Chapter Review in Chapter 2 Water, The Solvent Of Life

Chapter Review

KEY TERMS

Terms in bold are defined in the glossary.

Problems

1. Effect of Local Environment on Ionic Bond Strength The ATP-binding site of an enzyme is buried in the interior of the enzyme, in a hydrophobic environment. Suppose that the ionic interaction between enzyme and ATP took place at the surface of the enzyme, exposed to water. Would this enzyme-substrate interaction be stronger or would it be weaker? Why?
2. Biological Advantage of Weak Interactions The associations between biomolecules are often stabilized by hydrogen bonds, electrostatic interactions, the hydrophobic effect, and van der Waals interactions. How are weak interactions such as these advantageous to an organism?
3. Solubility of Ethanol in Water Ethane $mathml alt text870$ $left-parenthesis CH Subscript 3 Baseline CH Subscript 3 Baseline right-parenthesis$ and ethanol $mathml alt text871$ $left-parenthesis CH Subscript 3 Baseline CH Subscript 2 Baseline OH right-parenthesis$ differ in their molecular makeup by only one atom, yet ethanol is much more soluble in water than ethane. Describe the features of ethanol that make it more water soluble than ethane.
4. Calculation of pH from Hydrogen Ion Concentration What is the pH of a solution that has an $mathml alt text874$ $upper H Superscript plus$ concentration of
1. $mathml alt text875$ $1.75 times 10 Superscript minus Superscript 5 Baseline mol slash upper L$ ;
2. $mathml alt text876$ $6.50 times 10 Superscript minus Superscript 10 Baseline mol slash upper L$ ;
3. $mathml alt text877$ $1.0 times 10 Superscript negative 4 Baseline mol slash upper L$ ;
4. $mathml alt text878$ $1.50 times 10 Superscript negative 5 Baseline mol slash upper L$ ?
5. Calculation of Hydrogen Ion Concentration from pH What is the $mathml alt text880$ $upper H Superscript plus$ concentration of a solution with pH of
1. 3.82;
2. 6.52;
3. 11.11?
6. Acidity of Gastric HCl A technician in a hospital laboratory obtained a 10.0 mL sample of gastric juice from a patient several hours after a meal and titrated the sample with 0.1 m NaOH to neutrality. The neutralization of gastric HCl required 7.2 mL of NaOH. The patient’s stomach contained no ingested food or drink at the time of sample harvest. Therefore, assume that no buffers were present. What was the pH of the gastric juice?
7. Calculation of the pH of a Strong Acid or Base
1. Write out the acid dissociation reaction for hydrochloric acid.
2. Calculate the pH of a solution of $mathml alt text894$ $5 times 10 Superscript negative 4 Baseline upper M$ hydrochloric acid at $mathml alt text895$ $25 degree upper C$ .
3. Write out the acid dissociation reaction for sodium hydroxide.
4. Calculate the pH of a solution of $mathml alt text897$ $7 times 10 Superscript minus Superscript 5 Baseline upper M$ sodium hydroxide at $mathml alt text898$ $25 degree upper C$ .
8. Calculation of pH from Concentration of Strong Acid Calculate the pH of a solution prepared by diluting 3.0 mL of 2.5 m HCl to a final volume of 100 mL with $mathml alt text904$ $upper H Subscript 2 Baseline upper O$ .
9. Measurement of Acetylcholine Levels by pH Changes You have a 15 mL sample of acetylcholine (a neurotransmitter) with an unknown concentration and a pH of 7.65. You incubate this sample with the enzyme acetylcholinesterase to convert all of the acetylcholine to choline and acetic acid. The acetic acid dissociates to yield acetate and hydrogen ions.

From left to right, acetylcholine has C H 3 bonded to C that is double bonded to O above and further bonded to O that is bonded to C H 2 bonded to C H 2 bonded to N plus that is bonded to 3 C H 3 groups. This reacts with H 2 O to produce choline and acetate. Acetate consists of the left half of the chain up to the O, which becomes O minus. Choline consists of the right half of the chain except for O, which becomes O H. There is an additional H plus present.

