CH2OH-, OH-"). The most acidic proton on CH3OH is the hydroxyl proton, so CH2OH is the incorrect conjugate

Which one of the following statements concerning 50 mL of basic solution composed of 0.2 M NaH2PO4 and 0.2 M Na2HPO4 is correct? Question 6 Answer Choices A. The concentration of Na+ is 0.4 M. B. The pH of the solution will not change with the addition of 20 mL of water. C. The addition of 1 mL of 0.1 M HCl to the reaction will result in a pH of 2. D. With the addition of ammonia, a precipitate should be formed

B. The correct statement concerning 50 mL of basic solution composed of 0.2 M NaH2PO4 and 0.2 M Na2HPO4 is the pH of the solution will not change with the addition of 20 mL of water. This solution is a buffer. The pH of a buffer does not change with the addition or removal of water.

What is the expression for the acid dissociation constant (Ka) of sulfurous acid?

B. The expression for the acid dissociation constant (Ka) of sulfurous acid is Ka = . The acid dissociation constant, like any equilibrium constant, is the product of the concentrations of products divided by the product of the concentrations of reactants, raised to the powers of their coefficients from the balanced chemical equation. The dissociation reaction of sulfurous acid (H2SO3) should show loss of only one proton (H2SO3 H+ + HSO3-), so the Ka expression must reflect that. Two choices are incorrect because the reactant is HSO3-, and not H2SO3. Finally, since the ratio of the three species is 1:1:1, the H+ concentration should not be squared in the equilibrium expression.

Which one of the following graphs best represents the titration of a weak base with HCl?

B. The graph that best represents the titration of a weak base with HCl is: Graphs A and C can be eliminated right away since we begin with a base; the initial pH must be greater than 7. Since the equivalence point of the titration curve in D occurs below pH 7, this must be the curve for a weak base titrated with strong acid. (A strong base titrated with a strong acid would have its equivalence point at pH 7, like the curve in B.) The graphs have been shown below for reference.

Which of the following combinations will form a solution with the greatest buffering capacity? Question 5 Answer Choices A. 0.2 M HCl and 0.2 M NaCl B. 0.2 M CH3CO2H and 0.2 M NaCH3CO2 C. 0.02 M CH3CO2H and 0.02 M NaCH3CO2 D. 0.02 M HCl and 0.02 M NaCl

B. 0.2 M CH3CO2H and 0.2 M NaCH3CO2

What is the pH of a solution with a hydronium (H3O+) ion concentration of 3 × 10-1 M? Question 1 Answer Choices A. 2.0 B. 0.5 C. 1.5 D. 1.0

B. 0.5

In the formation of CO2 through the reaction of O2 and C(s), graphite acts as a(n): Question 12 Answer Choices A. oxidizing agent. B. reducing agent. C. catalyst. D. intermediate.

B. Carbon (graphite) begins as a solid with an oxidation state of 0 and ends as carbon dioxide with an oxidation state of +4. This reaction is therefore an oxidation of carbon, which indicates that it was functioning as a reducing agent (choice B is correct, choice A is not). Graphite is a starting reagent (so choice D can be eliminated) and is consumed in the process which means it cannot be a catalyst (eliminate choice C).

What is the relationship between the mass of an atom and the sum of the masses of its constituent particles? Question 28 Answer Choices A. The mass of the atom is greater. B. The sum of the masses of the constituent particles is greater. C. Both masses will be equal. D. It depends on the number of particles comprising the atom.

B. From a stability perspective, the bound atom has to have a lower energy compared to the constituent particles (or else nature would prefer free protons and neutrons moving in space). From Einstein's equation, E = mc2, this difference in energy translates into a mass defect in the bound atom (albeit a very small difference). In other words, the mass of the atom is smaller than the sum of the masses of its constituent atoms because the energy of the bound atom is smaller than the individual particles. Thus, choice B is correct.

In a fully charged lead-acid battery, metallic Pb is one of the two electrode materials. Which of the following redox reactions is possible during recharge of a spent lead-acid battery? Question 7 Answer Choices A. Pb → PbSO4 + 2e− B. PbSO4 + 2e− → Pb C. Pb → PbO2 + 4e− D. Pb(H2O)4 + 2e− → Pb

B. If Pb0 is an electrode material in the fully charged battery, then it will be oxidized during discharge since anionic Pb is not feasible. The recharge reaction must then be the reduction of a cationic Pb species to Pb0 (choice B is the correct answer). Eliminate choices A and C as they do not produce Pb0, but describe the oxidation of Pb0, which would take place during discharge. Choice D is not an electrically balanced equation as both lead species are in the 0 oxidation state.

