LS1a Fall 2014 Practice Problem Set #1 I. Basic Concept Questions 1. Consider a hypothetical reaction carried out at constant temperature and pressure: The Gibbs free energy curve of this reaction is depicted below: G A+B 50% Products, 50% Reactants C+D a. (1 point) Provide as much information as you can about the value of G rxn (e.g., >1, <1, =1, >0, <0, =0, etc ). b. (1 point) Provide as much information as you can about the value of Keq (e.g., >1, <1, =1, >0, <0, =0, etc ). c. (2 points) Label the point on the x-axis at which ΔG is zero. Provide as much information as you can about the value of the ratio of [products]/[reactants] at this point. (e.g., >1, <1, =1, >0, <0, =0, >Keq, <Keq, =Keq, > G rxn, < G rxn, = G rxn, ect ). d. (1 point) What is the sign of ΔG when the system diagrammed above contains 70% C+D and 30% A+B? 1
e. (1 point) Cells can drive thermodynamically unfavorable reactions by coupling them with favorable reactions. Draw a Gibbs free energy curve of a reaction such that it could drive reaction if the two reactions were coupled: 2. Shown below is a free energy diagram for the reaction A + B C + D a. On the diagram, clearly label ΔG rxn as well as the position of equilibrium on the x- axis. Also, indicate with arrows the direction in which the reaction will proceed spontaneously at the two points (1 and 2) highlighted on the curve. G 1 2 A + B C + D b. Is ΔG rxn for this reaction positive or negative? Is the reaction (as written) thermodynamically favorable or unfavorable under standard state conditions? c. What can you say about the magnitude of Keq for this reaction? 2
d. What is the sign (or the value) of ΔG at equilbrium? 3. Humans metabolize consumed ethanol into Acetyl-CoA, which can then be used to generate cellular energy. This process occurs in multiple steps as shown below: a. The first reaction in the pathway and its G rxn is shown below: Based on the value of G rxn, is the Keq for this reaction greater than 1, less than 1, or equal to 1 under standard state conditions? b. Whether or not this reaction can proceed towards the right under non-standard conditions depends on where the reaction lies with respect to equilibrium. What do the magnitude of G at a particular state and the sign of G at a particular state tell us about the reaction at that particular state and its relative relationship to reaching equilibrium? c. The second step of the reaction and its G rxn is shown below: Based on the value of G rxn, is the Keq for this reaction greater than 1, less than 1, or equal to 1 under standard state conditions? d. In order to determine G at a particular state one needs to take into account the concentrations of reactants and products as well as the reaction temperature. Consider the following situation inside a cell: 0.08% blood alcohol level [Ethanol] = 1.73 x 10-2 M [Acetaldehyde] = 20 x 10-5 M physiological temperature = 310 K R = 8.3 J/mol K Under these conditions and at physiological temperature, calculate the G for this particular state of the first reaction. Show your work for this calculation. 4. Consider the reaction written below. 3
a. Write the expression for the equilibrium constant, Keq. b. When the above reaction reaches equilibrium, does the reaction mixture contain mostly products or mostly reactants? Briefly explain. c. Write the expression for the equilibrium constant for the reaction in the reverse direction (right to left), 1/Keq. d. In which direction would the reaction shift if you increased the concentration of HCO3 -? Briefly explain. 5. Shown below is the structure of the non-drowsy anti-histamine Fexofenadine: In this structure, two ionizable groups ( COOH and NH + ) are labeled with their respective pka values. Both groups are shown in their protonated states. a. Determine the ratio of protonated to deprotonated carboxyl groups ( COOH) present in this molecule when it is placed into a solution of constant ph = 5.5. Show your work for this calculation and express your answer as [protonated]/[deprotonated]=. 4