At the end of the incubation period, you measure the pH again and find that it has decreased to 6.87. Assuming there was no buffer in the assay mixture, determine the number of nanomoles of acetylcholine in the original 15 mL sample.
10. Relationship Between $mathml alt text912$ $bold p bold-italic upper K Subscript bold a Baseline$ and pH Which aqueous solution has the lowest pH: 0.1 m hydrofluoric acid $mathml alt text915$ $left-parenthesis p upper K Subscript a Baseline equals 3.20 right-parenthesis$ ; 0.1 m acetic acid $mathml alt text917$ $left-parenthesis p upper K Subscript a Baseline equals 4.86 right-parenthesis$ ; 0.1 m formic acid $mathml alt text919$ $left-parenthesis p upper K Subscript a Baseline equals 3.75 right-parenthesis$ ; or 0.1 m lactic acid $mathml alt text921$ $left-parenthesis p upper K Subscript a Baseline equals 7.86 right-parenthesis$ ?
11. Properties of Strong and Weak Acids Classify each acid or property as representing a strong acid or a weak acid:
1. hydrochloric acid;
2. acetic acid;
3. strong tendency to dissociate protons;
4. larger $mathml alt text922$ $upper K Subscript a$ ;
5. partially dissociates into ions;
6. larger $mathml alt text923$ $p upper K Subscript a Baseline$ .
12. Simulated Vinegar One way to make vinegar is to prepare a solution of acetic acid, the sole acid component of vinegar, at the proper pH (see Fig. 2-14) and add appropriate flavoring agents. Acetic acid is a liquid at $mathml alt text925$ $25 degree upper C$ , with a relative molecular mass $mathml alt text926$ $left-parenthesis upper M Subscript r Baseline right-parenthesis$ of 60, density of 1.049 g/mL, and acid dissociation constant $mathml alt text929$ $left-parenthesis upper K Subscript a Baseline right-parenthesis$ of $mathml alt text930$ $1.7 times 10 Superscript negative 5 Baseline upper M$ . Calculate the volume of acetic acid needed to produce 1 L of simulated vinegar from distilled water (see Fig. 2-15).
13. Identifying Conjugate Bases Write the conjugate base for each acid:
1. $mathml alt text932$ $upper H Subscript 3 Baseline PO Subscript 4$
2. $mathml alt text933$ $upper H Subscript 2 Baseline CO Subscript 3$
3. $mathml alt text934$ $CH Subscript 3 Baseline COOH$
4. $mathml alt text935$ $CH Subscript 3 Baseline NH Subscript 3 Superscript plus$
14. Calculation of the pH of a Mixture of a Weak Acid and Its Conjugate Base Calculate the pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid $mathml alt text938$ $left-parenthesis p upper K Subscript a Baseline equals 4.76 right-parenthesis$ of
1. 2:1;
2. 1:3;
3. 5:1;
4. 1:1;
5. 1:10.
15. Effect of pH on Solubility The strongly polar, hydrogen-bonding properties of water make it an excellent solvent for ionic (charged) species. By contrast, nonionized, nonpolar organic molecules, such as benzene, are relatively insoluble in water. In principle, the aqueous solubility of any organic acid or base can be increased by converting the molecules to charged species. For example, the solubility of benzoic acid in water is low. Adding sodium bicarbonate to a mixture of water and benzoic acid raises the pH and deprotonates the benzoic acid to form benzoate ion, which is quite soluble in water.

Benzoic acid has a benzene ring bonded to C that is double bonded to O and bonded to O with a highlighted H. Text reads, p K subscript a is approximately 5. Benzoate ion has the same structure except that the highlighted H is gone and the O H has become O minus.

Categorize the given compounds based on whether they are more soluble in an aqueous solution of 0.1 m NaOH or 0.1 m HCl. (The dissociable protons are shown in red.)

Part a shows the quinoline ion as a six-membered benzene ring fused by its right side bond with a six-membered ring that has double bonds at the top right and lower right sides and N plus with a highlighted H substituted for C at the top vertex. Part b shows m-cresol, which is a benzene ring with O bonded to highlighted H meta to a bond to something else. Part c shows 2-(methylthio)pyridine ion as a benzene ring with N plus with a bonded highlighted H substituted for carbon at the top vertex and a bond to S that begins a chain at the ortho position.
16. Treatment of Poison Ivy Rash Urushiol, the component of poison ivy that is responsible for the characteristic itchy rash, is a mixture of catechols substituted with various long-chain alkyl groups.

Urushiol has a benzene ring with O H at the top vertex, O with highlighted H at the top right vertex, and a chain extending from the lower right vertex. The chain has n C H 2s and ends with C H 3. Text reads, p K subscript a equals approximately 8.

Which of these treatments would be most effective at removing catechols from the surface of the skin after exposure to poison ivy? Justify your choice.
1. Wash the area with cold water.
2. Wash the area with dilute vinegar or lemon juice.
3. Wash the area with soap and water.
4. Wash the area with soap, water, and baking soda (sodium bicarbonate).
17. pH and Drug Absorption Aspirin is a weak acid with a $mathml alt text949$ $p upper K Subscript a Baseline$ of 3.5 (the ionizable H is shown in red):

A benzene ring has two chains positioned ortho to each other. One chain has O bonded to C that is double bonded to O and single bonded to C H 3. The other has C double bonded to O and further bonded to O bonded to a highlighted H.