Which of the following salts would yield acidic aqueous solutions when dissolved in water? NH4ClO4 AgNO3 NaC2H3O2 Question 24 Answer Choices A. I only B. I and II only C. III only D. I and III only

B. If the cation of a salt is the conjugate acid of a weak base, it will make a solution acidic. Similarly, if the anion of a salt is the conjugate base of a weak acid, it will make a solution basic. Ammonium perchlorate, NH4ClO4, contains the conjugate base (ClO4-) of a strong acid, which will have no reaction with water, and a conjugate acid (NH4+) of the weak base, NH3, which should make the solution acidic. Roman numeral I is therefore true (eliminate choice C). Silver nitrate, AgNO3, also contains the conjugate base (NO3-) of a strong acid, and its cation is reactive with water to form an insoluble hydroxide. Since Ag+ pulls hydroxide out of the solution, it lowers the pH of the solution by producing excess hydrogen ions (Ag+ + H2O → AgOH + H+). Roman numeral II is therefore also true (eliminate choices A and D). The anion in sodium acetate (C2H3O2-) is the conjugate base of a weak acid and will therefore make the solution basic according to the following hydrolysis equation: C2H3O2- + H2O → HC2H3O2 + OH-. The Na+ ion is not reactive with water since it is from the strong base NaOH.

Which of the following can be deduced from the following reaction? 2 Fe2+ + H2O2 + 2 H+ → 2 Fe3+ + 2 H2O E°= 1.0 V Question 17 Answer Choices A. The reaction is nonspontaneous. B. At equilibrium, the products are favored. C. The products are less stable than the reactants. D. The oxidation half-reaction is more spontaneous than the reduction half-reaction.

B. Reaction potentials provide much of the same information as ΔG and Keq. As this reaction has a positive potential, it is spontaneous and the products must therefore be more stable than the reactants and favored at equilibrium (choice B is correct while choices A and C are eliminated). The overall reaction cannot directly tell us which of the two half-reactions is more spontaneous (choice D is incorrect).

Given the titration curve for H2S below, at what pH would an equal concentration of HS- and H2S be observed? A. 4.0 B. 7.0 C. 9.5 D. 12

B. The concentration of an acid and its conjugate base are equal at half-equivalence points (where pH of the solution = pKa of the acid). Half-equivalence points can be determined by first identifying each equivalence point in a titration curve (each vertical portion of the trace). Half-equivalence points are at the middle of the flat portions of the graph directly to the left of each equivalence point. In this instance, we are being asked about the first proton dissociating from hydrosulfuric acid, so we are interested in the first half-equivalence point which can be found at a pH of 7.0 (choice B is correct).

Which of the following is a contributing factor to the relative acidity of HF compared to HCl? Question 8 Answer Choices A. HCl exhibits stronger hydrogen bonding. B. HF has a greater boiling point. C. F- is less stable in solution than Cl-. D. HCl has a shorter bond length.

C. A contributing factor to the relative acidity of HF compared to HCl is that F- is less stable in solution than Cl-. All hydrogen halide compounds are strong acids except for HF. There are several reasons for this. HF has significant hydrogen bonding between molecules which hinders the proton from contributing to acidity. None of the other hydrogen halides have hydrogen bonding. HF is the smallest of all the hydrogen halides, which increases the electrostatic interaction between hydrogen and fluorine according to Coulomb's law. HF does have a high boiling point due to its relatively strong intermolecular forces, but this does not directly affect acidity. Since HF is much less likely to dissociate into H+ and F- compared to the dissociation of HCl into H+ and Cl-, F- must be less stable as an ion in solution compared to Cl-.