b. Determine the ratio of protonated to deprotonated nitrogen atoms (-NH + ) present in this molecule when it is placed into a solution of constant ph = 5.5. Show your work for this calculation and express your answer as [protonated]/[deprotonated] =. c. Draw the specific structure of this molecule that is present in excess in a solution of ph = 5.5 (i.e., the structure that represents the dominant form at ph = 5.5). You only need to specifically draw out the relevant portion of the molecule that shows the two ionizable groups as shown below. d. Draw the structure that is present in excess at ph=1. e. Draw the structure that is present in excess at ph=11. 5
6. In order to facilitate digestion, your stomach maintains a strongly acidic environment through the production of hydrochloric acid. a) What is the ph range of your gastric juices if the concentration of HCl varies between 0.01M and 0.1 M? (You can assume the strong acid HCl completely dissociates in solution into H + and its conjugate base) b) Pepsin, a digestive enzyme that is utilized in your stomach, has two critical aspartate residues in its active site (we will learn much more about these aspartyl proteases later on in the course): In order for the enzyme to function at an optimal rate, one aspartate must be protonated; the other must remain deprotonated. The pka value of these special aspartates is about 1.7. i. At what ph would half of the aspartates be protonated (Hint: Use the Henderson-Hasselbach equation)? ii. How does the acidity of your stomach (refer to part a) enable this enzyme to function optimally? 7. Shown below is the amino acid lysine. The approximate pka values for the molecule s functional groups are indicated: sidechain 6
Draw the dominant charged species of this molecule at each given ph: ph 1 ph 6 ph 10 ph 12 8. The structure of nicotine is drawn below along with the pk a values of its two ionizable groups shown in their protonated states. a. Draw the structure of nicotine that is present in excess in a solution of constant ph = 5. b. Draw the structure of nicotine that is present in excess in a solution of constant ph = 11. c. Consider the ionizable group with a pk a of 8.0. For this group only, indicate the relative ratio of protonated/deprotonated species at each ph value listed below. Protonated Deprotonated ph = 6 ph = 7 ph = 8 ph = 9 ph = 10 7
9. Consider the following balanced dissociation reaction for a monoprotic acid in water: a. Calculate the pk a value for this acid. b. Write the expression for the acid dissociation constant (Ka) for the above reaction in terms of the concentrations of molecules involved. c. An amount of lactic acid is dissolved in water and the reaction is allowed to reach equilibrium. The concentration of the protonated acid at equilibrium is found to be 2.5 M. What is the ph of the solution? Show your calculations. [Hint: One molecule of lactate is generated for each molecule of H3O + that is generated; therefore, the concentration of lactate equals the concentration of H3O + at equilibrium. You may also disregard the auto-ionization of water.] d. Briefly compare and contrast ph and pka. Describe in your own words what each quantity measures. e. Draw a Gibbs free energy curve for the dissociation of lactic acid to lactate and hydronium ion. 8
G Reactant 50% Products/Reactant Products f. If you change the ph of the solution to 9, which species would be predominant at this ph? How would this change the product to reactant ratio at equilibrium? Draw a Gibbs free energy curve that would represent the same reaction at ph=9. Briefly explain your answer. G Reactant 50% Products/Reactant Products 9