Aspirin is absorbed into the blood through the cells lining the stomach and the small intestine. Absorption requires passage through the plasma membrane. The polarity of the molecule determines the absorption rate: charged and highly polar molecules pass slowly, whereas neutral hydrophobic molecules pass rapidly. The pH of the stomach contents is about 1.5, and the pH of the contents of the small intestine is about 6. Based on this information, is more aspirin absorbed into the bloodstream from the stomach or from the small intestine? Clearly justify your choice.
18. Calculation of pH from Molar Concentrations The $mathml alt text956$ $p upper K Subscript a Baseline$ of $mathml alt text957$ $NH Subscript 4 Superscript plus Baseline slash NH Subscript 3$ is 9.25. Calculate the pH of a solution containing $mathml alt text960$ $0.12 upper M NH Subscript 4 Baseline Cl$ and 0.03 m NaOH.
19. Calculation of pH after Titration of Weak Acid A compound has a $mathml alt text963$ $p upper K Subscript a Baseline$ of 7.4. You add 100 mL of a 1.0 m solution of this compound at pH 8.0 to 30 mL of 1.0 m hydrochloric acid. What is the pH of the resulting solution?
20. Properties of a Buffer The amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a $mathml alt text971$ $p upper K Subscript a Baseline$ of 9.6, can exist either in the protonated form $mathml alt text973$ $left-parenthesis em-dash NH Subscript 3 Superscript plus Baseline right-parenthesis$ or as the free base $mathml alt text974$ $left-parenthesis em-dash NH Subscript 2 Baseline right-parenthesis$ , because of the reversible equilibrium

$R — {NH}_{3}^{+} ⇌ R — {NH}_{2} + H^{+}$ $upper R em-dash NH Subscript 3 Superscript plus Baseline right harpoon over left harpoon upper R em-dash NH Subscript 2 Baseline plus upper H Superscript plus$
1. In what pH range can glycine be used as an effective buffer due to its amino group?
2. In a 0.1 m solution of glycine at pH 9.0, what fraction of glycine has its amino group in the $mathml alt text979$ $em-dash NH Subscript 3 Superscript plus$ form?
3. How much 5 m KOH must be added to 1.0 L of 0.1 m glycine at pH 9.0 to bring its pH to exactly 10.0?
4. When 99% of the glycine is in its $mathml alt text987$ $em-dash NH Subscript 3 Superscript plus$ form, what is the numerical relation between the pH of the solution and the $mathml alt text989$ $p upper K Subscript a Baseline$ of the amino group?
21. Calculation of the $mathml alt text990$ $bold p bold-italic upper K Subscript bold a Baseline$ of an Ionizable Group by Titration Suppose a biochemist has 10 mL of a 1.0 m solution of a compound with two ionizable groups at a pH of 8.00. She adds 10.0 mL of 1.00 m HCl, which changes the pH to 3.20. The $mathml alt text999$ $p upper K Subscript a Baseline$ value of one of the groups $mathml alt text1000$ $left-parenthesis p upper K 1 right-parenthesis$ is 3.8 and it is known that $mathml alt text1001$ $p upper K 2$ is between 7 and 10. What is the exact value of $mathml alt text1004$ $p upper K 2$ ?
22. Calculation of the pH of a Solution of a Polyprotic Acid The amino acid histidine has ionizable groups with $mathml alt text1006$ $p upper K Subscript a Baseline$ values of 1.8, 6.0, and 9.2, as shown (His = imidazole group). A biochemist makes up 100 mL of a 0.10 m solution of histidine at a pH of 5.40. She then adds 40 mL of 0.10 m HCl. What is the pH of the resulting solution?