Chlorosulfonic acid (HSO3Cl) is stronger than sulfuric acid when solvated in water, yet it is far less acidic in common organic solvents. When 1.16 g of HSO3Cl is dissolved in 10 mL of hexane, spectroscopic methods determine the [SO3Cl-] to be 2.2 × 10-3 M. Which of the following is closest to the pKa of chlorosulfonic acid in hexane? Question 11 Answer Choices A. 6.2 B. 2.8 C. 5.4 D. 0.2

C. When 1.16 g of HSO3Cl is dissolved in 10 mL of hexane, spectroscopic methods determine the concentration of [SO3Cl-] to be 2.2 × 10-3 M. The pKa of chlorosulfonic acid in hexane is closest to 5.4. The molecular weight of HSO3Cl is ~116 g/mol, so the initial concentration for HSO3Cl in 10 mL of hexane is 1 M. If [SO3Cl-] is 2.2 × 10-3 M, then the Ka ([SO3Cl-][H+]/[HSO3Cl]) in question can be expressed as (2.2 × 10-3)2/(1 - 2.2 × 10-3). The denominator can be safely simplified to 1, giving Ka = ~4 × 10-6. Since -log (4 × 10-6) is between 5 and 6, 5.4 is the only possible answer.

Which of the following is an amphoteric species? Question 2 Answer Choices A. H+ B. S2- C. HS- D. H2S

C. HS- is an amphoteric species. An amphoteric species is one that can act as either an acid or a base. The dissociation of hydrogen sulfide (H2S) is shown below: H2S <--> H+ + HS- <--> H+ + S2- Of these species, H2S is deprotonated in the reaction, and therefore acts as an acid. S2- is the final deprotonated product, and is therefore the final base produced by the full dissociation. However, since HS- can be protonated or deprotonated, it is amphoteric.

The Ka of a buffer is 4.5 × 10-4. If the concentration of undissociated weak acid is equal to the concentration of the conjugate base, the pH of this buffer system is between: Question 16 Answer Choices A. 5 and 6. B. 2 and 3. C. 3 and 4. D. 4 and 5.

C. The Ka of a buffer is 4.5 × 10-4. If the concentration of undissociated weak acid is equal to the concentration of the conjugate base, the pH of this buffer system is between 3 and 4. The pH of a buffer can be calculated using the Henderson-Hasselbalch equation, pH = pKa + log ([conjugate base]/[conjugate acid]) When the concentrations of the acid and base are equal, the fraction on the right-hand side is 1; since log 1 = 0, we have pH = pKa. If the Ka of the acid is 4.5 × 10-4, which is between 10-4 and 10-3, the pKa (and, therefore, the pH) is between 3 and 4.

Which one of the following best approximates the pH of a solution when 99.9% of the acid in a pH 1 solution is neutralized? Question 5 Answer Choices A. 7 B. 1 C. 4 D. 6.7

C. The pH of a solution when 99.9% of the acid in a pH 1 solution is neutralized is 4. The neutralization of 99.9% of acid tells us that for every one thousand H+ originally in solution, only one remains. A change in [H+] by a factor of 103 corresponds to a change of 3 pH units. If the initial pH was 1.0, then the final pH must be 4.

Peroxide concentration in a sample is often determined by redox titration with I−. The following equation describes the relevant reaction: 2 H+ + H2O2 + 2 I− → I2 + 2 H2O What was the initial concentration of peroxide in a 100 mL sample if 4.2 mL of a 1 M solution of NaI was required to reach the endpoint? Question 11 Answer Choices A. 0.0042 M B. 0.042 M C. 0.021 M D. 0.0021 M

C. 4.2 mL of a 1 M solution of NaI contains 0.0042 moles of I−. Since two moles of I− are required for each peroxide molecule, there must have initially been 0.0021 mol of peroxide (eliminate choices A and B). In 100 mL this amounts to a concentration of 0.021 M (eliminate choice D).

The normal range for the pH of blood is between 7.35-7.45. In the absence of blood's normal bicarbonate buffer system, which pair of compounds (in an equimolar mixture) would buffer blood at the same pH? Question 15 Answer Choices A. HCO2H (pKa = 3.7) and HCO2- B. HF (pKa = 4.8) and F- C. HSO3- (pKa = 7.2) and SO32- D. H3BO3 (pKa = 9.3) and H2BO3-

C. Each of these conjugate pairs will help maintain the pH of a solution containing them around the listed pKa values of the acids. Buffers are most effective one point on either side of their pKa. The buffer system with the acid pKa nearest the physiological pH of blood is HSO3- and SO32- (choice C is correct).