II. Applied Concept Questions 10. Glycolysis is a 10-step cellular process by which the sugar glucose is broken down to capture chemical potential energy from food. The initial reactant (or starting material ) is one molecule of glucose. The cell uses the energy released by glycolysis to create high energy molecules, such as ATP, which can fuel other cellular reactions. In the second step of glycolysis, glucose-6-phosphate is converted into fructose-6- phosphate, as shown below. The G rxn for this step is +1.7 kj/mol. Step 2: G rxn = +1.7 kj/mol a. Write expressions for the equilibrium constants for steps 2 of glycolysis shown above in terms of the molar concentrations of reactants and products. b. At a temperature of 298 K, determine the ratio (products over reactants) between these two molecules once they reach equilibrium. Which molecule does the reaction favor? (R = 8.31 J mol -1 K -1 ) The next step (step #3) of glycolysis uses fructose-6-phosphate ( F6P ) as a reactant to produce fructose-1,6-bisphosphate ( Fru1,6bP ), as shown below. The G rxn for this step is -14.2 kj/mol. Step 3: G rxn = -14.2 kj/mol 10
c. How does the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate affect the relative ratio of glucose-6-phosphate to fructose-6-phosphate at equilibrium? d. Assume you start with two batches of cells growing under equal conditions. You then add an inhibitor that stops the conversion of F6P to Fru1,6bP to one of the two batches of cells, and you allow the other batch to grow normally. Which batch of cells do you predict will have more glucose-6-phosphate? Briefly explain your reasoning. e. Consider a cell in which the concentrations of fructose-6-phosphate, fructose-1,6- bisphosphate, ATP, and ADP are 0.01 mm, 3.5 mm, 2.25 mm, and 0.25 mm, respectively. Calculate ΔG at 37 C (310 K) for the synthesis of fructose-1,6- bisphosphate. Based on your results, is the reaction thermodynamically favored in this cell? (R = 8.314x10-3 kj mol -1 K -1 ) 11
11. You are given a mystery solution of multiple molecules of unknown concentrations in an aqueous solution. Shown below is a list of all of the molecules that might be present in the solution along with their approximate pka values. Molecule pka (approximate) Carbonic Acid 6 and 10 Phosphoric Acid 2, 7, and 12 TAPS 8.5 Acetic Acid 4.5 Triethyl Amine 11 MES 6 HEPES 3 and 7 Glycylglycine 3 and 8 You analyze the solution using a titration experiment and obtain the data shown below: a. Which of the molecules shown in the table above are present in the solution? How do you know? Assume all molecules in the solution are present at a concentration that allows detection by this method. 12
b. Of the molecules you identified in part (a), which is the most concentrated in the mystery solution? c. If 1 L of the mystery solution was used for the experiment and the titration was conducted with 0.2 M NaOH, then what is the concentration of molecule you identified in part (b)? 12. Shown below are several reactions describing the dissolution of various solids ( s ) in water, along with reaction describing the autoionization of water. Each reaction is shown with either their equilibrium constants or ΔG rxn values at 25 C. Keq ΔG rxn (kj mol -1 ) 51.6 2x10-8 55.63 38.59 1x10-14 79.9 a. Fill in the missing values for Keq and ΔG rxn in the boxes above. Show calculations to support your answers. (R = 8.314x10-3 kj mol -1 K -1 ) b. A set of coupled reactions can be summed together and simplified to describe an overall net reaction. ΔG rxn of the overall net reaction can be expressed as the sum of the ΔG rxn values of the component reactions. The example of CO2 dissolution coupled to carbonic acid dissociation from lecture is shown below: Using this strategy, draw a Gibbs free energy curve for the reaction shown below at 25 C. Briefly explain why you drew the diagram as you did. 13
c. If AgNO3 is dissolved in unbuffered, pure water, would you expect the ph of the solution to increase, decrease, or stay the same? Briefly explain your answer. d. AgOH(s) is added to unbuffered, pure water and found to dissolve poorly. Which of the following, when added in equal molar amounts to the AgOH/water mixture, will lead to the greatest concentration of Ag + (aq) at equilibrium? Briefly explain your answer, and include an explanation for why the other choices lead to less Ag + (aq) at equilibrium than the choice you selected. i. HCl (pka << 0) ii. HNO3 (pka << 0) iii. NaOH iv. NaCl 14