On the left, histidine has C H bonded to N H 3 plus to the left, C O O with a highlighted H above, and C H 2 below that is further bonded to C at the top left vertex of a five-membered ring. The ring has N H substituted for C at the top vertex, C bonded to H at the bottom left vertex and right vertex, N H plus substituted for C at the bottom right vertex, and double bonds on the left side and right bottom side. The ionizable group is C O O H that becomes C O O minus. The p K 1 shown above arrows showing a reversible reaction is 1.8. This produces a similar molecule that has lost the highlighted H to produce C O O minus on the end opposite the ring. The H of N H plus at the lower right vertex of the ring is highlighted. The next ionizable group his H i s H plus that becomes H i s. The p K 2 shown above the arrows for the reversible reaction is 6.2. The product has N instead of N H plus at the lower right vertex of the ring, and N H 3 has a highlighted H. The next ionizable group is N H 3 plus that becomes N H 2. This produces a similar molecule except that the highlighted H is lost and N H 3 plus becomes N H 2.
23. Calculation of Original pH from Final pH after Titration A biochemist has 100 mL of a 0.100 m solution of a weak acid with a $mathml alt text1022$ $p upper K Subscript a Baseline$ of 6.3. He adds 6.0 mL of 1.0 m HCl, which changes the pH to 5.7. What was the pH of the original solution?
24. Preparation of a Phosphate Buffer Phosphoric acid $mathml alt text1029$ $left-parenthesis upper H Subscript 3 Baseline PO Subscript 4 Baseline right-parenthesis$ , a triprotic acid, has three $mathml alt text1030$ $p upper K Subscript a Baseline$ values: 2.14, 6.86, and 12.4. What molar ratio of $mathml alt text1034$ $HPO Subscript 4 Superscript 2 minus$ to $mathml alt text1035$ $upper H Subscript 2 Baseline PO Subscript 4 Superscript bold minus$ in solution would produce a pH of 7.0? Hint: Only one of the $mathml alt text1038$ $p upper K Subscript a Baseline$ values is relevant here.
25. Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in commercial pH meters gives an electrical response proportional to the concentration of hydrogen ion. Converting these responses to a pH reading requires calibration of the electrode against standard solutions of known $mathml alt text1042$ $upper H Superscript plus$ concentration. Preparation of the pH 7.00 standard buffer uses dihydrogen phosphate $mathml alt text1044$ $left-parenthesis NaH Subscript 2 Baseline PO Subscript 4 Baseline bullet upper H Subscript 2 Baseline upper O$ ; FW 138) and disodium hydrogen phosphate $mathml alt text1045$ $left-parenthesis Na Subscript 2 Baseline HPO Subscript 4 Baseline$ ; FW 142). Phosphoric acid $mathml alt text1046$ $left-parenthesis upper H Subscript 3 Baseline PO Subscript 4 Baseline right-parenthesis$ , a triprotic acid, has three $mathml alt text1047$ $p upper K Subscript a Baseline$ values: 2.14, 6.86, and 12.4. Calculate the weight in grams of sodium dihydrogen phosphate and disodium hydrogen phosphate needed to prepare 1.0 L of a standard buffer with a total phosphate concentration of 0.10 m (see Fig. 2-15).
26. Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a $mathml alt text1054$ $p upper K Subscript a Baseline$ of 6.00, calculate the ratio of conjugate base to acid at a pH of 5.00.
27. Preparation of Buffer of Known pH and Strength You have 0.10 m solutions of acetic acid $mathml alt text1060$ $left-parenthesis p upper K Subscript a Baseline equals 4.76 right-parenthesis$ and sodium acetate. If you wanted to prepare 1.0 L of 0.10 m acetate buffer of pH 4.00, how many milliliters of acetic acid and sodium acetate would you mix together?
28. Choice of Weak Acid for a Buffer Determine whether each weak acid would best buffer at pH 3.0, at pH 5.0, or at pH 9.0:
1. formic acid $mathml alt text1068$ $left-parenthesis p upper K Subscript a Baseline equals 3.8 right-parenthesis$ ;
2. acetic acid $mathml alt text1069$ $left-parenthesis p upper K Subscript a Baseline equals 4.76 right-parenthesis$ ;
3. ammonium $mathml alt text1070$ $left-parenthesis p upper K Subscript a Baseline equals 9.25 right-parenthesis$ ;
4. boric acid $mathml alt text1071$ $left-parenthesis p upper K Subscript a Baseline equals 9.24 right-parenthesis$ ;
5. chloroacetic acid $mathml alt text1072$ $left-parenthesis p upper K Subscript a Baseline equals 2.87 right-parenthesis$ ;
6. hydrazoic acid $mathml alt text1073$ $left-parenthesis p upper K Subscript a Baseline equals 4.6 right-parenthesis$ . Briefly justify your answer.
29. Working with Buffers A buffer contains 0.010 mol of lactic acid $mathml alt text1074$ $left-parenthesis p upper K Subscript a Baseline equals 3.86 right-parenthesis$ and 0.050 mol of sodium lactate per liter.
1. Calculate the pH of the buffer.
2. Calculate the change in pH after adding 5.0 mL of 0.5 m HCl to 1 L of the buffer.
3. What pH change would you expect if you added the same quantity of HCl to 1 L of pure water?
30. Use of Molar Concentrations to Calculate pH What is the pH of a solution that contains 0.20 m sodium acetate and 0.60 m acetic acid $mathml alt text1088$ $left-parenthesis p upper K Subscript a Baseline equals 4.76 right-parenthesis$ ?
31. Preparation of an Acetate Buffer Calculate the concentrations of acetic acid $mathml alt text1089$ $left-parenthesis p upper K Subscript a Baseline equals 4.76 right-parenthesis$ and sodium acetate necessary to prepare a 0.2 m buffer solution at pH 5.0.
32. pH of Insect Defensive Secretion You have been observing an insect that defends itself from enemies by secreting a caustic liquid. Analysis of the liquid shows it to have a total concentration of formate plus formic acid $mathml alt text1093$ $left-parenthesis upper K Subscript a Baseline equals 1.8 times 10 Superscript negative 4 Baseline right-parenthesis$ of 1.45 m. Further analysis reveals that the concentration of formate ion is 0.015 m. What is the pH of the secretion?