Which of the following best characterizes electron flow in an electrolytic cell? Question 9 Answer Choices A. Electrons flow from anode to cathode due to the intrinsic potential difference between the electrode materials. B. Electrons flow from cathode to anode due to the intrinsic potential difference between the electrode materials. C. Electrons flow from anode to cathode due to the driving force provided by an external source. D. Electrons flow from cathode to anode due to the electrostatic attraction of the respective poles.

C. Electrons always flow from anode to cathode (eliminate choices B and D) in galvanic or electrolytic cells. In galvanic cells, this flow of electrons occurs spontaneously as electrons flow down their potential gradient. For an electrolytic cell (which we are being asked about here), the intrinsic potential difference between the materials that constitute the anode and cathode is negative, indicating the given reaction is nonspontaneous and the flow of electrons from anode to cathode requires assistance from an external source, such as a battery (choice A is wrong and choice C is correct).

In which of the following redox reactions are exactly six moles of electrons transferred? Question 20 Answer Choices A. 5 SnCl2 + 2 KMnO4 + 16 HCl → 2 MnCl2 + 5 SnCl4 + 8 H2O + 2 KCl B. 2 FeCl3 + Ti(s) → TiCl2 + 2 FeCl2 C. 3 Pd(s) + Na2Cr2O7 + 14 HNO3 → 3 Pd(NO3)2 + 2 Cr(NO3)3 + 7 H2O + 2 NaNO3 D. 2 H2SO3 + H2O2 → H2S2O6 + 2 H2O

C. For any redox reaction, the easiest way to determine the total number of moles of transferred electrons is to determine the change in oxidation states of a given element over the course of the reaction, then multiply by the total number of moles of the element required in the reaction. Each reaction will have two such substances (one that gains electrons and one that loses them), and either substance will work. Choice A shows 5 mol Sn changing from a +2 to a +4 oxidation state, and 2 mol Mn7+ changing to Mn2+, both 10 mol e- transfers (eliminate choice A). Choice B shows 2 mol Fe going from a +3 to a +2 oxidation state, and 1 mole of Ti0 going to Ti2+, both 2 mol e- transfers (eliminate choice B). Choice C shows 3 mol Pb0 going to Pb2+, and 2 mol Cr6+ going to Cr3+, both 6 mol e- transfers. Choice D shows 2 mol S change from a +4 to a +5 oxidation state, while 2 mol O changes from -1 to -2 oxidation states, both 2 mol e- transfers (eliminate choice D).

Question 26 Based on the titration curve below, which of the following is the most likely analyte? A. HF (pKa = 3.2) B. H2SO3 (pKa = 1.8) C. NaOCl (pKb = 6.5) D. NaC2H3O2 (pKb = 9.3)

C. Here we are presented with a titration curve where we know neither the titrant (solution of known composition and concentration) nor the analyte (solution being analyzed). Since the solution begins in the basic range, we can conclude the analyte to be an basic (choices A and B are incorrect). Furthermore, at the equivalence point (the most vertical section of the graph) the pH of the solution is below seven, indicating that a strong acid was mixed with a weak base. The half-equivalence point is approximately 7.5, and should match the pKa of the conjugate acid of our basic analyte. After determining the pKa values of the conjugate acids from the listed pKb values for choices C and D, we find that NaOCl would have a half equivalence point nearest 7.5 (choice C is correct).

A battery stops functioning shortly after its initial use. What observation could explain the sudden failure of a battery? Question 13 Answer Choices A. Increased solid formation at the cathode B. Etching visible on the anode C. Fracture of the anode/cathode barrier D. A slight color change in one of the chambers

C. Mixing of the anode and cathode chambers in a galvanic cell would result in rapid battery death. When the reactants are separated, electrons are forced to follow a specified path to provide useful electrical energy. Breaking this barrier means electrons quickly transfer from reducing agent to oxidizing agent directly, making choice C the best answer. Solid formation at the cathode, etching visible on the anode, and color changes may be part of normal battery function, so we cannot be certain they resulted in battery death (eliminate choices A, B and D).