13. Dimethyl phosphoric acid is structurally similar to the phosphates in the backbone of DNA. Shown below is the deprotonation reaction of dimethyl phosphoric acid: a. Write out the equation for the acidity constant (Ka) of dimethyl phosphoric acid (use the abbreviations A and B for your equation). b. Experimentally, the Ka is determined to be 0.01. What is the pka value? c. Determine the ratio of the protonated to deprotonated forms of this molecule at each given ph (in the format of 1:n or n:1, where n is a number). I. ph 1: A (protonated) : B (deprotonated) II. ph 5: III. ph 9: IV. ph 13: 15
Answers 1. a. G rxn > 0 b. Keq < 1 c. G = 0 at the lowest energy point, which is the same point along the x-axis where [P]/[R]=Keq. d. Positive e. 2 a. The curve should be drawn to show a negative G rxn, and the lowest point on the curve must be to the right of the 50% point on the x-axis. G 1 Grxn 2 A + B Eq. C + D 16
b. ΔG rxn is negative because G for the products is lower than G for the reactants. Under standard state conditions this reaction will be thermodynamically favorable. c. Since the products are favored over the reactants, Keq will be greater than 1. d. At equilibrium, ΔG is equal to 0. 3a. Keq will be far less than one, indicating that the reactants are favored over the products. b. The G value at a particular state describes how far away from equilibrium the system is at that state (since G=0 at equilibrium). The sign of G at a particular state tells us the direction in which a reaction will spontaneously proceed to achieve equilibrium. If G at a particular state of the reaction is less than zero (negative), then the reaction (as written) will spontaneously proceed to the right to convert reactants to product to reach to equilibrium. If G at a particular state of the reaction is greater than zero (positive), the reaction will spontaneously proceed to the left to convert products back into reactants to reach equilibrium. The magnitude of G at a particular state provides us with a measure of how far away from equilibrium the reaction lies at that state. The greater the magnitude, the farther away the reaction lies from equilibrium. c. Keq will be far greater than one, indicating that the products are favored over the reactants. d. G = G rxn + RT ln ([Acetaldehyde]/[Ethanol]) G = 47,200 J/mol + (8.3 J/mol K)(310 K) ln [(20 x 10-5 )/(1.73 x 10-2 )] G = 47,200 J/mol + (8.3 J/mol K)(310 K) ln (1.16 x 10-2 ) G = 47,200 J/mol + (-11476 J/mol) G = 35,724 J/mol (using rounded numbers) 4a. Keq = [SO4 2- aq][hco3 - aq]/[hso4 - aq][co3 2- aq] b. The reaction will contain mostly products at equilibrium since Keq >1. c. 1/Keq = [HSO4 - aq][co3 2- aq]/ [SO4 2- aq][hco3 - aq] d. If we increase the concentration of HCO3 - aq, the reaction will shift to the left in order to restore the equilibrium ratio of products over reactants. If the concentration of HCO3 - aq is increased, the concentrations of both HSO4 - aq and CO3 2- will increase (and the concentration of SO4 2- aq will decrease) so that the ratio of products over reactants will once again equal Keq. 5a. pka = ph + log [HA]/[A - ] 4.5 = 5.5 + log [HA]/[A - ] -1 = log [HA]/[A - ] 10-1 = [HA]/[A - ] Thus, [protonated]/[deprotonated] = [HA]/[A - ] = 1/10 b. pka = ph + log [HA - ]/[A - ] 9.5 = 5.5 + log [HA]/[A - ] 4 =log [HA]/[A - ] 10 4 = [HA]/[A - ] Thus, [protonated]/[deprotonated] = 10,000/1 17
c. d. e. 6a. ph= -log [H + ] ph= -log [0.1M] = 1; ph= -log [0.01M] = 2 ph is between 1 and 2 b. i. pka= ph + log [HA] [A - ] So if [HA]=[A-] pka= ph +log (1) pka= ph + 0 pka= ph Therefore at ph=1.7 half of the aspartates will be protonated. ii. The ph of your stomach is between 1 and 2, approximately equal to the pka of asparate. So, it varies closely around the ph where half the residues would be protonated and half deprotonated. 7. ph 1 ph 6 ph10 ph12 18