33. Calculation of $mathml alt text1097$ $bold p bold-italic upper K Subscript bold a Baseline$ An unknown compound, X, is thought to have a carboxyl group with a $mathml alt text1098$ $p upper K Subscript a Baseline$ of 2.0 and another ionizable group with a $mathml alt text1100$ $p upper K Subscript a Baseline$ between 5 and 8. When 75 mL of 0.1 m NaOH is added to 100 mL of a 0.1 m solution of X at pH 2.0, the pH increases to 6.72. Calculate the $mathml alt text1110$ $p upper K Subscript a Baseline$ of the second ionizable group of X.
34. Ionized Forms of Amino Acids at Different pH Levels Glycine is a diprotic acid that can undergo two dissociation reactions, one for the $mathml alt text1112$ $alpha$ -amino group $mathml alt text1113$ $left-parenthesis em-dash NH Subscript 3 Superscript plus Baseline right-parenthesis$ and the other for the carboxyl $mathml alt text1114$ $left-parenthesis em-dash COOH right-parenthesis$ group. Therefore, it has two $mathml alt text1115$ $p upper K Subscript a Baseline$ values. The carboxyl group has a $mathml alt text1116$ $p upper K 1$ of 2.34 and the α-amino group has a $mathml alt text1119$ $p upper K 2$ of 9.60. Glycine can exist in fully deprotonated $mathml alt text1121$ $left-parenthesis NH Subscript 2 Baseline em-dash CH Subscript 2 Baseline em-dash COO Superscript minus Baseline right-parenthesis$ , fully protonated $mathml alt text1122$ $left-parenthesis Superscript plus Baseline NH Subscript 3 Baseline em-dash CH Subscript 2 Baseline em-dash COOH right-parenthesis$ , or zwitterionic form $mathml alt text1123$ $left-parenthesis Superscript plus Baseline NH Subscript 3 Baseline em-dash CH Subscript 2 Baseline em-dash COO Superscript minus Baseline right-parenthesis$ . Determine which form of glycine would be present in the highest concentration in a solution of
1. pH 1.0;
2. pH 6.0;
3. pH 7.0;
4. pH 8.0;
5. pH 11.9. Explain your answers in terms of pH relative to the two $mathml alt text1130$ $p upper K Subscript a Baseline$ values.
35. Control of Blood pH by Respiratory Rate
1. The partial pressure of $CO Subscript 2$ $CO Subscript 2$ $mathml alt text1133$ $left-parenthesis pCO Subscript 2 Baseline right-parenthesis$ in the lungs can be varied rapidly by the rate and depth of breathing. For example, a common remedy to alleviate hiccups is to increase the concentration of $mathml alt text1134$ $CO Subscript 2$ in the lungs. This can be achieved by holding one’s breath, by very slow and shallow breathing (hypoventilation), or by breathing in and out of a paper bag. Under such conditions, $mathml alt text1135$ $pCO Subscript 2$ in the air space of the lungs rises above normal. How would increasing $mathml alt text1136$ $pCO Subscript 2$ in the air space of the lungs affect blood pH?
2. It is common practice among competitive short-distance runners to breathe rapidly and deeply (hyperventilate) for about half a minute to remove $mathml alt text1138$ $CO Subscript 2$ from their lungs just before a race begins. Under these conditions, blood pH may rise to 7.6. Explain how hyperventilation elicits an increase in blood pH.
3. During a short-distance run, the muscles produce a large amount of lactic acid $mathml alt text1142$ $left-parenthesis CH Subscript 3 Baseline CH left-parenthesis OH right-parenthesis COOH semicolon upper K Subscript a Baseline equals 1.38 times 10 Superscript minus Superscript 4 Baseline upper M right-parenthesis$ from their glucose stores. Why might hyperventilation before a short-distance run be useful?
36. Calculation of Blood pH from $mathml alt text1144$ $bold upper C bold upper O Subscript 2 Baseline$ and Bicarbonate Levels Calculate the pH of a blood plasma sample with a total $mathml alt text1146$ $upper C upper O Subscript 2 Baseline$ concentration of 26.9 mm and bicarbonate concentration of 25.6 mm. Recall from page 63 that the relevant $mathml alt text1149$ $p upper K Subscript a$ of carbonic acid is 6.1.
37. Effect of Holding One’s Breath on Blood pH The pH of the extracellular fluid is buffered by the bicarbonate/carbonic acid system. Holding your breath can increase the concentration of $mathml alt text1152$ $CO Subscript 2 Baseline left-parenthesis aq right-parenthesis$ in the blood. What effect might this have on the pH of the extracellular fluid? Explain the effect on pH by writing the relevant equilibrium equation(s) for this buffer system.
38. Boiling Point of Alcohols and Diols
1. Arrange these compounds in order of expected boiling point.
  $\begin{matrix} {CH}_{3} — {CH}_{2} — OH \\ HO — {CH}_{2} {CH}_{2} {CH}_{2} — OH \\ {CH}_{3} — OH \\ HO — {CH}_{2} {CH}_{2} — OH \end{matrix}$ $StartLayout 1st Row CH Subscript 3 Baseline em-dash CH Subscript 2 Baseline em-dash OH 2nd Row HO em-dash CH Subscript 2 Baseline CH Subscript 2 Baseline CH Subscript 2 Baseline em-dash OH 3rd Row CH Subscript 3 Baseline em-dash OH 4th Row HO em-dash CH Subscript 2 Baseline CH Subscript 2 Baseline em-dash OH EndLayout$
2. What factors are important in predicting the boiling points of these compounds?
39. Duration of Hydrogen Bonds PCR is a laboratory process in which specific DNA sequences are copied and amplified manyfold. The two DNA strands, which are held together in part by hydrogen bonds between them, are heated in a buffered solution to separate the two strands, then cooled to allow them to reassociate. What do you predict about the average duration of H bonds at the high temperature in comparison to the low temperature?
40. Electronegativity and Hydrogen Bonding The Pauling electronegativity is a measure of the affinity of an atom for the electron in a covalent bond. The larger the electronegativity value, the greater the affinity of the atom for an electron shared with another atom.