Question 22 Which of the following would exhibit a color change only at the end of the titration of 20 milliliters of 0.75 M NaOH into 60 milliliters of 0.25 M HClO4? Question 22 Answer Choices A.Thymol blue (pKa = 1.7) B. Methyl orange (pKa = 3.5) C. Bromothymol blue (pKa = 7.1) D. Phenolphthalein (pKa = 9.7)

C. The number of moles of the acid and base can be determined by multiplying the molarity and the volume. Thus 0.75 M NaOH * 0.020 L gives 0.150 mol NaOH and 0.25 M HClO4 * 0.060 L gives 0.150 mol HClO4. The neutralization reaction here involves equal quantities of a strong acid and strong base resulting in a neutral solution (pH = 7). The appropriate indicator will not change color until all the acid has been neutralized by the addition of the full amount of base. An indicator's given color is dictated by its protonation state, since it is itself a weak acid. When the solution pH approaches that of the pKa of the indicator, the protonation state of the compound changes and the color of the indicator changes. Therefore, we will select an indicator with a p?Ka nearest the pH we wish to see a color change (choice C is correct).

Twenty milliliters of a 8.4 x 10-3 M solution of HBr is added to sixty milliliters of water. What is the final pH of the solution? Question 16 Answer Choices A. 0.6 B. 1.8 C. 2.7 D. 3.3

C. This problem requires us to perform a dilution before calculating the pH. Using C1V1 = C2V2, we find our final concentration to be 2.1 x 10-3 M. Since we are dealing with a strong acid, the concentration of hydronium ion is the same as the molarity. To find the pH, we need only find -log [H+] and find our answer to be between a pH of 2 and 3 after appropriate rounding to 1 x 10-2 M and 1 x 10-3 M (choice C is correct).

Mixing which of the following with an equal volume of 1 M HCl would generate a final solution with pH 7? Question 29 Answer Choices A. 1 M H2SO4 B. 1 M Mg(OH)2 C. 1 M KOH D. 1 M CH3COOH

C. This question is asking about a neutralization reaction. If equal quantities of the acid and base in question are mixed and generate a pH of 7, the two species have to be a strong acid and a strong base (if it were a weak acid and a strong base, the result would have been a basic solution, while if it were a strong acid and a weak base, the result would have been an acidic solution). Choices A and D are acids, so should be eliminated. Between choices B and C, only KOH is a strong base, and moreover has one equivalent of hydroxide to neutralize the hydrogen ions, making choice C correct.

A mixture of acidic wastes is found to be 2 M H2SO4, 0.5 M HI, and 3 M HNO3. What volume of 5 M NaOH solution will be required, per liter of acidic waste, to completely neutralize the acid? Question 12 Answer Choices A. 3.0 L B. 1.5 mL C. 1.1 L D. 1.5 L

D. A mixture of acidic wastes is found to be 2 M H2SO4, 0.5 M HI, and 3 M HNO3. It will require 1.5 L of a 5 M solution of NaOH to completely neutralize 1 L of the acidic waste in question. The easiest way to answer this question is to determine the molarity of H+ in the acidic wastes. Accounting for all dissociable protons there are 7.5 moles of H+ per liter of solution. Remember, there are two acidic protons on H2SO4. It is then the case that 7.5 moles = 5 MNaOH x ? L solution, resulting in 1.5 L of NaOH solution being necessary for complete neutralization.

A titration of which of the following aqueous hydrogen halides with NaOH will show an equivalence point above pH = 7? Question 10 Answer Choices A. HBr (aq) B. HCl (aq) C. HI (aq) D. HF (aq)

D. A titration of HF with NaOH will show an equivalence point above pH = 7. HF is the only weak acid of the hydrogen halide series, thus, when titrated with a strong base, it has a basic equivalence point. All the other choices have equivalence points at or very near pH = 7.

Ba(NO2)2 is a basic salt due to the interaction of the nitrite ion with water according to the following reaction: NO2- + H2O <--> HNO2 + OH- Why does the equilibrium shown favor the left side of the reaction? Question 5 Answer Choices A. H2O is a stronger acid than HNO2. B. H2O is a weaker base than OH−. C. NO2− is a stronger base than OH−. D. NO2− is a weaker base than OH−.

D. Ba(NO2)2 is a basic salt due to the interaction of the nitrite ion with water according to the following reaction: NO2- + H2O <--> HNO2 + OH- The equilibrium shown favors the left side of the reaction because NO2− is a weaker base than OH−. Acid/base equilibria always favor the weaker acid/base pair. Since NO2- has a negative charge and no proton to donate, it must be the base, making water the acid. If the equilibrium favors the left hand side of the reaction, NO2- and H2O must be the weakest base and acid, respectively, eliminating these choices. While the choice, "H2O is a weaker base than OH−" is a true statement, water is not acting as a base in this particular reaction so this choice is not a good justification.