8. a. b. c. Protonated Deprotonated ph = 6 100 1 ph = 7 10 1 ph = 8 1 1 ph = 9 1 10 ph = 10 1 100 9. a. pk a = -log(ka)= 4 b. Ka = [Lactate][H3O + ]/[Lactic Acid] (water should not be included in the denominator) c. Ka = [Lactate][H3O + ]/[Lactic Acid] Ka * [Lactic Acid]= [Lactate][H3O + ] Since one molecule of lactate is generated for each molecule of H3O + that is generated, [Lactate]=[H3O + ]. Ka * [Lactic Acid]= [H3O + ][H3O + ] (1x10-4 )(2.5 ) = [H3O + ] 2 [(1x10-4 )(2.5)] = [H3O + ] [H3O + ] = 0.0158 M ph = -log ([H3O + ]) = -log(0.0158) = 1.8 d. The ph is a property of a solution; it represents the total concentration of H + or H3O + present in solution. e. pka is a property of an acid (or acidic group); it provides a measure of the strength of an acid or acidic group, or its tendency to give up proton. 19
G Reactant 50% Products/Reactant Products The shape of the curve should indicate that it is unfavorable, and the equilibrium point being to the left of the 50% marker. f. G Reactant 50% Products/Reactant Products The graph should be the same as in part (e), even though at ph 9, lactate, the deprotonated form of the acid (i.e., the conjugate base or A - ), will predominate. Even though the ratio of protonated to deprotonated acid changes, the value of the ratio of [products]/[reactants] at equilibrium does not change. The energy diagram is the same because ph does not affect Keq. 10. a. Step 2: K eq = [fructose 6 phosphate] [glucose 6 phosphate] or [F6P] [G6P] b. At equilibrium, the ratio of [F6P]/[G6P] is equal to K eq. We can therefore solve for K eq using the following equation: G rxn = RTln(K eq ) ( G rxn )/RT = ln(k eq ) 20
K eq = e ( G rxn/rt) K eq = e ( 1,700 J mol)/(8.31 J mol 1 K 1 )(298 K) K eq = 0.5 = [F6P]/[G6P] at equilibrium, such that glucose-6-phosphate is favored by the reaction at this temperature c. The ratio is not affected. While step 3 decreases the concentration of fructose- 6-phosphate, the Keq of step 2 remains unaffected. The depletion of fructose- 6-phosphate will drive the reaction in step 2 such that more glucose-6- phosphate will be converted into fructose-6-phosphate to restore equilibrium via Le Chatelier s Principle. However the relative ratio between glucose-6-phosphate and fructose-6- phosphate remains the same once equilibrium is re-established as dictated by the value of K eq (i.e., 2 G6P for every 1 F6P). This is basically the same question as 3f. d. We would expect to see more G6P in the batch of cells in which the inhibitor was added. We can think of this in two (equivalent) ways: 1) In the cells with the inhibitor added, stopping the conversion of F6P to Fru1,6bP stops the depletion of F6P in the cells. With more F6P present in the cells, the concentration of G6P also increases in order to maintain equilibrium. 2) In the cells lacking the inhibitor, F6P continues to be depleted through its conversion to Fru1,6bP. This conversion is thermodynamically very favorable, such that there is less F6P in equilibrium with G6P. This will cause the concentration of G6P to also decrease in order to maintain equilibrium. e. G = G rxn + RT ln([products]/[reactants]) G = (-14.2 kj/mol) + (8.314x10-3 kj mol -1 K -1 )(310 K)ln[(0.00025 M)(0.0035 M)/(0.00225 M)/(0.00001 M)] G = (-14.2 kj/mol) + (8.314x10-3 kj mol -1 K -1 )(310 K)ln[38.889] G = (-14.2 kj/mol) + (8.314x10-3 kj mol -1 K -1 )(310 K)(3.6607) G = (-14.2 kj/mol) + (9.430 kj/mol) G = -4.77 kj/mol The formation of fructose-1,6-bisphosphate is thermodynamically favored because G is negative. 11a. The solution contains carbonic acid, MES, and glycylglycine. It has a pka at 8, and glycylglycine is the only molecule on the chart with this property; therefore glycylglycine must be in the sample. It takes the same amount of NaOH to reach equivalence point following pka 8 as it does to reach the equivalence point of pka 3, such that the amount of glycylglycine is sufficient to explain both the plateaus at pka 3 and 8 (i.e., there is the same number of moles of both the pka 3 and pka 8 acidic groups, which equals the number of moles of glycylglycine). The sample contains a pka at 10, and carbonic acid is the only molecule that has this property; therefore the sample must contain carbonic acid. Carbonic acid also has a pka at 6, but it takes about half as much NaOH to reach the equivalence point of pka 10 as it does to reach the equivalence point of pka 6, 21