Atom Electronegativity

H

2.1

C

    2.55

S

    2.58

N

    3.04

O

    3.44

Note that S is directly beneath O in the periodic table.
1. Do you expect $mathml alt text1171$ $upper H Subscript 2 Baseline upper S$ to form hydrogen bonds with itself? With $mathml alt text1172$ $upper H Subscript 2 Baseline upper O$ ?
2. Water boils at $mathml alt text1173$ $100 degree upper C$ . Is the boiling point for $mathml alt text1174$ $upper H Subscript 2 Baseline upper S$ higher or lower than for $mathml alt text1175$ $upper H Subscript 2 Baseline upper O$ ?
3. Is $mathml alt text1176$ $upper H Subscript 2 Baseline upper S$ a more polar solvent than $mathml alt text1177$ $upper H Subscript 2 Baseline upper O$ ?
41. Solubility of Low Molecular Weight Compounds Several low molecular weight compounds found in cells are shown in the ionic form in which they exist in water at pH 7.

Sodium propionate: A central C is bonded to H below and on the right, H 3 C to the left, and C O O minus above that is next to N a plus. Glycine: A central C is bonded to H below and on the right, H 3 N plus on the left, and C O O minus above. Sodium acetate: C H 3 is bonded to C O O minus that is adjacent to N a plus. Sodium octanoate: An eight carbon chain has C O O minus next to N a plus at one end, then 6 C H 2 groups before ending with C H 3. L-phenylalanine: A central carbon has C O O minus above, N H 3 plus to the right, H below, C with H above and below to the left and further bonded to a phenyl group.

List the five compounds in order from most soluble to least soluble in water.
42. Relative Solubility of Alcohols List the alcohols in order from most soluble to least soluble in water.