Which of the following compounds is a weak electrolyte? Question 4 Answer Choices A. BaCl2 B. CH3OH C. KOH D. CH3COOH

D. CH3COOH is a weak electrolyte. Strong electrolytes dissociate into ions when dissolved in water, yielding a solution that conducts electricity. Weak electrolytes partially dissociate in solution. Both the strong base KOH and salt BaCl2 are strong electrolytes because they are ionic compounds, and can therefore be eliminated. Answer choices CH3COOH and CH3OH are covalently bonded molecular compounds, but only CH3COOH partially dissociates due to the slightly acidic nature of the COOH group. CH3OH does not dissociate to any appreciable extent and is a non-electrolyte.

The addition of 1 g of which of the following compounds to distilled water would cause the least change in pH? Question 10 Answer Choices A. LiH B. HCl C. LiOH D. CsOH

D. The addition of 1 g of CsOH to distilled water would cause the least change in pH. Since the question posits that 1 g of each strong electrolyte (which undergoes complete dissociation) is added to distilled water, the compound with the highest molecular weight will yield the lowest molar concentration of either H+ or OH-. Cesium hydroxide (MW = ~150) is far heavier than any of the other compounds in question and is therefore the correct answer. LiH is a strong base, and when added to water will undergo the reaction LiH + H2O ? LiOH + H2. Hence the resultant pH from 1 g of LiH will be just slightly higher than 1 g of LiOH.

Calculate the concentration of F- ions in a 2 M solution of hydrogen fluoride, HF. (The Ka of HF is 6.8 × 10-4.) Question 14 Answer Choices A. 1.2 × 10-2 M B. 1.8 × 10-2 M C. 1.4 × 10-3 M D. 3.7 × 10-2 M

D. The concentration of F- ions in a 2 M solution of hydrogen fluoride, HF is 3.7 × 10-2 M. (The Ka of HF is 6.8 × 10-4.) The dissociation of HF is given by HF(aq) → H+(aq) + F-(aq) Let [H+] = [F-] = x. Although the actual equilibrium concentration of HF is (2 - x) M due to dissociation, we assume that x is very small (compared to 2) and ignore the difference. That is, we take [HF] = 2 M at equilibrium. Since the equilibrium expression for this reaction is Ka = [H+][F-]/[HF], we have Ka = x2/2, so x2 = 2Ka = 2(6.8 × 10-4) = 13.6 × 10-4 which implies that x = 3.7 × 10-2. (Note that our assumption that x is small compared to 2 was reasonable; x is less than 2% of 2.)

When designing experiments to study the action of pharmacological agents at physiological pH (~7.3) for use in human blood, which of the following conjugate pairs would serve as the best buffer solution? Question 11 Answer Choices A. CH3CO2H/CH3CO2-; pKa = 4.8 B. H2CO3/HCO3-; pKa = 6.4 C. HPO42-/ PO43-; pKa = 12.1 D. H2PO4-/HPO42- ; pKa = 7.1

D. When designing experiments to study the action of pharmacological agents at physiological pH (~7.3) for use in human blood, H2PO4-/HPO42- would serve as the best buffer solution. A buffer solution composed of an acid with a pKa near the pH of interest along with its conjugate base allows for effective buffering against changes brought on by the action of either added acid or base. The closest pKa to 7.3 is that of the H2PO4-/HPO42- conjugate pair.

A titration is performed on 100 mL of a 1 M solution of nitrous acid (Ka = 4 × 10-4) using a 2 M NaOH solution. Upon addition of 25 mL NaOH, the pH of the solution will be closest to: Question 4 Answer Choices A. 5.4 B. 1.4 C. 7.4 D. 3.4

D. 3.4

Phosphorus-32 has a half-life of 14 days. If a sample of 32P originally gave a reading of 100 milliCuries, what is the reading after 49 days? Question 23 Answer Choices A. 25 mCi B. 12.5 mCi C. 8.8 mCi D. 6.3 mCi

D. It is best to approach half-life questions by determining how many half-lives it takes to reach the total time allowed for decay and cutting the activity in half that many times. 1 half-life = 14 days = 50 milliCuries → 2 half-lives = 28 days = 25 milliCuries → 3 half-lives = 42 days = 12.5 milliCuries → 4 half-lives = 56 days = 6.25 milliCuries In this case 49 days is between 3 and 4 half-lives, making choice D the best answer (3.5 half-lives = 49 days = 8.84 milliCuries).