such that the amount of carbonic acid in solution is not sufficient to explain the length of the pka 6 plateau. Therefore there must be something else in the solution with a pka of 6, and MES is the only possible molecule. b. Glycylglycine requires 5 ml of NaOH to titrate. c. The most concentrated species requires ~5 ml 0.2 M NaOH to titrate. Therefore there are ~1x10-3 moles of glycylglycine in the solution. This is in a volume 1 L, so the concentration is about ~1.0 mm. 12. a. Keq ΔG rxn (kj mol -1 ) 51.6-9.77 1.77x10-10 55.63 2x10-8 43.92 1.72x10-7 38.59 1x10-14 79.9 ΔG rxn = -RT*ln(Keq) ΔG rxn = -(8.314 J mol -1 K -1 ) (298 K) ln(keq) For Reaction 1: ΔG rxn = -(8.314x10-3 kj mol -1 K -1 ) (298 K) ln(51.6) = -9.77 kj mol -1 For Reaction 3: ΔG rxn = (8.314x10-3 kj mol -1 K -1 ) (298 K) ln(2x10-8 ) = 43.9 kj mol -1 Keq = e^[-δg rxn/(rt)] = e^[-δg rxn/(8.314x10-3 kj mol -1 K -1 *298 K)] For Reaction 2: Keq = e^[-55.63 kj/mol/(8.314x10-3 kj mol -1 K -1 *298 K)] = 1.77x10-10 For Reaction 4: Keq = e^[-38.59 kj/mol/(8.314x10-3 kj mol -1 K -1 *298 K)] = 1.72x10-7 22
b. Break into three component reactions and sum their ΔG rxn values: 1.) AgNO3 Ag + + NO3 - ΔG rxn = -9.77 kj mol -1 2.) Cl - + Ag + AgCl ΔG rxn = -55.63 kj mol -1 3.) CuCl Cu + + Cl - ΔG rxn = +38.59 kj mol -1 ΔG rxn = (38.59 9.77 55.63) kj mol -1 = -26.81 kj mol -1. Since ΔG rxn is negative, the overall net reaction is favorable and the equilibrium (minimum of the graph) must be on the right side of the graph. c. The ph would decrease. Silver nitrate dissociates to release silver ions when dissolved in water (first reaction in chart). These ions would react with hydroxide ions to form silver hydroxide, a solid that would precipitate out of solution (which thermodynamically favored; third reaction in chart). This would decrease the amount of hydroxide in the solution, which would shift the equilibrium between water, hydroxide, and hydronium to create more hydronium and hydroxide (last reaction in chart). This results in an increase in hydronium concentration and a decrease in ph. Or: d. HNO3 will lead to the greatest silver ion concentration. Both HCl and HNO3 dissociate in water to release H3O + (since their pka values are small). The increase in H3O + concentration shifts the equilibrium between water, H3O +, and OH - towards water, which decreases the amount of OH - in the solution. The decreased OH - concentration shifts the AgOH/OH - /Ag + 23
equilibrium away from AgOH, which increases the concentration of Ag + in the solution. While both HNO3 and HCl generate hydronium to affect the Ag + in this way, they differ in how their conjugate base (either NO3 - or Cl - ) affects the Ag + concentration. The addition of either NO3 - or Cl - would shift the Ag + equilibrium away from Ag + and towards either AgNO3 of AgCl. Since roughly equal molar amounts of NO3 - and Cl - are being added in each situation, there would be a greater equilibrium concentration of Ag + when HNO3 is added than when HCl is added because the equilibrium constant for formation of AgNO3 from Ag + and NO3 - is much, much lower than the equilibrium constant for the formation of AgCl from Ag + and Cl -. NaOH would shift the equilibrium towards AgOH by increasing OH - concentration. NaCl increase the Cl - concentration to shift the equilibrium away from Ag + and towards AgCl. 13a. Ka = [B][H + ]/[A]; or in water, Ka = [B][H3O + ]/[A] b. pka = -log Ka = -log 0.01 = -log 10-2 = - (-2 log 10) = 2 c. ph 1: 10:1 ph 5: 1: 1000 ph 9: 1: 10 7 ph 13: 1: 10 11 d. The phosphates in the DNA backbone are structurally analogous to dimethyl phosphoric acid and are likely to have similar pka values. At ph 7.4, the deprotonated, or negatively charged state will be favored by 10 5 over the protonated state. The picture shown above would thus have to be altered to appear as shown below: As the phosphate groups are deprotonated, no hydrogen bonds can form as proposed in the structure. In addition, since like charges repel, this arrangement would be extremely energetically unfavorable and is not likely to occur naturally. 24