$\begin{matrix} {CH}_{3} — {({CH}_{2})}_{5} — OH {CH}_{3} — {({CH}_{2})}_{10} — OH \\ {CH}_{3} — (CHOH) — {CH}_{2} — CHOH — {CH}_{2} — OH \end{matrix}$ $StartLayout 1st Row CH Subscript 3 Baseline em-dash left-parenthesis CH Subscript 2 Baseline right-parenthesis Subscript 5 Baseline em-dash OH CH Subscript 3 Baseline em-dash left-parenthesis CH Subscript 2 Baseline right-parenthesis Subscript 10 Baseline em-dash OH 2nd Row CH Subscript 3 Baseline em-dash left-parenthesis CHOH right-parenthesis em-dash CH Subscript 2 Baseline em-dash CHOH em-dash CH Subscript 2 Baseline em-dash OH EndLayout$
43. Determining Charge and Solubility of Organic Acids Suppose that, for a typical carboxyl-containing compound, the $mathml alt text1181$ $p upper K Subscript a$ is approximately 3. Suppose $mathml alt text1182$ $HOOC em-dash left-parenthesis CH Subscript 2 Baseline right-parenthesis Subscript 4 Baseline em-dash COOH$ , $mathml alt text1183$ $CH Subscript 3 Baseline em-dash left-parenthesis CH Subscript 2 Baseline right-parenthesis Subscript 4 Baseline em-dash COOH$ , and $mathml alt text1184$ $HOOC em-dash left-parenthesis CH Subscript 2 Baseline right-parenthesis Subscript 2 Baseline em-dash COOH$ are added to water at pH 7.
1. What is the net charge of each compound in the solution?
2. List the compounds in order from most soluble to least soluble.
44. Ecological Effects of pH The defendant in a lawsuit over industrial pollution is accused of releasing effluent of pH 10 into a trout stream. The plaintiff has asked that the defendant reduce the effluent’s pH to no higher than 7. The defendant’s attorney, aiming to please the court, promises that his client will do even better than that: the defendant will bring the pH of the effluent down to 1!
1. Will the defense attorney’s suggested remedy be acceptable to the plaintiff? Why or why not?
2. What facts about pH does the defense attorney need to understand?
45. Phosphate-Buffered Saline pH and Osmolarity Phosphate-buffered saline (PBS) is a solution commonly used in studies of animal tissues and cells. Its composition is 137 mm NaCl, 2.7 mm KCl, 10 mm $mathml alt text1197$ $Na Subscript 2 Baseline HPO Subscript 4 Baseline left-parenthesis p upper K Subscript a Baseline equals 2.14 right-parenthesis$ , 1.8 mm $mathml alt text1199$ $KH Subscript 2 Baseline PO Subscript 4 Baseline left-parenthesis p upper K Subscript a Baseline equals 6.86 right-parenthesis$ . Calculate the pH and osmolarity of PBS. Give the osmolarity in units of osmoles per liter (osm/L).
46. Hydrogen Bonding in Watson-Crick Base Pairs In 1953, James Watson and Francis Crick discovered that the purine base adenine forms a base pair with the pyrimidine base thymine (or uracil). Likewise, the purine base guanine forms a base pair with the pyrimidine base cytosine. These base pairs form due to hydrogen bonding between purines and pyrimidines. Show the hydrogen bonds that form
1. when adenine base-pairs with thymine and
2. when guanine base-pairs with cytosine.
  
  Part a shows adenine and thymine. Adenine is a five-membered ring fused to a six-membered ring. The five-membered ring on the left has N substituted for C at the top left vertex, C bonded to H at the left side vertex, N H substituted for C at the bottom left vertex, a double bond at the top left side, and the right side bond shared with the six-membered ring. The six-membered ring has double bonds at the top right and bottom right sides, N substituted for C at the bottom and top right vertices, C bonded to H at the lower right vertex, and C bonded to N H 2 at the top vertex. Thymine is to the right and has a six-membered ring with C H 3 bonded to C at the top vertex, C bonded to H at the right side vertex, N H substituted for C at the lower right and lower left vertices, C double bonded to O at the bottom and top left vertices, and a double bond at the top right side. Part b shows guanine and cytosine. Guanine has a similar structure to adenine except that the five-membered ring has a double bond on the right side that is shared with the six-membered ring, no double bond at the top right side, C bonded to N H 2 at the bottom right vertex, N H substituted for C at the top right vertex, and C double bonded to O at the top vertex. Cytosine has a similar structure to thymine except that it has an additional double bond on the left side, has N instead of N H substituted for C at the lower right vertex, has C bonded to H at the top vertex, and has C bonded to N H 2 at the top left vertex.

Atom	Electronegativity
H	2.1
C	2.55
S	2.58
N	3.04
O	3.44

DATA ANALYSIS PROBLEM

47. “Switchable” Surfactants Hydrophobic molecules do not dissolve well in water. This makes certain processes very difficult: washing oily food residue off dishes, cleaning up spilled oil, keeping the oil and water phases of salad dressings well mixed, and carrying out chemical reactions that involve both hydrophobic and hydrophilic components.