A radioactive nucleus decays by emitting a positron particle. What is true about the resulting daughter nucleus? Question 2 Answer Choices A. Its mass number increases. B. Its mass number decreases. C. Its atomic number increases. D. Its atomic number decreases.

D. Beta particles, both positive and negative, cause no change in mass number (eliminate choices A and B). However, positive beta particles (positrons) essentially convert one of the protons into a neutron, thus decreasing the atomic number during the process and making choice D correct.

Which of the following is the conjugate species formed when sodium acetate is added to water? A. Na+ B. H3O+ C. CH3COO- D. CH3COOH

D. When sodium acetate is added to water, acetate removes a proton from water to generate acetic acid and a hydroxide ion. Given that acetate is accepting a proton, it is acting as a base and the conjugate acid formed would be acetic acid, making choice D correct.

When a sample of a weak base is completely neutralized with a strong acid, the pH at the equivalence point will be: Question 2 Answer Choices A. indeterminable given the information provided. B. basic. C. neutral. D. acidic.

D. acidic.

What is the pOH of a mixture made by adding 150 mL of 0.20 M sodium hydroxide to 50 mL of 0.20 M hydrochloric acid? Question 27 Answer Choices A. 13.0 B. 12.5 C. 1.5 D. 1.0

D. From the values in the question, there is an excess of the strong base, sodium hydroxide, so the solution must be basic. This should give a pH value higher than 7.0 and a pOH value LOWER than 7.0 (eliminate choices A and B). pOH is defined as -log[OH-], so find the molar concentration of NaOH. The moles of hydroxide are: M x V = (0.20 M)(150 x 10-3 L) = 30 x 10-3 mol NaOH The moles of acid added are: M x V = (0.20 M)(50 x 10-3 L) = 10 x 10-3 mol HCl This leaves 20 x 10-3 mol of base in excess. The final concentration of base will be: so pOH = -log(0.1) = 1.0

Given the standard reduction potentials below, if a galvanic electrochemical cell were prepared using these two half-reactions, what would the values of E°cell and ΔG° be, respectively? [AuCl4]- + 3 e− → Au(s) + 4 Cl- E° = 0.930 V Br2(aq) + 2 e− → 2 Br- E° = 1.087 V Question 33 Answer Choices A. 2.017 V and 1,170 kJ B. 1.401 V and -45.5 kJ C. -0.157 V and 45.5 kJ D. 0.157 V and -90.9 kJ

D. Galvanic cells are spontaneous electrochemical cells, so the E°cell must be positive (eliminate choice C); ΔG° must be negative for a spontaneous process (eliminate choice A). To create a positive standard cell voltage, the gold half-reaction should be reversed and become an oxidation half-reaction. This will make the E°cell = +1.087 + (-0.930 V) = +0.157 V (eliminate choice A, choice D is correct). Note that to calculate ΔG°, apply the equation ΔG° = -nFE°cell. The total number of electrons transferred, n, will be 6 based on the balanced equation from the half reactions: (2 Au(s) + 8 Cl- + 3 Br2(aq) → 6 Br- + 2 [AuCl4]-). This makes ΔG° = -(6 mol e-)(96,500 C/mol e-)(+0.157 V) ≈ -(6)(100,000)(0.16) ≈ -96,000 J, or just less than -96 kJ via estimation.

Which of following acids would dissociate to the greatest degree? Question 32 Answer Choices A. HCN (Ka = 6.2 x 10-10) B. H3BO3 (Ka = 5.4 x 10-10) C. HNO2 (Ka = 4.0 x 10-4) D. H2SO3 (Ka = 1.4 x 10-2)

D. Given the Ka, which is an equilibrium constant, we can find the compound with the greatest degree of dissociation by looking for the largest Ka (answer D is correct). Had this question listed pKas, we would be looking for the smallest pKa as that would correlate with the largest Ka and the greatest degree of dissociation.