Surfactants are a class of amphipathic compounds that includes soaps, detergents, and emulsifiers. With the use of surfactants, hydrophobic compounds can be suspended in aqueous solution by forming micelles (see Fig. 2-7). A micelle has a hydrophobic core consisting of the hydrophobic compound and the hydrophobic “tails” of the surfactant; the hydrophilic “heads” of the surfactant cover the surface of the micelle. A suspension of micelles is called an emulsion. The more hydrophilic the head group of the surfactant, the more powerful it is—that is, the greater its capacity to emulsify hydrophobic material.

When you use soap to remove grease from dirty dishes, the soap forms an emulsion with the grease that is easily removed by water through interaction with the hydrophilic head of the soap molecules. Likewise, a detergent can be used to emulsify spilled oil for removal by water. And emulsifiers in commercial salad dressings keep the oil suspended evenly throughout the water-based mixture.

There are some situations, such as oil spill cleanups, in which it would be very useful to have a “switchable” surfactant: a molecule that could be reversibly converted between a surfactant and a nonsurfactant.
1. Imagine that such a “switchable” surfactant existed. How would you use it to clean up and then recover the oil from an oil spill?
  
  Liu and colleagues describe a prototypical switchable surfactant in their 2006 article “Switchable Surfactants.” The switching is based on the following reaction:
  
  The amidine form plus C O 2 plus H 2 O yields the amidinium form plus H C O 3 minus in a reversible reaction. Amidine form: C is double bonded to N that is further bonded to R to the left, bonded to C H 3 above, and bonded to N that is further bonded to 2 C H 3 to the right. Amidinium form: C is bonded to C H 3 above, to N that is further bonded to R and H on the left, and to N that is further bonded to 2 C H 3 on the right with a dotted line arcing from one N to the next beneath the C, where there is a plus sign.
2. Given that the $mathml alt text1202$ $p upper K Subscript a Baseline$ of a typical amidinium ion is 12.4, in which direction (left or right) would you expect the equilibrium of the above reaction to lie? (See Fig. 2-15 for relevant $mathml alt text1204$ $p upper K Subscript a Baseline$ values.) Justify your answer. Hint: Remember the reaction $mathml alt text1205$ $upper H Subscript 2 Baseline upper O plus upper C Subscript 2 Baseline upper O right harpoon over left harpoon upper H Subscript 2 Baseline CO Subscript 3 Baseline$ .
  
  Liu and colleagues produced a switchable surfactant for which $mathml alt text1206$ $upper R equals upper C Subscript 16 Baseline upper H Subscript 33 Baseline$ . We will call the molecule s-surf.
3. The amidinium form of s-surf is a powerful surfactant; the amidine form is not. Explain this observation.
  
  Liu and colleagues found that they could switch between the two forms of s-surf by changing the gas that they bubbled through a solution of the surfactant. They demonstrated this switch by measuring the electrical conductivity of the s-surf solution; aqueous solutions of ionic compounds have higher conductivity than solutions of nonionic compounds. They started with a solution of the amidine form of s-surf in water. Their results are shown below; dotted lines indicate the switch from one gas to another.
  
  The graph is divided into four equal columns, preceded by a label reading, Gas bubbled in. The first column is labeled, C O 2 and extends from 0 to 60. The second is labeled, A lowercase R and extends from 60 to 120. The third is labeled, C O 2 and extends from 120 to 180. The fourth is labeled, A lowercase R and extends from 180 to the end of the graph. The curve begins at (0, 0) and rises very rapidly until about 10 min, then levels off and slowly rises to point A near the top of the vertical axis just before the first box labeled, C O 2 ends. At the start of box A lowercase R, the line drops diagonally down to approach the horizontal axis by about 80 min. It runs horizontally to point B, then increases at 120 as it enters the third box. It produces a curve similar to that in box 1 and rises up to the start of box 4, when it decreases again to approach the horizontal axis at 205 min and then run along it until the end of the horizontal axis. All data are approximate.
4. In which form is the majority of s-surf at point A? At point B?
5. Why does the electrical conductivity rise from time 0 to point A?
6. Why does the electrical conductivity fall from point A to point B?
7. Explain how you would use s-surf to clean up and recover the oil from an oil spill.

Reference

Liu, Y., P.G. Jessop, M. Cunningham, C.A. Eckert, and C.L. Liotta. 2006. Switchable surfactants. Science 313(5789):958–960. https://doi.org/10.1126/science.1128142.