In a fully charged NiCad battery, the anodic and cathodic reactions are as follows: Cd + 2 OH- → Cd(OH)2 + 2e- (anode)2 NiO(OH) + 2 H2O + 2e- → 2 Ni(OH)2 + 2 OH- (cathode) Which of the following statements is FALSE? Question 30 Answer Choices A. Ni2+ is oxidized as the battery is recharged. B. Cd2+ is reduced as the battery is recharged. C. Cd is oxidized as the battery is discharged. D. Ni2+ is reduced as the battery is discharged.

D. In discharge Cd is converted from Cd0 to Cd2+, indicating a loss of electrons, or oxidation (eliminate choice C). That means during recharge the reverse process will happen, so Cd2+ is reduced (eliminate choice B). During discharge, the NiO(OH) is converted to Ni(OH)2, which involves a conversion of Ni from the +3 oxidation state to the +2 state, or a reduction (choice D is the false statement). Therefore Ni2+ is oxidized during recharge (eliminate choice A).

A potassium chloride salt bridge connects the two halves of an electrochemical cell. Which of the following best describes the path the salt bridge ions follow? Question 10 Answer Choices A. Potassium flows to the anode, and chloride flows to the cathode. B. Potassium flows to the anode, and chloride flows to the anode. C. Potassium flows to the cathode, and chloride flows to the cathode. D. Potassium flows to the cathode, and chloride flows to the anode.

D. Salt bridges are responsible for neutralizing charge imbalance that results from the reaction occurring in a galvanic cell (or two-chamber electrolytic cell), so the cations and anions must travel to opposite electrodes. This eliminates choices B and C. Given that electrons flow from the anode to the cathode, anions from the salt bridge must travel to the anode and cations from the salt bridge must travel to the cathode to balance the change in charge (choice D is correct).

Which of the following would result in a change in the electrical potential of a galvanic cell consisting of solid silver and copper electrodes in solutions of their respective ions? Question 19 Answer Choices A. Increasing the amount of solid silver anode in a reaction chamber B. Changing the identity of the salt in the salt bridge C. Changing the wire to another conductive metal D. Increasing the temperature of the cell

D. The Nernst equation, which expresses the electrical potential of a cell, can be written as: Thus, changing the temperature will affect E for the cell in question (and also Keq and ΔG - choice D is correct). Similarly, changes in reagent concentrations (affecting Q) would also affect the electrical potential. The remaining factors are not components of the Nernst equation and will not affect electrical potential. Specifically, changing the amount of solid silver anode (this is not part of the Q expression as it is a solid), the identity of the salt bridge, or the wire connecting the anode/cathode would have limited impact on potential.

The oxidation states of sulfur in SO42- and SO32- are (respectively): Question 4 Answer Choices A. +8 and +6. B. +8 and +4. C. +6 and +6. D. +6 and +4.

D. The oxidation states of sulfur in sulfate and sulfite can be determined using the typical oxidation state rules. Oxygen maintains an oxidation state of -2, totaling -8 and -6, respectively, while the overall anions have a charge of -2. This means that sulfur must have charges of +6 and +4, respectively, to cancel out the negative charge and result in an overall -2 for the anion, making choice D correct.

Which of the following best describes the reduction half-reaction in the following reaction? Cu(s) + 2 NO3-(aq) + 4 H+(aq) → Cu2+(aq) + 2 NO2(g) + 2 H2O(l) Question 25 Answer Choices A. Cu(s) + NO3-(aq) → Cu2+(aq) + NO2(g) B. Cu(s) → Cu2+(aq) + 2 e- C. 4 H+(aq) + O2(g) + 4 e- → 2 H2O(l) D. 2 NO3-(aq) + 2 e- + 4 H+(aq) → 2 NO2(g) +2 H2O(l)

D. This question is asking for a reduction half-reaction which therefore must involve only accepting electrons (choices A and B are wrong). The reaction described here involves the reduction of nitrogen in nitrate and the oxidation of copper metal to copper(II) cation. While answer choice C describes the oxidation of oxygen, this reaction is not a part of the overall reaction as oxygen gas is not involved (choice C is wrong). The reduction of nitrate to nitrogen dioxide gas describes the reduction half-reaction in the overall reaction (choice D is correct).